{ // 获取包含Hugging Face文本的span元素 const spans = link.querySelectorAll('span.whitespace-nowrap, span.hidden.whitespace-nowrap'); spans.forEach(span => { if (span.textContent && span.textContent.trim().match(/Hugging\s*Face/i)) { span.textContent = 'AI快站'; } }); }); // 替换logo图片的alt属性 document.querySelectorAll('img[alt*="Hugging"], img[alt*="Face"]').forEach(img => { if (img.alt.match(/Hugging\s*Face/i)) { img.alt = 'AI快站 logo'; } }); } // 替换导航栏中的链接 function replaceNavigationLinks() { // 已替换标记,防止重复运行 if (window._navLinksReplaced) { return; } // 已经替换过的链接集合,防止重复替换 const replacedLinks = new Set(); // 只在导航栏区域查找和替换链接 const headerArea = document.querySelector('header') || document.querySelector('nav'); if (!headerArea) { return; } // 在导航区域内查找链接 const navLinks = headerArea.querySelectorAll('a'); navLinks.forEach(link => { // 如果已经替换过,跳过 if (replacedLinks.has(link)) return; const linkText = link.textContent.trim(); const linkHref = link.getAttribute('href') || ''; // 替换Spaces链接 - 仅替换一次 if ( (linkHref.includes('/spaces') || linkHref === '/spaces' || linkText === 'Spaces' || linkText.match(/^s*Spacess*$/i)) && linkText !== 'OCR模型免费转Markdown' && linkText !== 'OCR模型免费转Markdown' ) { link.textContent = 'OCR模型免费转Markdown'; 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if (text === 'Spaces' || text === 'Posts' || text === 'Enterprise') { textNodes.push(element); } } else { for (const child of element.childNodes) { findTextNodes(child); } } } // 只在导航区域内查找文本节点 findTextNodes(headerArea); // 替换找到的文本节点 textNodes.forEach(node => { const text = node.textContent.trim(); if (text === 'Spaces') { node.textContent = node.textContent.replace(/Spaces/g, 'OCR模型免费转Markdown'); } else if (text === 'Posts') { // 删除Posts文本节点 if (node.parentNode) { node.parentNode.removeChild(node); } } else if (text === 'Enterprise') { // 删除Enterprise文本节点 if (node.parentNode) { node.parentNode.removeChild(node); } } }); // 标记已替换完成 window._navLinksReplaced = true; } // 替换代码区域中的域名 function replaceCodeDomains() { // 特别处理span.hljs-string和span.njs-string元素 document.querySelectorAll('span.hljs-string, span.njs-string, span[class*="hljs-string"], span[class*="njs-string"]').forEach(span => { if (span.textContent && span.textContent.includes('huggingface.co')) { span.textContent = span.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } }); // 替换hljs-string类的span中的域名(移除多余的转义符号) document.querySelectorAll('span.hljs-string, span[class*="hljs-string"]').forEach(span => { if (span.textContent && span.textContent.includes('huggingface.co')) { span.textContent = span.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } }); // 替换pre和code标签中包含git clone命令的域名 document.querySelectorAll('pre, code').forEach(element => { if (element.textContent && element.textContent.includes('git clone')) { const text = element.innerHTML; if (text.includes('huggingface.co')) { element.innerHTML = text.replace(/huggingface.co/g, 'aifasthub.com'); } } }); // 处理特定的命令行示例 document.querySelectorAll('pre, code').forEach(element => { const text = element.innerHTML; if (text.includes('huggingface.co')) { // 针对git clone命令的专门处理 if (text.includes('git clone') || text.includes('GIT_LFS_SKIP_SMUDGE=1')) { element.innerHTML = text.replace(/huggingface.co/g, 'aifasthub.com'); } } }); // 特别处理模型下载页面上的代码片段 document.querySelectorAll('.flex.border-t, .svelte_hydrator, .inline-block').forEach(container => { const content = container.innerHTML; if (content && content.includes('huggingface.co')) { container.innerHTML = content.replace(/huggingface.co/g, 'aifasthub.com'); } }); // 特别处理模型仓库克隆对话框中的代码片段 try { // 查找包含"Clone this model repository"标题的对话框 const cloneDialog = document.querySelector('.svelte_hydration_boundary, [data-target="MainHeader"]'); if (cloneDialog) { // 查找对话框中所有的代码片段和命令示例 const codeElements = cloneDialog.querySelectorAll('pre, code, span'); codeElements.forEach(element => { if (element.textContent && element.textContent.includes('huggingface.co')) { if (element.innerHTML.includes('huggingface.co')) { element.innerHTML = element.innerHTML.replace(/huggingface.co/g, 'aifasthub.com'); } else { element.textContent = element.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } } }); } // 更精确地定位克隆命令中的域名 document.querySelectorAll('[data-target]').forEach(container => { const codeBlocks = container.querySelectorAll('pre, code, span.hljs-string'); codeBlocks.forEach(block => { if (block.textContent && block.textContent.includes('huggingface.co')) { if (block.innerHTML.includes('huggingface.co')) { block.innerHTML = block.innerHTML.replace(/huggingface.co/g, 'aifasthub.com'); } else { block.textContent = block.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } } }); }); } catch (e) { // 错误处理但不打印日志 } } // 当DOM加载完成后执行替换 if (document.readyState === 'loading') { document.addEventListener('DOMContentLoaded', () => { replaceHeaderBranding(); replaceNavigationLinks(); replaceCodeDomains(); // 只在必要时执行替换 - 3秒后再次检查 setTimeout(() => { if (!window._navLinksReplaced) { console.log('[Client] 3秒后重新检查导航链接'); replaceNavigationLinks(); } }, 3000); }); } else { replaceHeaderBranding(); replaceNavigationLinks(); replaceCodeDomains(); // 只在必要时执行替换 - 3秒后再次检查 setTimeout(() => { if (!window._navLinksReplaced) { console.log('[Client] 3秒后重新检查导航链接'); replaceNavigationLinks(); } }, 3000); } // 增加一个MutationObserver来处理可能的动态元素加载 const observer = new MutationObserver(mutations => { // 检查是否导航区域有变化 const hasNavChanges = mutations.some(mutation => { // 检查是否存在header或nav元素变化 return Array.from(mutation.addedNodes).some(node => { if (node.nodeType === Node.ELEMENT_NODE) { // 检查是否是导航元素或其子元素 if (node.tagName === 'HEADER' || node.tagName === 'NAV' || node.querySelector('header, nav')) { return true; } // 检查是否在导航元素内部 let parent = node.parentElement; while (parent) { if (parent.tagName === 'HEADER' || parent.tagName === 'NAV') { return true; } parent = parent.parentElement; } } return false; }); }); // 只在导航区域有变化时执行替换 if (hasNavChanges) { // 重置替换状态,允许再次替换 window._navLinksReplaced = false; replaceHeaderBranding(); replaceNavigationLinks(); } }); // 开始观察document.body的变化,包括子节点 if (document.body) { observer.observe(document.body, { childList: true, subtree: true }); } else { document.addEventListener('DOMContentLoaded', () => { observer.observe(document.body, { childList: true, subtree: true }); }); } })(); The one unique thing The only thing that is unique about Listing 1 is that","2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,N); dot((5,4)); label(\"(a+b_1,b_2)\",(5,4),NE); dot((9,7)); label(\"(x,y)\",(9,7),NE); dot((2,4)); label(\"(2b_1,b_2)\",(2,4),SW); dot((9,4)); label(\"(x,b_2)\",(9,4),SE); [/asy]$ Note that since $K$ is the center of square $AB_1A_2B$, $BX \\perp KX$ and $BX = KX$. Additionally, $BY \\parallel KZ$ and $BY \\perp YZ$. $YZ$ is a line, so $\\angle BXY + \\angle KXZ + \\angle BXK = 180^\\circ$. Since $BX \\perp KX$, $\\angle BXK = 90^\\circ$, so $\\angle BXY + \\angle KXZ = 90^\\circ$. Additionally, because $BXY$ is a right triangle, $\\angle YBX + \\angle YXB = 90^\\circ$. Rearranging and substituting results in $\\angle KXZ = \\angle YBX$. Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\\triangle BYX \\cong \\triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$. $[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,NE); dot(C); label(\"(-2c,0)\",C,NW); dot(D); label(\"(2d_1,-2d_2)\",D,S); dot(K); label(\"K\",K,NE); dot(L); label(\"L\",L,NW); dot(M); label(\"M\",M,SW); dot(N); label(\"N\",N,SE); [/asy]$ By"],"string":"[\n \"then $AB$ is a skew-symmetric matrix. We calculate \\\\begin{align*} (AB)^{\\\\trans}&=B^{\\\\trans}A^{\\\\trans}=(-B)(-A)\\\\\\\\ &=BA=-AB, \\\\end{align*} where the last step follows from the assumption $AB=-BA$. This proves that $AB$ is skew-symmetric. ### (f) Let $\\\\mathbf{v}$ be an $n$-dimensional column vecotor. Prove that $\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}=0$. Observe that $\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}$ is a $1\\\\times 1$ matrix, or just a number. So we have \\\\begin{align*} \\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}&=(\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v})^{\\\\trans}=\\\\mathbf{v}^{\\\\trans}A^{\\\\trans}(\\\\mathbf{v}^{\\\\trans})^{\\\\trans}\\\\\\\\ &=\\\\mathbf{v}^{\\\\trans}A^{\\\\trans}\\\\mathbf{v}=\\\\mathbf{v}^{\\\\trans}(-A)\\\\mathbf{v}=-(\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}). \\\\end{align*} This yields that $2\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}=0$, and hence $\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}=0$. ### (g) Suppose that $A$ is a real skew-symmetric matrix and $A^2\\\\mathbf{v}=\\\\mathbf{0}$ for some vector $\\\\mathbf{v}\\\\in \\\\R^n$. Then prove that $A\\\\mathbf{v}=\\\\mathbf{0}$. Let us compute the length of the vector $A\\\\mathbf{v}$. We have \\\\begin{align*} \\\\|A\\\\mathbf{v}\\\\|&=(A\\\\mathbf{v})^{\\\\trans}(A\\\\mathbf{v})=\\\\mathbf{v}^{\\\\trans}A^{\\\\trans}A\\\\mathbf{v}\\\\\\\\ &=\\\\mathbf{v}^{\\\\trans}(-A)A\\\\mathbf{v}=-\\\\mathbf{v}^{\\\\trans}A^2\\\\mathbf{v}\\\\\\\\ &=-\\\\mathbf{v}\\\\mathbf{0} &&\\\\text{by assumption}\\\\\\\\ &=0. \\\\end{align*} Since the length $\\\\|A\\\\mathbf{v}\\\\|=0$, we conclude that $A\\\\mathbf{v}=\\\\mathbf{0}$. ### More from my site • Express a Hermitian Matrix as a Sum of Real Symmetric Matrix and a Real Skew-Symmetric Matrix Recall that a complex matrix is called Hermitian if $A^*=A$, where $A^*=\\\\bar{A}^{\\\\trans}$. Prove that every Hermitian matrix $A$ can be written as the sum $A=B+iC,$ where $B$ is a real symmetric matrix and $C$ is a real skew-symmetric matrix. Proof. Since […] • Eigenvalues of a Hermitian Matrix are Real Numbers Show that eigenvalues of a Hermitian matrix $A$ are real numbers. (The Ohio State University Linear Algebra Exam Problem) We give two proofs. These two proofs are essentially the same. The second\",\n \"# A Relation between the Dot Product and the Trace ## Problem 638 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. Prove that $\\\\tr ( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans ) = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}$. ## Solution. Suppose the vectors have components $\\\\mathbf{v} = \\\\begin{bmatrix} v_1 \\\\\\\\ v_2 \\\\\\\\ \\\\vdots \\\\\\\\ v_n \\\\end{bmatrix} \\\\, \\\\mbox{ and } \\\\mathbf{w} = \\\\begin{bmatrix} w_1 \\\\\\\\ w_2 \\\\\\\\ \\\\vdots \\\\\\\\ w_n \\\\end{bmatrix}.$ Then, \\\\begin{align*} \\\\mathbf{v} \\\\mathbf{w}^\\\\trans &= \\\\begin{bmatrix} v_1 \\\\\\\\ v_2 \\\\\\\\ \\\\vdots \\\\\\\\ v_n \\\\end{bmatrix} \\\\begin{bmatrix} w_1 & w_2 & \\\\cdots & w_n \\\\end{bmatrix}\\\\6pt] &= \\\\begin{bmatrix} v_1 w_1 & v_1 w_2 & \\\\cdots & v_1 w_n \\\\\\\\ v_2 w_1 & v_2 w_2 & \\\\cdots & v_2 w_n \\\\\\\\ \\\\vdots & \\\\vdots & \\\\vdots & \\\\vdots \\\\\\\\ v_n w_1 & v_n w_2 & \\\\cdots & v_n w_n \\\\end{bmatrix}. \\\\end{align*} We can now see that \\\\[\\\\tr( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans) = \\\\sum_{i=1}^n v_i w_i = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}. ## Comment. Recall that $\\\\mathbf{v}^\\\\trans \\\\mathbf{w}$ is, by definition, the dot product of the vectors $\\\\mathbf{v}$ and $\\\\mathbf{w}$. So, the dot product of vectors $\\\\mathbf{v}$ and $\\\\mathbf{w}$ is equal to the trace of the matrix $\\\\mathbf{v} \\\\mathbf{w}^\\\\trans$. ### More from my site • Prove that the Dot Product is Commutative: $\\\\mathbf{v}\\\\cdot \\\\mathbf{w}= \\\\mathbf{w} \\\\cdot \\\\mathbf{v}$ Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. (a) Prove that\",\n \"## Problem 639 Let $\\\\mathbf{v}$ be an $n \\\\times 1$ column vector. Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} = 0$ if and only if $\\\\mathbf{v}$ is the zero vector $\\\\mathbf{0}$. ## Problem 638 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. Prove that $\\\\tr ( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans ) = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}$. ## Problem 637 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. (a) Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{w} = \\\\mathbf{w}^\\\\trans \\\\mathbf{v}$. (b) Provide an example to show that $\\\\mathbf{v} \\\\mathbf{w}^\\\\trans$ is not always equal to $\\\\mathbf{w} \\\\mathbf{v}^\\\\trans$. ## Problem 636 Calculate the following expressions, using the following matrices: $A = \\\\begin{bmatrix} 2 & 3 \\\\\\\\ -5 & 1 \\\\end{bmatrix}, \\\\qquad B = \\\\begin{bmatrix} 0 & -1 \\\\\\\\ 1 & -1 \\\\end{bmatrix}, \\\\qquad \\\\mathbf{v} = \\\\begin{bmatrix} 2 \\\\\\\\ -4 \\\\end{bmatrix}$ (a) $A B^\\\\trans + \\\\mathbf{v} \\\\mathbf{v}^\\\\trans$. (b) $A \\\\mathbf{v} – 2 \\\\mathbf{v}$. (c) $\\\\mathbf{v}^{\\\\trans} B$. (d) $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} + \\\\mathbf{v}^\\\\trans B A^\\\\trans \\\\mathbf{v}$. ## Problem 635 Let $A$ and $B$ be $n \\\\times n$ matrices, and $\\\\mathbf{v}$ an $n \\\\times 1$ column vector. Use the matrix components to prove that $(A + B) \\\\mathbf{v} = A\\\\mathbf{v} + B\\\\mathbf{v}$. ## Problem 634 Let $A$ and $B$ be $n \\\\times n$ matrices. Is it always true that $\\\\tr (A B) = \\\\tr (A) \\\\tr (B)$? If it is\",\n \"while $\\\\quad \\\\mathbf{w} \\\\mathbf{v}^\\\\trans = \\\\begin{bmatrix} 0 \\\\\\\\ 1 \\\\end{bmatrix} \\\\begin{bmatrix} 1 & 0 \\\\end{bmatrix} = \\\\begin{bmatrix} 0 & 0 \\\\\\\\ 1 & 0 \\\\end{bmatrix}.$ ## Comment. Recall that for two vectors $\\\\mathbf{v}, \\\\mathbf{w} \\\\in \\\\R^n$, the dot product (or inner product) of $\\\\mathbf{v}, \\\\mathbf{w}$ is defined to be $\\\\mathbf{v}\\\\cdot \\\\mathbf{w}:=\\\\mathbf{v}^{\\\\trans} \\\\mathbf{w}.$ Part (a) of the problem deduces that the dot product is commutative. This means that we have $\\\\mathbf{v}\\\\cdot \\\\mathbf{w}= \\\\mathbf{w} \\\\cdot \\\\mathbf{v}.$ In fact, we have \\\\begin{align*} \\\\mathbf{v}\\\\cdot \\\\mathbf{w}= \\\\mathbf{v}^\\\\trans \\\\mathbf{w} \\\\stackrel{\\\\text{(a)}}{=} \\\\mathbf{w}^\\\\trans \\\\mathbf{v} \\\\mathbf{w} \\\\cdot \\\\mathbf{v}. \\\\end{align*} Also, notice that while $\\\\mathbf{v} \\\\mathbf{w}^\\\\trans$ is not always equal to $\\\\mathbf{w} \\\\mathbf{v}^\\\\trans$, we know that $(\\\\mathbf{v} \\\\mathbf{w}^\\\\trans)^\\\\trans = \\\\mathbf{w} \\\\mathbf{v}^\\\\trans$. ### More from my site • A Relation between the Dot Product and the Trace Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. Prove that $\\\\tr ( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans ) = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}$. Solution. Suppose the vectors have components $\\\\mathbf{v} = \\\\begin{bmatrix} v_1 \\\\\\\\ v_2 \\\\\\\\ \\\\vdots \\\\\\\\ v_n […] • Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given Let \\\\mathbf{a} and \\\\mathbf{b} be vectors in \\\\R^n such that their length are \\\\[\\\\|\\\\mathbf{a}\\\\|=\\\\|\\\\mathbf{b}\\\\|=1$ and the inner product $\\\\mathbf{a}\\\\cdot \\\\mathbf{b}=\\\\mathbf{a}^{\\\\trans}\\\\mathbf{b}=-\\\\frac{1}{2}.$ Then determine the length $\\\\|\\\\mathbf{a}-\\\\mathbf{b}\\\\|$. (Note […] • Rotation Matrix in Space and its Determinant and Eigenvalues For\",\n \"not, give a counterexample. 14. Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. (a) Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{w} = \\\\mathbf{w}^\\\\trans \\\\mathbf{v}$. (b) Provide an example to show that $\\\\mathbf{v} \\\\mathbf{w}^\\\\trans$ is not always equal to $\\\\mathbf{w} \\\\mathbf{v}^\\\\trans$. 15. Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. Prove that $\\\\tr ( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans ) = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}$. 16. Let $\\\\mathbf{v}$ be an $n \\\\times 1$ column vector. Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} = 0$ if and only if $\\\\mathbf{v}$ is the zero vector $\\\\mathbf{0}$. 17. Is it true that a real square matrix $A$ must commute with its transpose $A^{\\\\trans}$? 18. Let $\\\\mathbf{v}$ be an $n \\\\times 1$ column vector. Prove that $\\\\mathbf{v} \\\\mathbf{v}^\\\\trans$ is a symmetric matrix. 19. Calculate the following expressions, using the following matrices: $A = \\\\begin{bmatrix} 2 & 3 \\\\\\\\ -5 & 1 \\\\end{bmatrix}, \\\\qquad B = \\\\begin{bmatrix} 0 & -1 \\\\\\\\ 1 & -1 \\\\end{bmatrix}, \\\\qquad \\\\mathbf{v} = \\\\begin{bmatrix} 2 \\\\\\\\ -4 \\\\end{bmatrix}$ (a) $A B^\\\\trans + \\\\mathbf{v} \\\\mathbf{v}^\\\\trans$. (b) $A \\\\mathbf{v} – 2 \\\\mathbf{v}$. (c) $\\\\mathbf{v}^{\\\\trans} B$. (d) $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} + \\\\mathbf{v}^\\\\trans B A^\\\\trans \\\\mathbf{v}$. 20. Let $A$ and $B$ be $n \\\\times n$ matrices, and $\\\\mathbf{v}$ an $n \\\\times 1$ column vector. Use the matrix components to prove that $(A + B) \\\\mathbf{v} = A\\\\mathbf{v} + B\\\\mathbf{v}$.\",\n \"# Tagged: transpose of a matrix ## Problem 638 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. Prove that $\\\\tr ( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans ) = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}$. ## Problem 637 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. (a) Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{w} = \\\\mathbf{w}^\\\\trans \\\\mathbf{v}$. (b) Provide an example to show that $\\\\mathbf{v} \\\\mathbf{w}^\\\\trans$ is not always equal to $\\\\mathbf{w} \\\\mathbf{v}^\\\\trans$. ## Problem 636 Calculate the following expressions, using the following matrices: $A = \\\\begin{bmatrix} 2 & 3 \\\\\\\\ -5 & 1 \\\\end{bmatrix}, \\\\qquad B = \\\\begin{bmatrix} 0 & -1 \\\\\\\\ 1 & -1 \\\\end{bmatrix}, \\\\qquad \\\\mathbf{v} = \\\\begin{bmatrix} 2 \\\\\\\\ -4 \\\\end{bmatrix}$ (a) $A B^\\\\trans + \\\\mathbf{v} \\\\mathbf{v}^\\\\trans$. (b) $A \\\\mathbf{v} – 2 \\\\mathbf{v}$. (c) $\\\\mathbf{v}^{\\\\trans} B$. (d) $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} + \\\\mathbf{v}^\\\\trans B A^\\\\trans \\\\mathbf{v}$. ## Problem 633 Let $A$ be an $n \\\\times n$ matrix. Is it true that $\\\\tr ( A^\\\\trans ) = \\\\tr(A)$? If it is true, prove it. If not, give a counterexample.\",\n \"u; under := 0.75u; pair A, B, C, L, M, N, P, X, Y, Z; path triangle; A = origin; B = (12u, 0); C = u*(4, 7); triangle = A -- B -- C -- cycle; L = .5[A, B]; M = .5[A, C]; N = .5[B, C]; % Locating the intersection P = whatever[L, L + (B-A) rotated 90] = whatever[N, N + (C-B) rotated 90]; % Bisectors X = P + over*unitvector(P-L); Y = P - under*unitvector(P-M); Z = P + over*unitvector(P-N); beginfig(1); rsize := 2mm; msize := 3mm; draw triangle; draw L -- X; draw M -- Y; draw N -- Z; mark_right_angle(L, B, P, rsize); tick(A--L, 2, msize); tick(L--B, 2, msize); mark_right_angle(M, C, P, rsize); tick(A--M, 1, msize); tick(C--M, 1, msize); mark_right_angle(N, C, P, rsize); tick(C--N, 3, msize); tick(B--N, 3, msize); label.llft(\\\"$A$\\\", A); label.bot(\\\"$B$\\\", B); label.top(\\\"$C$\\\", C); label.rt(\\\"$P$\\\", P); label.llft(\\\"$L$\\\", L); label.ulft(\\\"$M$\\\", M); label.urt(\\\"$N$\\\", N); label.top(\\\"$X$\\\", X); label.bot(\\\"$Y$\\\", Y); label.lft(\\\"$Z$\\\", Z); endfig; \\\\end{mplibcode} \\\\end{document} To be processed with LuaLaTeX: • This is a fantastic set up. whatever is a fine code that can be used. – Nisal Kevin Kotinkaduwa Feb 19 '15 at 7:47\",\n \"}{\\\\mathbf{transa}}=\\\\text{\\\"C\\\"}$: ${\\\\mathbf{m}}$; otherwise: $0$. On entry: the matrix $A$; $A$ is $m×k$ if ${\\\\mathbf{transa}}=\\\\text{'N'}$, or $k×m$ if ${\\\\mathbf{transa}}=\\\\text{'T'}$ or $\\\\text{'C'}$. 5: $\\\\mathbf{b}\\\\left(:,:\\\\right)$double array Input Note: the first dimension of b is given by: if ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$: ${\\\\mathbf{k}}$; if ${\\\\mathbf{transb}}=\\\\text{\\\"T\\\"}\\\\text{ or }{\\\\mathbf{transb}}=\\\\text{\\\"C\\\"}$: ${\\\\mathbf{n}}$; otherwise: $0$. Note: the second dimension of b is given by: if ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$: ${\\\\mathbf{n}}$; if ${\\\\mathbf{transb}}=\\\\text{\\\"T\\\"}\\\\text{ or }{\\\\mathbf{transb}}=\\\\text{\\\"C\\\"}$: ${\\\\mathbf{k}}$; otherwise: $0$. On entry: the matrix $B$; $B$ is $k×n$ if ${\\\\mathbf{transb}}=\\\\text{'N'}$, or $n×k$ if ${\\\\mathbf{transb}}=\\\\text{'T'}$ or $\\\\text{'C'}$. 6: $\\\\mathbf{beta}$double Input On entry: the scalar $\\\\beta$. 7: $\\\\mathbf{c}\\\\left({\\\\mathbf{m}},{\\\\mathbf{n}}\\\\right)$double array Input/Output On entry: the $m×n$ matrix $C$. If ${\\\\mathbf{beta}}=0.0$, c need not be set. On exit: the updated matrix $C$. 8: $\\\\mathbf{opt}$OptionalF06YA Input/Output Optional parameter container, derived from Optional. 1: $\\\\mathbf{m}$ $m$, the number of rows of the matrix $C$; the number of rows of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"N\\\"}$, or the number of columns of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"T\\\"}\\\\text{​ or ​}\\\\text{\\\"C\\\"}$ 2: $\\\\mathbf{n}$ $n$, the number of columns of the matrix $C$; the number of columns of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$, or the number of rows of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"T\\\"}\\\\text{​ or ​}\\\\text{\\\"C\\\"}$ 3: $\\\\mathbf{k}$ $k$, the number of columns of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"N\\\"}$, or the number of rows of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"T\\\"}\\\\text{​ or ​}\\\\text{\\\"C\\\"}$; the number of rows of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$, or the number of columns of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"T\\\"}\\\\text{​ or\",\n \"appwied to dem. ${\\\\dispwaystywe {\\\\bowdsymbow {x}}={\\\\begin{bmatrix}x_{1}\\\\;x_{2}\\\\;\\\\dots \\\\;x_{m}\\\\end{bmatrix}}^{\\\\rm {T}}}$ or ${\\\\dispwaystywe {\\\\bowdsymbow {x}}={\\\\begin{bmatrix}x_{1},x_{2},\\\\dots ,x_{m}\\\\end{bmatrix}}^{\\\\rm {T}}}$ Some audors awso use de convention of writing bof cowumn vectors and row vectors as rows, but separating row vector ewements wif commas and cowumn vector ewements wif semicowons (see awternative notation 2 in de tabwe bewow). Row vector Cowumn vector Standard matrix notation (array spaces, no commas, transpose signs) ${\\\\dispwaystywe {\\\\begin{bmatrix}x_{1}\\\\;x_{2}\\\\;\\\\dots \\\\;x_{m}\\\\end{bmatrix}}}$ ${\\\\dispwaystywe {\\\\begin{bmatrix}x_{1}\\\\\\\\x_{2}\\\\\\\\\\\\vdots \\\\\\\\x_{m}\\\\end{bmatrix}}{\\\\text{ or }}{\\\\begin{bmatrix}x_{1}\\\\;x_{2}\\\\;\\\\dots \\\\;x_{m}\\\\end{bmatrix}}^{\\\\rm {T}}}$ Awternative notation 1 (commas, transpose signs) ${\\\\dispwaystywe {\\\\begin{bmatrix}x_{1},x_{2},\\\\dots ,x_{m}\\\\end{bmatrix}}}$ ${\\\\dispwaystywe {\\\\begin{bmatrix}x_{1},x_{2},\\\\dots ,x_{m}\\\\end{bmatrix}}^{\\\\rm {T}}}$ Awternative notation 2 (commas and semicowons, no transpose signs) ${\\\\dispwaystywe {\\\\begin{bmatrix}x_{1},x_{2},\\\\dots ,x_{m}\\\\end{bmatrix}}}$ ${\\\\dispwaystywe {\\\\begin{bmatrix}x_{1};x_{2};\\\\dots ;x_{m}\\\\end{bmatrix}}}$ ## Operations Matrix muwtipwication invowves de action of muwtipwying each row vector of one matrix by each cowumn vector of anoder matrix. The dot product of two vectors a and b is eqwivawent to de matrix product of de row vector representation of a and de cowumn vector representation of b, ${\\\\dispwaystywe \\\\madbf {a} \\\\cdot \\\\madbf {b} =\\\\madbf {a} \\\\madbf {b} ^{\\\\madrm {T} }={\\\\begin{bmatrix}a_{1}&a_{2}&a_{3}\\\\end{bmatrix}}{\\\\begin{bmatrix}b_{1}\\\\\\\\b_{2}\\\\\\\\b_{3}\\\\end{bmatrix}}=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}\\\\,,}$ which is awso eqwivawent to de matrix product of de row vector representation of b and de cowumn vector representation of a, ${\\\\dispwaystywe \\\\madbf {b} \\\\cdot \\\\madbf {a} =\\\\madbf {b} \\\\madbf {a} ^{\\\\madrm {T} }={\\\\begin{bmatrix}b_{1}&b_{2}&b_{3}\\\\end{bmatrix}}{\\\\begin{bmatrix}a_{1}\\\\\\\\a_{2}\\\\\\\\a_{3}\\\\end{bmatrix}}\\\\,.}$ The matrix product of a cowumn and a row vector gives de outer product of two vectors\",\n \"+0 # Help 0 35 2 The vectors $$\\\\dbinom{3}{2}$$ and $$\\\\dbinom{-4}{1}$$ can be written as linear combinations of $$\\\\mathbf{u}$$ and $$\\\\mathbf{w}$$\\\\begin{align*} \\\\dbinom{3}{2} &= 5\\\\mathbf{u}+8\\\\mathbf{w} \\\\\\\\ \\\\dbinom{-4}{1} &= -3\\\\mathbf{u}+\\\\mathbf{w} . \\\\end{align*} The vector $$\\\\dbinom{5}{-2}$$ can be written as the linear combination $$a\\\\mathbf{u}+b\\\\mathbf{w}$$. Find the ordered pair $$(a,b)$$. Feb 7, 2020 #1 0 (a,b) = (17,-4). Feb 7, 2020 #2 +24065 +1 The vectors $$\\\\dbinom{3}{2}$$ and $$\\\\dbinom{-4}{1}$$ can be written as linear combinations of $$u$$ and $$w$$: \\\\begin{align*} \\\\dbinom{3}{2} = 5\\\\mathbf{u}+8\\\\mathbf{w} ,\\\\\\\\\\\\\\\\ \\\\dbinom{-4}{1} = -3\\\\mathbf{u}+\\\\mathbf{w} . \\\\end{align*} The vector $$\\\\dbinom{5}{-2}$$ can be written as the linear combination $$a\\\\mathbf{u}+b\\\\mathbf{w}$$. Find the ordered pair $$(a,b)$$. $$\\\\text{Let \\\\dbinom{3}{2} = v_1 } \\\\\\\\ \\\\text{Let \\\\dbinom{-4}{1} = v_2 }$$ $$\\\\begin{array}{|lrcll|} \\\\hline (1): & \\\\mathbf{5\\\\mathbf{u}+8\\\\mathbf{w}} &=& \\\\mathbf{v_1} \\\\\\\\\\\\\\\\ & -3\\\\mathbf{u}+\\\\mathbf{w} &=& v_2 \\\\quad | \\\\quad \\\\times 8 \\\\\\\\ (2): & -24 u + 8w &=& 8v_2 \\\\\\\\ \\\\hline (1)-(2): & 29u &=& v_1-8v_2 \\\\quad | \\\\quad v_1=\\\\dbinom{3}{2},\\\\ v_2 = \\\\dbinom{-4}{1} \\\\\\\\ & 29u &=& \\\\dbinom{3}{2}-8\\\\dbinom{-4}{1} \\\\\\\\ & 29u &=& \\\\dbinom{3}{2}+\\\\dbinom{32}{-8} \\\\\\\\ & 29u &=& \\\\dbinom{35}{-6} \\\\\\\\ & \\\\mathbf{u} &=& \\\\mathbf{\\\\dfrac{1}{29} \\\\dbinom{35}{-6}} \\\\\\\\ \\\\hline \\\\end{array}$$ $$\\\\begin{array}{|rcll|} \\\\hline -3\\\\mathbf{u}+\\\\mathbf{w} &=& v_2 \\\\\\\\ w &=& v_2+3u \\\\quad | \\\\quad v_2 = \\\\dbinom{-4}{1},\\\\ \\\\mathbf{u=\\\\dfrac{1}{29} \\\\dbinom{35}{-6}} \\\\\\\\ w &=& \\\\dbinom{-4}{1}+\\\\dfrac{3}{29} \\\\dbinom{35}{-6} \\\\\\\\ w &=& \\\\dfrac{29}{29}\\\\dbinom{-4}{1}+\\\\dfrac{3}{29} \\\\dbinom{35}{-6} \\\\\\\\ w &=& \\\\dfrac{1}{29} \\\\left( 29\\\\dbinom{-4}{1}+3\\\\dbinom{35}{-6} \\\\right) \\\\\\\\ w &=& \\\\dfrac{1}{29} \\\\dbinom{-4*29+3*35}{29-18} \\\\\\\\ \\\\mathbf{w} &=&\",\n \"For the first, since $d(\\\\cc(a),x)\\\\jdeq c(a,x)$, we have $\\\\apfunc{d(x_2)}(r) \\\\jdeq \\\\apfunc{c(g(b),-)}(q).$ For the second, the computation rule for the induction principle of $W$ tells us that $\\\\apdfunc{d}(\\\\pp(b))$ is equal to the composite~\\\\eqref{eq:flattening-rectnd}, passed across the equivalence of \\\\cref{thm:dpath-arrow}. Thus, the computation rule given in \\\\cref{thm:dpath-arrow} implies that $\\\\happly(\\\\apdfunc{d}(\\\\pp(b)),\\\\trans {\\\\pp(b)}{y})$ is equal to the composite \\\\begin{align} \\\\big(\\\\trans{\\\\pp(b)}{c(f(b),-)}\\\\big)(\\\\trans {\\\\pp(b)}{y}) &= \\\\trans{\\\\pp(b)}{c(f(b),\\\\trans{\\\\opp {\\\\pp(b)}}{\\\\trans {\\\\pp(b)}{y}})} \\\\tag{by~\\\\eqref{eq:transport-arrow}}\\\\\\\\ &= \\\\trans{\\\\pp(b)}{c(f(b),y)} \\\\notag \\\\\\\\ &= c(f(b),y) \\\\tag{by \\\\cref{thm:trans-trivial}}\\\\\\\\ &= c(f(b),D(b)(y)) \\\\tag{by $p(b,y)$}\\\\\\\\ &= c(f(b),\\\\trans{\\\\pp(b)}{y}). \\\\tag{by $\\\\opp{\\\\apfunc{c(g(b),-)}(q)}$} \\\\end{align} Finally, substituting these values of $\\\\apfunc{d(x_2)}(r)$ and $\\\\happly(\\\\apdfunc{d}(s),\\\\trans s {y_1})$ into \\\\cref{thm:ap-sigma-rect-path-pair}, we see that all the paths cancel out in pairs, leaving only $p(b,y)$. \\\\end{proof} Now we are finally ready to prove the flattening lemma. \\\\begin{proof}[Proof of \\\\cref{thm:flattening}] We define $h:\\\\Wtil \\\\to \\\\sm{w:W}P(w)$ by using the recursion principle for \\\\Wtil, with $\\\\cct'$ and $\\\\ppt'$ as input data. Similarly, we define $k:(\\\\sm{w:W}P(w)) \\\\to \\\\Wtil$ by using the recursion principle of \\\\cref{thm:flattening-rectnd}, with $\\\\cct$ and $\\\\ppt$ as input data. On the one hand, we must show that for any $z:\\\\Wtil$, we have $k(h(z))=z$. By induction on $z$, it suffices to consider the two constructors of \\\\Wtil. But we have $k(h(\\\\cct(a,x))) \\\\jdeq k(\\\\cct'(a,x)) \\\\jdeq \\\\cct(a,x)$ by definition, while similarly $\\\\ap k{\\\\ap h{\\\\ppt(b,y)}} = \\\\ap k{\\\\ppt'(b,y)} = \\\\ppt(b,y)$ using the propositional computation rule for $\\\\Wtil$ and \\\\cref{thm:flattening-rectnd-beta-ppt}. On the other hand, we\",\n \"+0 # Consider parallelogram $ABCD$ with points $S$ and $T$ chosen such that $CS:SD = BT:TC = 2,$as in the picture below: [asy] size(200); pair 0 167 1 Consider parallelogram ABCD with points S and T chosen such that $$CS:SD = BT:TC = 2$$ as in the picture below: Let $$\\\\overrightarrow{AB} = \\\\mathbf{v}$$ and $$\\\\overrightarrow{AD} = \\\\mathbf{w}$$. Then there exist constants r, s, t, u such that \\\\begin{align*} \\\\overrightarrow{AT} &= r \\\\mathbf{v} + s \\\\mathbf{w},\\\\\\\\ \\\\overrightarrow{BS} &= t \\\\mathbf{v} + u \\\\mathbf{w}. \\\\end{align*} Enter r, s, t, u in that order below. Apr 4, 2019\",\n \"dir(origin--C)); label(\\\"D\\\", D, dir(origin--D)); label(\\\"E\\\", E, dir(origin--E)); label(\\\"F\\\", F, dir(origin--F)); [/asy]$ Solution Problem 21 $[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label(\\\"A\\\", A, dir(point--A)); label(\\\"B\\\", B, dir(point--B)); label(\\\"C\\\", C, dir(point--C)); label(\\\"D\\\", D, dir(point--D)); label(\\\"E\\\", E, dir(point--E)); [/asy]$ Solution Problem 28 $[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label(\\\"A\\\", A, NE); label(\\\"B\\\", B, NW); label(\\\"C\\\", C, S); label(\\\"D\\\", D, E); label(\\\"4\\\", (4,2,0), SW); label(\\\"4\\\", (2,4,0), SE); label(\\\"3\\\", (0, 4, 1.5), E); [/asy]$ Solution\",\n \"# Tagged: transpose ## Problem 713 Determine bases for $\\\\calN(A)$ and $\\\\calN(A^{T}A)$ when $A= \\\\begin{bmatrix} 1 & 2 & 1 \\\\\\\\ 1 & 1 & 3 \\\\\\\\ 0 & 0 & 0 \\\\end{bmatrix} .$ Then, determine the ranks and nullities of the matrices $A$ and $A^{\\\\trans}A$. ## Problem 640 Let $\\\\mathbf{v}$ be an $n \\\\times 1$ column vector. Prove that $\\\\mathbf{v} \\\\mathbf{v}^\\\\trans$ is a symmetric matrix. ## Problem 639 Let $\\\\mathbf{v}$ be an $n \\\\times 1$ column vector. Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} = 0$ if and only if $\\\\mathbf{v}$ is the zero vector $\\\\mathbf{0}$. ## Problem 638 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. Prove that $\\\\tr ( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans ) = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}$. ## Problem 637 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. (a) Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{w} = \\\\mathbf{w}^\\\\trans \\\\mathbf{v}$. (b) Provide an example to show that $\\\\mathbf{v} \\\\mathbf{w}^\\\\trans$ is not always equal to $\\\\mathbf{w} \\\\mathbf{v}^\\\\trans$. ## Problem 636 Calculate the following expressions, using the following matrices: $A = \\\\begin{bmatrix} 2 & 3 \\\\\\\\ -5 & 1 \\\\end{bmatrix}, \\\\qquad B = \\\\begin{bmatrix} 0 & -1 \\\\\\\\ 1 & -1 \\\\end{bmatrix}, \\\\qquad \\\\mathbf{v} = \\\\begin{bmatrix} 2 \\\\\\\\ -4 \\\\end{bmatrix}$ (a) $A B^\\\\trans + \\\\mathbf{v} \\\\mathbf{v}^\\\\trans$. (b) $A \\\\mathbf{v} – 2 \\\\mathbf{v}$. (c) $\\\\mathbf{v}^{\\\\trans} B$. (d) $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} + \\\\mathbf{v}^\\\\trans\",\n \"pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } for(int i = 0; i < 2; ++i) /* left */ fillrect((s/2^(ceil(i/2)),s/2^(floor(i/2)))); for(int i = 0; i < n; ++i) /* right */ fillrect(shiftR,shiftR + (s/2^(ceil(i/2)),s/2^(floor(i/2)))); label(\\\"\\\\frac 12\\\",(s*3/4,s/2),sm); label(\\\"\\\\cdots\\\",(s*1/4,s/2),sm); label(\\\"\\\\frac 12\\\",shiftR+(s*3/4,s/2),sm); label(\\\"\\\\cdots\\\",shiftR+(s*1/4,s/2),sm); label(\\\"\\\\frac 14\\\",shiftR+(s*1/4,s*3/4),sm); label(\\\"\\\\frac 18\\\",shiftR+(s*3/8,s/4),sm); htick((0,-1), (s,-1)); htick(shiftR + (0,-1), shiftR + (s,-1)); label(\\\"1\\\",(s/2,-1),S,sm); label(\\\"1\\\",shiftR+(s/2,-1),S,sm); [/asy]$ The infinite geometric series $\\\\frac 12 + \\\\frac {1}{2^2} + \\\\frac {1}{2^3} + \\\\cdots = 1$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 4; real h = 2; pen colors[] = {rgb(0.8,0,0),rgb(0,0.8,0)}; void drawTriGrid(real s){ for(int i = 0; i < 4; ++i){ draw( (-s*3/2,s*(3/2 - i)) -- (s*3/2,s*(3/2 - i)), linetype(\\\"2 2\\\")); draw( (s*(3/2 - i),-s*3/2) -- (s*(3/2 - i),s*3/2), linetype(\\\"2 2\\\")); } } void fillrect(pair A, pair B, pen p){ filldraw(A--(A.x,B.y)--B--(B.x,A.y)--cycle, p, linewidth(1)); } for(int i = 0; i < n; ++i) { fillrect( ((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) , ((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[0]); fillrect(-((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) ,-((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[1]); fillrect( (-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) , (h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[0]); fillrect(-(-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) ,-(h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[1]); drawTriGrid(h/3^i); } [/asy]$ The infinite geometric series $\\\\frac 13 + \\\\frac {1}{3^2} + \\\\frac {1}{3^3} + \\\\cdots = \\\\frac 12$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 10; real h = 6; pen colors[] = {rgb(0.9,0,0),rgb(0,0.9,0),rgb(0,0,0.9)}; pair shiftR = (h+3,0); void drawEquilaterals(pair A, real s){ filldraw(A--A+s*expi(2*pi/3)--A+(-s,0)--cycle,colors[0]); filldraw(A--A+s*expi(2*pi/3)--A+s*expi(1*pi/3)--cycle,colors[1]); filldraw(A--A+s*expi(1*pi/3)--A+(s,0)--cycle,colors[2]); } for(int i = 0; i < n; ++i) drawEquilaterals(shiftR + (0,h-h/(2^i)\",\n \"x_5 \\\\\\\\ 0 \\\\\\\\ -x_5 \\\\\\\\ x_5 \\\\end{bmatrix}=x_3\\\\begin{bmatrix} 1 \\\\\\\\ -2 \\\\\\\\ 1 \\\\\\\\ 0 \\\\\\\\ 0 \\\\end{bmatrix}+x_5\\\\begin{bmatrix} 2 \\\\\\\\ 1 \\\\\\\\ 0 \\\\\\\\ -1\\\\\\\\ 1 \\\\end{bmatrix}. \\\\end{align*} Therefore, the vector form for the general solution is \\\\[\\\\mathbf{x}=x_3\\\\begin{bmatrix} 1 \\\\\\\\ -2 \\\\\\\\ 1 \\\\\\\\ 0 \\\\\\\\ 0 \\\\end{bmatrix}+x_5\\\\begin{bmatrix} 2 \\\\\\\\ 1 \\\\\\\\ 0 \\\\\\\\ -1\\\\\\\\ 1 \\\\end{bmatrix}. (b) Matrix product and transpose First of all, note that by the property of the transpose we have $(A-B)^{\\\\trans}=A^{\\\\trans}-B^{\\\\trans}.$ Hence we can simply the middle part: \\\\begin{align*} A^{\\\\trans}-(A-B)^{\\\\trans}&= A^{\\\\trans}-(A^{\\\\trans}-B^{\\\\trans})\\\\\\\\ &=A^{\\\\trans}-A^{\\\\trans}+B^{\\\\trans}=B^{\\\\trans}. \\\\end{align*} Thus the expression becomes $\\\\mathbf{v}^{\\\\trans}\\\\left( A^{\\\\trans}-(A-B)^{\\\\trans}\\\\right)C=\\\\mathbf{v}^{\\\\trans}B^{\\\\trans}C.$ The transposes of $\\\\mathbf{v}$ and $B$ are $\\\\mathbf{v}^{\\\\trans}=\\\\begin{bmatrix} 0 & 1 & 0 \\\\\\\\ \\\\end{bmatrix} \\\\text{ and } B^{\\\\trans}=\\\\begin{bmatrix} 1 & 0 \\\\\\\\ 0 & 1 \\\\\\\\ 1 &0 \\\\end{bmatrix}.$ Thus we have \\\\begin{align*} &\\\\mathbf{v}^{\\\\trans}\\\\left( A^{\\\\trans}-(A-B)^{\\\\trans}\\\\right)C=\\\\mathbf{v}^{\\\\trans}B^{\\\\trans}C\\\\\\\\ &=\\\\begin{bmatrix} 0 & 1 & 0 \\\\\\\\ \\\\end{bmatrix} \\\\begin{bmatrix} 1 & 0 \\\\\\\\ 0 & 1 \\\\\\\\ 1 &0 \\\\end{bmatrix} \\\\begin{bmatrix} 1 & 2\\\\\\\\ 0& 6 \\\\end{bmatrix}\\\\\\\\ &=\\\\begin{bmatrix} 0 & 1 \\\\end{bmatrix}\\\\begin{bmatrix} 1 & 2\\\\\\\\ 0& 6 \\\\end{bmatrix} =\\\\begin{bmatrix} 0 & 6 \\\\end{bmatrix}. \\\\end{align*} In conclusion, we have obtained $\\\\mathbf{v}^{\\\\trans}\\\\left( A^{\\\\trans}-(A-B)^{\\\\trans}\\\\right)C=\\\\begin{bmatrix} 0 & 6 \\\\end{bmatrix}.$ Comment. These are Quiz 2 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017. List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017 There\",\n \"y} \\\\qquad \\\\text{and} \\\\qquad \\\\alpha A {\\\\mathbf x} = A ( \\\\alpha {\\\\mathbf x}), \\\\end{equation*} where \\\\begin{equation*} {\\\\mathbf x} = \\\\begin{pmatrix} x_1 \\\\\\\\ x_2 \\\\\\\\ \\\\vdots \\\\\\\\ x_n \\\\end{pmatrix}. \\\\end{equation*} We will often abbreviate the matrix $A$ by writing $(a_{ij})\\\\text{.}$ Conversely, if $T : {\\\\mathbb R}^n \\\\rightarrow {\\\\mathbb R}^m$ is a linear map, we can associate a matrix $A$ with $T$ by considering what $T$ does to the vectors \\\\begin{align*} {\\\\mathbf e}_1 & = (1, 0, \\\\ldots, 0)^\\\\transpose\\\\\\\\ {\\\\mathbf e}_2 & = (0, 1, \\\\ldots, 0)^\\\\transpose\\\\\\\\ & \\\\vdots & \\\\\\\\ {\\\\mathbf e}_n & = (0, 0, \\\\ldots, 1)^\\\\transpose. \\\\end{align*} We can write any vector ${\\\\mathbf x} = (x_1, \\\\ldots, x_n)^\\\\transpose$ as \\\\begin{equation*} x_1 {\\\\mathbf e}_1 + x_2 {\\\\mathbf e}_2 + \\\\cdots + x_n {\\\\mathbf e}_n. \\\\end{equation*} Consequently, if \\\\begin{align*} T({\\\\mathbf e}_1) & = (a_{11}, a_{21}, \\\\ldots, a_{m1})^\\\\transpose,\\\\\\\\ T({\\\\mathbf e}_2) & = (a_{12}, a_{22}, \\\\ldots, a_{m2})^\\\\transpose,\\\\\\\\ & \\\\vdots & \\\\\\\\ T({\\\\mathbf e}_n) & = (a_{1n}, a_{2n}, \\\\ldots, a_{mn})^\\\\transpose, \\\\end{align*} then \\\\begin{align*} T({\\\\mathbf x} ) & = T(x_1 {\\\\mathbf e}_1 + x_2 {\\\\mathbf e}_2 + \\\\cdots + x_n {\\\\mathbf e}_n)\\\\\\\\ & = x_1 T({\\\\mathbf e}_1) + x_2 T({\\\\mathbf e}_2) + \\\\cdots + x_n T({\\\\mathbf e}_n)\\\\\\\\ & = \\\\left( \\\\sum_{k=1}^{n} a_{1k} x_k, \\\\ldots, \\\\sum_{k=1}^{n} a_{mk} x_k \\\\right)^\\\\transpose\\\\\\\\ & = A {\\\\mathbf x}. \\\\end{align*} ###### Example12.1 If we let $T : {\\\\mathbb\",\n \"\\\\mathbf{e}_{49}$, but we can see that the 49 encoding vectors are divided into 7 rows of 7 vectors each. Now the idea is to use only one row of 7 encoding vectors and expand those vectors to the other 6 rows, essentially reusing their values. Because it is discouraged to have the same values for different encoding vectors, each vector of dimension (a.k.a height) config.hidden_size=4 is cut into the lower encoding vector $\\\\mathbf{e}_\\\\text{down}$ of size $1$ and $\\\\mathbf{e}_\\\\text{up}$ of size $3$, so that the lower part can be expanded along the row dimension and the upper part can be expanded along the column dimension. Let's visualize for more clarity. We can see that we have cut the embedding vectors into $\\\\mathbf{e}_\\\\text{down}$ (in blue) and $\\\\mathbf{e}_\\\\text{up}$ (in yellow). Now for the \\\"sub\\\"-vectors $\\\\mathbf{E}_\\\\text{down} = \\\\left[\\\\mathbf{e}_{\\\\text{down},1}, \\\\ldots, \\\\mathbf{e}_{\\\\text{down},49}\\\\right]$ only the first row, a.k.a. the width in the graphic, of $7$ is kept and expanded along the column dimension, a.k.a. the depth of the graphic. Inversely, for the \\\"sub\\\"-vectors $\\\\mathbf{E}_\\\\text{up} = \\\\left[\\\\mathbf{e}_{\\\\text{up},1}, \\\\ldots, \\\\mathbf{e}_{\\\\text{up},49}\\\\right]$ only the first column of $7$ is kept and expanded along the row dimension. The resulting embedding vectors $\\\\mathbf{e'}_{i}$ then correspond to $\\\\mathbf{e'}_{i} = \\\\left[ \\\\left[\\\\mathbf{e}_{\\\\text{down, } i \\\\% n_\\\\text{max}^1}\\\\right]^T, \\\\left[\\\\mathbf{e}_{\\\\text{up, } \\\\left \\\\lfloor{\\\\frac{i}{{n}^2_{\\\\text{max}}}}\\\\right \\\\rfloor} \\\\right]^T \\\\right]^T$ whereas $n_\\\\text{max}^1 = 7$ and $n_\\\\text{max}^2 = 7$ in our\",\n \"var Aa = 0; //Do not modify any of the following code. var bAng = Ba*Math.PI/180; var aAng = Aa*Math.PI/180; //Compute horizontal and vertical components // for each vector. var Bx = Bm*Math.cos(bAng); var By = Bm*Math.sin(bAng); var Ax = Am*Math.cos(aAng); var Ay = Am*Math.sin(aAng); //Compute sum of vectors var Cx = Bx + Ax; var Cy = By + Ay; var Ca = getAngle(Cx,Cy); var Cm = Math.sqrt(Cx*Cx + Cy*Cy); //Compute difference between vectors var Dx = Bx - Ax; var Dy = By - Ay; var Da = getAngle(Dx,Dy); var Dm = Math.sqrt(Dx*Dx + Dy*Dy); document.write(\\\"Bm = \\\" + Bm.toFixed(2) + \\\"
\\\"); document.write(\\\"Ba = \\\" + Ba.toFixed(2) + \\\" deg
\\\"); document.write(\\\"Am = \\\" + Am.toFixed(2) + \\\"
\\\"); document.write(\\\"Aa = \\\" + Aa.toFixed(2) + \\\" deg
\\\"); document.write(\\\"Bx = \\\" + Bx.toFixed(2) + \\\"
\\\"); document.write(\\\"By = \\\" + By.toFixed(2) + \\\"
\\\"); document.write(\\\"Ax = \\\" + Ax.toFixed(2) + \\\"
\\\"); document.write(\\\"Ay = \\\" + Ay.toFixed(2) + \\\"
\\\"); document.write(\\\"Cx = \\\" + Cx.toFixed(2) + \\\"
\\\"); document.write(\\\"Cy = \\\" + Cy.toFixed(2) + \\\"
\\\"); document.write(\\\"Ca = \\\" + Ca.toFixed(2) + \\\" deg
\\\"); document.write(\\\"Cm = \\\" + Cm.toFixed(2) + \\\"
\\\"); document.write(\\\"Da = \\\" + Da.toFixed(2) + \\\" deg
\\\"); document.write(\\\"Dm = \\\" + Dm.toFixed(2) + \\\"
\\\"); document.write(\\\"End Script\\\"); The one unique thing The only thing that is unique about Listing 1 is that\",\n \"2\\\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\\\"(2a,0)\\\",A,NE); dot(B); label(\\\"(2b_1,2b_2)\\\",B,N); dot((5,4)); label(\\\"(a+b_1,b_2)\\\",(5,4),NE); dot((9,7)); label(\\\"(x,y)\\\",(9,7),NE); dot((2,4)); label(\\\"(2b_1,b_2)\\\",(2,4),SW); dot((9,4)); label(\\\"(x,b_2)\\\",(9,4),SE); [/asy]$ Note that since $K$ is the center of square $AB_1A_2B$, $BX \\\\perp KX$ and $BX = KX$. Additionally, $BY \\\\parallel KZ$ and $BY \\\\perp YZ$. $YZ$ is a line, so $\\\\angle BXY + \\\\angle KXZ + \\\\angle BXK = 180^\\\\circ$. Since $BX \\\\perp KX$, $\\\\angle BXK = 90^\\\\circ$, so $\\\\angle BXY + \\\\angle KXZ = 90^\\\\circ$. Additionally, because $BXY$ is a right triangle, $\\\\angle YBX + \\\\angle YXB = 90^\\\\circ$. Rearranging and substituting results in $\\\\angle KXZ = \\\\angle YBX$. Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\\\\triangle BYX \\\\cong \\\\triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$. $[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\\\"2 2\\\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label(\\\"(2a,0)\\\",A,NE); dot(B); label(\\\"(2b_1,2b_2)\\\",B,NE); dot(C); label(\\\"(-2c,0)\\\",C,NW); dot(D); label(\\\"(2d_1,-2d_2)\\\",D,S); dot(K); label(\\\"K\\\",K,NE); dot(L); label(\\\"L\\\",L,NW); dot(M); label(\\\"M\\\",M,SW); dot(N); label(\\\"N\\\",N,SE); [/asy]$ By\"\n]"},"ps_retrievals":{"kind":"list like","value":["not, give a counterexample. 14. Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. (a) Prove that $\\mathbf{v}^\\trans \\mathbf{w} = \\mathbf{w}^\\trans \\mathbf{v}$. (b) Provide an example to show that $\\mathbf{v} \\mathbf{w}^\\trans$ is not always equal to $\\mathbf{w} \\mathbf{v}^\\trans$. 15. Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. Prove that $\\tr ( \\mathbf{v} \\mathbf{w}^\\trans ) = \\mathbf{v}^\\trans \\mathbf{w}$. 16. Let $\\mathbf{v}$ be an $n \\times 1$ column vector. Prove that $\\mathbf{v}^\\trans \\mathbf{v} = 0$ if and only if $\\mathbf{v}$ is the zero vector $\\mathbf{0}$. 17. Is it true that a real square matrix $A$ must commute with its transpose $A^{\\trans}$? 18. Let $\\mathbf{v}$ be an $n \\times 1$ column vector. Prove that $\\mathbf{v} \\mathbf{v}^\\trans$ is a symmetric matrix. 19. Calculate the following expressions, using the following matrices: $A = \\begin{bmatrix} 2 & 3 \\\\ -5 & 1 \\end{bmatrix}, \\qquad B = \\begin{bmatrix} 0 & -1 \\\\ 1 & -1 \\end{bmatrix}, \\qquad \\mathbf{v} = \\begin{bmatrix} 2 \\\\ -4 \\end{bmatrix}$ (a) $A B^\\trans + \\mathbf{v} \\mathbf{v}^\\trans$. (b) $A \\mathbf{v} – 2 \\mathbf{v}$. (c) $\\mathbf{v}^{\\trans} B$. (d) $\\mathbf{v}^\\trans \\mathbf{v} + \\mathbf{v}^\\trans B A^\\trans \\mathbf{v}$. 20. Let $A$ and $B$ be $n \\times n$ matrices, and $\\mathbf{v}$ an $n \\times 1$ column vector. Use the matrix components to prove that $(A + B) \\mathbf{v} = A\\mathbf{v} + B\\mathbf{v}$.","## Problem 639 Let $\\mathbf{v}$ be an $n \\times 1$ column vector. Prove that $\\mathbf{v}^\\trans \\mathbf{v} = 0$ if and only if $\\mathbf{v}$ is the zero vector $\\mathbf{0}$. ## Problem 638 Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. Prove that $\\tr ( \\mathbf{v} \\mathbf{w}^\\trans ) = \\mathbf{v}^\\trans \\mathbf{w}$. ## Problem 637 Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. (a) Prove that $\\mathbf{v}^\\trans \\mathbf{w} = \\mathbf{w}^\\trans \\mathbf{v}$. (b) Provide an example to show that $\\mathbf{v} \\mathbf{w}^\\trans$ is not always equal to $\\mathbf{w} \\mathbf{v}^\\trans$. ## Problem 636 Calculate the following expressions, using the following matrices: $A = \\begin{bmatrix} 2 & 3 \\\\ -5 & 1 \\end{bmatrix}, \\qquad B = \\begin{bmatrix} 0 & -1 \\\\ 1 & -1 \\end{bmatrix}, \\qquad \\mathbf{v} = \\begin{bmatrix} 2 \\\\ -4 \\end{bmatrix}$ (a) $A B^\\trans + \\mathbf{v} \\mathbf{v}^\\trans$. (b) $A \\mathbf{v} – 2 \\mathbf{v}$. (c) $\\mathbf{v}^{\\trans} B$. (d) $\\mathbf{v}^\\trans \\mathbf{v} + \\mathbf{v}^\\trans B A^\\trans \\mathbf{v}$. ## Problem 635 Let $A$ and $B$ be $n \\times n$ matrices, and $\\mathbf{v}$ an $n \\times 1$ column vector. Use the matrix components to prove that $(A + B) \\mathbf{v} = A\\mathbf{v} + B\\mathbf{v}$. ## Problem 634 Let $A$ and $B$ be $n \\times n$ matrices. Is it always true that $\\tr (A B) = \\tr (A) \\tr (B)$? If it is","then $AB$ is a skew-symmetric matrix. We calculate \\begin{align*} (AB)^{\\trans}&=B^{\\trans}A^{\\trans}=(-B)(-A)\\\\ &=BA=-AB, \\end{align*} where the last step follows from the assumption $AB=-BA$. This proves that $AB$ is skew-symmetric. ### (f) Let $\\mathbf{v}$ be an $n$-dimensional column vecotor. Prove that $\\mathbf{v}^{\\trans}A\\mathbf{v}=0$. Observe that $\\mathbf{v}^{\\trans}A\\mathbf{v}$ is a $1\\times 1$ matrix, or just a number. So we have \\begin{align*} \\mathbf{v}^{\\trans}A\\mathbf{v}&=(\\mathbf{v}^{\\trans}A\\mathbf{v})^{\\trans}=\\mathbf{v}^{\\trans}A^{\\trans}(\\mathbf{v}^{\\trans})^{\\trans}\\\\ &=\\mathbf{v}^{\\trans}A^{\\trans}\\mathbf{v}=\\mathbf{v}^{\\trans}(-A)\\mathbf{v}=-(\\mathbf{v}^{\\trans}A\\mathbf{v}). \\end{align*} This yields that $2\\mathbf{v}^{\\trans}A\\mathbf{v}=0$, and hence $\\mathbf{v}^{\\trans}A\\mathbf{v}=0$. ### (g) Suppose that $A$ is a real skew-symmetric matrix and $A^2\\mathbf{v}=\\mathbf{0}$ for some vector $\\mathbf{v}\\in \\R^n$. Then prove that $A\\mathbf{v}=\\mathbf{0}$. Let us compute the length of the vector $A\\mathbf{v}$. We have \\begin{align*} \\|A\\mathbf{v}\\|&=(A\\mathbf{v})^{\\trans}(A\\mathbf{v})=\\mathbf{v}^{\\trans}A^{\\trans}A\\mathbf{v}\\\\ &=\\mathbf{v}^{\\trans}(-A)A\\mathbf{v}=-\\mathbf{v}^{\\trans}A^2\\mathbf{v}\\\\ &=-\\mathbf{v}\\mathbf{0} &&\\text{by assumption}\\\\ &=0. \\end{align*} Since the length $\\|A\\mathbf{v}\\|=0$, we conclude that $A\\mathbf{v}=\\mathbf{0}$. ### More from my site • Express a Hermitian Matrix as a Sum of Real Symmetric Matrix and a Real Skew-Symmetric Matrix Recall that a complex matrix is called Hermitian if $A^*=A$, where $A^*=\\bar{A}^{\\trans}$. Prove that every Hermitian matrix $A$ can be written as the sum $A=B+iC,$ where $B$ is a real symmetric matrix and $C$ is a real skew-symmetric matrix. Proof. Since […] • Eigenvalues of a Hermitian Matrix are Real Numbers Show that eigenvalues of a Hermitian matrix $A$ are real numbers. (The Ohio State University Linear Algebra Exam Problem) We give two proofs. These two proofs are essentially the same. The second","# Tagged: transpose of a matrix ## Problem 638 Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. Prove that $\\tr ( \\mathbf{v} \\mathbf{w}^\\trans ) = \\mathbf{v}^\\trans \\mathbf{w}$. ## Problem 637 Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. (a) Prove that $\\mathbf{v}^\\trans \\mathbf{w} = \\mathbf{w}^\\trans \\mathbf{v}$. (b) Provide an example to show that $\\mathbf{v} \\mathbf{w}^\\trans$ is not always equal to $\\mathbf{w} \\mathbf{v}^\\trans$. ## Problem 636 Calculate the following expressions, using the following matrices: $A = \\begin{bmatrix} 2 & 3 \\\\ -5 & 1 \\end{bmatrix}, \\qquad B = \\begin{bmatrix} 0 & -1 \\\\ 1 & -1 \\end{bmatrix}, \\qquad \\mathbf{v} = \\begin{bmatrix} 2 \\\\ -4 \\end{bmatrix}$ (a) $A B^\\trans + \\mathbf{v} \\mathbf{v}^\\trans$. (b) $A \\mathbf{v} – 2 \\mathbf{v}$. (c) $\\mathbf{v}^{\\trans} B$. (d) $\\mathbf{v}^\\trans \\mathbf{v} + \\mathbf{v}^\\trans B A^\\trans \\mathbf{v}$. ## Problem 633 Let $A$ be an $n \\times n$ matrix. Is it true that $\\tr ( A^\\trans ) = \\tr(A)$? If it is true, prove it. If not, give a counterexample.","# Tagged: transpose ## Problem 713 Determine bases for $\\calN(A)$ and $\\calN(A^{T}A)$ when $A= \\begin{bmatrix} 1 & 2 & 1 \\\\ 1 & 1 & 3 \\\\ 0 & 0 & 0 \\end{bmatrix} .$ Then, determine the ranks and nullities of the matrices $A$ and $A^{\\trans}A$. ## Problem 640 Let $\\mathbf{v}$ be an $n \\times 1$ column vector. Prove that $\\mathbf{v} \\mathbf{v}^\\trans$ is a symmetric matrix. ## Problem 639 Let $\\mathbf{v}$ be an $n \\times 1$ column vector. Prove that $\\mathbf{v}^\\trans \\mathbf{v} = 0$ if and only if $\\mathbf{v}$ is the zero vector $\\mathbf{0}$. ## Problem 638 Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. Prove that $\\tr ( \\mathbf{v} \\mathbf{w}^\\trans ) = \\mathbf{v}^\\trans \\mathbf{w}$. ## Problem 637 Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. (a) Prove that $\\mathbf{v}^\\trans \\mathbf{w} = \\mathbf{w}^\\trans \\mathbf{v}$. (b) Provide an example to show that $\\mathbf{v} \\mathbf{w}^\\trans$ is not always equal to $\\mathbf{w} \\mathbf{v}^\\trans$. ## Problem 636 Calculate the following expressions, using the following matrices: $A = \\begin{bmatrix} 2 & 3 \\\\ -5 & 1 \\end{bmatrix}, \\qquad B = \\begin{bmatrix} 0 & -1 \\\\ 1 & -1 \\end{bmatrix}, \\qquad \\mathbf{v} = \\begin{bmatrix} 2 \\\\ -4 \\end{bmatrix}$ (a) $A B^\\trans + \\mathbf{v} \\mathbf{v}^\\trans$. (b) $A \\mathbf{v} – 2 \\mathbf{v}$. (c) $\\mathbf{v}^{\\trans} B$. (d) $\\mathbf{v}^\\trans \\mathbf{v} + \\mathbf{v}^\\trans","}{\\mathbf{transa}}=\\text{\"C\"}$: ${\\mathbf{m}}$; otherwise: $0$. On entry: the matrix $A$; $A$ is $m×k$ if ${\\mathbf{transa}}=\\text{'N'}$, or $k×m$ if ${\\mathbf{transa}}=\\text{'T'}$ or $\\text{'C'}$. 5: $\\mathbf{b}\\left(:,:\\right)$double array Input Note: the first dimension of b is given by: if ${\\mathbf{transb}}=\\text{\"N\"}$: ${\\mathbf{k}}$; if ${\\mathbf{transb}}=\\text{\"T\"}\\text{ or }{\\mathbf{transb}}=\\text{\"C\"}$: ${\\mathbf{n}}$; otherwise: $0$. Note: the second dimension of b is given by: if ${\\mathbf{transb}}=\\text{\"N\"}$: ${\\mathbf{n}}$; if ${\\mathbf{transb}}=\\text{\"T\"}\\text{ or }{\\mathbf{transb}}=\\text{\"C\"}$: ${\\mathbf{k}}$; otherwise: $0$. On entry: the matrix $B$; $B$ is $k×n$ if ${\\mathbf{transb}}=\\text{'N'}$, or $n×k$ if ${\\mathbf{transb}}=\\text{'T'}$ or $\\text{'C'}$. 6: $\\mathbf{beta}$double Input On entry: the scalar $\\beta$. 7: $\\mathbf{c}\\left({\\mathbf{m}},{\\mathbf{n}}\\right)$double array Input/Output On entry: the $m×n$ matrix $C$. If ${\\mathbf{beta}}=0.0$, c need not be set. On exit: the updated matrix $C$. 8: $\\mathbf{opt}$OptionalF06YA Input/Output Optional parameter container, derived from Optional. 1: $\\mathbf{m}$ $m$, the number of rows of the matrix $C$; the number of rows of $A$ if ${\\mathbf{transa}}=\\text{\"N\"}$, or the number of columns of $A$ if ${\\mathbf{transa}}=\\text{\"T\"}\\text{​ or ​}\\text{\"C\"}$ 2: $\\mathbf{n}$ $n$, the number of columns of the matrix $C$; the number of columns of $B$ if ${\\mathbf{transb}}=\\text{\"N\"}$, or the number of rows of $B$ if ${\\mathbf{transb}}=\\text{\"T\"}\\text{​ or ​}\\text{\"C\"}$ 3: $\\mathbf{k}$ $k$, the number of columns of $A$ if ${\\mathbf{transa}}=\\text{\"N\"}$, or the number of rows of $A$ if ${\\mathbf{transa}}=\\text{\"T\"}\\text{​ or ​}\\text{\"C\"}$; the number of rows of $B$ if ${\\mathbf{transb}}=\\text{\"N\"}$, or the number of columns of $B$ if ${\\mathbf{transb}}=\\text{\"T\"}\\text{​ or","while $\\quad \\mathbf{w} \\mathbf{v}^\\trans = \\begin{bmatrix} 0 \\\\ 1 \\end{bmatrix} \\begin{bmatrix} 1 & 0 \\end{bmatrix} = \\begin{bmatrix} 0 & 0 \\\\ 1 & 0 \\end{bmatrix}.$ ## Comment. Recall that for two vectors $\\mathbf{v}, \\mathbf{w} \\in \\R^n$, the dot product (or inner product) of $\\mathbf{v}, \\mathbf{w}$ is defined to be $\\mathbf{v}\\cdot \\mathbf{w}:=\\mathbf{v}^{\\trans} \\mathbf{w}.$ Part (a) of the problem deduces that the dot product is commutative. This means that we have $\\mathbf{v}\\cdot \\mathbf{w}= \\mathbf{w} \\cdot \\mathbf{v}.$ In fact, we have \\begin{align*} \\mathbf{v}\\cdot \\mathbf{w}= \\mathbf{v}^\\trans \\mathbf{w} \\stackrel{\\text{(a)}}{=} \\mathbf{w}^\\trans \\mathbf{v} \\mathbf{w} \\cdot \\mathbf{v}. \\end{align*} Also, notice that while $\\mathbf{v} \\mathbf{w}^\\trans$ is not always equal to $\\mathbf{w} \\mathbf{v}^\\trans$, we know that $(\\mathbf{v} \\mathbf{w}^\\trans)^\\trans = \\mathbf{w} \\mathbf{v}^\\trans$. ### More from my site • A Relation between the Dot Product and the Trace Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. Prove that $\\tr ( \\mathbf{v} \\mathbf{w}^\\trans ) = \\mathbf{v}^\\trans \\mathbf{w}$. Solution. Suppose the vectors have components $\\mathbf{v} = \\begin{bmatrix} v_1 \\\\ v_2 \\\\ \\vdots \\\\ v_n […] • Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given Let \\mathbf{a} and \\mathbf{b} be vectors in \\R^n such that their length are \\[\\|\\mathbf{a}\\|=\\|\\mathbf{b}\\|=1$ and the inner product $\\mathbf{a}\\cdot \\mathbf{b}=\\mathbf{a}^{\\trans}\\mathbf{b}=-\\frac{1}{2}.$ Then determine the length $\\|\\mathbf{a}-\\mathbf{b}\\|$. (Note […] • Rotation Matrix in Space and its Determinant and Eigenvalues For","[/asy]$ $\\textbf{(D)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(0,0)*cat); dot((0,1.5)); [/asy]$ $\\textbf{(E)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(0,1)*cat); dot((1.5,0)); [/asy]$ ## Problem 24 A ship travels from point A to point B along a semicircular path, centered at Island X. Then it travels along a straight path from B to C. Which of these graphs best shows the ship's distance from Island X as it moves along its course? $[asy]size(150); pair X=origin, A=(-5,0), B=(5,0), C=(0,5); draw(Arc(X, 5, 180, 360)^^B--C); dot(X); label(\"X\", X, NE); label(\"C\", C, N); label(\"B\", B, E); label(\"A\", A, W);[/asy]$ $\\textbf{(A)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\"distance traveled\", (8,0), S); label(rotate(90)*\"distance to X\", (0,8), W); draw(Arc((4,10), 4, 0, 180)^^(8,10)--(16,12)); [/asy]$ $\\textbf{(B)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\"distance traveled\", (8,0), S); label(rotate(90)*\"distance to X\", (0,8), W); draw(Arc((12,10), 4, 180, 360)^^(0,10)--(8,10)); [/asy]$ $\\textbf{(C)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\"distance traveled\", (8,0), S); label(rotate(90)*\"distance to X\", (0,8), W); draw((0,8)--(10,10)--(16,8)); [/asy]$ $\\textbf{(D)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\"distance traveled\", (8,0), S); label(rotate(90)*\"distance to X\", (0,8), W); draw(Arc((12,10), 4, 0, 180)^^(0,10)--(8,10)); [/asy]$ $\\textbf{(E)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\"distance traveled\", (8,0), S); label(rotate(90)*\"distance to X\", (0,8), W); draw((0,6)--(6,6)--(16,10)); [/asy]$ ## Problem 25 In the figure, the area of square WXYZ is $25 \\text{cm}^2$. The four smaller squares have sides 1 cm long,","# Difference between revisions of \"2005 AMC 8 Problems/Problem 13\" ## Problem The area of polygon $ABCDEF$ is 52 with $AB=8$, $BC=9$ and $FA=5$. What is $DE+EF$? $[asy] pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4); draw(a--b--c--d--e--f--cycle); draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a); draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b); draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c); draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d); draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f); label(\"A\", a, NW); label(\"B\", b, NE); label(\"C\", c, SE); label(\"D\", d, SW); label(\"E\", e, SW); label(\"F\", f, SW); label(\"5\", (0,6.5), W); label(\"8\", (4,9), N); label(\"9\", (8, 4.5), E); [/asy]$ $\\textbf{(A)}\\ 7\\qquad\\textbf{(B)}\\ 8\\qquad\\textbf{(C)}\\ 9\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 11$ ## Solution Notice that $AF + DE = BC$, so $DE=4$. Let $O$ be the intersection of the extensions of $AF$ and $DC$, which makes rectangle $ABCO$. The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$. $$\\text{Area} = 52 = 8 \\cdot 9- EF \\cdot 4$$ Solving for the unknown, $EF=5$, therefore $DE+EF=4+5=\\boxed{\\textbf{(C)}\\ 9}$.","+0 # Consider parallelogram $ABCD$ with points $S$ and $T$ chosen such that $CS:SD = BT:TC = 2,$as in the picture below: [asy] size(200); pair 0 167 1 Consider parallelogram ABCD with points S and T chosen such that $$CS:SD = BT:TC = 2$$ as in the picture below: Let $$\\overrightarrow{AB} = \\mathbf{v}$$ and $$\\overrightarrow{AD} = \\mathbf{w}$$. Then there exist constants r, s, t, u such that \\begin{align*} \\overrightarrow{AT} &= r \\mathbf{v} + s \\mathbf{w},\\\\ \\overrightarrow{BS} &= t \\mathbf{v} + u \\mathbf{w}. \\end{align*} Enter r, s, t, u in that order below. Apr 4, 2019","Find a Basis of the Subspace of All Vectors that are Perpendicular to the Columns of the Matrix Problem 40 Find a basis for the subspace $W$ of all vectors in $\\R^4$ which are perpendicular to the columns of the matrix $A=\\begin{bmatrix} 11 & 12 & 13 & 14 \\\\ 21 &22 & 23 & 24 \\\\ 31 & 32 & 33 & 34 \\\\ 41 & 42 & 43 & 44 \\end{bmatrix}.$ (Harvard University Exam) Contents Hint. 1. Show that $W=\\calN(A^{\\trans})$. 2. Find a basis of $\\calN(A^{\\trans})$ by reducing the matrix $A^{\\trans}$. Solution. Let us write $A=[A_1 \\, A_2 \\, A_3 \\, A_4]$, where $A_i$ is the $i$-th column vector of $A$ for $i=1,2,3,4$. First we claim that a vector $\\mathbf{x}\\in \\R^4$ is perpendicular to all column vectors $A_i$ if and only if $\\mathbf{x} \\in \\calN(A^{\\trans})$. To see this, we compute \\begin{align*} A^{\\trans} \\mathbf{x} =\\begin{bmatrix} A_1^{\\trans} \\\\ A_2^{\\trans} \\\\ A_3^{\\trans} \\\\ A_4^{\\trans} \\end{bmatrix}\\mathbf{x} =\\begin{bmatrix} A_1^{\\trans}\\mathbf{x} \\\\ A_2^{\\trans} \\mathbf{x}\\\\ A_3^{\\trans} \\mathbf{x}\\\\ A_4^{\\trans} \\mathbf{x} \\end{bmatrix}. \\end{align*} From this equality the claim follows immediately. So we proved that $\\calN(A^{\\trans}) =W$. From this, we see that $W$ is actually a subspace of $\\R^4$. Thus, we need to find a basis for the null space of the transpose $A^{\\trans}$. We apply elementary row operations to $A^{\\trans}$ and obtain a reduced row echelon","y} \\qquad \\text{and} \\qquad \\alpha A {\\mathbf x} = A ( \\alpha {\\mathbf x}), \\end{equation*} where \\begin{equation*} {\\mathbf x} = \\begin{pmatrix} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n \\end{pmatrix}. \\end{equation*} We will often abbreviate the matrix $A$ by writing $(a_{ij})\\text{.}$ Conversely, if $T : {\\mathbb R}^n \\rightarrow {\\mathbb R}^m$ is a linear map, we can associate a matrix $A$ with $T$ by considering what $T$ does to the vectors \\begin{align*} {\\mathbf e}_1 & = (1, 0, \\ldots, 0)^\\transpose\\\\ {\\mathbf e}_2 & = (0, 1, \\ldots, 0)^\\transpose\\\\ & \\vdots & \\\\ {\\mathbf e}_n & = (0, 0, \\ldots, 1)^\\transpose. \\end{align*} We can write any vector ${\\mathbf x} = (x_1, \\ldots, x_n)^\\transpose$ as \\begin{equation*} x_1 {\\mathbf e}_1 + x_2 {\\mathbf e}_2 + \\cdots + x_n {\\mathbf e}_n. \\end{equation*} Consequently, if \\begin{align*} T({\\mathbf e}_1) & = (a_{11}, a_{21}, \\ldots, a_{m1})^\\transpose,\\\\ T({\\mathbf e}_2) & = (a_{12}, a_{22}, \\ldots, a_{m2})^\\transpose,\\\\ & \\vdots & \\\\ T({\\mathbf e}_n) & = (a_{1n}, a_{2n}, \\ldots, a_{mn})^\\transpose, \\end{align*} then \\begin{align*} T({\\mathbf x} ) & = T(x_1 {\\mathbf e}_1 + x_2 {\\mathbf e}_2 + \\cdots + x_n {\\mathbf e}_n)\\\\ & = x_1 T({\\mathbf e}_1) + x_2 T({\\mathbf e}_2) + \\cdots + x_n T({\\mathbf e}_n)\\\\ & = \\left( \\sum_{k=1}^{n} a_{1k} x_k, \\ldots, \\sum_{k=1}^{n} a_{mk} x_k \\right)^\\transpose\\\\ & = A {\\mathbf x}. \\end{align*} ###### Example12.1 If we let $T : {\\mathbb","${\\mathbf{transa}}=\\text{\"T\"}$ or $\\text{\"C\"}$ The operation involves ${A}^{\\mathrm{T}}$. Constraint: ${\\mathbf{transa}}=\\text{\"N\"}$, $\\text{\"T\"}$ or $\\text{\"C\"}$. transb Type: System..::..String On entry: specifies whether the operation involves $B$ or ${B}^{\\mathrm{T}}$. ${\\mathbf{transb}}=\\text{\"N\"}$ The operation involves $B$. ${\\mathbf{transb}}=\\text{\"T\"}$ or $\\text{\"C\"}$ The operation involves ${B}^{\\mathrm{T}}$. Constraint: ${\\mathbf{transb}}=\\text{\"N\"}$, $\\text{\"T\"}$ or $\\text{\"C\"}$. m Type: System..::..Int32 On entry: $m$, the number of rows of the matrix $C$; the number of rows of $A$ if ${\\mathbf{transa}}=\\text{\"N\"}$, or the number of columns of $A$ if ${\\mathbf{transa}}=\\text{\"T\"}$ or $\\text{\"C\"}$. Constraint: ${\\mathbf{m}}\\ge 0$. n Type: System..::..Int32 On entry: $n$, the number of columns of the matrix $C$; the number of columns of $B$ if ${\\mathbf{transb}}=\\text{\"N\"}$, or the number of rows of $B$ if ${\\mathbf{transb}}=\\text{\"T\"}$ or $\\text{\"C\"}$. Constraint: ${\\mathbf{n}}\\ge 0$. k Type: System..::..Int32 On entry: $k$, the number of columns of $A$ if ${\\mathbf{transa}}=\\text{\"N\"}$, or the number of rows of $A$ if ${\\mathbf{transa}}=\\text{\"T\"}$ or $\\text{\"C\"}$; the number of rows of $B$ if ${\\mathbf{transb}}=\\text{\"N\"}$, or the number of columns of $B$ if ${\\mathbf{transb}}=\\text{\"T\"}$ or $\\text{\"C\"}$. Constraint: ${\\mathbf{k}}\\ge 0$. alpha Type: System..::..Double On entry: the scalar $\\alpha$. a Type: array[,](,)[,][,] An array of size [dim1, dim2] Note: dim1 must satisfy the constraint: • if ${\\mathbf{transa}}=\\text{\"N\"}$, $\\mathrm{dim1}\\ge \\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,{\\mathbf{m}}\\right)$; • if ${\\mathbf{transa}}=\\text{\"T\"}$ or $\\text{\"C\"}$, $\\mathrm{dim1}\\ge \\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,{\\mathbf{k}}\\right)$. Note: the second dimension of the array a must be at least $\\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,{\\mathbf{k}}\\right)$ if ${\\mathbf{transa}}=\\text{\"N\"}$ and at least $\\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,{\\mathbf{m}}\\right)$ if ${\\mathbf{transa}}=\\text{\"T\"}$ or","and software for solving the generalized Sylvester equation and estimating the separation between regular matrix pairs ACM Trans. Math. Software 22 78–103 ## 5 Parameters 1: TRANS – CHARACTER(1)Input On entry: if ${\\mathbf{TRANS}}=\\text{'N'}$, solve the generalized Sylvester equation (1). If ${\\mathbf{TRANS}}=\\text{'T'}$, solve the ‘transposed’ system (2). Constraint: ${\\mathbf{TRANS}}=\\text{'N'}$ or $\\text{'T'}$. 2: IJOB – INTEGERInput On entry: specifies what kind of functionality is to be performed when ${\\mathbf{TRANS}}=\\text{'N'}$. ${\\mathbf{IJOB}}=0$ Solve (1) only. ${\\mathbf{IJOB}}=1$ The functionality of ${\\mathbf{IJOB}}=0$ and $3$. ${\\mathbf{IJOB}}=2$ The functionality of ${\\mathbf{IJOB}}=0$ and $4$. ${\\mathbf{IJOB}}=3$ Only an estimate of $\\mathrm{Dif}\\left[\\left(A,D\\right),\\left(B,E\\right)\\right]$ is computed based on the Frobenius norm. ${\\mathbf{IJOB}}=4$ Only an estimate of $\\mathrm{Dif}\\left[\\left(A,D\\right),\\left(B,E\\right)\\right]$ is computed based on the $1$-norm. If ${\\mathbf{TRANS}}=\\text{'T'}$, IJOB is not referenced. Constraint: if ${\\mathbf{TRANS}}=\\text{'N'}$, $0\\le {\\mathbf{IJOB}}\\le 4$. 3: M – INTEGERInput On entry: $m$, the order of the matrices $A$ and $D$, and the row dimension of the matrices $C$, $F$, $R$ and $L$. Constraint: ${\\mathbf{M}}\\ge 0$. 4: N – INTEGERInput On entry: $n$, the order of the matrices $B$ and $E$, and the column dimension of the matrices $C$, $F$, $R$ and $L$. Constraint: ${\\mathbf{N}}\\ge 0$. 5: A(LDA,$*$) – REAL (KIND=nag_wp) arrayInput Note: the second dimension of the array A must be at least $\\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,{\\mathbf{M}}\\right)$. On entry: the upper quasi-triangular matrix $A$. 6: LDA – INTEGERInput On entry: the first dimension of","sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7)); /* draw(rightanglemark((0,0),a*expi(rot1),(c,0))); */ filldraw(sq1, rgb(0.95,1,0.95)); filldraw(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); label(\"a\",a/2*expi(rot1),SE,sm); label(\"b\",a/2*expi(rot1)+(c/2,0),SW,sm); label(\"c\",(c/2,-c),S,sm); [/asy]$ COMING: The last proof of the Pythagorean Theorem we shall present on this page, this one by dissection. $[asy] defaultpen(linewidth(0.7)+fontsize(10)); unitsize(15); real a = 3.6, b = 4.8, c = (a^2 + b^2)^.5; pair shiftR = (a+b+2,0); pen sm = fontsize(10), heavy = linewidth(1); void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = 0.5){ /* Arrow option resizes */ fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); } real s1 = 5, s2 = 7, s3 = 8; // triangle side lengths pen c1 = rgb(0.5,0.5,1), c2 = rgb(0.5,1,0.8), c3 = rgb(0.5,1,0.5); // color pens pair shiftR = (s1+1,0); // distance between two diagrams pair A=(0,0), B = (s1,0), C = intersectionpoints(Circle(A, s3), Circle(B, s2))[0], I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,A,C), F = foot(I,A,B); // draw left diagram filldraw(A--I--B--cycle, c1, heavy); filldraw(B--I--C--cycle, c2, heavy); filldraw(A--I--C--cycle, c3, heavy); dot(I); draw(incircle(A,B,C), heavy); draw(I--D, linetype(\"2 2\")); draw(I--E, linetype(\"2 2\")); draw(I--F, linetype(\"2 2\")); label(\"a\",(A+B)/2,S); label(\"b\",(B+C)/2,NE); label(\"c\",(A+C)/2,NW); label(\"r\",(I+F)/2,W); draw(rightanglemark(I,F,A)); draw(rightanglemark(I,D,B)); draw(rightanglemark(I,E,C)); pair reflectC = C + 2*(foot(C,A,I) - C), reflectI = (reflectC.x - I.x, I.y); // draw right diagram filldraw(shift(shiftR)*(A--I--B--cycle), c1, heavy);","\\mathbf{e}_{49}$, but we can see that the 49 encoding vectors are divided into 7 rows of 7 vectors each. Now the idea is to use only one row of 7 encoding vectors and expand those vectors to the other 6 rows, essentially reusing their values. Because it is discouraged to have the same values for different encoding vectors, each vector of dimension (a.k.a height) config.hidden_size=4 is cut into the lower encoding vector $\\mathbf{e}_\\text{down}$ of size $1$ and $\\mathbf{e}_\\text{up}$ of size $3$, so that the lower part can be expanded along the row dimension and the upper part can be expanded along the column dimension. Let's visualize for more clarity. We can see that we have cut the embedding vectors into $\\mathbf{e}_\\text{down}$ (in blue) and $\\mathbf{e}_\\text{up}$ (in yellow). Now for the \"sub\"-vectors $\\mathbf{E}_\\text{down} = \\left[\\mathbf{e}_{\\text{down},1}, \\ldots, \\mathbf{e}_{\\text{down},49}\\right]$ only the first row, a.k.a. the width in the graphic, of $7$ is kept and expanded along the column dimension, a.k.a. the depth of the graphic. Inversely, for the \"sub\"-vectors $\\mathbf{E}_\\text{up} = \\left[\\mathbf{e}_{\\text{up},1}, \\ldots, \\mathbf{e}_{\\text{up},49}\\right]$ only the first column of $7$ is kept and expanded along the row dimension. The resulting embedding vectors $\\mathbf{e'}_{i}$ then correspond to $\\mathbf{e'}_{i} = \\left[ \\left[\\mathbf{e}_{\\text{down, } i \\% n_\\text{max}^1}\\right]^T, \\left[\\mathbf{e}_{\\text{up, } \\left \\lfloor{\\frac{i}{{n}^2_{\\text{max}}}}\\right \\rfloor} \\right]^T \\right]^T$ whereas $n_\\text{max}^1 = 7$ and $n_\\text{max}^2 = 7$ in our","adjacent side vectors $\\overrightarrow{(a,b)}, \\overrightarrow{(c,d)}$ is given by $\\overrightarrow{(a,b)} \\times \\overrightarrow{(c,d)} = ad-bc$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); real r = 3.5; // radius pair shiftL = (-2.5*r,0); // distance between 2 diagrams /* returns the vertex of the interior equilateral triangle with one edge shared with the dodecagon */ pair dodecagonPt(int i) { return r*dir(i*360/12) + rotate(60)*(r*(dir((i+1)*360/12) - dir(i*360/12))); } /* left diagram */ path dodecagon = shiftL+(r,0)--shiftL+r*dir(30); for(int i = 1; i < 12; ++i) dodecagon = dodecagon--shiftL+r*dir(i*30); dodecagon = dodecagon--cycle; filldraw(dodecagon, rgb(0.5,1,0.5)); draw(Circle(shiftL, r), linetype(\"2 2\")); dot((0,0)); draw(shiftL--shiftL+(r,0)); label(\"R\",shiftL+(r/2,0),S); /* right diagram */ for(int i = 0; i < 9; ++i) { filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.8,0.8,0)); if (i % 3 == 1) { filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir((i+1)*360/12)--cycle, rgb(0.8,0.8,0)); filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir(floor(i/3)*90)--cycle, rgb(0,0.8,0)); filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir((i+1)*360/12)--r*dir(floor(i/3)*90+90)--cycle, rgb(0,0.8,0)); } } for(int i = 9; i < 12; ++i) { filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype(\"2 2\")); filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype(\"2 2\")); filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(1,1,0.5), linetype(\"2 2\")); } [/asy]$ The area of a dodecagon is $3R^2$, where $R$ is the circumradius. $[asy] pathpen = linewidth(1); unitsize(15); pen dotted = linetype(\"2 4\"); path xaxis = (-3,0)--(3,0); pair A = (-2,2), B = (1.5,1.5), B3 = (-1.5,0), B2 = (B.x,-B.y), C2 = IP(xaxis, A--B2); D(xaxis,Arrows(8)); D(D(A)--D(C2)--D(B)); D(D(B2)--C2,dashed+linewidth(0.7)); D(A--D(B3)--B,dotted+linewidth(0.7)); D(B3--B2,dotted); MP(\"(a,b)\",A,W); MP(\"(c,d)\",B,E); MP(\"(c,-d)\",B2,E); [/asy]$ The smallest distance necessary to travel between $(a,b)$, the x-axis, and then $(c,d)$ for $b,d > 0$ is given","([shift=(nw)] -.5, .5); \\coordinate[label=right:$A_2$] (A2) at ([shift=(ne)] .5, .5); \\coordinate[label=right:$A_3$] (A3) at ([shift=(se)] .5, -.5); \\coordinate[label=left:$A_4$] (A4) at ([shift=(sw)] -.5, -.5); % Grid and outer layers of detector \\draw[black, very thick] (A1) rectangle (A3); \\draw[black, very thick] (sw) rectangle (ne); \\draw[step=5mm,black] (sw) grid (ne); % Anode wiring \\draw (A1) -- (nw); \\draw (A2) -- (ne); \\draw (A3) -- (se); \\draw (A4) -- (sw); %Colour in distorted areas \\draw[fill=white] (sw) parabola bend +(2, .6) (se) -- cycle; \\draw[fill=white] (nw) parabola bend +(2, -.6) (ne) -- cycle; \\begin{scope}[shift=(sw), rotate=90] \\draw[fill=white] (0,0) parabola bend +(2, -.6) ++(4, 0) -- cycle; \\end{scope} \\begin{scope}[shift=(se), rotate=90] \\draw[fill=white] (0,0) parabola bend +(2, .6) ++(4, 0) -- cycle; \\end{scope} \\end{tikzpicture} \\end{document} • Thank you jakun! It looks fantastic! Just one quick question. With the lines which end in '-- cycle'. Does that just cycle around the 3 points earlier to create a shape which is then filled in? – AlphaBetaGamma96 May 4 '17 at 18:37 • @AlphaBetaGamma96 \"The --cycle causes the current path to be closed (actually the current part of the current path) by smoothly joining the first and last point.\" see tikz documentation page 36/37. – jakun May 5 '17 at 5:22 You've used bend left/bend right earlier in the drawing, use that again, and it works fine. For example \\draw[fill=white] (8.5,0) to[bend right]","less. Let $S=\\{p_1(x), p_2(x), p_3(x), p_4(x)\\},$ where \\begin{align*} p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\\\ p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3. \\end{align*} (a) Find a basis $Q$ of the span $\\Span(S)$ consisting of polynomials in $S$. (b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$. (The Ohio State University, Linear Algebra Midterm) ## Problem 606 Let $V$ be a vector space and $B$ be a basis for $V$. Let $\\mathbf{w}_1, \\mathbf{w}_2, \\mathbf{w}_3, \\mathbf{w}_4, \\mathbf{w}_5$ be vectors in $V$. Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\\mathbf{w}_1, \\mathbf{w}_2, \\mathbf{w}_3, \\mathbf{w}_4, \\mathbf{w}_5$ with respect to the basis $B$. After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form $\\begin{bmatrix} 1 & 0 & 2 & 1 & 0 \\\\ 0 & 1 & 3 & 0 & 1 \\\\ 0 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 \\end{bmatrix}.$ (a) What is the dimension of $V$? (b) What is the dimension of $\\Span\\{\\mathbf{w}_1, \\mathbf{w}_2, \\mathbf{w}_3, \\mathbf{w}_4, \\mathbf{w}_5\\}$? (The Ohio State University, Linear Algebra Midterm) ## Problem 604 Let $A=\\begin{bmatrix} 1 & -1 & 0 & 0 \\\\ 0 &1 & 1 & 1 \\\\ 1 & -1 & 0 & 0","21. Let $A$ be a square matrix such that $A^{\\trans}A=A$, where $A^{\\trans}$ is the transpose matrix of $A$. Prove that $A$ is idempotent, that is, $A^2=A$. Also, prove that $A$ is a symmetric matrix. 22. Let $A$ and $B$ be $n \\times n$ real symmetric matrices. Prove the followings. (a) The product $AB$ is symmetric if and only if $AB=BA$. (b) If the product $AB$ is a diagonal matrix, then $AB=BA$. 23. A $2 \\times 2$ matrix has two parallel columns and $\\tr(A)=5$. Find $\\tr(A^2)$. 24. Let $I$ be the $n\\times n$ identity matrix, where $n$ is a positive integer. Prove that there are no $n\\times n$ matrices $X$ and $Y$ such that $XY-YX=I$. 25. Let $A=(a_{i j})$ and $B=(b_{i j})$ be $n\\times n$ real matrices for some $n \\in \\N$. (a) Express $\\tr(AB^{\\trans})$ in terms of the entries of the matrices $A$ and $B$. Here $B^{\\trans}$ is the transpose matrix of $B$. (b) Show that $\\tr(AA^{\\trans})$ is the sum of the square of the entries of $A$. (c) Show that if $A$ is nonzero symmetric matrix, then $\\tr(A^2)>0$. 26. For each of the following matrix $A$, prove that $\\mathbf{x}^{\\trans}A\\mathbf{x} \\geq 0$ for all vectors $\\mathbf{x}$ in $\\R^2$. Also, determine those vectors $\\mathbf{x}\\in \\R^2$ such that $\\mathbf{x}^{\\trans}A\\mathbf{x}=0$. (a) $A=\\begin{bmatrix} 4 & 2\\\\ 2& 1 \\end{bmatrix}$. (b) $A=\\begin{bmatrix} 2"],"string":"[\n \"not, give a counterexample. 14. Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. (a) Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{w} = \\\\mathbf{w}^\\\\trans \\\\mathbf{v}$. (b) Provide an example to show that $\\\\mathbf{v} \\\\mathbf{w}^\\\\trans$ is not always equal to $\\\\mathbf{w} \\\\mathbf{v}^\\\\trans$. 15. Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. Prove that $\\\\tr ( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans ) = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}$. 16. Let $\\\\mathbf{v}$ be an $n \\\\times 1$ column vector. Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} = 0$ if and only if $\\\\mathbf{v}$ is the zero vector $\\\\mathbf{0}$. 17. Is it true that a real square matrix $A$ must commute with its transpose $A^{\\\\trans}$? 18. Let $\\\\mathbf{v}$ be an $n \\\\times 1$ column vector. Prove that $\\\\mathbf{v} \\\\mathbf{v}^\\\\trans$ is a symmetric matrix. 19. Calculate the following expressions, using the following matrices: $A = \\\\begin{bmatrix} 2 & 3 \\\\\\\\ -5 & 1 \\\\end{bmatrix}, \\\\qquad B = \\\\begin{bmatrix} 0 & -1 \\\\\\\\ 1 & -1 \\\\end{bmatrix}, \\\\qquad \\\\mathbf{v} = \\\\begin{bmatrix} 2 \\\\\\\\ -4 \\\\end{bmatrix}$ (a) $A B^\\\\trans + \\\\mathbf{v} \\\\mathbf{v}^\\\\trans$. (b) $A \\\\mathbf{v} – 2 \\\\mathbf{v}$. (c) $\\\\mathbf{v}^{\\\\trans} B$. (d) $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} + \\\\mathbf{v}^\\\\trans B A^\\\\trans \\\\mathbf{v}$. 20. Let $A$ and $B$ be $n \\\\times n$ matrices, and $\\\\mathbf{v}$ an $n \\\\times 1$ column vector. Use the matrix components to prove that $(A + B) \\\\mathbf{v} = A\\\\mathbf{v} + B\\\\mathbf{v}$.\",\n \"## Problem 639 Let $\\\\mathbf{v}$ be an $n \\\\times 1$ column vector. Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} = 0$ if and only if $\\\\mathbf{v}$ is the zero vector $\\\\mathbf{0}$. ## Problem 638 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. Prove that $\\\\tr ( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans ) = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}$. ## Problem 637 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. (a) Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{w} = \\\\mathbf{w}^\\\\trans \\\\mathbf{v}$. (b) Provide an example to show that $\\\\mathbf{v} \\\\mathbf{w}^\\\\trans$ is not always equal to $\\\\mathbf{w} \\\\mathbf{v}^\\\\trans$. ## Problem 636 Calculate the following expressions, using the following matrices: $A = \\\\begin{bmatrix} 2 & 3 \\\\\\\\ -5 & 1 \\\\end{bmatrix}, \\\\qquad B = \\\\begin{bmatrix} 0 & -1 \\\\\\\\ 1 & -1 \\\\end{bmatrix}, \\\\qquad \\\\mathbf{v} = \\\\begin{bmatrix} 2 \\\\\\\\ -4 \\\\end{bmatrix}$ (a) $A B^\\\\trans + \\\\mathbf{v} \\\\mathbf{v}^\\\\trans$. (b) $A \\\\mathbf{v} – 2 \\\\mathbf{v}$. (c) $\\\\mathbf{v}^{\\\\trans} B$. (d) $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} + \\\\mathbf{v}^\\\\trans B A^\\\\trans \\\\mathbf{v}$. ## Problem 635 Let $A$ and $B$ be $n \\\\times n$ matrices, and $\\\\mathbf{v}$ an $n \\\\times 1$ column vector. Use the matrix components to prove that $(A + B) \\\\mathbf{v} = A\\\\mathbf{v} + B\\\\mathbf{v}$. ## Problem 634 Let $A$ and $B$ be $n \\\\times n$ matrices. Is it always true that $\\\\tr (A B) = \\\\tr (A) \\\\tr (B)$? If it is\",\n \"then $AB$ is a skew-symmetric matrix. We calculate \\\\begin{align*} (AB)^{\\\\trans}&=B^{\\\\trans}A^{\\\\trans}=(-B)(-A)\\\\\\\\ &=BA=-AB, \\\\end{align*} where the last step follows from the assumption $AB=-BA$. This proves that $AB$ is skew-symmetric. ### (f) Let $\\\\mathbf{v}$ be an $n$-dimensional column vecotor. Prove that $\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}=0$. Observe that $\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}$ is a $1\\\\times 1$ matrix, or just a number. So we have \\\\begin{align*} \\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}&=(\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v})^{\\\\trans}=\\\\mathbf{v}^{\\\\trans}A^{\\\\trans}(\\\\mathbf{v}^{\\\\trans})^{\\\\trans}\\\\\\\\ &=\\\\mathbf{v}^{\\\\trans}A^{\\\\trans}\\\\mathbf{v}=\\\\mathbf{v}^{\\\\trans}(-A)\\\\mathbf{v}=-(\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}). \\\\end{align*} This yields that $2\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}=0$, and hence $\\\\mathbf{v}^{\\\\trans}A\\\\mathbf{v}=0$. ### (g) Suppose that $A$ is a real skew-symmetric matrix and $A^2\\\\mathbf{v}=\\\\mathbf{0}$ for some vector $\\\\mathbf{v}\\\\in \\\\R^n$. Then prove that $A\\\\mathbf{v}=\\\\mathbf{0}$. Let us compute the length of the vector $A\\\\mathbf{v}$. We have \\\\begin{align*} \\\\|A\\\\mathbf{v}\\\\|&=(A\\\\mathbf{v})^{\\\\trans}(A\\\\mathbf{v})=\\\\mathbf{v}^{\\\\trans}A^{\\\\trans}A\\\\mathbf{v}\\\\\\\\ &=\\\\mathbf{v}^{\\\\trans}(-A)A\\\\mathbf{v}=-\\\\mathbf{v}^{\\\\trans}A^2\\\\mathbf{v}\\\\\\\\ &=-\\\\mathbf{v}\\\\mathbf{0} &&\\\\text{by assumption}\\\\\\\\ &=0. \\\\end{align*} Since the length $\\\\|A\\\\mathbf{v}\\\\|=0$, we conclude that $A\\\\mathbf{v}=\\\\mathbf{0}$. ### More from my site • Express a Hermitian Matrix as a Sum of Real Symmetric Matrix and a Real Skew-Symmetric Matrix Recall that a complex matrix is called Hermitian if $A^*=A$, where $A^*=\\\\bar{A}^{\\\\trans}$. Prove that every Hermitian matrix $A$ can be written as the sum $A=B+iC,$ where $B$ is a real symmetric matrix and $C$ is a real skew-symmetric matrix. Proof. Since […] • Eigenvalues of a Hermitian Matrix are Real Numbers Show that eigenvalues of a Hermitian matrix $A$ are real numbers. (The Ohio State University Linear Algebra Exam Problem) We give two proofs. These two proofs are essentially the same. The second\",\n \"# Tagged: transpose of a matrix ## Problem 638 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. Prove that $\\\\tr ( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans ) = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}$. ## Problem 637 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. (a) Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{w} = \\\\mathbf{w}^\\\\trans \\\\mathbf{v}$. (b) Provide an example to show that $\\\\mathbf{v} \\\\mathbf{w}^\\\\trans$ is not always equal to $\\\\mathbf{w} \\\\mathbf{v}^\\\\trans$. ## Problem 636 Calculate the following expressions, using the following matrices: $A = \\\\begin{bmatrix} 2 & 3 \\\\\\\\ -5 & 1 \\\\end{bmatrix}, \\\\qquad B = \\\\begin{bmatrix} 0 & -1 \\\\\\\\ 1 & -1 \\\\end{bmatrix}, \\\\qquad \\\\mathbf{v} = \\\\begin{bmatrix} 2 \\\\\\\\ -4 \\\\end{bmatrix}$ (a) $A B^\\\\trans + \\\\mathbf{v} \\\\mathbf{v}^\\\\trans$. (b) $A \\\\mathbf{v} – 2 \\\\mathbf{v}$. (c) $\\\\mathbf{v}^{\\\\trans} B$. (d) $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} + \\\\mathbf{v}^\\\\trans B A^\\\\trans \\\\mathbf{v}$. ## Problem 633 Let $A$ be an $n \\\\times n$ matrix. Is it true that $\\\\tr ( A^\\\\trans ) = \\\\tr(A)$? If it is true, prove it. If not, give a counterexample.\",\n \"# Tagged: transpose ## Problem 713 Determine bases for $\\\\calN(A)$ and $\\\\calN(A^{T}A)$ when $A= \\\\begin{bmatrix} 1 & 2 & 1 \\\\\\\\ 1 & 1 & 3 \\\\\\\\ 0 & 0 & 0 \\\\end{bmatrix} .$ Then, determine the ranks and nullities of the matrices $A$ and $A^{\\\\trans}A$. ## Problem 640 Let $\\\\mathbf{v}$ be an $n \\\\times 1$ column vector. Prove that $\\\\mathbf{v} \\\\mathbf{v}^\\\\trans$ is a symmetric matrix. ## Problem 639 Let $\\\\mathbf{v}$ be an $n \\\\times 1$ column vector. Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} = 0$ if and only if $\\\\mathbf{v}$ is the zero vector $\\\\mathbf{0}$. ## Problem 638 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. Prove that $\\\\tr ( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans ) = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}$. ## Problem 637 Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. (a) Prove that $\\\\mathbf{v}^\\\\trans \\\\mathbf{w} = \\\\mathbf{w}^\\\\trans \\\\mathbf{v}$. (b) Provide an example to show that $\\\\mathbf{v} \\\\mathbf{w}^\\\\trans$ is not always equal to $\\\\mathbf{w} \\\\mathbf{v}^\\\\trans$. ## Problem 636 Calculate the following expressions, using the following matrices: $A = \\\\begin{bmatrix} 2 & 3 \\\\\\\\ -5 & 1 \\\\end{bmatrix}, \\\\qquad B = \\\\begin{bmatrix} 0 & -1 \\\\\\\\ 1 & -1 \\\\end{bmatrix}, \\\\qquad \\\\mathbf{v} = \\\\begin{bmatrix} 2 \\\\\\\\ -4 \\\\end{bmatrix}$ (a) $A B^\\\\trans + \\\\mathbf{v} \\\\mathbf{v}^\\\\trans$. (b) $A \\\\mathbf{v} – 2 \\\\mathbf{v}$. (c) $\\\\mathbf{v}^{\\\\trans} B$. (d) $\\\\mathbf{v}^\\\\trans \\\\mathbf{v} + \\\\mathbf{v}^\\\\trans\",\n \"}{\\\\mathbf{transa}}=\\\\text{\\\"C\\\"}$: ${\\\\mathbf{m}}$; otherwise: $0$. On entry: the matrix $A$; $A$ is $m×k$ if ${\\\\mathbf{transa}}=\\\\text{'N'}$, or $k×m$ if ${\\\\mathbf{transa}}=\\\\text{'T'}$ or $\\\\text{'C'}$. 5: $\\\\mathbf{b}\\\\left(:,:\\\\right)$double array Input Note: the first dimension of b is given by: if ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$: ${\\\\mathbf{k}}$; if ${\\\\mathbf{transb}}=\\\\text{\\\"T\\\"}\\\\text{ or }{\\\\mathbf{transb}}=\\\\text{\\\"C\\\"}$: ${\\\\mathbf{n}}$; otherwise: $0$. Note: the second dimension of b is given by: if ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$: ${\\\\mathbf{n}}$; if ${\\\\mathbf{transb}}=\\\\text{\\\"T\\\"}\\\\text{ or }{\\\\mathbf{transb}}=\\\\text{\\\"C\\\"}$: ${\\\\mathbf{k}}$; otherwise: $0$. On entry: the matrix $B$; $B$ is $k×n$ if ${\\\\mathbf{transb}}=\\\\text{'N'}$, or $n×k$ if ${\\\\mathbf{transb}}=\\\\text{'T'}$ or $\\\\text{'C'}$. 6: $\\\\mathbf{beta}$double Input On entry: the scalar $\\\\beta$. 7: $\\\\mathbf{c}\\\\left({\\\\mathbf{m}},{\\\\mathbf{n}}\\\\right)$double array Input/Output On entry: the $m×n$ matrix $C$. If ${\\\\mathbf{beta}}=0.0$, c need not be set. On exit: the updated matrix $C$. 8: $\\\\mathbf{opt}$OptionalF06YA Input/Output Optional parameter container, derived from Optional. 1: $\\\\mathbf{m}$ $m$, the number of rows of the matrix $C$; the number of rows of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"N\\\"}$, or the number of columns of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"T\\\"}\\\\text{​ or ​}\\\\text{\\\"C\\\"}$ 2: $\\\\mathbf{n}$ $n$, the number of columns of the matrix $C$; the number of columns of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$, or the number of rows of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"T\\\"}\\\\text{​ or ​}\\\\text{\\\"C\\\"}$ 3: $\\\\mathbf{k}$ $k$, the number of columns of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"N\\\"}$, or the number of rows of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"T\\\"}\\\\text{​ or ​}\\\\text{\\\"C\\\"}$; the number of rows of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$, or the number of columns of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"T\\\"}\\\\text{​ or\",\n \"while $\\\\quad \\\\mathbf{w} \\\\mathbf{v}^\\\\trans = \\\\begin{bmatrix} 0 \\\\\\\\ 1 \\\\end{bmatrix} \\\\begin{bmatrix} 1 & 0 \\\\end{bmatrix} = \\\\begin{bmatrix} 0 & 0 \\\\\\\\ 1 & 0 \\\\end{bmatrix}.$ ## Comment. Recall that for two vectors $\\\\mathbf{v}, \\\\mathbf{w} \\\\in \\\\R^n$, the dot product (or inner product) of $\\\\mathbf{v}, \\\\mathbf{w}$ is defined to be $\\\\mathbf{v}\\\\cdot \\\\mathbf{w}:=\\\\mathbf{v}^{\\\\trans} \\\\mathbf{w}.$ Part (a) of the problem deduces that the dot product is commutative. This means that we have $\\\\mathbf{v}\\\\cdot \\\\mathbf{w}= \\\\mathbf{w} \\\\cdot \\\\mathbf{v}.$ In fact, we have \\\\begin{align*} \\\\mathbf{v}\\\\cdot \\\\mathbf{w}= \\\\mathbf{v}^\\\\trans \\\\mathbf{w} \\\\stackrel{\\\\text{(a)}}{=} \\\\mathbf{w}^\\\\trans \\\\mathbf{v} \\\\mathbf{w} \\\\cdot \\\\mathbf{v}. \\\\end{align*} Also, notice that while $\\\\mathbf{v} \\\\mathbf{w}^\\\\trans$ is not always equal to $\\\\mathbf{w} \\\\mathbf{v}^\\\\trans$, we know that $(\\\\mathbf{v} \\\\mathbf{w}^\\\\trans)^\\\\trans = \\\\mathbf{w} \\\\mathbf{v}^\\\\trans$. ### More from my site • A Relation between the Dot Product and the Trace Let $\\\\mathbf{v}$ and $\\\\mathbf{w}$ be two $n \\\\times 1$ column vectors. Prove that $\\\\tr ( \\\\mathbf{v} \\\\mathbf{w}^\\\\trans ) = \\\\mathbf{v}^\\\\trans \\\\mathbf{w}$. Solution. Suppose the vectors have components $\\\\mathbf{v} = \\\\begin{bmatrix} v_1 \\\\\\\\ v_2 \\\\\\\\ \\\\vdots \\\\\\\\ v_n […] • Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given Let \\\\mathbf{a} and \\\\mathbf{b} be vectors in \\\\R^n such that their length are \\\\[\\\\|\\\\mathbf{a}\\\\|=\\\\|\\\\mathbf{b}\\\\|=1$ and the inner product $\\\\mathbf{a}\\\\cdot \\\\mathbf{b}=\\\\mathbf{a}^{\\\\trans}\\\\mathbf{b}=-\\\\frac{1}{2}.$ Then determine the length $\\\\|\\\\mathbf{a}-\\\\mathbf{b}\\\\|$. (Note […] • Rotation Matrix in Space and its Determinant and Eigenvalues For\",\n \"[/asy]$ $\\\\textbf{(D)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(0,0)*cat); dot((0,1.5)); [/asy]$ $\\\\textbf{(E)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(0,1)*cat); dot((1.5,0)); [/asy]$ ## Problem 24 A ship travels from point A to point B along a semicircular path, centered at Island X. Then it travels along a straight path from B to C. Which of these graphs best shows the ship's distance from Island X as it moves along its course? $[asy]size(150); pair X=origin, A=(-5,0), B=(5,0), C=(0,5); draw(Arc(X, 5, 180, 360)^^B--C); dot(X); label(\\\"X\\\", X, NE); label(\\\"C\\\", C, N); label(\\\"B\\\", B, E); label(\\\"A\\\", A, W);[/asy]$ $\\\\textbf{(A)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\\\"distance traveled\\\", (8,0), S); label(rotate(90)*\\\"distance to X\\\", (0,8), W); draw(Arc((4,10), 4, 0, 180)^^(8,10)--(16,12)); [/asy]$ $\\\\textbf{(B)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\\\"distance traveled\\\", (8,0), S); label(rotate(90)*\\\"distance to X\\\", (0,8), W); draw(Arc((12,10), 4, 180, 360)^^(0,10)--(8,10)); [/asy]$ $\\\\textbf{(C)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\\\"distance traveled\\\", (8,0), S); label(rotate(90)*\\\"distance to X\\\", (0,8), W); draw((0,8)--(10,10)--(16,8)); [/asy]$ $\\\\textbf{(D)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\\\"distance traveled\\\", (8,0), S); label(rotate(90)*\\\"distance to X\\\", (0,8), W); draw(Arc((12,10), 4, 0, 180)^^(0,10)--(8,10)); [/asy]$ $\\\\textbf{(E)}$ $[asy] defaultpen(fontsize(7)); size(80); draw((0,16)--origin--(16,0), linewidth(0.9)); label(\\\"distance traveled\\\", (8,0), S); label(rotate(90)*\\\"distance to X\\\", (0,8), W); draw((0,6)--(6,6)--(16,10)); [/asy]$ ## Problem 25 In the figure, the area of square WXYZ is $25 \\\\text{cm}^2$. The four smaller squares have sides 1 cm long,\",\n \"# Difference between revisions of \\\"2005 AMC 8 Problems/Problem 13\\\" ## Problem The area of polygon $ABCDEF$ is 52 with $AB=8$, $BC=9$ and $FA=5$. What is $DE+EF$? $[asy] pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4); draw(a--b--c--d--e--f--cycle); draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a); draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b); draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c); draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d); draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f); label(\\\"A\\\", a, NW); label(\\\"B\\\", b, NE); label(\\\"C\\\", c, SE); label(\\\"D\\\", d, SW); label(\\\"E\\\", e, SW); label(\\\"F\\\", f, SW); label(\\\"5\\\", (0,6.5), W); label(\\\"8\\\", (4,9), N); label(\\\"9\\\", (8, 4.5), E); [/asy]$ $\\\\textbf{(A)}\\\\ 7\\\\qquad\\\\textbf{(B)}\\\\ 8\\\\qquad\\\\textbf{(C)}\\\\ 9\\\\qquad\\\\textbf{(D)}\\\\ 10\\\\qquad\\\\textbf{(E)}\\\\ 11$ ## Solution Notice that $AF + DE = BC$, so $DE=4$. Let $O$ be the intersection of the extensions of $AF$ and $DC$, which makes rectangle $ABCO$. The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$. $$\\\\text{Area} = 52 = 8 \\\\cdot 9- EF \\\\cdot 4$$ Solving for the unknown, $EF=5$, therefore $DE+EF=4+5=\\\\boxed{\\\\textbf{(C)}\\\\ 9}$.\",\n \"+0 # Consider parallelogram $ABCD$ with points $S$ and $T$ chosen such that $CS:SD = BT:TC = 2,$as in the picture below: [asy] size(200); pair 0 167 1 Consider parallelogram ABCD with points S and T chosen such that $$CS:SD = BT:TC = 2$$ as in the picture below: Let $$\\\\overrightarrow{AB} = \\\\mathbf{v}$$ and $$\\\\overrightarrow{AD} = \\\\mathbf{w}$$. Then there exist constants r, s, t, u such that \\\\begin{align*} \\\\overrightarrow{AT} &= r \\\\mathbf{v} + s \\\\mathbf{w},\\\\\\\\ \\\\overrightarrow{BS} &= t \\\\mathbf{v} + u \\\\mathbf{w}. \\\\end{align*} Enter r, s, t, u in that order below. Apr 4, 2019\",\n \"Find a Basis of the Subspace of All Vectors that are Perpendicular to the Columns of the Matrix Problem 40 Find a basis for the subspace $W$ of all vectors in $\\\\R^4$ which are perpendicular to the columns of the matrix $A=\\\\begin{bmatrix} 11 & 12 & 13 & 14 \\\\\\\\ 21 &22 & 23 & 24 \\\\\\\\ 31 & 32 & 33 & 34 \\\\\\\\ 41 & 42 & 43 & 44 \\\\end{bmatrix}.$ (Harvard University Exam) Contents Hint. 1. Show that $W=\\\\calN(A^{\\\\trans})$. 2. Find a basis of $\\\\calN(A^{\\\\trans})$ by reducing the matrix $A^{\\\\trans}$. Solution. Let us write $A=[A_1 \\\\, A_2 \\\\, A_3 \\\\, A_4]$, where $A_i$ is the $i$-th column vector of $A$ for $i=1,2,3,4$. First we claim that a vector $\\\\mathbf{x}\\\\in \\\\R^4$ is perpendicular to all column vectors $A_i$ if and only if $\\\\mathbf{x} \\\\in \\\\calN(A^{\\\\trans})$. To see this, we compute \\\\begin{align*} A^{\\\\trans} \\\\mathbf{x} =\\\\begin{bmatrix} A_1^{\\\\trans} \\\\\\\\ A_2^{\\\\trans} \\\\\\\\ A_3^{\\\\trans} \\\\\\\\ A_4^{\\\\trans} \\\\end{bmatrix}\\\\mathbf{x} =\\\\begin{bmatrix} A_1^{\\\\trans}\\\\mathbf{x} \\\\\\\\ A_2^{\\\\trans} \\\\mathbf{x}\\\\\\\\ A_3^{\\\\trans} \\\\mathbf{x}\\\\\\\\ A_4^{\\\\trans} \\\\mathbf{x} \\\\end{bmatrix}. \\\\end{align*} From this equality the claim follows immediately. So we proved that $\\\\calN(A^{\\\\trans}) =W$. From this, we see that $W$ is actually a subspace of $\\\\R^4$. Thus, we need to find a basis for the null space of the transpose $A^{\\\\trans}$. We apply elementary row operations to $A^{\\\\trans}$ and obtain a reduced row echelon\",\n \"y} \\\\qquad \\\\text{and} \\\\qquad \\\\alpha A {\\\\mathbf x} = A ( \\\\alpha {\\\\mathbf x}), \\\\end{equation*} where \\\\begin{equation*} {\\\\mathbf x} = \\\\begin{pmatrix} x_1 \\\\\\\\ x_2 \\\\\\\\ \\\\vdots \\\\\\\\ x_n \\\\end{pmatrix}. \\\\end{equation*} We will often abbreviate the matrix $A$ by writing $(a_{ij})\\\\text{.}$ Conversely, if $T : {\\\\mathbb R}^n \\\\rightarrow {\\\\mathbb R}^m$ is a linear map, we can associate a matrix $A$ with $T$ by considering what $T$ does to the vectors \\\\begin{align*} {\\\\mathbf e}_1 & = (1, 0, \\\\ldots, 0)^\\\\transpose\\\\\\\\ {\\\\mathbf e}_2 & = (0, 1, \\\\ldots, 0)^\\\\transpose\\\\\\\\ & \\\\vdots & \\\\\\\\ {\\\\mathbf e}_n & = (0, 0, \\\\ldots, 1)^\\\\transpose. \\\\end{align*} We can write any vector ${\\\\mathbf x} = (x_1, \\\\ldots, x_n)^\\\\transpose$ as \\\\begin{equation*} x_1 {\\\\mathbf e}_1 + x_2 {\\\\mathbf e}_2 + \\\\cdots + x_n {\\\\mathbf e}_n. \\\\end{equation*} Consequently, if \\\\begin{align*} T({\\\\mathbf e}_1) & = (a_{11}, a_{21}, \\\\ldots, a_{m1})^\\\\transpose,\\\\\\\\ T({\\\\mathbf e}_2) & = (a_{12}, a_{22}, \\\\ldots, a_{m2})^\\\\transpose,\\\\\\\\ & \\\\vdots & \\\\\\\\ T({\\\\mathbf e}_n) & = (a_{1n}, a_{2n}, \\\\ldots, a_{mn})^\\\\transpose, \\\\end{align*} then \\\\begin{align*} T({\\\\mathbf x} ) & = T(x_1 {\\\\mathbf e}_1 + x_2 {\\\\mathbf e}_2 + \\\\cdots + x_n {\\\\mathbf e}_n)\\\\\\\\ & = x_1 T({\\\\mathbf e}_1) + x_2 T({\\\\mathbf e}_2) + \\\\cdots + x_n T({\\\\mathbf e}_n)\\\\\\\\ & = \\\\left( \\\\sum_{k=1}^{n} a_{1k} x_k, \\\\ldots, \\\\sum_{k=1}^{n} a_{mk} x_k \\\\right)^\\\\transpose\\\\\\\\ & = A {\\\\mathbf x}. \\\\end{align*} ###### Example12.1 If we let $T : {\\\\mathbb\",\n \"${\\\\mathbf{transa}}=\\\\text{\\\"T\\\"}$ or $\\\\text{\\\"C\\\"}$ The operation involves ${A}^{\\\\mathrm{T}}$. Constraint: ${\\\\mathbf{transa}}=\\\\text{\\\"N\\\"}$, $\\\\text{\\\"T\\\"}$ or $\\\\text{\\\"C\\\"}$. transb Type: System..::..String On entry: specifies whether the operation involves $B$ or ${B}^{\\\\mathrm{T}}$. ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$ The operation involves $B$. ${\\\\mathbf{transb}}=\\\\text{\\\"T\\\"}$ or $\\\\text{\\\"C\\\"}$ The operation involves ${B}^{\\\\mathrm{T}}$. Constraint: ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$, $\\\\text{\\\"T\\\"}$ or $\\\\text{\\\"C\\\"}$. m Type: System..::..Int32 On entry: $m$, the number of rows of the matrix $C$; the number of rows of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"N\\\"}$, or the number of columns of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"T\\\"}$ or $\\\\text{\\\"C\\\"}$. Constraint: ${\\\\mathbf{m}}\\\\ge 0$. n Type: System..::..Int32 On entry: $n$, the number of columns of the matrix $C$; the number of columns of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$, or the number of rows of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"T\\\"}$ or $\\\\text{\\\"C\\\"}$. Constraint: ${\\\\mathbf{n}}\\\\ge 0$. k Type: System..::..Int32 On entry: $k$, the number of columns of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"N\\\"}$, or the number of rows of $A$ if ${\\\\mathbf{transa}}=\\\\text{\\\"T\\\"}$ or $\\\\text{\\\"C\\\"}$; the number of rows of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"N\\\"}$, or the number of columns of $B$ if ${\\\\mathbf{transb}}=\\\\text{\\\"T\\\"}$ or $\\\\text{\\\"C\\\"}$. Constraint: ${\\\\mathbf{k}}\\\\ge 0$. alpha Type: System..::..Double On entry: the scalar $\\\\alpha$. a Type: array[,](,)[,][,] An array of size [dim1, dim2] Note: dim1 must satisfy the constraint: • if ${\\\\mathbf{transa}}=\\\\text{\\\"N\\\"}$, $\\\\mathrm{dim1}\\\\ge \\\\mathrm{max}\\\\phantom{\\\\rule{0.125em}{0ex}}\\\\left(1,{\\\\mathbf{m}}\\\\right)$; • if ${\\\\mathbf{transa}}=\\\\text{\\\"T\\\"}$ or $\\\\text{\\\"C\\\"}$, $\\\\mathrm{dim1}\\\\ge \\\\mathrm{max}\\\\phantom{\\\\rule{0.125em}{0ex}}\\\\left(1,{\\\\mathbf{k}}\\\\right)$. Note: the second dimension of the array a must be at least $\\\\mathrm{max}\\\\phantom{\\\\rule{0.125em}{0ex}}\\\\left(1,{\\\\mathbf{k}}\\\\right)$ if ${\\\\mathbf{transa}}=\\\\text{\\\"N\\\"}$ and at least $\\\\mathrm{max}\\\\phantom{\\\\rule{0.125em}{0ex}}\\\\left(1,{\\\\mathbf{m}}\\\\right)$ if ${\\\\mathbf{transa}}=\\\\text{\\\"T\\\"}$ or\",\n \"and software for solving the generalized Sylvester equation and estimating the separation between regular matrix pairs ACM Trans. Math. Software 22 78–103 ## 5 Parameters 1: TRANS – CHARACTER(1)Input On entry: if ${\\\\mathbf{TRANS}}=\\\\text{'N'}$, solve the generalized Sylvester equation (1). If ${\\\\mathbf{TRANS}}=\\\\text{'T'}$, solve the ‘transposed’ system (2). Constraint: ${\\\\mathbf{TRANS}}=\\\\text{'N'}$ or $\\\\text{'T'}$. 2: IJOB – INTEGERInput On entry: specifies what kind of functionality is to be performed when ${\\\\mathbf{TRANS}}=\\\\text{'N'}$. ${\\\\mathbf{IJOB}}=0$ Solve (1) only. ${\\\\mathbf{IJOB}}=1$ The functionality of ${\\\\mathbf{IJOB}}=0$ and $3$. ${\\\\mathbf{IJOB}}=2$ The functionality of ${\\\\mathbf{IJOB}}=0$ and $4$. ${\\\\mathbf{IJOB}}=3$ Only an estimate of $\\\\mathrm{Dif}\\\\left[\\\\left(A,D\\\\right),\\\\left(B,E\\\\right)\\\\right]$ is computed based on the Frobenius norm. ${\\\\mathbf{IJOB}}=4$ Only an estimate of $\\\\mathrm{Dif}\\\\left[\\\\left(A,D\\\\right),\\\\left(B,E\\\\right)\\\\right]$ is computed based on the $1$-norm. If ${\\\\mathbf{TRANS}}=\\\\text{'T'}$, IJOB is not referenced. Constraint: if ${\\\\mathbf{TRANS}}=\\\\text{'N'}$, $0\\\\le {\\\\mathbf{IJOB}}\\\\le 4$. 3: M – INTEGERInput On entry: $m$, the order of the matrices $A$ and $D$, and the row dimension of the matrices $C$, $F$, $R$ and $L$. Constraint: ${\\\\mathbf{M}}\\\\ge 0$. 4: N – INTEGERInput On entry: $n$, the order of the matrices $B$ and $E$, and the column dimension of the matrices $C$, $F$, $R$ and $L$. Constraint: ${\\\\mathbf{N}}\\\\ge 0$. 5: A(LDA,$*$) – REAL (KIND=nag_wp) arrayInput Note: the second dimension of the array A must be at least $\\\\mathrm{max}\\\\phantom{\\\\rule{0.125em}{0ex}}\\\\left(1,{\\\\mathbf{M}}\\\\right)$. On entry: the upper quasi-triangular matrix $A$. 6: LDA – INTEGERInput On entry: the first dimension of\",\n \"sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7)); /* draw(rightanglemark((0,0),a*expi(rot1),(c,0))); */ filldraw(sq1, rgb(0.95,1,0.95)); filldraw(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); label(\\\"a\\\",a/2*expi(rot1),SE,sm); label(\\\"b\\\",a/2*expi(rot1)+(c/2,0),SW,sm); label(\\\"c\\\",(c/2,-c),S,sm); [/asy]$ COMING: The last proof of the Pythagorean Theorem we shall present on this page, this one by dissection. $[asy] defaultpen(linewidth(0.7)+fontsize(10)); unitsize(15); real a = 3.6, b = 4.8, c = (a^2 + b^2)^.5; pair shiftR = (a+b+2,0); pen sm = fontsize(10), heavy = linewidth(1); void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = 0.5){ /* Arrow option resizes */ fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); } real s1 = 5, s2 = 7, s3 = 8; // triangle side lengths pen c1 = rgb(0.5,0.5,1), c2 = rgb(0.5,1,0.8), c3 = rgb(0.5,1,0.5); // color pens pair shiftR = (s1+1,0); // distance between two diagrams pair A=(0,0), B = (s1,0), C = intersectionpoints(Circle(A, s3), Circle(B, s2))[0], I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,A,C), F = foot(I,A,B); // draw left diagram filldraw(A--I--B--cycle, c1, heavy); filldraw(B--I--C--cycle, c2, heavy); filldraw(A--I--C--cycle, c3, heavy); dot(I); draw(incircle(A,B,C), heavy); draw(I--D, linetype(\\\"2 2\\\")); draw(I--E, linetype(\\\"2 2\\\")); draw(I--F, linetype(\\\"2 2\\\")); label(\\\"a\\\",(A+B)/2,S); label(\\\"b\\\",(B+C)/2,NE); label(\\\"c\\\",(A+C)/2,NW); label(\\\"r\\\",(I+F)/2,W); draw(rightanglemark(I,F,A)); draw(rightanglemark(I,D,B)); draw(rightanglemark(I,E,C)); pair reflectC = C + 2*(foot(C,A,I) - C), reflectI = (reflectC.x - I.x, I.y); // draw right diagram filldraw(shift(shiftR)*(A--I--B--cycle), c1, heavy);\",\n \"\\\\mathbf{e}_{49}$, but we can see that the 49 encoding vectors are divided into 7 rows of 7 vectors each. Now the idea is to use only one row of 7 encoding vectors and expand those vectors to the other 6 rows, essentially reusing their values. Because it is discouraged to have the same values for different encoding vectors, each vector of dimension (a.k.a height) config.hidden_size=4 is cut into the lower encoding vector $\\\\mathbf{e}_\\\\text{down}$ of size $1$ and $\\\\mathbf{e}_\\\\text{up}$ of size $3$, so that the lower part can be expanded along the row dimension and the upper part can be expanded along the column dimension. Let's visualize for more clarity. We can see that we have cut the embedding vectors into $\\\\mathbf{e}_\\\\text{down}$ (in blue) and $\\\\mathbf{e}_\\\\text{up}$ (in yellow). Now for the \\\"sub\\\"-vectors $\\\\mathbf{E}_\\\\text{down} = \\\\left[\\\\mathbf{e}_{\\\\text{down},1}, \\\\ldots, \\\\mathbf{e}_{\\\\text{down},49}\\\\right]$ only the first row, a.k.a. the width in the graphic, of $7$ is kept and expanded along the column dimension, a.k.a. the depth of the graphic. Inversely, for the \\\"sub\\\"-vectors $\\\\mathbf{E}_\\\\text{up} = \\\\left[\\\\mathbf{e}_{\\\\text{up},1}, \\\\ldots, \\\\mathbf{e}_{\\\\text{up},49}\\\\right]$ only the first column of $7$ is kept and expanded along the row dimension. The resulting embedding vectors $\\\\mathbf{e'}_{i}$ then correspond to $\\\\mathbf{e'}_{i} = \\\\left[ \\\\left[\\\\mathbf{e}_{\\\\text{down, } i \\\\% n_\\\\text{max}^1}\\\\right]^T, \\\\left[\\\\mathbf{e}_{\\\\text{up, } \\\\left \\\\lfloor{\\\\frac{i}{{n}^2_{\\\\text{max}}}}\\\\right \\\\rfloor} \\\\right]^T \\\\right]^T$ whereas $n_\\\\text{max}^1 = 7$ and $n_\\\\text{max}^2 = 7$ in our\",\n \"adjacent side vectors $\\\\overrightarrow{(a,b)}, \\\\overrightarrow{(c,d)}$ is given by $\\\\overrightarrow{(a,b)} \\\\times \\\\overrightarrow{(c,d)} = ad-bc$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); real r = 3.5; // radius pair shiftL = (-2.5*r,0); // distance between 2 diagrams /* returns the vertex of the interior equilateral triangle with one edge shared with the dodecagon */ pair dodecagonPt(int i) { return r*dir(i*360/12) + rotate(60)*(r*(dir((i+1)*360/12) - dir(i*360/12))); } /* left diagram */ path dodecagon = shiftL+(r,0)--shiftL+r*dir(30); for(int i = 1; i < 12; ++i) dodecagon = dodecagon--shiftL+r*dir(i*30); dodecagon = dodecagon--cycle; filldraw(dodecagon, rgb(0.5,1,0.5)); draw(Circle(shiftL, r), linetype(\\\"2 2\\\")); dot((0,0)); draw(shiftL--shiftL+(r,0)); label(\\\"R\\\",shiftL+(r/2,0),S); /* right diagram */ for(int i = 0; i < 9; ++i) { filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.8,0.8,0)); if (i % 3 == 1) { filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir((i+1)*360/12)--cycle, rgb(0.8,0.8,0)); filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir(floor(i/3)*90)--cycle, rgb(0,0.8,0)); filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir((i+1)*360/12)--r*dir(floor(i/3)*90+90)--cycle, rgb(0,0.8,0)); } } for(int i = 9; i < 12; ++i) { filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype(\\\"2 2\\\")); filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype(\\\"2 2\\\")); filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(1,1,0.5), linetype(\\\"2 2\\\")); } [/asy]$ The area of a dodecagon is $3R^2$, where $R$ is the circumradius. $[asy] pathpen = linewidth(1); unitsize(15); pen dotted = linetype(\\\"2 4\\\"); path xaxis = (-3,0)--(3,0); pair A = (-2,2), B = (1.5,1.5), B3 = (-1.5,0), B2 = (B.x,-B.y), C2 = IP(xaxis, A--B2); D(xaxis,Arrows(8)); D(D(A)--D(C2)--D(B)); D(D(B2)--C2,dashed+linewidth(0.7)); D(A--D(B3)--B,dotted+linewidth(0.7)); D(B3--B2,dotted); MP(\\\"(a,b)\\\",A,W); MP(\\\"(c,d)\\\",B,E); MP(\\\"(c,-d)\\\",B2,E); [/asy]$ The smallest distance necessary to travel between $(a,b)$, the x-axis, and then $(c,d)$ for $b,d > 0$ is given\",\n \"([shift=(nw)] -.5, .5); \\\\coordinate[label=right:$A_2$] (A2) at ([shift=(ne)] .5, .5); \\\\coordinate[label=right:$A_3$] (A3) at ([shift=(se)] .5, -.5); \\\\coordinate[label=left:$A_4$] (A4) at ([shift=(sw)] -.5, -.5); % Grid and outer layers of detector \\\\draw[black, very thick] (A1) rectangle (A3); \\\\draw[black, very thick] (sw) rectangle (ne); \\\\draw[step=5mm,black] (sw) grid (ne); % Anode wiring \\\\draw (A1) -- (nw); \\\\draw (A2) -- (ne); \\\\draw (A3) -- (se); \\\\draw (A4) -- (sw); %Colour in distorted areas \\\\draw[fill=white] (sw) parabola bend +(2, .6) (se) -- cycle; \\\\draw[fill=white] (nw) parabola bend +(2, -.6) (ne) -- cycle; \\\\begin{scope}[shift=(sw), rotate=90] \\\\draw[fill=white] (0,0) parabola bend +(2, -.6) ++(4, 0) -- cycle; \\\\end{scope} \\\\begin{scope}[shift=(se), rotate=90] \\\\draw[fill=white] (0,0) parabola bend +(2, .6) ++(4, 0) -- cycle; \\\\end{scope} \\\\end{tikzpicture} \\\\end{document} • Thank you jakun! It looks fantastic! Just one quick question. With the lines which end in '-- cycle'. Does that just cycle around the 3 points earlier to create a shape which is then filled in? – AlphaBetaGamma96 May 4 '17 at 18:37 • @AlphaBetaGamma96 \\\"The --cycle causes the current path to be closed (actually the current part of the current path) by smoothly joining the first and last point.\\\" see tikz documentation page 36/37. – jakun May 5 '17 at 5:22 You've used bend left/bend right earlier in the drawing, use that again, and it works fine. For example \\\\draw[fill=white] (8.5,0) to[bend right]\",\n \"less. Let $S=\\\\{p_1(x), p_2(x), p_3(x), p_4(x)\\\\},$ where \\\\begin{align*} p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\\\\\\\ p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3. \\\\end{align*} (a) Find a basis $Q$ of the span $\\\\Span(S)$ consisting of polynomials in $S$. (b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$. (The Ohio State University, Linear Algebra Midterm) ## Problem 606 Let $V$ be a vector space and $B$ be a basis for $V$. Let $\\\\mathbf{w}_1, \\\\mathbf{w}_2, \\\\mathbf{w}_3, \\\\mathbf{w}_4, \\\\mathbf{w}_5$ be vectors in $V$. Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\\\\mathbf{w}_1, \\\\mathbf{w}_2, \\\\mathbf{w}_3, \\\\mathbf{w}_4, \\\\mathbf{w}_5$ with respect to the basis $B$. After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form $\\\\begin{bmatrix} 1 & 0 & 2 & 1 & 0 \\\\\\\\ 0 & 1 & 3 & 0 & 1 \\\\\\\\ 0 & 0 & 0 & 0 & 0 \\\\\\\\ 0 & 0 & 0 & 0 & 0 \\\\end{bmatrix}.$ (a) What is the dimension of $V$? (b) What is the dimension of $\\\\Span\\\\{\\\\mathbf{w}_1, \\\\mathbf{w}_2, \\\\mathbf{w}_3, \\\\mathbf{w}_4, \\\\mathbf{w}_5\\\\}$? (The Ohio State University, Linear Algebra Midterm) ## Problem 604 Let $A=\\\\begin{bmatrix} 1 & -1 & 0 & 0 \\\\\\\\ 0 &1 & 1 & 1 \\\\\\\\ 1 & -1 & 0 & 0\",\n \"21. Let $A$ be a square matrix such that $A^{\\\\trans}A=A$, where $A^{\\\\trans}$ is the transpose matrix of $A$. Prove that $A$ is idempotent, that is, $A^2=A$. Also, prove that $A$ is a symmetric matrix. 22. Let $A$ and $B$ be $n \\\\times n$ real symmetric matrices. Prove the followings. (a) The product $AB$ is symmetric if and only if $AB=BA$. (b) If the product $AB$ is a diagonal matrix, then $AB=BA$. 23. A $2 \\\\times 2$ matrix has two parallel columns and $\\\\tr(A)=5$. Find $\\\\tr(A^2)$. 24. Let $I$ be the $n\\\\times n$ identity matrix, where $n$ is a positive integer. Prove that there are no $n\\\\times n$ matrices $X$ and $Y$ such that $XY-YX=I$. 25. Let $A=(a_{i j})$ and $B=(b_{i j})$ be $n\\\\times n$ real matrices for some $n \\\\in \\\\N$. (a) Express $\\\\tr(AB^{\\\\trans})$ in terms of the entries of the matrices $A$ and $B$. Here $B^{\\\\trans}$ is the transpose matrix of $B$. (b) Show that $\\\\tr(AA^{\\\\trans})$ is the sum of the square of the entries of $A$. (c) Show that if $A$ is nonzero symmetric matrix, then $\\\\tr(A^2)>0$. 26. For each of the following matrix $A$, prove that $\\\\mathbf{x}^{\\\\trans}A\\\\mathbf{x} \\\\geq 0$ for all vectors $\\\\mathbf{x}$ in $\\\\R^2$. Also, determine those vectors $\\\\mathbf{x}\\\\in \\\\R^2$ such that $\\\\mathbf{x}^{\\\\trans}A\\\\mathbf{x}=0$. (a) $A=\\\\begin{bmatrix} 4 & 2\\\\\\\\ 2& 1 \\\\end{bmatrix}$. (b) $A=\\\\begin{bmatrix} 2\"\n]"}}},{"rowIdx":6934,"cells":{"problem":{"kind":"string","value":"Find all values of $a$ for which the points $(0,0,0),$ $(1,a,0),$ $(0,1,a),$ and $(a,0,1)$ are coplanar."},"level":{"kind":"string","value":"Level 5"},"type":{"kind":"string","value":"Precalculus"},"solution":{"kind":"string","value":"If the points $(0,0,0),$ $(1,a,0),$ $(0,1,a),$ and $(a,0,1)$ are coplanar, then the parallelepiped generated by the corresponding vectors $\\begin{pmatrix} 1 \\\\ a \\\\ 0 \\end{pmatrix},$ $\\begin{pmatrix} 0 \\\\ 1 \\\\ a \\end{pmatrix},$ and $\\begin{pmatrix} a \\\\ 0 \\\\ 1 \\end{pmatrix}$ has a volume of 0. Thus,\n\\[\\begin{vmatrix} 1 & 0 & a \\\\ a & 1 & 0 \\\\ 0 & a & 1 \\end{vmatrix} = 0.\\]Expanding the determinant, we get\n\\begin{align*}\n\\begin{vmatrix} 1 & 0 & a \\\\ a & 1 & 0 \\\\ 0 & a & 1 \\end{vmatrix} &= 1 \\begin{vmatrix} 1 & 0 \\\\ a & 1 \\end{vmatrix} + a \\begin{vmatrix} a & 1 \\\\ 0 & a \\end{vmatrix} \\\\\n&= 1((1)(1) - (0)(a)) + a((a)(a) - (1)(0)) \\\\\n&= a^3 + 1.\n\\end{align*}Then $a^3 + 1 = 0,$ so $a = \\boxed{-1}.$"},"answer":{"kind":"string","value":"-1"},"p_retrievals":{"kind":"list like","value":["Using vector method, show that the points Question: Using vector method, show that the points $A(4,5,1), B(0,-1,-1), C(3,9,4)$ and $D(-4,4,4)$ are coplanar. Solution:","# The vectors $\\hat{i}+3\\hat{j},5\\hat{k}$ and $\\lambda\\hat{i}-\\hat{j}$ are coplanar.Find the value of $\\lambda$. This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com","of all points. consider the example $$avg(avg(0,0,0),avg(1,1)) = 0.5 \\neq 0.4 = avg(0,0,0,1,1)$$.","Want to ask us a question? Click here Browse Questions Ad –1 vote # The vectors $\\hat{i}+3\\hat{j},5\\hat{k}$ and $\\lambda\\hat{i}-\\hat{j}$ are coplanar.Find the value of $\\lambda$. This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com Can you answer this question? 0 votes 2 answers 0 votes 0 answers 0 votes 0 answers 0 votes 0 answers 0 votes 0 answers 0 votes 0 answers 0 votes 0 answers","# The lines $\\frac{x-2}{1} = \\frac{y-3}{1} = \\frac{z-4}{-k}$ and $\\frac{x-1}{k} = \\frac{y-4}{2} = \\frac{z-5}{1}$ are coplanar if : ( A ) $k=1$ or $-1$ ( B ) $k=3$ or $-3$ ( C ) $k=0$ or $-1$ ( D ) $k=0$ or $-3$","from zero (unless $P_0 .. P_n$ are all precisely coplanar) and the solution sought for is, at last, the column of the $3 \\times 3$ right orthogonal matrix $V_L$ corresponding to the smallest singular value.","point$(3,1,2)\\$. ...","$-$ Position vector of A Position vector of C $-$ Position vector of B. are parallel vectors. But $B$ is a point common to them. Hence, the given points and $C$ are collinear. (iv) Given the points and . Then, Position vector of B $-$ Position vector of A Position vector of C $-$ Position vector of B are parallel vectors. But $B$ is a point common to them. Hence, the given points and $C$ are collinear. #### Question 3: If $\\stackrel{\\to }{a}$, $\\stackrel{\\to }{b}$, $\\stackrel{\\to }{c}$ are non-zero, non-coplanar vectors, prove that the following vectors are coplanar: (i) (ii) (i) The three vectors are coplanar if one of them is expressible as a linear combination of the other two . Let Solving first two of these equations, we get . Clearly, these values of x and y satisfies the third equation. Hence, the given vectors are coplanar. (ii) The three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let Solving first two of these equations, we get . These values of x and y does not satisfy the third equation. Hence, the given vectors are not coplanar. #### Question 4: Show that the four points having position vectors are coplanar. Let the given four points be and $S$","of all points}.$$ It is like asking for an equation describing all of $$\\mathbb R^3$$, as a subset of $$\\mathbb R^3$$.","For $x=10$ we get $y=0$ Exists in table (hence it is a complete square) So corresponding point is: $(10,0)$ So all the affine points are $(0,1)$ $(0,10)$ $(1,0)$ $(2,2)$$(2,9)$ $(9,2)$ $(9,9)$ $(10,0)$","# If the points $(1,2)$ and $(3,4)$ were to be on the same side of the line $3x -5y +a =0$, then: ( A ) $a < 7$ or $a > 11$ ( B ) $7 < a < 11$ ( C ) $a = 11$ ( D ) $a=7$","1 - 4 = 5$ $x^2 + 2x - 8 = 0$ $(x + 4)(x - 2) = 0$ $x = -4, 2$ So the two points are (-4, -1), and (2, -1).","1e4)) points(x[m],y[m], pch=19, cex=1/2, col=\"#0000e010\")","$$\\mathbb{Q}$$, you can find torsion points of order $$m-1=1,2,3,4,5,6,7,8,9,10$$, and $$12$$.","## prove coplanarity Here is the task that I need to solve: \"Let a,b,c are non-coplanar vectors and p,q,r are three vectors. Show that p,q,r are non-coplanar if and only if there exist numbers $x_i,y_i,z_i (i=1,2,3)$ where $a=x_1p + x_2q+x_3r, b=y_1p+y_2q+y_3r, c=z_1p+z_2q+z_3r$\" I know that I need to show it in both ways that is: If p,q,r are non-coplanar then there exist numbers $x_i,y_i,z_i (i=1,2,3)$ where $a=x_1p + x_2q+x_3r, b=y_1p+y_2q+y_3r, c=z_1p+z_2q+z_3r$ and If exist numbers $x_i,y_i,z_i (i=1,2,3)$ where $a=x_1p + x_2q+x_3r, b=y_1p+y_2q+y_3r, c=z_1p+z_2q+z_3r$ then p,q,r are non-coplanar. I will begin with the second case. I need to show that u*p+v*q+w*r=0 leads to u=v=w=0. I know the fact that m*a+n*b+l*c=0 leads to m=n=l=0 So that $m*(x_1p + x_2q+x_3r)+n*(y_1p+y_2q+y_3r)$+ $l*(z_1p+z_2q+z_3r)=0$ and $ p*(x_1*m+x_2*m+x_3*m)+q*(y_1*n+y_2*n+y_3*n)$ + $r*(z_1*l+z_2*l+z_3*l)=0 $ Only I do not know if its valid to take: $u=(x_1*m+x_2*m+x_3*m)$ $v=(y_1*n+y_2*n+y_3*n)$ $r=(z_1*l+z_2*l+z_3*l)$ In this case because of m=n=l=0 then u=v=r=0.","at two points in $[0, 1]$.","$0 < a < 1$. If $a = 1$ it defines the point (0, 0).","$0$. Find the location of: $A[3,4,5]$ $A[1,2,7]$ $A[4,3,9]$","find that, for instance, $$4.2$$ in the 7 point scale corresponds to a higher value than $$4.8$$ in the 9 point scale.","1)$, $(2, 0)$, $(1, -1)$, $(0, -2)$, $(-1, -1)$, $(-2, 0)$ as well as $(0, 0)$, for a total of $9$ moves. If the object makes $3$ moves, it can end up at $(0, 3)$, $(2, 1)$, $(1, 2)$, $(3, 0)$, $(2, -1)$, $(1, -2)$, $(0, -3)$, $(-1, -2)$, $(-2, -1)$, $(0, -3)$ etc. for a total of $16$ moves. At this point we can guess that for n moves, there are $(n + 1)^2$ different endpoints. Thus, for 10 moves, there are $11^2 = 121$ endpoints, and the answer is $\\boxed{B}$."],"string":"[\n \"Using vector method, show that the points Question: Using vector method, show that the points $A(4,5,1), B(0,-1,-1), C(3,9,4)$ and $D(-4,4,4)$ are coplanar. Solution:\",\n \"# The vectors $\\\\hat{i}+3\\\\hat{j},5\\\\hat{k}$ and $\\\\lambda\\\\hat{i}-\\\\hat{j}$ are coplanar.Find the value of $\\\\lambda$. This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com\",\n \"of all points. consider the example $$avg(avg(0,0,0),avg(1,1)) = 0.5 \\\\neq 0.4 = avg(0,0,0,1,1)$$.\",\n \"Want to ask us a question? Click here Browse Questions Ad –1 vote # The vectors $\\\\hat{i}+3\\\\hat{j},5\\\\hat{k}$ and $\\\\lambda\\\\hat{i}-\\\\hat{j}$ are coplanar.Find the value of $\\\\lambda$. This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com Can you answer this question? 0 votes 2 answers 0 votes 0 answers 0 votes 0 answers 0 votes 0 answers 0 votes 0 answers 0 votes 0 answers 0 votes 0 answers\",\n \"# The lines $\\\\frac{x-2}{1} = \\\\frac{y-3}{1} = \\\\frac{z-4}{-k}$ and $\\\\frac{x-1}{k} = \\\\frac{y-4}{2} = \\\\frac{z-5}{1}$ are coplanar if : ( A ) $k=1$ or $-1$ ( B ) $k=3$ or $-3$ ( C ) $k=0$ or $-1$ ( D ) $k=0$ or $-3$\",\n \"from zero (unless $P_0 .. P_n$ are all precisely coplanar) and the solution sought for is, at last, the column of the $3 \\\\times 3$ right orthogonal matrix $V_L$ corresponding to the smallest singular value.\",\n \"point$(3,1,2)\\\\$. ...\",\n \"$-$ Position vector of A Position vector of C $-$ Position vector of B. are parallel vectors. But $B$ is a point common to them. Hence, the given points and $C$ are collinear. (iv) Given the points and . Then, Position vector of B $-$ Position vector of A Position vector of C $-$ Position vector of B are parallel vectors. But $B$ is a point common to them. Hence, the given points and $C$ are collinear. #### Question 3: If $\\\\stackrel{\\\\to }{a}$, $\\\\stackrel{\\\\to }{b}$, $\\\\stackrel{\\\\to }{c}$ are non-zero, non-coplanar vectors, prove that the following vectors are coplanar: (i) (ii) (i) The three vectors are coplanar if one of them is expressible as a linear combination of the other two . Let Solving first two of these equations, we get . Clearly, these values of x and y satisfies the third equation. Hence, the given vectors are coplanar. (ii) The three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let Solving first two of these equations, we get . These values of x and y does not satisfy the third equation. Hence, the given vectors are not coplanar. #### Question 4: Show that the four points having position vectors are coplanar. Let the given four points be and $S$\",\n \"of all points}.$$ It is like asking for an equation describing all of $$\\\\mathbb R^3$$, as a subset of $$\\\\mathbb R^3$$.\",\n \"For $x=10$ we get $y=0$ Exists in table (hence it is a complete square) So corresponding point is: $(10,0)$ So all the affine points are $(0,1)$ $(0,10)$ $(1,0)$ $(2,2)$$(2,9)$ $(9,2)$ $(9,9)$ $(10,0)$\",\n \"# If the points $(1,2)$ and $(3,4)$ were to be on the same side of the line $3x -5y +a =0$, then: ( A ) $a < 7$ or $a > 11$ ( B ) $7 < a < 11$ ( C ) $a = 11$ ( D ) $a=7$\",\n \"1 - 4 = 5$ $x^2 + 2x - 8 = 0$ $(x + 4)(x - 2) = 0$ $x = -4, 2$ So the two points are (-4, -1), and (2, -1).\",\n \"1e4)) points(x[m],y[m], pch=19, cex=1/2, col=\\\"#0000e010\\\")\",\n \"$$\\\\mathbb{Q}$$, you can find torsion points of order $$m-1=1,2,3,4,5,6,7,8,9,10$$, and $$12$$.\",\n \"## prove coplanarity Here is the task that I need to solve: \\\"Let a,b,c are non-coplanar vectors and p,q,r are three vectors. Show that p,q,r are non-coplanar if and only if there exist numbers $x_i,y_i,z_i (i=1,2,3)$ where $a=x_1p + x_2q+x_3r, b=y_1p+y_2q+y_3r, c=z_1p+z_2q+z_3r$\\\" I know that I need to show it in both ways that is: If p,q,r are non-coplanar then there exist numbers $x_i,y_i,z_i (i=1,2,3)$ where $a=x_1p + x_2q+x_3r, b=y_1p+y_2q+y_3r, c=z_1p+z_2q+z_3r$ and If exist numbers $x_i,y_i,z_i (i=1,2,3)$ where $a=x_1p + x_2q+x_3r, b=y_1p+y_2q+y_3r, c=z_1p+z_2q+z_3r$ then p,q,r are non-coplanar. I will begin with the second case. I need to show that u*p+v*q+w*r=0 leads to u=v=w=0. I know the fact that m*a+n*b+l*c=0 leads to m=n=l=0 So that $m*(x_1p + x_2q+x_3r)+n*(y_1p+y_2q+y_3r)$+ $l*(z_1p+z_2q+z_3r)=0$ and $ p*(x_1*m+x_2*m+x_3*m)+q*(y_1*n+y_2*n+y_3*n)$ + $r*(z_1*l+z_2*l+z_3*l)=0 $ Only I do not know if its valid to take: $u=(x_1*m+x_2*m+x_3*m)$ $v=(y_1*n+y_2*n+y_3*n)$ $r=(z_1*l+z_2*l+z_3*l)$ In this case because of m=n=l=0 then u=v=r=0.\",\n \"at two points in $[0, 1]$.\",\n \"$0 < a < 1$. If $a = 1$ it defines the point (0, 0).\",\n \"$0$. Find the location of: $A[3,4,5]$ $A[1,2,7]$ $A[4,3,9]$\",\n \"find that, for instance, $$4.2$$ in the 7 point scale corresponds to a higher value than $$4.8$$ in the 9 point scale.\",\n \"1)$, $(2, 0)$, $(1, -1)$, $(0, -2)$, $(-1, -1)$, $(-2, 0)$ as well as $(0, 0)$, for a total of $9$ moves. If the object makes $3$ moves, it can end up at $(0, 3)$, $(2, 1)$, $(1, 2)$, $(3, 0)$, $(2, -1)$, $(1, -2)$, $(0, -3)$, $(-1, -2)$, $(-2, -1)$, $(0, -3)$ etc. for a total of $16$ moves. At this point we can guess that for n moves, there are $(n + 1)^2$ different endpoints. Thus, for 10 moves, there are $11^2 = 121$ endpoints, and the answer is $\\\\boxed{B}$.\"\n]"},"s_retrievals":{"kind":"list like","value":["## Change of variables, from dxdy to dudv Suppose we have a function $f(x, y)$ and we rewrite it as $f(u(x, y), v(x, y))$, i.e. perform a change of variables, then the caculation of area using double integral can be carried out with these new variables $u, v$. But we have to figure out the relation between $dudv$ and $dxdy$. Here is the thing, with the variables $x, y$, $dxdy$ is the area of the parallelogram between the vectors $(dx, 0)$ and $(0, dy)$, under the new variables, these vectors are transformed into two new vectors $(u_x dx, v_x dx)$, $(u_y dy, v_y dy)$, and $dudv$ is the area of the parallelogram between these two new vectors. $\\begin{array}{cc} (dx, 0) & (u_x dx, v_x dx) \\\\ (0, dy) & (u_y dy, v_y dy) \\end{array}$ To get the area of this parallelogram, we need the determinant of this matrix: \\begin{align*} M &= \\begin{pmatrix} u_x dx & u_y dy \\\\ v_x dx & v_y dy \\end{pmatrix} \\\\ &= \\begin{pmatrix} u_x & u_y \\\\ v_x & v_y \\end{pmatrix} \\begin{pmatrix} dx & 0 \\\\ 0 & dy \\end{pmatrix} \\\\ &= J \\begin{pmatrix} dx & 0 \\\\ 0 & dy \\end{pmatrix}, \\end{align*} where $J$ stands for the Jacobian matrix of $f$. So the determinant is \\begin{align*} \\det(M) &= \\det(J) dxdy \\end{align*} Remember in","$(1,n)$-entry are $-1$. The other entries are zero. ## Hint. 1. Calculate the first row cofactor expansion. 2. The determinant of a triangular matrix is the product of its diagonal entries. ## Solution. Apply the cofactor expansion corresponding to the first row. We obtain \\begin{align*} \\det(A)&= \\begin{vmatrix} 1 & 0 & \\dots & 0 & 0 & 0 \\\\ 1 & 1 & \\dots & 0 & 0 & 0 \\\\ \\vdots & \\vdots & \\dots & \\ddots & \\vdots & \\vdots \\\\ 0 & 0 &\\dots & 1 & 1 & 0\\\\ 0 & 0 &\\dots & 0 & 1 & 1 \\end{vmatrix} +(-1)^{n+1} \\begin{vmatrix} 1 & 1 & 0 & \\dots & 0 & 0 \\\\ 0 & 1 & 1 & \\dots & 0 & 0 \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\vdots \\\\ 0 & 0 & 0 &\\dots & 1 & 1 \\\\ 0 & 0 & 0 &\\dots & 0 & 1 \\end{vmatrix} \\end{align*} The two smaller (minor) $n-1 \\times n-1$ matrices are both triangular. The determinant of a triangular matrix is the product of its diagonal entries. Thus we see that \\begin{align*} \\det(A)&=1+(-1)^{n+1} \\\\ &= \\begin{cases} 2 & \\text{ if } n \\text{ is odd}\\\\ 0 & \\text{ if } n \\text{ is even}. \\end{cases} \\end{align*}","= [1,-2,1]$$. • Solve $$Uw = v_3$$: \\begin{align*} \\begin{bmatrix} 3 & 1 & -1 & 1 \\\\ -6 & -2 & 2 & -2 \\\\ 2 & 1 & 0 & 1 \\end{bmatrix} \\leadsto \\begin{bmatrix} 1 & 0 & -1 & 0 \\\\ 0 & 1 & 2 & 1 \\\\ 0 & 0 & 0 & 0 \\end{bmatrix} ,\\end{align*} so take $$v_3 = [0,1,0]$$. • Putting things together: \\begin{align*} A &= P^{-1}J P \\text{ where } \\\\ J = J_1(\\lambda = 2) \\oplus J_2(\\lambda = 1) &= \\begin{bmatrix} 2 & 0 & 0 \\\\ 0 & 1 & 1 \\\\ 0 & 0 & 1 \\end{bmatrix} \\\\ P = [v_1, v_2, v_3] &= \\begin{bmatrix} 1 & 1 & 0 \\\\ -2 & -2 & 1 \\\\ 0 & 1 & 0 \\end{bmatrix} .\\end{align*} • Write $$\\min_A(t) = (t-2)(t-1)^{\\ell_1}$$, then since $$\\min_A(t)$$ divides $$\\chi_A(t)$$ either $$\\ell_1 = 1, 2$$. • $$\\ell_1$$ is the size of the largest block corresponding to $$\\lambda = 1$$, which is size 2, so $$\\lambda_1=2$$. • Thus \\begin{align*} \\min_A(t) = (t-2)(t-1)^2 .\\end{align*} ## Fall 2020 #5#algebra/qual/work Consider the following matrix: \\begin{align*} B \\coloneqq \\begin{bmatrix} 1 & 3 & 3 \\\\ 2 & 2 & 3 \\\\ -1 & -2 & -2 \\end{bmatrix} .\\end{align*} • Find the minimal polynomial of $$B$$. • Find a $$3\\times","upon a pair of entries of which are not both zero. So, for example, when, the matrix can be \\begin{pmatrix} 0&-1&0\\\\ 1&0&0\\\\ 0&0&-1\\\\ \\end{pmatrix}. For the vector, the resulting simplex has vertices \\begin{pmatrix}1\\ 0\\ 1/\\surd2\\end{pmatrix}, \\begin{pmatrix}0\\ 1\\ -1/\\surd2\\end{pmatrix}, \\begin{pmatrix}-1\\ 0\\ 1/\\surd2\\end{pmatrix}, \\begin{pmatrix}0\\ -1\\ -1/\\surd2\\end{pmatrix}, each of which has distance 2 from the others. Geometric properties Volume The volume of an n-simplex in n-dimensional space with vertices (v0, ..., vn) is Volume= 1 n! \\left|\\det \\begin{pmatrix} v1-v0&&v2-v0&&&&vn-v0 \\end{pmatrix}\\right| where each column of the n × n determinant is a vector that points from vertex to another vertex .[6] This formula is particularly useful when v0 is the origin. The expression Volume= 1 n! T \\det\\left[ \\begin{pmatrix} v 0 T \\ v 0 \\\\vdots T \\end{pmatrix} \\begin{pmatrix} v \\ v 1-v 0&v2-v0&&vn-v 1/2 0 \\end{pmatrix} \\right] employs a Gram determinant and works even when the n-simplex's vertices are in a Euclidean space with more than n dimensions, e.g., a triangle in R3 . A more symmetric way to compute the volume of an n-simplex in Rn is Volume={1\\overn!}\\left|\\det \\begin{pmatrix} v0&v1&&vn\\\\ 1&1&&1 \\end{pmatrix}\\right|. Another common way of computing the volume of the simplex is via the Cayley–Menger determinant, which works even when the n-simplex's vertices are in a Euclidean space with more than n dimensions. Without the 1/n! it is the","\\\\ &\\; \\; \\; + (\\hat c - \\hat dx_1)y + \\hat d xy \\end{align*} It follows that this solution would uniquely determine the desired bilinear interpolant $$f(x,y)$$ at the original points: $$\\begin{pmatrix} a \\\\ b \\\\ c \\\\ d \\end{pmatrix} = \\begin{pmatrix} 1 & -x_1 & -y_1 & x_1y_1 \\\\ 0 & 1 & 0 & -y_1 \\\\ 0 & 0 & 1 & -x_1 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} \\hat a \\\\ \\hat b \\\\ \\hat c \\\\ \\hat d \\end{pmatrix}$$ In that sense we can assume without loss of generality that $$(x_1,y_1) = (0,0)$$, and we proceed to find the determinant of $$M$$ in this slightly special case: \\begin{align*} \\det M &= \\left| \\begin{array}{cccc} 1 & 0 & 0 & 0 \\\\ 1 & x_2 & y_2 & x_2 y_2 \\\\ 1 & x_3 & y_3 & x_3 y_3 \\\\ 1 & x_4 & y_4 & x_4 y_4 \\end{array} \\right| \\\\ &= \\left| \\begin{array}{ccc} x_2 & y_2 & x_2 y_2 \\\\ x_3 & y_3 & x_3 y_3 \\\\ x_4 & y_4 & x_4 y_4 \\end{array} \\right| \\\\ &= x_2x_4y_3(y_4-y_2) + x_3x_4y_2(y_3-y_4) + x_2x_3y_4(y_2-y_3) \\end{align*} With this expression we can identify some geometric conditions on the four points for which the matrix has zero determinant and is thus not invertible. One","\\hat j & \\hat k \\\\ 1 & -3 & 0 \\\\ 0 & -2 & 1 \\end{vmatrix} = (-3, -1, -2) \\begin{vmatrix} \\hat i & \\hat j & \\hat k \\\\ 1 & 0 & 1 \\\\ 1 & -3 & 1 \\end{vmatrix} = (3, 0, -3) \\begin{vmatrix} \\hat i & \\hat j & \\hat k \\\\ 4 & -2 & 5 \\\\ 0 & 2 & 1 \\end{vmatrix} = (-12, -4, 8)$ Then I find a point for each line: $P_r (0, 1, 3), P_s (2, 2, 2), P_t (-1, -1, 1)$ After that I do the determinant with the first row as the difference between the two points of the two lines, and on the second and third raw I put the directional vectors of the two lines. For $r$ and $s$: $\\begin{vmatrix} 0-2 & 1-2 & 3-2 \\\\ -3 & -1 & -2 \\\\ 3 & 0 & -3 \\end{vmatrix} = 12$ Non-coplanar. $\\begin{vmatrix} 2-(-1) & 2-(-1) & 2-1 \\\\ 3 & 0 & -3 \\\\ -12 & -4 & 8 \\\\ \\end{vmatrix} = -10$ Non-coplanar. $\\begin{vmatrix} 0-(-1) & 1-(-1) & 3-1 \\\\ -3 & -1 & -2 \\\\ -12 & -4 & 8 \\\\ \\end{vmatrix} = 80$ Non-coplanar. The thing is that all three of them are askew (non-coplanar). Could someone check it","with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ is $\\frac{1}{2} \\begin{vmatrix} x_1 &x_2 &x_3 \\\\ y_1 &y_2 &y_3 \\\\ 1 &1 &1 \\end{vmatrix}.$ 3. (Bittinger 1973) Prove that the area of a triangle with vertices at $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ whose coordinates are integers has an area of $N$ or $N/2$ for some positive integer $N$. ## Section III - Other Formulas for Determinants (This section is optional. Later sections do not depend on this material.) Determinants are a fount of interesting and amusing formulas. Here is one that is often seen in calculus classes and used to compute determinants by hand. ### 1 - Laplace's Expansion Example 1.1 In this permutation expansion \\begin{array}{rl} \\begin{vmatrix} t_{1,1} &t_{1,2} &t_{1,3} \\\\ t_{2,1} &t_{2,2} &t_{2,3} \\\\ t_{3,1} &t_{3,2} &t_{3,3} \\end{vmatrix} &=\\begin{align} &t_{1,1}t_{2,2}t_{3,3}\\begin{vmatrix} 1 &0 &0 \\\\ 0 &1 &0 \\\\ 0 &0 &1 \\end{vmatrix} +t_{1,1}t_{2,3}t_{3,2}\\begin{vmatrix} 1 &0 &0 \\\\ 0 &0 &1 \\\\ 0 &1 &0 \\end{vmatrix} \\\\ &\\quad +t_{1,2}t_{2,1}t_{3,3}\\begin{vmatrix} 0 &1 &0 \\\\ 1 &0 &0 \\\\ 0 &0 &1 \\end{vmatrix} +t_{1,2}t_{2,3}t_{3,1}\\begin{vmatrix} 0 &1 &0 \\\\ 0 &0 &1 \\\\ 1 &0 &0 \\end{vmatrix} \\\\ &\\quad +t_{1,3}t_{2,1}t_{3,2}\\begin{vmatrix} 0 &0 &1 \\\\ 1 &0 &0 \\\\ 0 &1 &0 \\end{vmatrix} +t_{1,3}t_{2,2}t_{3,1}\\begin{vmatrix} 0 &0 &1 \\\\ 0 &1 &0 \\\\ 1 &0 &0 \\end{vmatrix} \\end{align} \\end{array} we can, for instance, factor out the entries","trick on the second column gives $$\\begin{vmatrix} a_{11} & a_{12}\\\\ a_{21} & a_{22} \\end{vmatrix} = a_{11} \\left (a_{12} \\underbrace{\\begin{vmatrix} 1 & 1\\\\ 0 & 0 \\end{vmatrix}}_{0} + a_{22}\\underbrace{\\begin{vmatrix} 1 & 0\\\\ 0 & 1 \\end{vmatrix}}_{1} \\right )+ a_{21} \\left( a_{12}\\underbrace{\\begin{vmatrix} 0 & 1\\\\ 1 & 0 \\end{vmatrix}}_{-1} + a_{22} \\underbrace{\\begin{vmatrix} 0 & 0\\\\ 1& 1 \\end{vmatrix}}_{0} \\right).$$ (7.24) In the above equation, the first and the last determinants are zero because they have two identical columns. The second determinant is $1$ according to property (i) of the definition. The third determinant is $-1$ since $$\\begin{vmatrix} 0 & 1\\\\ 1 & 0 \\end{vmatrix} = - \\begin{vmatrix} 1 & 0\\\\ 0 & 1 \\end{vmatrix} = -1 ,$$ (7.25) where we used property $(v)$ to swap two columns and then property $(i)$. Thus we have showed that the definition of the determinant gives that $$\\begin{vmatrix} a_{11} & a_{12}\\\\ a_{21} & a_{22} \\end{vmatrix} = a_{11} a_{22} - a_{21} a_{12} ,$$ (7.26) which coincides with our earlier definition. In this example we calculated a $2 \\times 2$ determinant by expanding each column. By doing this for a general $n \\times n$ matrix, one obtains an explicit expression for the determinant. This expression is best expressed using the concept of a permutation of $n$ numbers. Definition 7.2: A permutation $p$ of $n$ elements is","& a_{32} & a_{33} & a_{34}\\\\a_{41} & a_{42} & a_{43} & a_{44}\\end{pmatrix}$ . . Find determinant. Can I assume you know the \"Cofactor\" method? $\\text{det }\\!\\!A \\;=\\;a_{11}\\begin{vmatrix}a_{22}&a_{23}&a_{24} \\\\ a_{32}&a_{33}&a_{34} \\\\ a_{42}&a_{43}&a_{44} \\end{vmatrix} - a_{12}\\begin{vmatrix}a_{21}&a_{23}&a_{24}\\\\a_{31}& a_{33}&a_{34}\\\\a_{41}&a_{43}&a_{44} \\end{vmatrix}$ . $+\\:a_{13}\\begin{vmatrix}a_{21}&a_{22}&a_{24}\\\\a_{3 1}&a_{32}&a_{34} \\\\ a_{41}&a_{42}&a_{44}\\end{vmatrix} - a_{14}\\begin{vmatrix}a_{21}& a_{22}&a_{23}\\\\a_{31}&a_{32}&a_{33}\\\\a_{41}&a_{42} &a_{43} \\end{vmatrix}$ . . . $= \\;a_{11}\\bigg( a_{22}\\begin{vmatrix}a_{33}&a_{34}\\\\a_{43}&a_{44}\\ end{vmatrix} - a_{23}\\begin{vmatrix}a_{32}&a_{34}\\\\a_{42}&a_{44}\\ end{vmatrix} + a_{24}\\begin{vmatrix}a_{32}&a_{33}\\\\a_{42}&a_{43}\\ end{vmatrix} \\bigg)$ . . . . . $-a_{12}\\bigg(a_{21}\\begin{vmatrix}a_{33}&a_{34}\\\\a_ {43}&a_{44}\\end{vmatrix} -a_{23}\\begin{vmatrix}a_{31}&a_{34}\\\\a_{41}&a_{44}\\ end{vmatrix} + a_{24}\\begin{vmatrix}a_{31}&a_{33}\\\\a_{41}&a_{43} \\end{vmatrix} \\bigg)$ . . . . . . . $+a_{13}\\bigg(a_{21}\\begin{vmatrix}a_{32}&a_{34}\\\\a _{42}&a_{44}\\end{vmatrix} -a_{22}\\begin{vmatrix}a_{31}&a_{34}\\\\a_{41}&a_{44}\\ end{vmatrix} + a_{24}\\begin{vmatrix}a_{31}&a_{32}\\\\a_{41}&a_{42}\\ end{vmatrix}\\bigg)$ . . . . . . . . . $-a_{14}\\bigg(a_{21}\\begin{vmatrix}a_{32}&a_{33}\\\\a_ {42}&a_{43} \\end{vmatrix} - a_{22}\\begin{vmatrix}a_{31}&a_{33}\\\\a_{41}&a_{43}\\ end{vmatrix} + a_{23}\\begin{vmatrix}a_{31}&a_{32}\\\\a_{41}&a_{42}\\ end{vmatrix} \\bigg)$ . . . and so on.","0 \\\\ 2x + 3y - 2z + 2 = 0.$$ Notice that, these are two equations in three variables, so you have a free variable say $z=t$, then we have $$x + 2y = 1-t \\\\ 2x + 3y = 2t-2.$$ Solving the last system gives $$\\left\\{ x=-7+7\\,t,y=4-4\\,t \\right\\} .$$ Then the parametrized equation of the line is given by $$(x,y,z)= (-7+7t, 4-4t,t)=(-7,4,0)+(7,-4,1)t .$$ The first plane has normal vector $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$$ and the second has normal vector $$\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $$v_1,v_2$$ is $$v_1\\times v_2$$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\times \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $$y=0$$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $$x=0, z=1$$, thus the line of intersection is $$\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$$ • Can you please elaborate as to what you did to find any random point on the line?I didn't quite understand – Karan Singh Apr 26 '16 at 9:01 • In your set of linear equations, where did that last '-1' (in equation 1) come from?. And where did that last '+2' (in equation 2) come from?. – loldrup May 21 '16 at 20:45 •","remaining uncovered $2 \\times 2$ matrix, then multiply that by $a_{1i}$. The determinant is the sum of those values, alternating addition and subtraction. Worked Examples Example 1 Find the determinant of the matrix $\\begin{pmatrix} 1 & 2 & 1\\\\ 0 & 3 & 4\\\\ 3 & 1 & 4 \\end{pmatrix}$. Solution \\begin{align} \\lvert \\mathbf{A} \\rvert &= 1\\begin{vmatrix} 3 & 4\\\\ 1 & 4 \\end{vmatrix} -2 \\begin{vmatrix} 0 & 4\\\\ 3 & 4 \\end{vmatrix} +1 \\begin{vmatrix} 0 & 3\\\\ 3 & 1 \\end{vmatrix}\\\\\\\\ &= 1(3\\times 4 - 4\\times 1) -2(0\\times 4 - 4\\times 3) +1(0 \\times 1 - 3\\times 3)\\\\ &= 1(12-4) -2(0-12) +1(0-9)\\\\ &= 8 +24-9\\\\ &=23 \\end{align} Example 2 Find the determinant of the matrix $\\begin{pmatrix} 1 & 0 & 3\\\\ -1 & -1 & -3\\\\ 0 & 0 & 6 \\end{pmatrix}$. Solution \\begin{align} \\lvert \\mathbf{A} \\rvert &= 1\\begin{vmatrix} -1 & -3\\\\ 0 & 6 \\end{vmatrix} -0 \\begin{vmatrix} -1 & -3\\\\ 0 & 6 \\end{vmatrix} +3 \\begin{vmatrix} -1 & -1\\\\ 0 & 0 \\end{vmatrix}\\\\\\\\ &= 1\\bigl((-1)\\times 6 - (-3)\\times 0\\bigr) -0(\\,\\dotso\\,) +3\\bigl((-1) \\times 0 - (-1)\\times 0\\bigr)\\\\ &= 1\\bigr((-6)-0\\bigl) -0 +3(0-0)\\\\ &=-6 +0\\\\ &=-6 \\end{align} Note: There was no need to work out the second $2 \\times 2$ determinant as it was being multiplied by zero. Video Example Hayley Bishop works out the determinant of the $3\\times","equal to the product of the elemnts on the diagonal. If you're not supposed to kniw that, just develop the determinant along the last column. You get $$(-1)^{n-1}(x+2)^{n-1}\\bigl((n-1)x-2\\bigr).$$ More generally: Given two numbers $$m$$ and $$x$$. Compute the determinant of the $$n\\times n$$-matrix whose entries on the main diagonal all equal $$m+x$$ and whose remaining $$n\\left(n-1\\right)$$ entries all equal $$m$$. Hint: The determinant is $$\\begin{vmatrix} m+x&m&m&\\dots &m &m\\\\ m&m+x&m&\\dots&m&m\\\\ m&m&m+x&\\dots &m&m\\\\ \\vdots & \\vdots &\\vdots &&\\vdots &\\vdots \\\\ m&m &m&\\dots &m+x& m \\\\ m&m&m&\\dots &m&m+x \\end{vmatrix}$$ Subtracting the 2nd row from the 1st, then the 3rd from the 2bd, &c., you obtain $$\\begin{vmatrix} x&-x&0&\\dots &0 &0\\\\ 0&x&-x&\\dots&0&0\\\\ 0&0&x&\\dots &0&0\\\\ \\vdots & \\vdots &\\vdots &&\\vdots &\\vdots \\\\ 0&0 &0&\\dots &x& -x\\\\ m&m&m&\\dots &m&m+x \\end{vmatrix} =x^{n-1}\\begin{vmatrix} 1&-1&0&\\dots &0 &0\\\\ 0&1&-1&\\dots&0&0\\\\ 0&0&1&\\dots &0&0\\\\ \\vdots & \\vdots &\\vdots &&\\vdots &\\vdots \\\\ 0&0 &0&\\dots &1& -1\\\\ m&m&m&\\dots &m&m+x \\end{vmatrix}$$ Can you prove inductively the latter determinant is equal to $$x+mn$$? You just have to play with the column vectors. Added: As @JeanMarie kindly reminded me, I (long ago) posted an answer to a similar problem which was simpler: First subtract the last row from each row above, to obtain $$\\begin{vmatrix} x&0 &0&\\dots &0 &-x\\\\ 0&x&0&\\dots&0&-x\\\\ 0&0&x&\\dots &0&-x\\\\ \\vdots & \\vdots &\\vdots &&\\vdots &\\vdots \\\\ 0&0 &0&\\dots &x& -x\\\\ m&m&m&\\dots &m&m+x \\end{vmatrix}$$ then add","-1 & 3 \\end{vmatrix}= 3\\begin{vmatrix} 2 & 1 \\\\ -1 & 3 \\end{vmatrix} - 0\\begin{vmatrix} 1 & -2 \\\\ -1 & 3 \\end{vmatrix} + 7\\begin{vmatrix} 1 & -2 \\\\ 2 & 1 \\end{vmatrix}=3(2\\cdot 3-1(-1))+7(1\\cdot 1-(-2)2)=56.$ Proposition. Let $$A$$ be an $$n\\times n$$ matrix, and let \\$1in. Then by expansion along the $$i$$-th row, we get $\\det(A)=\\sum_{j=1}^n(-1)^{i+j}a_{ij}\\det(A_{ij}).$ Let $$1\\le j\\le n$$. By expansion along the $$j$$-th column, we get $\\det(A)=\\sum_{i=1}^n(-1)^{i+j}a_{ij}\\det(A_{ij}).$ That is, we can compute the determinant by adding up the cofactors in any row or column. This gives $$2n$$ formulas for the determinant. Out of these $$2n$$ choices, we want to calculate the determinant along the row or column with the most zeros, since a zero entry makes the corresponding cofactor zero, and thus we don't have to calculate the determinant of the corresponding submatrix. Example. Let $A=\\begin{pmatrix} 2 & 0 & -1 \\\\ 1 & -2 & 2 \\\\ 3 & 0 & 8 \\end{pmatrix}.$ Using the definition, and expanding along the first column, we get $\\det(A)=2\\begin{vmatrix}-2 & 2 \\\\ 0 & 8 \\end{vmatrix}-1\\begin{vmatrix} 0 & -1 \\\\ 0 & 8 \\end{vmatrix} + 3 \\begin{vmatrix} 0 & -1 \\\\ -2 & 2 \\end{vmatrix} = 2((-2)8-2\\cdot0)-1(0\\cdot 8-(-1)0)+3(0\\cdot 2-(-1)(-2))=-38.$ Noticing that the second column has two zeros, we get $\\det(A)=-0\\det(A_{12})+(-2)\\begin{vmatrix}2 & -1 \\\\ 3 & 8 \\end{vmatrix}-0\\det(A_{32})=(-2)(2\\cdot 8-(-1)3)=-38.$ ##","allow us to stop writing $$\\det$$ over and over again: $\\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} := \\det\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$ Using the additivity of the bilinearity of the determinant on the first vector we can split this into: $\\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} = \\begin{vmatrix} a & b \\\\ 0 & d \\end{vmatrix} + \\begin{vmatrix} 0 & b \\\\ c & d \\end{vmatrix}$ And on the second vector...: $$\\label{additivity} ... = \\begin{vmatrix} a & b \\\\ 0 & 0 \\end{vmatrix} + \\begin{vmatrix} a & 0 \\\\ 0 & d \\end{vmatrix} + \\begin{vmatrix} 0 & b \\\\ c & 0 \\end{vmatrix} + \\begin{vmatrix} 0 & 0 \\\\ c & d \\end{vmatrix}$$ Then we can use homogeneity to extract the scalars: $$\\label{homogeneity} ... = ab\\begin{vmatrix} 1 & 1 \\\\ 0 & 0 \\end{vmatrix} + ad\\begin{vmatrix} 1 & 0 \\\\ 0 & 1 \\end{vmatrix} + bc\\begin{vmatrix} 0 & 1 \\\\ 1 & 0 \\end{vmatrix} + cd\\begin{vmatrix} 0 & 0 \\\\ 1 & 1 \\end{vmatrix}$$ Lastly we use that the determinant is alternating together with the determinant of an identity matrix being 1: $$\\label{final} ... = ab \\cdot 0 + ad\\cdot 1 + bc\\cdot -1 + cd\\cdot 0$$ And we conclude that the determinant of a general $$2\\times2$$ matrix","relation $P_k(a_0, \\dots, a_k) = a_0 P_{k-1}(a_1, \\dots, a_k) + P_{k-2}(a_2, \\dots, a_k).$ Initially, $$r_0 = \\frac{a_0}{1}$$ and $$r_1 = \\frac{a_0 a_1 + 1}{a_1}$$, thus \\begin{align}P_0(a_0)&=a_0,\\\\ P_1(a_0, a_1) &= a_0 a_1 + 1.\\end{align} For consistency, it is convenient to define $$P_{-1} = 1$$ and $$P_{-2}=0$$ and formally say that $$r_{-1} = \\frac{1}{0}$$ and $$r_{-2}=\\frac{0}{1}$$. From numerical analysis, it is known that the determinant of an arbitrary tridiagonal matrix $T_k = \\det \\begin{bmatrix} a_0 & b_0 & 0 & \\dots & 0 \\\\ c_0 & a_1 & b_1 & \\dots & 0 \\\\ 0 & c_1 & a_2 & . & \\vdots \\\\ \\vdots & \\vdots & . & \\ddots & c_{k-1} \\\\ 0 & 0 & \\dots & b_{k-1} & a_k \\end{bmatrix}$ can be computed recursively as $$T_k = a_k T_{k-1} - b_{k-1} c_{k-1} T_{k-2}$$. Comparing it to $$P_k$$, we get a direct expression $P_k = \\det \\begin{bmatrix} x_k & 1 & 0 & \\dots & 0 \\\\ -1 & x_{k-1} & 1 & \\dots & 0 \\\\ 0 & -1 & x_2 & . & \\vdots \\\\ \\vdots & \\vdots & . & \\ddots & 1 \\\\ 0 & 0 & \\dots & -1 & x_0 \\end{bmatrix}_{\\textstyle .}$ This polynomial is also known as the continuant due to its close relation with continued fraction. The continuant won't","points would be when $x'=0$ and $y'=0$: \\begin{align}0=(1+x)\\sin(y)\\end{align} \\begin{align}0=1-x-\\cos(y)\\end{align} From $(1)$, $\\sin(y)$ would be $0$ when $y=n\\pi$ where $n=0, 1, 2, 3,...$. Since there is a $\\cos(y)$ in $(2)$, the previous equation can be split into two cases: $y=2n\\pi$ where $n=0, 1, 2, 3,....$ and $y=n\\pi$ where $n=1, 3, 5,....$. Then in case 1 $\\cos(y)=1$ and in case 2 $cos(y)=-1$. Plugging both of these into $(2)$ gives the critical points: \\begin{align}(0, 2n\\pi) \\quad where \\quad n=0, 1, 2, 3,...\\end{align} \\begin{align}(2, n\\pi) \\quad where \\quad n=1, 3, 5,...\\end{align} The Jacobian matrix would be: \\begin{align}J=\\begin{pmatrix}F_{x}&F_{y}\\\\G_{x}&G_{y}\\end{pmatrix}=\\begin{pmatrix}\\sin(y)&-\\cos(y)(1+x)\\\\-1&\\sin(y)\\end{pmatrix}\\end{align} At $(0, 2n\\pi)$: \\begin{align}J=\\begin{pmatrix}0&1\\\\-1&0\\end{pmatrix}\\end{align} This gives eigenvalues $r_{1}=i$ and $r_{2}=-i$. Since $r_{1}, r_{2}=\\lambda\\pm i\\mu$ where $\\lambda=0$ the nonlinear system would be an indeterminate center or spiral point. At $(2, n\\pi)$: \\begin{align}J=\\begin{pmatrix}0&-3\\\\-1&0\\end{pmatrix}\\end{align} The eigenvalues would be $r_{1}=\\sqrt{3}$ and $r_{2}=-\\sqrt{3}$. Since $r_{2}<0$ are two distinct lines, then the coordinates for the point of intersection $[p_1, p_2, p_3]$ are given by: (2) \\begin{align} \\quad [p_1,p_2, p_3] = \\left [ \\begin{vmatrix} a_2 & a_3\\\\ b_2 & b_3 \\end{vmatrix}, \\begin{vmatrix} a_3 & a_1\\\\ b_3 & b_1 \\end{vmatrix}, \\begin{vmatrix} a_1 & a_2\\\\ b_1 & b_2 \\end{vmatrix} \\right ] \\end{align} Let's look at an example. ## Example 1 Consider the projective plane $\\mathbb{P}^2(\\mathbb{R})$. Find the point of intersection of the lines $<1, 1, 4>$ and $<2, 3, 1>$. Applying the formula above yields: (3) \\begin{align} \\quad [p_1,p_2, p_3] &= \\left [ \\begin{vmatrix} 1 & 4\\\\ 3 & 1 \\end{vmatrix}, \\begin{vmatrix} 4 & 1\\\\ 1 & 2 \\end{vmatrix}, \\begin{vmatrix} 1 & 1\\\\ 2 & 3 \\end{vmatrix} \\right ] \\\\ \\quad [p_1, p_2, p_3] &= [1 - 12, 8 - 1, 3 - 2] \\\\ \\quad [p_1, p_2, p_3] &= [-11, 7, 1] \\end{align} Similarly, we can verify that $[-11, 7, 1]$ lies on both $<1, 1, 4>$ and $<2, 3, 1>$ by verifyin that $[-11, 7, 1] \\cdot <1, 1, 4> = 0$ and $[-11, 7, 1] \\cdot <2, 3, 1> = 0$. We have that: (4) \\begin{align} \\quad [-11, 7, 1] \\cdot <1, 1, 4> = (-11)(1) + (7)(1) + (1)(4) = 0 \\end{align} And also: (5) \\begin{align} \\quad [-11, 7,","& a_{32} & a_{33} & a_{34}\\\\a_{41} & a_{42} & a_{43} & a_{44}\\end{pmatrix}$ . . Find determinant. Can I assume you know the \"Cofactor\" method? $\\text{det }\\!\\!A \\;=\\;a_{11}\\begin{vmatrix}a_{22}&a_{23}&a_{24} \\\\ a_{32}&a_{33}&a_{34} \\\\ a_{42}&a_{43}&a_{44} \\end{vmatrix} - a_{12}\\begin{vmatrix}a_{21}&a_{23}&a_{24}\\\\a_{31}& a_{33}&a_{34}\\\\a_{41}&a_{43}&a_{44} \\end{vmatrix}$ . $+\\:a_{13}\\begin{vmatrix}a_{21}&a_{22}&a_{24}\\\\a_{3 1}&a_{32}&a_{34} \\\\ a_{41}&a_{42}&a_{44}\\end{vmatrix} - a_{14}\\begin{vmatrix}a_{21}& a_{22}&a_{23}\\\\a_{31}&a_{32}&a_{33}\\\\a_{41}&a_{42} &a_{43} \\end{vmatrix}$ . . . $= \\;a_{11}\\bigg( a_{22}\\begin{vmatrix}a_{33}&a_{34}\\\\a_{43}&a_{44}\\ end{vmatrix} - a_{23}\\begin{vmatrix}a_{32}&a_{34}\\\\a_{42}&a_{44}\\ end{vmatrix} + a_{24}\\begin{vmatrix}a_{32}&a_{33}\\\\a_{42}&a_{43}\\ end{vmatrix} \\bigg)$ . . . . . $-a_{12}\\bigg(a_{21}\\begin{vmatrix}a_{33}&a_{34}\\\\a_ {43}&a_{44}\\end{vmatrix} -a_{23}\\begin{vmatrix}a_{31}&a_{34}\\\\a_{41}&a_{44}\\ end{vmatrix} + a_{24}\\begin{vmatrix}a_{31}&a_{33}\\\\a_{41}&a_{43} \\end{vmatrix} \\bigg)$ . . . . . . . $+a_{13}\\bigg(a_{21}\\begin{vmatrix}a_{32}&a_{34}\\\\a _{42}&a_{44}\\end{vmatrix} -a_{22}\\begin{vmatrix}a_{31}&a_{34}\\\\a_{41}&a_{44}\\ end{vmatrix} + a_{24}\\begin{vmatrix}a_{31}&a_{32}\\\\a_{41}&a_{42}\\ end{vmatrix}\\bigg)$ . . . . . . . . . $-a_{14}\\bigg(a_{21}\\begin{vmatrix}a_{32}&a_{33}\\\\a_ {42}&a_{43} \\end{vmatrix} - a_{22}\\begin{vmatrix}a_{31}&a_{33}\\\\a_{41}&a_{43}\\ end{vmatrix} + a_{23}\\begin{vmatrix}a_{31}&a_{32}\\\\a_{41}&a_{42}\\ end{vmatrix} \\bigg)$ . . . and so on.","but $(A-I)v_2\\ne 0$. Since $(A-I)^2=0$, you can take any vector in $\\Bbb{R}^3\\setminus Ker(A-I)$. Then $v_1=(A-I)v_2$ will be an eigenvector of $A$. Note that $Im(A-I)$ is one-dimensional generated by $\\begin{pmatrix} 1\\\\ 1\\\\ -1\\end{pmatrix}$, so $v_1$ will always be a multiple of $\\begin{pmatrix} 1\\\\ 1\\\\ -1\\end{pmatrix}$. Note also that $v_2$ cannot be in $Ker(A-I)$, since in that case $(A-I)v_2=0$. In order to complete the basis, take another vector $v_3$ in $Ker(A-I)$ which is not in the subspace generated by $v_1$. If you want to search directly for $v_1,v_2$ with $v_1=(A-I)v_2$, you have to note that $v_1$ is in the image of $A-I$, which is one dimensional, so it is a multiple of $\\begin{pmatrix} 1\\\\ 1\\\\ -1\\end{pmatrix}$. So in your example you can take $v_2=\\begin{pmatrix} 1\\\\ 0\\\\ 0\\end{pmatrix}$, then $v_1=(A-I)v_2=\\begin{pmatrix} 1\\\\ 1\\\\ -1\\end{pmatrix}$, and take $v_3$ another vector of $Ker(A-I)$, for example $v_3=\\begin{pmatrix} 2\\\\ -1\\\\ 0\\end{pmatrix}$. Then the matrix of $A$ with respect to the basis $\\{v_1,v_2,v_3\\}$ will be $\\begin{pmatrix} 1&1&0\\\\ 0&1&0\\\\0&0& 1\\end{pmatrix}$, since $A(v_1)=v_1$, $A(v_2)=v_1+v_2$ and $Av_3=v_3$.","\\end{vmatrix} = \\begin{vmatrix} \\ldots & 0\\ \\vc{a}_1 & \\ldots \\end{vmatrix} = 0 \\ \\begin{vmatrix} \\ldots & \\vc{a}_1 & \\ldots \\end{vmatrix} = 0.$$ (7.13) Property $(v)$ can be proved as follows. Form the determinant $$\\begin{vmatrix} \\ldots & \\vc{a}_i + \\vc{a}_j & \\ldots & \\vc{a}_i + \\vc{a}_j & \\ldots \\end{vmatrix} ,$$ (7.14) which is zero since it has two identical columns. Using the linearity of each column we get $$\\begin{array}{ll} 0 = & \\begin{vmatrix} \\ldots & \\vc{a}_i + \\vc{a}_j & \\ldots & \\vc{a}_i + \\vc{a}_j & \\ldots \\end{vmatrix} = \\\\ & \\begin{vmatrix} \\ldots & \\vc{a}_i & \\ldots & \\vc{a}_i & \\ldots \\end{vmatrix} + \\begin{vmatrix} \\ldots & \\vc{a}_i & \\ldots & \\vc{a}_j & \\ldots \\end{vmatrix} + \\\\ + & \\begin{vmatrix} \\ldots & \\vc{a}_j & \\ldots & \\vc{a}_i & \\ldots \\end{vmatrix} + \\begin{vmatrix} \\ldots & \\vc{a}_j & \\ldots & \\vc{a}_j & \\ldots \\end{vmatrix} . \\end{array}$$ (7.15) The first and last of these terms are zero, since they have two identical columns. We thus have $$0 = \\begin{vmatrix} \\ldots & \\vc{a}_i & \\ldots & \\vc{a}_j & \\ldots \\end{vmatrix} + \\begin{vmatrix} \\ldots & \\vc{a}_j & \\ldots & \\vc{a}_i & \\ldots \\end{vmatrix} ,$$ (7.16) which proves the statement. Property $(vi)$ can be proved as follows. Using the linearity we get $$\\begin{array}{ll} & \\begin{vmatrix} \\ldots & \\vc{a}_i + \\lambda \\vc{a}_j & \\ldots & \\vc{a}_j & \\ldots","to know that it's equal to -1, just do some matrix determinants. $$(1,0,0)\\times(0,0,1) = \\left( \\begin{vmatrix}0&0\\\\0&1\\end{vmatrix}, \\begin{vmatrix}0&1\\\\1&0\\end{vmatrix}, \\begin{vmatrix}1&0\\\\0&0\\end{vmatrix} \\right) = (0,-1,0)$$","15 '15 at 0:32 Note that $F(1,1,1) + F(1,2,3) = (2,6,12)$ and $F(2,3,4) = (1,8,27)$ and $(1,1,1) + (1,2,3) = (2,3,4)$. Then such $F$ never exist . If $F$ were linear, then $$F(1,1,1)+F(1,2,3) = F((1,1,1)+(1,2,3)) = F(2,3,4) = (1,8,27).$$ But $F(1,1,1)+F(1,2,3) = (1,2,3)+(1,4,9) = (2,6,12)$. The vectors $v_1=(1,1,1)$, $v_2=(1,2,3)$ and $v_3=(2,3,4)$ are not linearly independent; Gaussian elimination gives \\begin{align} \\begin{bmatrix} 1 & 1 & 2 \\\\ 1 & 2 & 3 \\\\ 1 & 3 & 4 \\end{bmatrix} &\\to \\begin{bmatrix} 1 & 1 & 2 \\\\ 0 & 1 & 1 \\\\ 0 & 2 & 2 \\end{bmatrix} \\\\&\\to \\begin{bmatrix} 1 & 1 & 2 \\\\ 0 & 1 & 1 \\\\ 0 & 0 & 0 \\end{bmatrix} \\\\&\\to \\begin{bmatrix} 1 & 0 & 1 \\\\ 0 & 1 & 1 \\\\ 0 & 0 & 0 \\end{bmatrix} \\end{align} which says that $v_3=v_1+v_2$. Since $(1,2,3)+(1,4,9)=(2,6,12)\\ne(1,8,27)$ no linear map exists with the requested property. Had the problem said $F(2,3,4)=(2,6,12)$, then infinitely many linear maps would exist: complete $\\{(1,1,1),(1,2,3)\\}$ to a basis and send the third basis vector to any vector. In the RREF above, the dominant columns (those with a leading $1$) are the first and the second column. The coefficients in the third column give exactly the coefficients to use for getting it as a linear combination of","{$M_4$} to an object with many points, then first compute {$M = M_4 M_3 M_2 M_1$} and then apply {$M$} to all the points. This is faster. ## 2.2 Definition 1. In 3D, the homogeneous point {$(x_h, y_h, z_h, w)$} corresponds to the Cartesian point {$(x_h/w, y_h/w, z_h/w)$} 2. There are an infinite number of homogeneous ways to represent each Cartesian point. E.g., (1,1,1,1), (2,2,2,2) and (3,3,3,3) all correspond to the Cartesian point (1,1,1). 3. For finite points, w is not 0. 1. Translation is a matrix multiply. To translate by {$(d_x,d_y,d_z)$} multiply by this matrix: {$$\\left( \\begin{array}{cccc} 1 &0&0&d_x\\\\ 0&1&0&d_y\\\\ 0&0&1&d_z \\\\ 0&0&0&1 \\end{array} \\right)$$} For example, translate by (1,2,3): {$$M = \\left(\\begin{array}{cccc} 1 & 0 & 0 & 1\\\\ 0 & 1 & 0 & 2\\\\ 0 & 0 & 1 & 3\\\\ 0 & 0 & 0 & 1\\end{array} \\right)$$} 2. To translate p=(5,5,5), make it homogeneous: (5,5,5,1) and compute 3. {$$\\begin{pmatrix} 6\\\\7\\\\8\\\\1\\end{pmatrix} = \\left(\\begin{array}{cccc} 1 & 0 & 0 & 1\\\\ 0 & 1 & 0 & 2\\\\ 0 & 0 & 1 & 3\\\\ 0 & 0 & 0 & 1\\end{array} \\right) \\begin{pmatrix}5\\\\5\\\\5\\\\1\\end{pmatrix}$$} 4. (10,10,10,2) is the same homogeneous point, and gives the same Cartesian answer: {$$\\begin{pmatrix} 12\\\\14 \\\\16\\\\2\\end{pmatrix} = \\left(\\begin{array}{cccc} 1 & 0 & 0 & 1\\\\ 0 & 1 & 0","& a_{33} \\end{vmatrix} - a_{12} \\begin{vmatrix} a_{21} & a_{23} \\\\ a_{31} & a_{33} \\end{vmatrix} + a_{13} \\begin{vmatrix} a_{21} & a_{22} \\\\ a_{31} & a_{32} \\end{vmatrix}$ Or another expansion: $\\begin{vmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \\end{vmatrix} = -a_{12} \\begin{vmatrix} a_{21} & a_{23} \\\\ a_{31} & a_{33} \\end{vmatrix} + a_{22} \\begin{vmatrix} a_{11} & a_{13} \\\\ a_{31} & a_{33} \\end{vmatrix} - a_{32} \\begin{vmatrix} a_{11} & a_{13} \\\\ a_{21} & a_{23} \\end{vmatrix}$ $\\begin{vmatrix} 1 & 2 & 0 \\\\ 4 & 1 & 0 \\end{vmatrix} = 1 \\begin{vmatrix} 1 & 0 \\\\ -1 & 3 \\end{vmatrix} - 2 \\begin{vmatrix} 4 & 0\\\\ 1 & 3 \\end{vmatrix} + 0 \\begin{vmatrix} 4 & 1 \\\\ 1 & -1 \\end{vmatrix}$ $= 1 (3-0) - 2 (12-0) = -21$ $\\begin{vmatrix} 0 & 0 & 0 & 2 \\\\ 1 & 0 & 0 & 3 \\\\ 0 & 1 & 0 & 2 \\\\ 0 & 0 & 1 & 3 \\end{vmatrix} = (-1)^{1+4} 2 \\begin{vmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{vmatrix} = -2$ $\\begin{vmatrix} 5 & 4 & 3 \\\\ 0 & -1 & 2 \\\\ 0 & 0 & 6 \\end{vmatrix} = 5 \\begin{vmatrix} -1 & 2 \\\\ 0 &","# Find SVD of a matrix Let A be a matrix, $$A= \\begin{pmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 2 \\\\ \\end{pmatrix}$$ and it's SVD, $$A=USV^t$$. Let $$U^1$$ be the first column of $$U$$ and $$V^1$$ the first column of $$V$$. There are a bunch of options, but the correct answer is: \"The singular values of $$A$$ are $$\\sigma _1 = 2$$ and $$\\sigma _2 =1$$, $$U^1=(0,1)$$ and $$V^1=(0,0,1)$$\" And I think I get to the same matrix they're talking about, but in a different order? 1. I found $$A^tA$$=$$\\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\\\ 0 & 2 \\\\ \\end{pmatrix}\\begin{pmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 2 \\\\ \\end{pmatrix}=\\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 4 \\\\ \\end{pmatrix}$$ 2. I found the eigenvectors and eigenvalues of $$A^tA$$, $$\\lambda_1=0$$, $$\\lambda_2=1$$, $$\\lambda_3=4$$. And their eigenvectors are $$S_0=[(1,0,0)]$$, $$S_1=[(0,1,0])$$, $$S_4=[(0,0,1)]$$. I know I don't have to orthonormalize them, because they form the canonical basis. So far, I have the singular values, which are $$\\sigma _1 = 1$$ and $$\\sigma _2 =2$$, $$S=\\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 2 & 0 \\\\ \\end{pmatrix}$$,$$V=V^t=\\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0","upon a pair of entries of which are not both zero. So, for example, when, the matrix can be \\begin{pmatrix} 0&-1&0\\\\ 1&0&0\\\\ 0&0&-1\\\\ \\end{pmatrix}. For the vector, the resulting simplex has vertices \\begin{pmatrix}1\\ 0\\ 1/\\surd2\\end{pmatrix}, \\begin{pmatrix}0\\ 1\\ -1/\\surd2\\end{pmatrix}, \\begin{pmatrix}-1\\ 0\\ 1/\\surd2\\end{pmatrix}, \\begin{pmatrix}0\\ -1\\ -1/\\surd2\\end{pmatrix}, each of which has distance 2 from the others. Geometric properties Volume The volume of an n-simplex in n-dimensional space with vertices (v0, ..., vn) is Volume= 1 n! \\left|\\det \\begin{pmatrix} v1-v0&&v2-v0&&&&vn-v0 \\end{pmatrix}\\right| where each column of the n × n determinant is a vector that points from vertex to another vertex .[6] This formula is particularly useful when v0 is the origin. The expression Volume= 1 n! T \\det\\left[ \\begin{pmatrix} v 0 T \\ v 0 \\\\vdots T \\end{pmatrix} \\begin{pmatrix} v \\ v 1-v 0&v2-v0&&vn-v 1/2 0 \\end{pmatrix} \\right] employs a Gram determinant and works even when the n-simplex's vertices are in a Euclidean space with more than n dimensions, e.g., a triangle in R3 . A more symmetric way to compute the volume of an n-simplex in Rn is Volume={1\\overn!}\\left|\\det \\begin{pmatrix} v0&v1&&vn\\\\ 1&1&&1 \\end{pmatrix}\\right|. Another common way of computing the volume of the simplex is via the Cayley–Menger determinant, which works even when the n-simplex's vertices are in a Euclidean space with more than n dimensions. Without the 1/n! it is the"],"string":"[\n \"5 \\\\end{pmatrix}$$ var v = [1.0, 2.0] var s = 3.0 var vsresult = [Double](count : v.count, repeatedValue : 0.0) vDSP_vsaddD(v, 1, &s, &vsresult, 1, vDSP_Length(v.count)) vsresult // returns [4.0, 5.0] $$\\\\begin{pmatrix} 1 \\\\\\\\ 2 \\\\end{pmatrix} \\\\times 3 = \\\\begin{pmatrix} 3 \\\\\\\\ 6 \\\\end{pmatrix}$$ vDSP_vsmulD(v, 1, &s, &vsresult, 1, vDSP_Length(v.count)) vsresult // returns [3.0, 6.0] $$\\\\begin{pmatrix} 1 \\\\\\\\ 2 \\\\end{pmatrix} \\\\div 3 = \\\\begin{pmatrix} 1/3 \\\\\\\\ 2/3 \\\\end{pmatrix}$$ vDSP_vsdivD(v, 1, &s, &vsresult, 1, vDSP_Length(v.count)) vsresult // returns [0.333333333333333, 0.666666666666667] ## Vector & Vector Vector-vector operations pose no challenge to Accelerate and the associated functions look like vDSP_v***D(vector_1, 1, vector_2, 1, &result, 1, length_of_vector) Here are a few worked-out examples: $$\\\\begin{pmatrix} 2 \\\\\\\\ 5 \\\\end{pmatrix} + \\\\begin{pmatrix} 3 \\\\\\\\ 4 \\\\end{pmatrix} = \\\\begin{pmatrix} 5 \\\\\\\\ 9 \\\\end{pmatrix}$$ var v1 = [2.0, 5.0] var v2 = [3.0, 4.0] var vvresult = [Double](count : 2, repeatedValue : 0.0) vDSP_vaddD(v1, 1, v2, 1, &vvresult, 1, vDSP_Length(v1.count)) vvresult // returns [5.0, 9.0] $$\\\\begin{pmatrix} 2 \\\\\\\\ 5 \\\\end{pmatrix} \\\\begin{pmatrix} 3 \\\\\\\\ 4 \\\\end{pmatrix} = \\\\begin{pmatrix} 6 \\\\\\\\ 20 \\\\end{pmatrix}$$ vDSP_vmulD(v1, 1, v2, 1, &vvresult, 1, vDSP_Length(v1.count)) vvresult // returns [6.0, 20.0] $$\\\\begin{pmatrix} 3 \\\\\\\\ 4 \\\\end{pmatrix} {\\\\bigg/} \\\\begin{pmatrix} 2 \\\\\\\\ 5 \\\\end{pmatrix} = \\\\begin{pmatrix} 1.5 \\\\\\\\ 0.8 \\\\end{pmatrix}$$ vDSP_vdivD(v1, 1, v2, 1, &vvresult, 1, vDSP_Length(v1.count)) vvresult // returns [1.5, 0.8] ## Dot Product $$\\\\begin{pmatrix}\",\n \"\\\\hat j & \\\\hat k \\\\\\\\ 1 & -3 & 0 \\\\\\\\ 0 & -2 & 1 \\\\end{vmatrix} = (-3, -1, -2) \\\\begin{vmatrix} \\\\hat i & \\\\hat j & \\\\hat k \\\\\\\\ 1 & 0 & 1 \\\\\\\\ 1 & -3 & 1 \\\\end{vmatrix} = (3, 0, -3) \\\\begin{vmatrix} \\\\hat i & \\\\hat j & \\\\hat k \\\\\\\\ 4 & -2 & 5 \\\\\\\\ 0 & 2 & 1 \\\\end{vmatrix} = (-12, -4, 8)$ Then I find a point for each line: $P_r (0, 1, 3), P_s (2, 2, 2), P_t (-1, -1, 1)$ After that I do the determinant with the first row as the difference between the two points of the two lines, and on the second and third raw I put the directional vectors of the two lines. For $r$ and $s$: $\\\\begin{vmatrix} 0-2 & 1-2 & 3-2 \\\\\\\\ -3 & -1 & -2 \\\\\\\\ 3 & 0 & -3 \\\\end{vmatrix} = 12$ Non-coplanar. $\\\\begin{vmatrix} 2-(-1) & 2-(-1) & 2-1 \\\\\\\\ 3 & 0 & -3 \\\\\\\\ -12 & -4 & 8 \\\\\\\\ \\\\end{vmatrix} = -10$ Non-coplanar. $\\\\begin{vmatrix} 0-(-1) & 1-(-1) & 3-1 \\\\\\\\ -3 & -1 & -2 \\\\\\\\ -12 & -4 & 8 \\\\\\\\ \\\\end{vmatrix} = 80$ Non-coplanar. The thing is that all three of them are askew (non-coplanar). Could someone check it\",\n \"& 0 & 0 & 0 & 0 \\\\\\\\ 0 & 1 & 0 & 0 & 0 \\\\\\\\ 0 & 0 & 2 & 0 & 0 \\\\\\\\ 0 & 0 & 0 & 3 & 0 \\\\\\\\ 0 & 0 & 0 & 0 & 4 \\\\end{pmatrix} + \\\\begin{pmatrix} 1 & -1 & 1 & -1 & 1 \\\\\\\\ 0 & 1 & -2 & 3 & -4 \\\\\\\\ 0 & 0 & 1 & -3 & 6 \\\\\\\\ 0 & 0 & 0 & 1 & -4 \\\\\\\\ 0 & 0 & 0 & 0 & 1 \\\\end{pmatrix}$$ and find $$\\\\begin{pmatrix} 1 & -1 & 3 & -1 & 1\\\\\\\\ 0 & 0 & -2 & 9 & -4\\\\\\\\ 0 & 0 & -1 & -3 & 18\\\\\\\\ 0 & 0 & 0 & -2 & -4\\\\\\\\ 0 & 0 & 0 & 0 & -3\\\\\\\\ \\\\end{pmatrix}$$ The RREF is $$\\\\begin{pmatrix} 1 & -1 & 0 & 0 & 0 \\\\\\\\ 0 & 0 & 1 & 0 & 0 \\\\\\\\ 0 & 0 & 0 & 1 & 0 \\\\\\\\ 0 & 0 & 0 & 0 & 1 \\\\\\\\ 0 & 0 & 0 & 0 & 0 \\\\end{pmatrix}$$ and a basis of the null space consists of the single vector $$\\\\begin{pmatrix} 1\",\n \"$(1,n)$-entry are $-1$. The other entries are zero. ## Hint. 1. Calculate the first row cofactor expansion. 2. The determinant of a triangular matrix is the product of its diagonal entries. ## Solution. Apply the cofactor expansion corresponding to the first row. We obtain \\\\begin{align*} \\\\det(A)&= \\\\begin{vmatrix} 1 & 0 & \\\\dots & 0 & 0 & 0 \\\\\\\\ 1 & 1 & \\\\dots & 0 & 0 & 0 \\\\\\\\ \\\\vdots & \\\\vdots & \\\\dots & \\\\ddots & \\\\vdots & \\\\vdots \\\\\\\\ 0 & 0 &\\\\dots & 1 & 1 & 0\\\\\\\\ 0 & 0 &\\\\dots & 0 & 1 & 1 \\\\end{vmatrix} +(-1)^{n+1} \\\\begin{vmatrix} 1 & 1 & 0 & \\\\dots & 0 & 0 \\\\\\\\ 0 & 1 & 1 & \\\\dots & 0 & 0 \\\\\\\\ \\\\vdots & \\\\vdots & \\\\vdots & \\\\ddots & \\\\vdots & \\\\vdots \\\\\\\\ 0 & 0 & 0 &\\\\dots & 1 & 1 \\\\\\\\ 0 & 0 & 0 &\\\\dots & 0 & 1 \\\\end{vmatrix} \\\\end{align*} The two smaller (minor) $n-1 \\\\times n-1$ matrices are both triangular. The determinant of a triangular matrix is the product of its diagonal entries. Thus we see that \\\\begin{align*} \\\\det(A)&=1+(-1)^{n+1} \\\\\\\\ &= \\\\begin{cases} 2 & \\\\text{ if } n \\\\text{ is odd}\\\\\\\\ 0 & \\\\text{ if } n \\\\text{ is even}. \\\\end{cases} \\\\end{align*}\",\n \"## Change of variables, from dxdy to dudv Suppose we have a function $f(x, y)$ and we rewrite it as $f(u(x, y), v(x, y))$, i.e. perform a change of variables, then the caculation of area using double integral can be carried out with these new variables $u, v$. But we have to figure out the relation between $dudv$ and $dxdy$. Here is the thing, with the variables $x, y$, $dxdy$ is the area of the parallelogram between the vectors $(dx, 0)$ and $(0, dy)$, under the new variables, these vectors are transformed into two new vectors $(u_x dx, v_x dx)$, $(u_y dy, v_y dy)$, and $dudv$ is the area of the parallelogram between these two new vectors. $\\\\begin{array}{cc} (dx, 0) & (u_x dx, v_x dx) \\\\\\\\ (0, dy) & (u_y dy, v_y dy) \\\\end{array}$ To get the area of this parallelogram, we need the determinant of this matrix: \\\\begin{align*} M &= \\\\begin{pmatrix} u_x dx & u_y dy \\\\\\\\ v_x dx & v_y dy \\\\end{pmatrix} \\\\\\\\ &= \\\\begin{pmatrix} u_x & u_y \\\\\\\\ v_x & v_y \\\\end{pmatrix} \\\\begin{pmatrix} dx & 0 \\\\\\\\ 0 & dy \\\\end{pmatrix} \\\\\\\\ &= J \\\\begin{pmatrix} dx & 0 \\\\\\\\ 0 & dy \\\\end{pmatrix}, \\\\end{align*} where $J$ stands for the Jacobian matrix of $f$. So the determinant is \\\\begin{align*} \\\\det(M) &= \\\\det(J) dxdy \\\\end{align*} Remember in\",\n \"= [1,-2,1]$$. • Solve $$Uw = v_3$$: \\\\begin{align*} \\\\begin{bmatrix} 3 & 1 & -1 & 1 \\\\\\\\ -6 & -2 & 2 & -2 \\\\\\\\ 2 & 1 & 0 & 1 \\\\end{bmatrix} \\\\leadsto \\\\begin{bmatrix} 1 & 0 & -1 & 0 \\\\\\\\ 0 & 1 & 2 & 1 \\\\\\\\ 0 & 0 & 0 & 0 \\\\end{bmatrix} ,\\\\end{align*} so take $$v_3 = [0,1,0]$$. • Putting things together: \\\\begin{align*} A &= P^{-1}J P \\\\text{ where } \\\\\\\\ J = J_1(\\\\lambda = 2) \\\\oplus J_2(\\\\lambda = 1) &= \\\\begin{bmatrix} 2 & 0 & 0 \\\\\\\\ 0 & 1 & 1 \\\\\\\\ 0 & 0 & 1 \\\\end{bmatrix} \\\\\\\\ P = [v_1, v_2, v_3] &= \\\\begin{bmatrix} 1 & 1 & 0 \\\\\\\\ -2 & -2 & 1 \\\\\\\\ 0 & 1 & 0 \\\\end{bmatrix} .\\\\end{align*} • Write $$\\\\min_A(t) = (t-2)(t-1)^{\\\\ell_1}$$, then since $$\\\\min_A(t)$$ divides $$\\\\chi_A(t)$$ either $$\\\\ell_1 = 1, 2$$. • $$\\\\ell_1$$ is the size of the largest block corresponding to $$\\\\lambda = 1$$, which is size 2, so $$\\\\lambda_1=2$$. • Thus \\\\begin{align*} \\\\min_A(t) = (t-2)(t-1)^2 .\\\\end{align*} ## Fall 2020 #5#algebra/qual/work Consider the following matrix: \\\\begin{align*} B \\\\coloneqq \\\\begin{bmatrix} 1 & 3 & 3 \\\\\\\\ 2 & 2 & 3 \\\\\\\\ -1 & -2 & -2 \\\\end{bmatrix} .\\\\end{align*} • Find the minimal polynomial of $$B$$. • Find a $$3\\\\times\",\n \"0 & 0 & 0 & 2\\\\\\\\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 \\\\end{pmatrix} \\\\xrightarrow[~]{rref} \\\\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 1\\\\\\\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 1\\\\\\\\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0\\\\\\\\ 0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 0\\\\\\\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 & -1 & 1\\\\\\\\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 \\\\end{pmatrix}$ What a mess! But we can still interpret the solution. From row 3 we have $v_3+v_7+v_8+v_{10} = 0$, then by non-negativity they are all zero. By removing column 3,7,8,10 and row 3 we instantly get $v_2 = v_5 = v_6 = 1, v_4 = 0$. Now the solution space is given by: $v = (0,1,0,0,1,1,0,0,1,0)^t + t(1,0,0,0,0,0,0,0,-1,0)^t$ where $t \\\\in\",\n \"## e Show that $\\\\alpha_1 - \\\\alpha_2$ is estimable and give its least squares estimator. Take $\\\\lambda^T = [0, 1,-1, 0, 0, 0]$ such that $\\\\lambda^T b = \\\\alpha_1 - \\\\alpha_2.$ Now we need to find $a$ such that $a^T X = \\\\lambda ^T.$ $[a_1, a_2, a_3, a_4, a_5, a_6] \\\\begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 0 \\\\\\\\ 1 & 1 & 0 & 0 & 1 & 0 \\\\\\\\ 1 & 1 & 0 & 0 & 0 & 1 \\\\\\\\ 1 & 0 & 1 & 1 & 0 & 0 \\\\\\\\ 1 & 0 & 1 & 0 & 1 & 0 \\\\\\\\ 1 & 0 & 1 & 0 & 0 & 1 \\\\end{bmatrix} = [0, 1, -1, 0, 0, 0]$ \\\\begin{align} a_1 + a_2 + a_3 + a_4 + a_5 + a_6 & = 0 \\\\\\\\ a_1 + a_2 + a_3 & = 1 \\\\\\\\ a_4 + a_5 + a_6 & = -1 \\\\\\\\ a_1 + a_4 & = 0 \\\\\\\\ a_2 + a_5 & = 0 \\\\\\\\ a_3 + a_6 & = 0. \\\\end{align} Now fix $a_5$ and $a_6$. \\\\begin{align} a_2 & = -a_5 \\\\\\\\ a_3 & = -a_6 \\\\\\\\ a_4 & = -1 + a_5 + a_6 \\\\\\\\ a_1 & = 1 - a_5 - a_6\",\n \"1 & 1 & 1 & 1 & 1 \\\\\\\\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\\\\\\\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\\\\\\\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\\\\\\\ \\\\end{pmatrix}$ ## Example 2 The RM(2,3) code is generated by the set: $\\\\{ v_0, v_1, v_2, v_3, v_1 \\\\wedge v_2, v_1 \\\\wedge v_3, v_2 \\\\wedge v_3 \\\\}$ or more explicitly by the rows of the matrix: $\\\\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\\\\\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\\\\\\\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\\\\\\\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\\\\\\\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\\\\\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\\\\\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\\\\\ \\\\end{pmatrix}$ ## Properties The following properties hold: 1 The set of\",\n \"\\\\\\\\ &\\\\; \\\\; \\\\; + (\\\\hat c - \\\\hat dx_1)y + \\\\hat d xy \\\\end{align*} It follows that this solution would uniquely determine the desired bilinear interpolant $$f(x,y)$$ at the original points: $$\\\\begin{pmatrix} a \\\\\\\\ b \\\\\\\\ c \\\\\\\\ d \\\\end{pmatrix} = \\\\begin{pmatrix} 1 & -x_1 & -y_1 & x_1y_1 \\\\\\\\ 0 & 1 & 0 & -y_1 \\\\\\\\ 0 & 0 & 1 & -x_1 \\\\\\\\ 0 & 0 & 0 & 1 \\\\end{pmatrix} \\\\begin{pmatrix} \\\\hat a \\\\\\\\ \\\\hat b \\\\\\\\ \\\\hat c \\\\\\\\ \\\\hat d \\\\end{pmatrix}$$ In that sense we can assume without loss of generality that $$(x_1,y_1) = (0,0)$$, and we proceed to find the determinant of $$M$$ in this slightly special case: \\\\begin{align*} \\\\det M &= \\\\left| \\\\begin{array}{cccc} 1 & 0 & 0 & 0 \\\\\\\\ 1 & x_2 & y_2 & x_2 y_2 \\\\\\\\ 1 & x_3 & y_3 & x_3 y_3 \\\\\\\\ 1 & x_4 & y_4 & x_4 y_4 \\\\end{array} \\\\right| \\\\\\\\ &= \\\\left| \\\\begin{array}{ccc} x_2 & y_2 & x_2 y_2 \\\\\\\\ x_3 & y_3 & x_3 y_3 \\\\\\\\ x_4 & y_4 & x_4 y_4 \\\\end{array} \\\\right| \\\\\\\\ &= x_2x_4y_3(y_4-y_2) + x_3x_4y_2(y_3-y_4) + x_2x_3y_4(y_2-y_3) \\\\end{align*} With this expression we can identify some geometric conditions on the four points for which the matrix has zero determinant and is thus not invertible. One\",\n \"and $$ are two distinct lines, then the coordinates for the point of intersection $[p_1, p_2, p_3]$ are given by: (2) \\\\begin{align} \\\\quad [p_1,p_2, p_3] = \\\\left [ \\\\begin{vmatrix} a_2 & a_3\\\\\\\\ b_2 & b_3 \\\\end{vmatrix}, \\\\begin{vmatrix} a_3 & a_1\\\\\\\\ b_3 & b_1 \\\\end{vmatrix}, \\\\begin{vmatrix} a_1 & a_2\\\\\\\\ b_1 & b_2 \\\\end{vmatrix} \\\\right ] \\\\end{align} Let's look at an example. ## Example 1 Consider the projective plane $\\\\mathbb{P}^2(\\\\mathbb{R})$. Find the point of intersection of the lines $<1, 1, 4>$ and $<2, 3, 1>$. Applying the formula above yields: (3) \\\\begin{align} \\\\quad [p_1,p_2, p_3] &= \\\\left [ \\\\begin{vmatrix} 1 & 4\\\\\\\\ 3 & 1 \\\\end{vmatrix}, \\\\begin{vmatrix} 4 & 1\\\\\\\\ 1 & 2 \\\\end{vmatrix}, \\\\begin{vmatrix} 1 & 1\\\\\\\\ 2 & 3 \\\\end{vmatrix} \\\\right ] \\\\\\\\ \\\\quad [p_1, p_2, p_3] &= [1 - 12, 8 - 1, 3 - 2] \\\\\\\\ \\\\quad [p_1, p_2, p_3] &= [-11, 7, 1] \\\\end{align} Similarly, we can verify that $[-11, 7, 1]$ lies on both $<1, 1, 4>$ and $<2, 3, 1>$ by verifyin that $[-11, 7, 1] \\\\cdot <1, 1, 4> = 0$ and $[-11, 7, 1] \\\\cdot <2, 3, 1> = 0$. We have that: (4) \\\\begin{align} \\\\quad [-11, 7, 1] \\\\cdot <1, 1, 4> = (-11)(1) + (7)(1) + (1)(4) = 0 \\\\end{align} And also: (5) \\\\begin{align} \\\\quad [-11, 7,\",\n \"& a_{32} & a_{33} & a_{34}\\\\\\\\a_{41} & a_{42} & a_{43} & a_{44}\\\\end{pmatrix}$ . . Find determinant. Can I assume you know the \\\"Cofactor\\\" method? $\\\\text{det }\\\\!\\\\!A \\\\;=\\\\;a_{11}\\\\begin{vmatrix}a_{22}&a_{23}&a_{24} \\\\\\\\ a_{32}&a_{33}&a_{34} \\\\\\\\ a_{42}&a_{43}&a_{44} \\\\end{vmatrix} - a_{12}\\\\begin{vmatrix}a_{21}&a_{23}&a_{24}\\\\\\\\a_{31}& a_{33}&a_{34}\\\\\\\\a_{41}&a_{43}&a_{44} \\\\end{vmatrix}$ . $+\\\\:a_{13}\\\\begin{vmatrix}a_{21}&a_{22}&a_{24}\\\\\\\\a_{3 1}&a_{32}&a_{34} \\\\\\\\ a_{41}&a_{42}&a_{44}\\\\end{vmatrix} - a_{14}\\\\begin{vmatrix}a_{21}& a_{22}&a_{23}\\\\\\\\a_{31}&a_{32}&a_{33}\\\\\\\\a_{41}&a_{42} &a_{43} \\\\end{vmatrix}$ . . . $= \\\\;a_{11}\\\\bigg( a_{22}\\\\begin{vmatrix}a_{33}&a_{34}\\\\\\\\a_{43}&a_{44}\\\\ end{vmatrix} - a_{23}\\\\begin{vmatrix}a_{32}&a_{34}\\\\\\\\a_{42}&a_{44}\\\\ end{vmatrix} + a_{24}\\\\begin{vmatrix}a_{32}&a_{33}\\\\\\\\a_{42}&a_{43}\\\\ end{vmatrix} \\\\bigg)$ . . . . . $-a_{12}\\\\bigg(a_{21}\\\\begin{vmatrix}a_{33}&a_{34}\\\\\\\\a_ {43}&a_{44}\\\\end{vmatrix} -a_{23}\\\\begin{vmatrix}a_{31}&a_{34}\\\\\\\\a_{41}&a_{44}\\\\ end{vmatrix} + a_{24}\\\\begin{vmatrix}a_{31}&a_{33}\\\\\\\\a_{41}&a_{43} \\\\end{vmatrix} \\\\bigg)$ . . . . . . . $+a_{13}\\\\bigg(a_{21}\\\\begin{vmatrix}a_{32}&a_{34}\\\\\\\\a _{42}&a_{44}\\\\end{vmatrix} -a_{22}\\\\begin{vmatrix}a_{31}&a_{34}\\\\\\\\a_{41}&a_{44}\\\\ end{vmatrix} + a_{24}\\\\begin{vmatrix}a_{31}&a_{32}\\\\\\\\a_{41}&a_{42}\\\\ end{vmatrix}\\\\bigg)$ . . . . . . . . . $-a_{14}\\\\bigg(a_{21}\\\\begin{vmatrix}a_{32}&a_{33}\\\\\\\\a_ {42}&a_{43} \\\\end{vmatrix} - a_{22}\\\\begin{vmatrix}a_{31}&a_{33}\\\\\\\\a_{41}&a_{43}\\\\ end{vmatrix} + a_{23}\\\\begin{vmatrix}a_{31}&a_{32}\\\\\\\\a_{41}&a_{42}\\\\ end{vmatrix} \\\\bigg)$ . . . and so on.\",\n \"but $(A-I)v_2\\\\ne 0$. Since $(A-I)^2=0$, you can take any vector in $\\\\Bbb{R}^3\\\\setminus Ker(A-I)$. Then $v_1=(A-I)v_2$ will be an eigenvector of $A$. Note that $Im(A-I)$ is one-dimensional generated by $\\\\begin{pmatrix} 1\\\\\\\\ 1\\\\\\\\ -1\\\\end{pmatrix}$, so $v_1$ will always be a multiple of $\\\\begin{pmatrix} 1\\\\\\\\ 1\\\\\\\\ -1\\\\end{pmatrix}$. Note also that $v_2$ cannot be in $Ker(A-I)$, since in that case $(A-I)v_2=0$. In order to complete the basis, take another vector $v_3$ in $Ker(A-I)$ which is not in the subspace generated by $v_1$. If you want to search directly for $v_1,v_2$ with $v_1=(A-I)v_2$, you have to note that $v_1$ is in the image of $A-I$, which is one dimensional, so it is a multiple of $\\\\begin{pmatrix} 1\\\\\\\\ 1\\\\\\\\ -1\\\\end{pmatrix}$. So in your example you can take $v_2=\\\\begin{pmatrix} 1\\\\\\\\ 0\\\\\\\\ 0\\\\end{pmatrix}$, then $v_1=(A-I)v_2=\\\\begin{pmatrix} 1\\\\\\\\ 1\\\\\\\\ -1\\\\end{pmatrix}$, and take $v_3$ another vector of $Ker(A-I)$, for example $v_3=\\\\begin{pmatrix} 2\\\\\\\\ -1\\\\\\\\ 0\\\\end{pmatrix}$. Then the matrix of $A$ with respect to the basis $\\\\{v_1,v_2,v_3\\\\}$ will be $\\\\begin{pmatrix} 1&1&0\\\\\\\\ 0&1&0\\\\\\\\0&0& 1\\\\end{pmatrix}$, since $A(v_1)=v_1$, $A(v_2)=v_1+v_2$ and $Av_3=v_3$.\",\n \"\\\\end{vmatrix} = \\\\begin{vmatrix} \\\\ldots & 0\\\\ \\\\vc{a}_1 & \\\\ldots \\\\end{vmatrix} = 0 \\\\ \\\\begin{vmatrix} \\\\ldots & \\\\vc{a}_1 & \\\\ldots \\\\end{vmatrix} = 0.$$ (7.13) Property $(v)$ can be proved as follows. Form the determinant $$\\\\begin{vmatrix} \\\\ldots & \\\\vc{a}_i + \\\\vc{a}_j & \\\\ldots & \\\\vc{a}_i + \\\\vc{a}_j & \\\\ldots \\\\end{vmatrix} ,$$ (7.14) which is zero since it has two identical columns. Using the linearity of each column we get $$\\\\begin{array}{ll} 0 = & \\\\begin{vmatrix} \\\\ldots & \\\\vc{a}_i + \\\\vc{a}_j & \\\\ldots & \\\\vc{a}_i + \\\\vc{a}_j & \\\\ldots \\\\end{vmatrix} = \\\\\\\\ & \\\\begin{vmatrix} \\\\ldots & \\\\vc{a}_i & \\\\ldots & \\\\vc{a}_i & \\\\ldots \\\\end{vmatrix} + \\\\begin{vmatrix} \\\\ldots & \\\\vc{a}_i & \\\\ldots & \\\\vc{a}_j & \\\\ldots \\\\end{vmatrix} + \\\\\\\\ + & \\\\begin{vmatrix} \\\\ldots & \\\\vc{a}_j & \\\\ldots & \\\\vc{a}_i & \\\\ldots \\\\end{vmatrix} + \\\\begin{vmatrix} \\\\ldots & \\\\vc{a}_j & \\\\ldots & \\\\vc{a}_j & \\\\ldots \\\\end{vmatrix} . \\\\end{array}$$ (7.15) The first and last of these terms are zero, since they have two identical columns. We thus have $$0 = \\\\begin{vmatrix} \\\\ldots & \\\\vc{a}_i & \\\\ldots & \\\\vc{a}_j & \\\\ldots \\\\end{vmatrix} + \\\\begin{vmatrix} \\\\ldots & \\\\vc{a}_j & \\\\ldots & \\\\vc{a}_i & \\\\ldots \\\\end{vmatrix} ,$$ (7.16) which proves the statement. Property $(vi)$ can be proved as follows. Using the linearity we get $$\\\\begin{array}{ll} & \\\\begin{vmatrix} \\\\ldots & \\\\vc{a}_i + \\\\lambda \\\\vc{a}_j & \\\\ldots & \\\\vc{a}_j & \\\\ldots\",\n \"to know that it's equal to -1, just do some matrix determinants. $$(1,0,0)\\\\times(0,0,1) = \\\\left( \\\\begin{vmatrix}0&0\\\\\\\\0&1\\\\end{vmatrix}, \\\\begin{vmatrix}0&1\\\\\\\\1&0\\\\end{vmatrix}, \\\\begin{vmatrix}1&0\\\\\\\\0&0\\\\end{vmatrix} \\\\right) = (0,-1,0)$$\",\n \"15 '15 at 0:32 Note that $F(1,1,1) + F(1,2,3) = (2,6,12)$ and $F(2,3,4) = (1,8,27)$ and $(1,1,1) + (1,2,3) = (2,3,4)$. Then such $F$ never exist . If $F$ were linear, then $$F(1,1,1)+F(1,2,3) = F((1,1,1)+(1,2,3)) = F(2,3,4) = (1,8,27).$$ But $F(1,1,1)+F(1,2,3) = (1,2,3)+(1,4,9) = (2,6,12)$. The vectors $v_1=(1,1,1)$, $v_2=(1,2,3)$ and $v_3=(2,3,4)$ are not linearly independent; Gaussian elimination gives \\\\begin{align} \\\\begin{bmatrix} 1 & 1 & 2 \\\\\\\\ 1 & 2 & 3 \\\\\\\\ 1 & 3 & 4 \\\\end{bmatrix} &\\\\to \\\\begin{bmatrix} 1 & 1 & 2 \\\\\\\\ 0 & 1 & 1 \\\\\\\\ 0 & 2 & 2 \\\\end{bmatrix} \\\\\\\\&\\\\to \\\\begin{bmatrix} 1 & 1 & 2 \\\\\\\\ 0 & 1 & 1 \\\\\\\\ 0 & 0 & 0 \\\\end{bmatrix} \\\\\\\\&\\\\to \\\\begin{bmatrix} 1 & 0 & 1 \\\\\\\\ 0 & 1 & 1 \\\\\\\\ 0 & 0 & 0 \\\\end{bmatrix} \\\\end{align} which says that $v_3=v_1+v_2$. Since $(1,2,3)+(1,4,9)=(2,6,12)\\\\ne(1,8,27)$ no linear map exists with the requested property. Had the problem said $F(2,3,4)=(2,6,12)$, then infinitely many linear maps would exist: complete $\\\\{(1,1,1),(1,2,3)\\\\}$ to a basis and send the third basis vector to any vector. In the RREF above, the dominant columns (those with a leading $1$) are the first and the second column. The coefficients in the third column give exactly the coefficients to use for getting it as a linear combination of\",\n \"{$M_4$} to an object with many points, then first compute {$M = M_4 M_3 M_2 M_1$} and then apply {$M$} to all the points. This is faster. ## 2.2 Definition 1. In 3D, the homogeneous point {$(x_h, y_h, z_h, w)$} corresponds to the Cartesian point {$(x_h/w, y_h/w, z_h/w)$} 2. There are an infinite number of homogeneous ways to represent each Cartesian point. E.g., (1,1,1,1), (2,2,2,2) and (3,3,3,3) all correspond to the Cartesian point (1,1,1). 3. For finite points, w is not 0. 1. Translation is a matrix multiply. To translate by {$(d_x,d_y,d_z)$} multiply by this matrix: {$$\\\\left( \\\\begin{array}{cccc} 1 &0&0&d_x\\\\\\\\ 0&1&0&d_y\\\\\\\\ 0&0&1&d_z \\\\\\\\ 0&0&0&1 \\\\end{array} \\\\right)$$} For example, translate by (1,2,3): {$$M = \\\\left(\\\\begin{array}{cccc} 1 & 0 & 0 & 1\\\\\\\\ 0 & 1 & 0 & 2\\\\\\\\ 0 & 0 & 1 & 3\\\\\\\\ 0 & 0 & 0 & 1\\\\end{array} \\\\right)$$} 2. To translate p=(5,5,5), make it homogeneous: (5,5,5,1) and compute 3. {$$\\\\begin{pmatrix} 6\\\\\\\\7\\\\\\\\8\\\\\\\\1\\\\end{pmatrix} = \\\\left(\\\\begin{array}{cccc} 1 & 0 & 0 & 1\\\\\\\\ 0 & 1 & 0 & 2\\\\\\\\ 0 & 0 & 1 & 3\\\\\\\\ 0 & 0 & 0 & 1\\\\end{array} \\\\right) \\\\begin{pmatrix}5\\\\\\\\5\\\\\\\\5\\\\\\\\1\\\\end{pmatrix}$$} 4. (10,10,10,2) is the same homogeneous point, and gives the same Cartesian answer: {$$\\\\begin{pmatrix} 12\\\\\\\\14 \\\\\\\\16\\\\\\\\2\\\\end{pmatrix} = \\\\left(\\\\begin{array}{cccc} 1 & 0 & 0 & 1\\\\\\\\ 0 & 1 & 0\",\n \"& a_{33} \\\\end{vmatrix} - a_{12} \\\\begin{vmatrix} a_{21} & a_{23} \\\\\\\\ a_{31} & a_{33} \\\\end{vmatrix} + a_{13} \\\\begin{vmatrix} a_{21} & a_{22} \\\\\\\\ a_{31} & a_{32} \\\\end{vmatrix}$ Or another expansion: $\\\\begin{vmatrix} a_{11} & a_{12} & a_{13} \\\\\\\\ a_{21} & a_{22} & a_{23} \\\\\\\\ a_{31} & a_{32} & a_{33} \\\\end{vmatrix} = -a_{12} \\\\begin{vmatrix} a_{21} & a_{23} \\\\\\\\ a_{31} & a_{33} \\\\end{vmatrix} + a_{22} \\\\begin{vmatrix} a_{11} & a_{13} \\\\\\\\ a_{31} & a_{33} \\\\end{vmatrix} - a_{32} \\\\begin{vmatrix} a_{11} & a_{13} \\\\\\\\ a_{21} & a_{23} \\\\end{vmatrix}$ $\\\\begin{vmatrix} 1 & 2 & 0 \\\\\\\\ 4 & 1 & 0 \\\\end{vmatrix} = 1 \\\\begin{vmatrix} 1 & 0 \\\\\\\\ -1 & 3 \\\\end{vmatrix} - 2 \\\\begin{vmatrix} 4 & 0\\\\\\\\ 1 & 3 \\\\end{vmatrix} + 0 \\\\begin{vmatrix} 4 & 1 \\\\\\\\ 1 & -1 \\\\end{vmatrix}$ $= 1 (3-0) - 2 (12-0) = -21$ $\\\\begin{vmatrix} 0 & 0 & 0 & 2 \\\\\\\\ 1 & 0 & 0 & 3 \\\\\\\\ 0 & 1 & 0 & 2 \\\\\\\\ 0 & 0 & 1 & 3 \\\\end{vmatrix} = (-1)^{1+4} 2 \\\\begin{vmatrix} 1 & 0 & 0 \\\\\\\\ 0 & 1 & 0 \\\\\\\\ 0 & 0 & 1 \\\\end{vmatrix} = -2$ $\\\\begin{vmatrix} 5 & 4 & 3 \\\\\\\\ 0 & -1 & 2 \\\\\\\\ 0 & 0 & 6 \\\\end{vmatrix} = 5 \\\\begin{vmatrix} -1 & 2 \\\\\\\\ 0 &\",\n \"# Find SVD of a matrix Let A be a matrix, $$A= \\\\begin{pmatrix} 0 & 1 & 0 \\\\\\\\ 0 & 0 & 2 \\\\\\\\ \\\\end{pmatrix}$$ and it's SVD, $$A=USV^t$$. Let $$U^1$$ be the first column of $$U$$ and $$V^1$$ the first column of $$V$$. There are a bunch of options, but the correct answer is: \\\"The singular values of $$A$$ are $$\\\\sigma _1 = 2$$ and $$\\\\sigma _2 =1$$, $$U^1=(0,1)$$ and $$V^1=(0,0,1)$$\\\" And I think I get to the same matrix they're talking about, but in a different order? 1. I found $$A^tA$$=$$\\\\begin{pmatrix} 0 & 0 \\\\\\\\ 1 & 0 \\\\\\\\ 0 & 2 \\\\\\\\ \\\\end{pmatrix}\\\\begin{pmatrix} 0 & 1 & 0 \\\\\\\\ 0 & 0 & 2 \\\\\\\\ \\\\end{pmatrix}=\\\\begin{pmatrix} 0 & 0 & 0 \\\\\\\\ 0 & 1 & 0 \\\\\\\\ 0 & 0 & 4 \\\\\\\\ \\\\end{pmatrix}$$ 2. I found the eigenvectors and eigenvalues of $$A^tA$$, $$\\\\lambda_1=0$$, $$\\\\lambda_2=1$$, $$\\\\lambda_3=4$$. And their eigenvectors are $$S_0=[(1,0,0)]$$, $$S_1=[(0,1,0])$$, $$S_4=[(0,0,1)]$$. I know I don't have to orthonormalize them, because they form the canonical basis. So far, I have the singular values, which are $$\\\\sigma _1 = 1$$ and $$\\\\sigma _2 =2$$, $$S=\\\\begin{pmatrix} 1 & 0 & 0 \\\\\\\\ 0 & 2 & 0 \\\\\\\\ \\\\end{pmatrix}$$,$$V=V^t=\\\\begin{pmatrix} 1 & 0 & 0 \\\\\\\\ 0 & 1 & 0 \\\\\\\\ 0 & 0\",\n \"upon a pair of entries of which are not both zero. So, for example, when, the matrix can be \\\\begin{pmatrix} 0&-1&0\\\\\\\\ 1&0&0\\\\\\\\ 0&0&-1\\\\\\\\ \\\\end{pmatrix}. For the vector, the resulting simplex has vertices \\\\begin{pmatrix}1\\\\ 0\\\\ 1/\\\\surd2\\\\end{pmatrix}, \\\\begin{pmatrix}0\\\\ 1\\\\ -1/\\\\surd2\\\\end{pmatrix}, \\\\begin{pmatrix}-1\\\\ 0\\\\ 1/\\\\surd2\\\\end{pmatrix}, \\\\begin{pmatrix}0\\\\ -1\\\\ -1/\\\\surd2\\\\end{pmatrix}, each of which has distance 2 from the others. Geometric properties Volume The volume of an n-simplex in n-dimensional space with vertices (v0, ..., vn) is Volume= 1 n! \\\\left|\\\\det \\\\begin{pmatrix} v1-v0&&v2-v0&&&&vn-v0 \\\\end{pmatrix}\\\\right| where each column of the n × n determinant is a vector that points from vertex to another vertex .[6] This formula is particularly useful when v0 is the origin. The expression Volume= 1 n! T \\\\det\\\\left[ \\\\begin{pmatrix} v 0 T \\\\ v 0 \\\\\\\\vdots T \\\\end{pmatrix} \\\\begin{pmatrix} v \\\\ v 1-v 0&v2-v0&&vn-v 1/2 0 \\\\end{pmatrix} \\\\right] employs a Gram determinant and works even when the n-simplex's vertices are in a Euclidean space with more than n dimensions, e.g., a triangle in R3 . A more symmetric way to compute the volume of an n-simplex in Rn is Volume={1\\\\overn!}\\\\left|\\\\det \\\\begin{pmatrix} v0&v1&&vn\\\\\\\\ 1&1&&1 \\\\end{pmatrix}\\\\right|. Another common way of computing the volume of the simplex is via the Cayley–Menger determinant, which works even when the n-simplex's vertices are in a Euclidean space with more than n dimensions. Without the 1/n! it is the\"\n]"}}},{"rowIdx":6935,"cells":{"problem":{"kind":"string","value":"Compute\n\\[\\begin{vmatrix} 7 & 3 \\\\ -1 & 2 \\end{vmatrix}.\\]"},"level":{"kind":"string","value":"Level 1"},"type":{"kind":"string","value":"Precalculus"},"solution":{"kind":"string","value":"We have that\n\\[\\begin{vmatrix} 7 & 3 \\\\ -1 & 2 \\end{vmatrix} = (7)(2) - (3)(-1) = \\boxed{17}.\\]"},"answer":{"kind":"string","value":"17"},"p_retrievals":{"kind":"list like","value":["4 & 2 \\\\\\\\ \\end{vmatrix}$$","\\end{vmatrix}\\:\\:\\:$ is $2$ $2e$ $2 \\pi$ $2i$","\\begin{vmatrix} -1 & -2 & 1\\\\ -3 & -2 & 1\\\\ 2& 2 & 1\\\\ \\end{vmatrix}$$ Δ = $$\\frac{1}{2}$$[-1(-2 - 2) + 2(-3 - 2) + 1(-6 + 4)] ⇒ Δ = 4 sq. unit","could someone help me out please $\\begin{vmatrix}7 & 11 & 13 & 17\\\\ 11 & 13 & 17 & 13\\\\ 13 & 17 & 13 & 11\\\\ 17 & 13 & 11 & 7\\end{vmatrix}=72^2$","2 & 5 \\end{vmatrix} \\begin{vmatrix} 11 \\\\ 15 \\end{vmatrix} = \\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix}$ and continue encryption as follows: $\\begin{vmatrix} 7 \\\\ 8 \\end{vmatrix}, \\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix} \\to \\begin{vmatrix} H \\\\ I \\end{vmatrix}, \\begin{vmatrix} A \\\\ T \\end{vmatrix}$ The matrix K is invertible, hence K − 1 exists such that KK − 1 = K − 1K = I2. To implement decrypting, we compute $K^{-1} = 9^{-1} \\begin{vmatrix} 5 & 23 \\\\ 24 & 3 \\end{vmatrix} = 3 \\begin{vmatrix} 5 & 23 \\\\ 24 & 3 \\end{vmatrix} = \\begin{vmatrix} 15 & 17 \\\\ 20 & 9 \\end{vmatrix}$ $HIAT \\to \\begin{vmatrix} H \\\\ I \\end{vmatrix}, \\begin{vmatrix} A \\\\ T \\end{vmatrix} \\to \\begin{vmatrix} 7 \\\\ 8 \\end{vmatrix}, \\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix}$ Then we compute $\\begin{vmatrix} 15 & 17 \\\\ 20 & 9 \\end{vmatrix}\\begin{vmatrix} 7 \\\\ 8 \\end{vmatrix} = \\begin{vmatrix} 7 \\\\ 4 \\end{vmatrix},$ $\\begin{vmatrix} 15 & 17 \\\\ 20 & 9 \\end{vmatrix}\\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix} = \\begin{vmatrix} 11 \\\\ 15 \\end{vmatrix}$ Therefore $\\begin{vmatrix} 7 \\\\ 4 \\end{vmatrix}, \\begin{vmatrix} 11 \\\\ 15 \\end{vmatrix} \\to \\begin{vmatrix} H \\\\ E \\end{vmatrix}, \\begin{vmatrix} L \\\\ P \\end{vmatrix} \\to HELP$. ## Security Unfortunately, the basic Hill cipher is vulnerable to a known-plaintext attack because it is completely linear. An opponent who intercepts n2 plaintext/ciphertext character pairs can set up a","& 4 & 7\\end{vmatrix} = 2\\begin{vmatrix}5 & 2 & 3 \\\\ 1 & 0 & 0 \\\\ 3 & 4 & 7\\end{vmatrix}$$ Where we have first factored out a $2$ from row $2$ and then subtracted row $1$ from row $2$. Now we expand along row $2$ to get $$2\\begin{vmatrix}5 & 2 & 3 \\\\ 1 &0 & 0 \\\\ 3 & 4 & 7\\end{vmatrix} = 2(-1)^{2+1}\\begin{vmatrix} 2 & 3 \\\\ 4 & 7\\end{vmatrix} = -2\\begin{vmatrix} 2 & 3 \\\\ 0 & 1\\end{vmatrix}$$ where in the last step we subtract twice row $1$ from row $2$. Now we simply multiply the diagonal entries to get determinant equal to $-4$.","and 9x - 5y + 8 =0 We have $\\begin{vmatrix} 2 & -3 & 5\\\\ 3 & 4 & -7\\\\ 9 & -5 & 8\\end{vmatrix}$ = 2(32 - 35) - (-3)(24 + 63) + 5(-15 - 36) = 2(-3) + 3(87) + 5(-51) = - 6 + 261 -255 = 0 Therefore, the given three straight lines are concurrent. The Straight Line","## Intermediate Algebra for College Students (7th Edition) We will have to re-arrange the given equations as: $x+2y=3$ and $4x+8y=12$ Need to apply Cramer's Rule. $x=\\dfrac{D_x}{D}$; $y=\\dfrac{D_y}{D}$ Now, $D=\\begin{vmatrix}1&2\\\\4&8\\end{vmatrix}=8-8=0$; $D_x=\\begin{vmatrix}3&2\\\\12&8\\end{vmatrix}=0$; $D_y=\\begin{vmatrix}1&3\\\\4&12\\end{vmatrix}=0$ Hence, we get same lines all are having the common point.","Comment Share Q) # Using properties of determinants, evaluate $A=\\begin{vmatrix}43 & 1 & 6\\\\35 & 7 & 4\\\\17 & 3 &2\\end{vmatrix}$ $\\begin{array}{1 1} 7 \\\\ 0 \\\\ -7 \\\\ 1 \\end{array}$","C_z \\end{vmatrix}$ 6. Aug 25, 2016 ### RJLiberator Ah. Okay guys, thank you for your help. The information I was unaware of was that it was a 3x3 determinant. I know calculated brute force both sides of 1.16 and arrived at the result.","Comment Share Q) # Solve for $x$ if $\\begin{vmatrix} 1 & 4 & 4 \\\\ 1 & -2 & 1 \\\\ 1 & 2x & x^2 \\end{vmatrix} =0$ $\\begin{array}{1 1} (A) x=2,-1 \\\\(B) x=-2,1 \\\\ (C) x=-2,-1 \\\\ none\\;of \\;the \\;above \\end{array}$","$|A| =3\\begin{vmatrix} 1 & -2 \\\\[0.3em] 3 & 1\\end{vmatrix}$ $+4\\begin{vmatrix} 1 & -2 \\\\[0.3em] 2 & 1 \\end{vmatrix}$ $+5\\begin{vmatrix} 1 & 1 \\\\[0.3em] 2 & 3 \\end{vmatrix}$$\\\\$ $=3(1+6)+4(1+4)+5(3-2) \\\\ =3(7)+4(5)+5(1)\\\\ 21+20+5=46$","0\\end{vmatrix}} \\vec{j} + {\\begin{vmatrix}-8 & 0\\\\ 0 & -9\\end{vmatrix}} \\vec{k} \\right\\}$ $= <-3 , 0 , 7> \\cdot <0 , 0 , 72> = 0 + 0 + 7\\cdot72 = 504$ The volume is $V = |504| = 504$ unit3","+ \\begin{vmatrix} 1&1\\\\ -4 & 2\\\\ \\end{vmatrix} &=& 0 \\\\\\\\ k*(5-6)+2+4 &=& 0\\\\ -k &=& -6 \\\\ \\mathbf{k} &=& \\mathbf{6} \\\\ \\hline \\end{array}$$ Aug 3, 2019","Note that $$\\det \\begin{vmatrix} 1 & 1 & 1 \\\\ 1 & -1 & 1 \\\\ -1 & 3 & -1 \\\\ \\end{vmatrix} =0$$ so the three $v_i$ are linearly dependent and therefor cannot span the original subspace","# Determinant equation Algebra Level 3 $\\large \\begin{vmatrix} x^3&20^3&17^3 \\\\ x&20&17 \\\\ 1&1&1 \\end{vmatrix} = 0$ Find the sum of all possible values of $$x$$ satisfy the equation above . ×","& -8 & 0 & 3 \\\\ 0 & 5 & 0 & -6 \\\\ \\end{vmatrix} = 2 \\cdot \\begin{vmatrix} 0 & 3 & -4 \\\\ -5 & -8 & 3 \\\\ 0 & 5 & -6 \\\\ \\end{vmatrix} = 2 \\cdot -(-5) \\cdot \\begin{vmatrix} 3 & -4 \\\\ 5 & -6 \\\\ \\end{vmatrix} = 2 \\cdot -(-5) \\cdot (-18 -(-20)) = 20.$$ I expanded along the third column in the first case and then along the first column, because these are the ones with the most zeros, so it saves a lot of computations. I am assuming that's what's in your book. If you want to expand along other columns, just keep track of the appropriate minors without making any computation mistakes (I guess that's the hard part ; the only trick is to make it slowly and be careful). So here we go : $$\\det A = \\begin{vmatrix} 5 & -7 & 2 & 2 \\\\ 0 & 3 & 0 & -4 \\\\ -5 & -8 & 0 & 3 \\\\ 0 & 5 & 0 & -6 \\\\ \\end{vmatrix} = 5 \\begin{vmatrix} 3 & 0 & -4 \\\\ -8 & 0 & 3 \\\\ 5 & 0 & -6 \\\\ \\end{vmatrix} -(-7) \\begin{vmatrix} 0 & 0 & -4 \\\\ -5 & 0 & 3","(n+3)^2 & (n+4)^2 \\\\ \\end{vmatrix} =-4(-1)(-1)(2) = -8$","\\\\ x_2 & y_2 & z_2 & 1 \\\\ x_3 & y_3 & z_3 & 1 \\\\ \\end{vmatrix}$$","&7 \\\\ 3& 8& 8\\\\ 5& 9&9 \\end{vmatrix}$ [$\\dpi{100} \\because$ Here two columns are identical ] $\\dpi{100} = 0$ Therefore we have the value of determinant = 0. Exams Articles Questions"],"string":"[\n \"4 & 2 \\\\\\\\\\\\\\\\ \\\\end{vmatrix}$$\",\n \"\\\\end{vmatrix}\\\\:\\\\:\\\\:$ is $2$ $2e$ $2 \\\\pi$ $2i$\",\n \"\\\\begin{vmatrix} -1 & -2 & 1\\\\\\\\ -3 & -2 & 1\\\\\\\\ 2& 2 & 1\\\\\\\\ \\\\end{vmatrix}$$ Δ = $$\\\\frac{1}{2}$$[-1(-2 - 2) + 2(-3 - 2) + 1(-6 + 4)] ⇒ Δ = 4 sq. unit\",\n \"could someone help me out please $\\\\begin{vmatrix}7 & 11 & 13 & 17\\\\\\\\ 11 & 13 & 17 & 13\\\\\\\\ 13 & 17 & 13 & 11\\\\\\\\ 17 & 13 & 11 & 7\\\\end{vmatrix}=72^2$\",\n \"2 & 5 \\\\end{vmatrix} \\\\begin{vmatrix} 11 \\\\\\\\ 15 \\\\end{vmatrix} = \\\\begin{vmatrix} 0 \\\\\\\\ 19 \\\\end{vmatrix}$ and continue encryption as follows: $\\\\begin{vmatrix} 7 \\\\\\\\ 8 \\\\end{vmatrix}, \\\\begin{vmatrix} 0 \\\\\\\\ 19 \\\\end{vmatrix} \\\\to \\\\begin{vmatrix} H \\\\\\\\ I \\\\end{vmatrix}, \\\\begin{vmatrix} A \\\\\\\\ T \\\\end{vmatrix}$ The matrix K is invertible, hence K − 1 exists such that KK − 1 = K − 1K = I2. To implement decrypting, we compute $K^{-1} = 9^{-1} \\\\begin{vmatrix} 5 & 23 \\\\\\\\ 24 & 3 \\\\end{vmatrix} = 3 \\\\begin{vmatrix} 5 & 23 \\\\\\\\ 24 & 3 \\\\end{vmatrix} = \\\\begin{vmatrix} 15 & 17 \\\\\\\\ 20 & 9 \\\\end{vmatrix}$ $HIAT \\\\to \\\\begin{vmatrix} H \\\\\\\\ I \\\\end{vmatrix}, \\\\begin{vmatrix} A \\\\\\\\ T \\\\end{vmatrix} \\\\to \\\\begin{vmatrix} 7 \\\\\\\\ 8 \\\\end{vmatrix}, \\\\begin{vmatrix} 0 \\\\\\\\ 19 \\\\end{vmatrix}$ Then we compute $\\\\begin{vmatrix} 15 & 17 \\\\\\\\ 20 & 9 \\\\end{vmatrix}\\\\begin{vmatrix} 7 \\\\\\\\ 8 \\\\end{vmatrix} = \\\\begin{vmatrix} 7 \\\\\\\\ 4 \\\\end{vmatrix},$ $\\\\begin{vmatrix} 15 & 17 \\\\\\\\ 20 & 9 \\\\end{vmatrix}\\\\begin{vmatrix} 0 \\\\\\\\ 19 \\\\end{vmatrix} = \\\\begin{vmatrix} 11 \\\\\\\\ 15 \\\\end{vmatrix}$ Therefore $\\\\begin{vmatrix} 7 \\\\\\\\ 4 \\\\end{vmatrix}, \\\\begin{vmatrix} 11 \\\\\\\\ 15 \\\\end{vmatrix} \\\\to \\\\begin{vmatrix} H \\\\\\\\ E \\\\end{vmatrix}, \\\\begin{vmatrix} L \\\\\\\\ P \\\\end{vmatrix} \\\\to HELP$. ## Security Unfortunately, the basic Hill cipher is vulnerable to a known-plaintext attack because it is completely linear. An opponent who intercepts n2 plaintext/ciphertext character pairs can set up a\",\n \"& 4 & 7\\\\end{vmatrix} = 2\\\\begin{vmatrix}5 & 2 & 3 \\\\\\\\ 1 & 0 & 0 \\\\\\\\ 3 & 4 & 7\\\\end{vmatrix}$$ Where we have first factored out a $2$ from row $2$ and then subtracted row $1$ from row $2$. Now we expand along row $2$ to get $$2\\\\begin{vmatrix}5 & 2 & 3 \\\\\\\\ 1 &0 & 0 \\\\\\\\ 3 & 4 & 7\\\\end{vmatrix} = 2(-1)^{2+1}\\\\begin{vmatrix} 2 & 3 \\\\\\\\ 4 & 7\\\\end{vmatrix} = -2\\\\begin{vmatrix} 2 & 3 \\\\\\\\ 0 & 1\\\\end{vmatrix}$$ where in the last step we subtract twice row $1$ from row $2$. Now we simply multiply the diagonal entries to get determinant equal to $-4$.\",\n \"and 9x - 5y + 8 =0 We have $\\\\begin{vmatrix} 2 & -3 & 5\\\\\\\\ 3 & 4 & -7\\\\\\\\ 9 & -5 & 8\\\\end{vmatrix}$ = 2(32 - 35) - (-3)(24 + 63) + 5(-15 - 36) = 2(-3) + 3(87) + 5(-51) = - 6 + 261 -255 = 0 Therefore, the given three straight lines are concurrent. The Straight Line\",\n \"## Intermediate Algebra for College Students (7th Edition) We will have to re-arrange the given equations as: $x+2y=3$ and $4x+8y=12$ Need to apply Cramer's Rule. $x=\\\\dfrac{D_x}{D}$; $y=\\\\dfrac{D_y}{D}$ Now, $D=\\\\begin{vmatrix}1&2\\\\\\\\4&8\\\\end{vmatrix}=8-8=0$; $D_x=\\\\begin{vmatrix}3&2\\\\\\\\12&8\\\\end{vmatrix}=0$; $D_y=\\\\begin{vmatrix}1&3\\\\\\\\4&12\\\\end{vmatrix}=0$ Hence, we get same lines all are having the common point.\",\n \"Comment Share Q) # Using properties of determinants, evaluate $A=\\\\begin{vmatrix}43 & 1 & 6\\\\\\\\35 & 7 & 4\\\\\\\\17 & 3 &2\\\\end{vmatrix}$ $\\\\begin{array}{1 1} 7 \\\\\\\\ 0 \\\\\\\\ -7 \\\\\\\\ 1 \\\\end{array}$\",\n \"C_z \\\\end{vmatrix}$ 6. Aug 25, 2016 ### RJLiberator Ah. Okay guys, thank you for your help. The information I was unaware of was that it was a 3x3 determinant. I know calculated brute force both sides of 1.16 and arrived at the result.\",\n \"Comment Share Q) # Solve for $x$ if $\\\\begin{vmatrix} 1 & 4 & 4 \\\\\\\\ 1 & -2 & 1 \\\\\\\\ 1 & 2x & x^2 \\\\end{vmatrix} =0$ $\\\\begin{array}{1 1} (A) x=2,-1 \\\\\\\\(B) x=-2,1 \\\\\\\\ (C) x=-2,-1 \\\\\\\\ none\\\\;of \\\\;the \\\\;above \\\\end{array}$\",\n \"$|A| =3\\\\begin{vmatrix} 1 & -2 \\\\\\\\[0.3em] 3 & 1\\\\end{vmatrix}$ $+4\\\\begin{vmatrix} 1 & -2 \\\\\\\\[0.3em] 2 & 1 \\\\end{vmatrix}$ $+5\\\\begin{vmatrix} 1 & 1 \\\\\\\\[0.3em] 2 & 3 \\\\end{vmatrix}$$\\\\\\\\$ $=3(1+6)+4(1+4)+5(3-2) \\\\\\\\ =3(7)+4(5)+5(1)\\\\\\\\ 21+20+5=46$\",\n \"0\\\\end{vmatrix}} \\\\vec{j} + {\\\\begin{vmatrix}-8 & 0\\\\\\\\ 0 & -9\\\\end{vmatrix}} \\\\vec{k} \\\\right\\\\}$ $= <-3 , 0 , 7> \\\\cdot <0 , 0 , 72> = 0 + 0 + 7\\\\cdot72 = 504$ The volume is $V = |504| = 504$ unit3\",\n \"+ \\\\begin{vmatrix} 1&1\\\\\\\\ -4 & 2\\\\\\\\ \\\\end{vmatrix} &=& 0 \\\\\\\\\\\\\\\\ k*(5-6)+2+4 &=& 0\\\\\\\\ -k &=& -6 \\\\\\\\ \\\\mathbf{k} &=& \\\\mathbf{6} \\\\\\\\ \\\\hline \\\\end{array}$$ Aug 3, 2019\",\n \"Note that $$\\\\det \\\\begin{vmatrix} 1 & 1 & 1 \\\\\\\\ 1 & -1 & 1 \\\\\\\\ -1 & 3 & -1 \\\\\\\\ \\\\end{vmatrix} =0$$ so the three $v_i$ are linearly dependent and therefor cannot span the original subspace\",\n \"# Determinant equation Algebra Level 3 $\\\\large \\\\begin{vmatrix} x^3&20^3&17^3 \\\\\\\\ x&20&17 \\\\\\\\ 1&1&1 \\\\end{vmatrix} = 0$ Find the sum of all possible values of $$x$$ satisfy the equation above . ×\",\n \"& -8 & 0 & 3 \\\\\\\\ 0 & 5 & 0 & -6 \\\\\\\\ \\\\end{vmatrix} = 2 \\\\cdot \\\\begin{vmatrix} 0 & 3 & -4 \\\\\\\\ -5 & -8 & 3 \\\\\\\\ 0 & 5 & -6 \\\\\\\\ \\\\end{vmatrix} = 2 \\\\cdot -(-5) \\\\cdot \\\\begin{vmatrix} 3 & -4 \\\\\\\\ 5 & -6 \\\\\\\\ \\\\end{vmatrix} = 2 \\\\cdot -(-5) \\\\cdot (-18 -(-20)) = 20.$$ I expanded along the third column in the first case and then along the first column, because these are the ones with the most zeros, so it saves a lot of computations. I am assuming that's what's in your book. If you want to expand along other columns, just keep track of the appropriate minors without making any computation mistakes (I guess that's the hard part ; the only trick is to make it slowly and be careful). So here we go : $$\\\\det A = \\\\begin{vmatrix} 5 & -7 & 2 & 2 \\\\\\\\ 0 & 3 & 0 & -4 \\\\\\\\ -5 & -8 & 0 & 3 \\\\\\\\ 0 & 5 & 0 & -6 \\\\\\\\ \\\\end{vmatrix} = 5 \\\\begin{vmatrix} 3 & 0 & -4 \\\\\\\\ -8 & 0 & 3 \\\\\\\\ 5 & 0 & -6 \\\\\\\\ \\\\end{vmatrix} -(-7) \\\\begin{vmatrix} 0 & 0 & -4 \\\\\\\\ -5 & 0 & 3\",\n \"(n+3)^2 & (n+4)^2 \\\\\\\\ \\\\end{vmatrix} =-4(-1)(-1)(2) = -8$\",\n \"\\\\\\\\ x_2 & y_2 & z_2 & 1 \\\\\\\\ x_3 & y_3 & z_3 & 1 \\\\\\\\ \\\\end{vmatrix}$$\",\n \"&7 \\\\\\\\ 3& 8& 8\\\\\\\\ 5& 9&9 \\\\end{vmatrix}$ [$\\\\dpi{100} \\\\because$ Here two columns are identical ] $\\\\dpi{100} = 0$ Therefore we have the value of determinant = 0. Exams Articles Questions\"\n]"},"s_retrievals":{"kind":"list like","value":["0\\end{vmatrix}} \\vec{j} + {\\begin{vmatrix}-8 & 0\\\\ 0 & -9\\end{vmatrix}} \\vec{k} \\right\\}$ $= <-3 , 0 , 7> \\cdot <0 , 0 , 72> = 0 + 0 + 7\\cdot72 = 504$ The volume is $V = |504| = 504$ unit3 Updated: 20 April 2017 (A Dendane)","could someone help me out please $\\begin{vmatrix}7 & 11 & 13 & 17\\\\ 11 & 13 & 17 & 13\\\\ 13 & 17 & 13 & 11\\\\ 17 & 13 & 11 & 7\\end{vmatrix}=72^2$","$|A| =3\\begin{vmatrix} 1 & -2 \\\\[0.3em] 3 & 1\\end{vmatrix}$ $+4\\begin{vmatrix} 1 & -2 \\\\[0.3em] 2 & 1 \\end{vmatrix}$ $+5\\begin{vmatrix} 1 & 1 \\\\[0.3em] 2 & 3 \\end{vmatrix}$$\\\\$ $=3(1+6)+4(1+4)+5(3-2) \\\\ =3(7)+4(5)+5(1)\\\\ 21+20+5=46$","0\\end{vmatrix}} \\vec{j} + {\\begin{vmatrix}-8 & 0\\\\ 0 & -9\\end{vmatrix}} \\vec{k} \\right\\}$ $= <-3 , 0 , 7> \\cdot <0 , 0 , 72> = 0 + 0 + 7\\cdot72 = 504$ The volume is $V = |504| = 504$ unit3","\\begin{vmatrix} -1 & -2 & 1\\\\ -3 & -2 & 1\\\\ 2& 2 & 1\\\\ \\end{vmatrix}$$ Δ = $$\\frac{1}{2}$$[-1(-2 - 2) + 2(-3 - 2) + 1(-6 + 4)] ⇒ Δ = 4 sq. unit","\\\\ 0 & 0 & 1 & -\\nicefrac{3}{17} \\\\ 0 & 0 & 43 & 12 \\end{vmatrix*} \\begin{matrix*}[l] \\\\ \\\\ \\leftarrow \\nicefrac{1}{17}r_3 \\ \\text{\\scriptsize (factor out 17)} \\\\ \\\\ \\end{matrix*} \\\\ &= -17\\begin{vmatrix*}[r] 1 & -3 & 5 & -12 \\\\ 0 & 1 & -12 & -3 \\\\ 0 & 0 & 1 & -\\nicefrac{3}{17} \\\\ 0 & 0 & 0 & \\nicefrac{333}{17} \\end{vmatrix*}","(n+3)^2 & (n+4)^2 \\\\ \\end{vmatrix} =-4(-1)(-1)(2) = -8$","Browse Questions # If $\\begin{vmatrix} 3x & 7 \\\\ 2 & 4 \\end{vmatrix}$= 10, find x. Toolbox: • The determinant value of a $2\\times 2$ matrix is $|A|=a_{11}\\times a_{22}-a_{21}\\times a_{12}$ $\\begin{vmatrix}3x & 7\\\\2 & 4\\end{vmatrix}$=10 Now let us evaluate the value of the determinant $3x\\times 4-7\\times 2=10.$ 12x-14=10. 12x=10+14 12x=24 x=$\\frac{24}{12}=2$ Hence the value of x=2.","4 & 2 \\\\\\\\ \\end{vmatrix}$$","the questions are solved in a way that does not require a calculator. Calculate the determinants of these matrices. For the $3 * 3$ and $4 * 4$ matrices, use at least two methods as applicable. Show all work. (1.) $\\begin{vmatrix} -3 & 3 \\\\[3ex] -1 & -2 \\end{vmatrix}$ $\\begin{vmatrix} -3 & 3 \\\\[3ex] -1 & -2 \\end{vmatrix} \\\\[3ex]$ $= (-3)(-2) - (-1)(3) \\\\[3ex] = 6 - (-3) \\\\[3ex] = 6 + 3 \\\\[3ex] = 9$ (2.) $\\begin{vmatrix} 7 & \\sqrt{7} \\\\[3ex] \\sqrt{7} & 7 \\end{vmatrix}$ $\\begin{vmatrix} 7 & \\sqrt{7} \\\\[3ex] \\sqrt{7} & 7 \\end{vmatrix} \\\\[3ex]$ $= (7)(7) - (\\sqrt{7})(\\sqrt{7}) \\\\[3ex] = 49 - 7 \\\\[3ex] = 42$ (3.) $\\begin{vmatrix} 3x^7 & -12 \\\\[3ex] x^7 & x^5 \\end{vmatrix}$ $\\begin{vmatrix} 3x^7 & -12 \\\\[3ex] x^7 & x^5 \\end{vmatrix} \\\\[3ex]$ $= (3x^7)(x^5) - (x^7)(-12) \\\\[3ex] = 3x^{12} - (-12x^7) \\\\[3ex] = 3x^{12} + 12x^7$ (4.) $\\begin{vmatrix} p & c \\\\[3ex] -d & -e \\end{vmatrix}$ $\\begin{vmatrix} p & c \\\\[3ex] -d & -e \\end{vmatrix} \\\\[3ex]$ $= (p)(-e) - (-d)(c) \\\\[3ex] = -ep - (-cd) \\\\[3ex] = -ep + cd \\\\[3ex] = cd - ep$ (5.) $\\begin{vmatrix} 2 & -1 & 6 \\\\[3ex] 4 & -6 & 0 \\\\[3ex] 5 & -2 & 6 \\end{vmatrix}$ First Method: Formula To solve it faster, we shall use the second row because it contains a","\\times 7 = \\boxed{70} \\end{gather*}$$","& 3 & 8\\\\4 & 3 & 2\\end {vmatrix}$$ $$det⁡ A=1( 4\\begin{vmatrix} 3 & 2 \\\\9 & 6\\end{vmatrix} – 3\\begin{vmatrix}4 & 2 \\\\4 & 6\\end{vmatrix}+8\\begin{vmatrix}4 & 3 \\\\4 & 9\\end{vmatrix}) -2( 8\\begin{vmatrix} 3 & 2 \\\\9 & 6\\end{vmatrix} – 7\\begin{vmatrix}4 & 2 \\\\4 & 6\\end{vmatrix}+2\\begin{vmatrix}4 & 3 \\\\4 & 9\\end{vmatrix}) +1( 8\\begin{vmatrix}3 & 8 \\\\9 & 6\\end{vmatrix} – 7\\begin{vmatrix}4 & 8 \\\\4 & 6\\end{vmatrix}+2\\begin{vmatrix}4 & 3 \\\\4 & 9\\end{vmatrix}) -1( 8\\begin{vmatrix} 3 & 8 \\\\3 & 2\\end{vmatrix} – 7\\begin{vmatrix}4 & 8 \\\\4 & 6\\end{vmatrix}+2\\begin{vmatrix}4 & 3 \\\\4 & 3\\end{vmatrix})$$ $$det⁡ A = 1[4(18-18)-3(24-8)+ 8(36-12)]-2[ 8(18-18)-7(24-8)+ 2(36-12)]+ 1[ 8(18-72)-7(24-32)+ 2(36-12)] -1[8(6-24)-7(8-32)+ 2(12-12)]$$ $$det⁡ A = 1[4(0)-3(16)+ 8(24)]-2[ 8(0)-7(16)+ 2(24)]+ 1[ 8(-54)-7(-8)+ 2(24)]-1[8(-18)-7(-24)+ 2(0)]$$ $$det⁡ A = 1[0-48+192]-2[0-112+48]+ 1[ -432+56+48]-1[-144+168+0]$$ $$det⁡ A = 1[144]-2[-64]+ 1[-328]-1[24]$$ $$det⁡ A = 144+128-328- 24$$ $$det⁡ A = -80$$ #### Expand Along the Row: For the calculations of matrix A = (aij)4×4 from expansion of row is determined by the following formula: $$det A = \\begin{vmatrix} a & b & c & d\\\\e & f & g &h \\\\i & j & k & l \\\\ m & n & o & p \\end{vmatrix} \\\\$$ $$det⁡ A= a\\begin{vmatrix} f & g & h\\\\j & k & l\\\\n & o & p\\end{vmatrix} – b\\begin{vmatrix}e & g & h\\\\i & k & l\\\\ m & o &","-4 & -1 & 0\\end{vmatrix}$ but ive no clue how to do this im pretty sure we've never been told could someone help me out please $\\begin{vmatrix}7 & 11 & 13 & 17\\\\ 11 & 13 & 17 & 13\\\\ 13 & 17 & 13 & 11\\\\ 17 & 13 & 11 & 7\\end{vmatrix}=72^2$","26\\\\ -3 & 5 & -16\\\\ \\end{vmatrix} = \\begin {vmatrix} 2 & 4 & 13 – 5 \\times 2\\\\ 5 & 1 & 26 – 5 \\times 5\\\\ -3 & 5 & -16 – 5 \\times (-3)\\\\ \\end{vmatrix} = \\begin {vmatrix} 2 & 4 & 3\\\\ 5 & 1 & 1\\\\ -3 & 5 & -1\\\\ \\end{vmatrix}$ Now expanding the determinant we get $\\begin {vmatrix} 2 & 4 & 3\\\\ 5 & 1 & 1\\\\ -3 & 5 & -1\\\\ \\end{vmatrix} = 2 \\times \\begin {vmatrix} 1 & 1\\\\ 5 & -1\\\\ \\end{vmatrix} – 4 \\times \\begin {vmatrix} 5 & 1\\\\ -3 & -1\\\\ \\end{vmatrix} + 3 \\times \\begin {vmatrix} 5 & 1\\\\ -3 & 5\\\\ \\end{vmatrix}$ $= 2 \\times (1 \\times (-1) – 1 \\times 5) – 4 \\times (5 \\times (-1) – 1 \\times (-3)) + 3 \\times (5 \\times 5 – 1 \\times (-3))$ $= 2 \\times (-1 – 5) – 4 \\times (-5 + 3) + 3 \\times (25 + 3)$ $= 2 \\times (-6) – 4 \\times (-2) + 3 \\times 28$ $= -12 + 8 + 84 = 80$ ### 7. Triangular Property of Determinant The triangular property of the determinant states that if the elements above or below the main diagonal are equal to zero, then the value of the determinant","and 9x - 5y + 8 =0 We have $\\begin{vmatrix} 2 & -3 & 5\\\\ 3 & 4 & -7\\\\ 9 & -5 & 8\\end{vmatrix}$ = 2(32 - 35) - (-3)(24 + 63) + 5(-15 - 36) = 2(-3) + 3(87) + 5(-51) = - 6 + 261 -255 = 0 Therefore, the given three straight lines are concurrent. The Straight Line","n & n+1 \\ 2 & 3 & … & n & n & n+1 \\ … & … & … & … & … & … \\ 2 & n & … & n-1 & n & n+1 \\ n & 3 & … & n-1 & n & n+1 end{vmatrix}$$ The closest one to mine from StackExchange was this one. But I didn’t manage to link these two determinants. No matter how I transform it, nothing worked for me. Any ideas?","0 & 2 \\end{vmatrix} + 3 \\begin{vmatrix} x & 1 \\\\ 0 & 3 \\end{vmatrix} = 132$$ $$2(2 - 12) + 5(2x) + 3(3x) = 132$$ $$-20 + 10x + 9x = 132$$ $$19x = 152$$ $$x = 8$$","\\\\ { – 7 } & { – 2 } \\end{vmatrix}$ $D_{ y } = \\begin{vmatrix} 2 & { 7 } \\\\ { 3 } & { – 7 } \\end{vmatrix}$ Recall the formula to evaluate a $2 \\times 2$ determinant: For a $2 \\timess 2$ matrix — $A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$ The determinant is calculated as — $det( A ) = | A | = \\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} = ad – bc$ Let’s calculate the determinants: $D = \\begin{vmatrix} 2 & 1 \\\\ { 3 } & { – 2 } \\end{vmatrix} = ( 2 )( – 2 ) – ( 1 )( 3 ) = – 4 – 3 = – 7$ $D_{ x } = \\begin{vmatrix} 7 & 1 \\\\ { – 7 } & { – 2 } \\end{vmatrix} = ( 7 )( – 2 ) – ( 1 )( – 7 ) = – 14 – ( – 7 ) = – 14 + 7 = – 7$ $D_{ y } = \\begin{vmatrix} 2 & { 7 } \\\\ { 3 } & { – 7 } \\end{vmatrix} = ( 2 )( – 7 ) – ( 7 )( 3 ) = -14 – 21 = – 35$ Now, we","Note that $$\\det \\begin{vmatrix} 1 & 1 & 1 \\\\ 1 & -1 & 1 \\\\ -1 & 3 & -1 \\\\ \\end{vmatrix} =0$$ so the three $v_i$ are linearly dependent and therefor cannot span the original subspace","\\times \\begin {vmatrix} 0 & 5\\\\ 0 & 0\\\\ \\end{vmatrix}$ $= 3 \\times (5 \\times 4 – 7 \\times 0) – 0 – 2 \\times (0 \\times 0 – 5 \\times 0)$ $= 3 \\times (20 – 0) – 0 – 2 \\times (0 – 0) = 3 \\times 20 – 0 – 2 \\times 0 = 3 \\times 20 – 0 – 0 = 60$. Example 2: Find the value of $\\begin {vmatrix} 1 & 0 & 0\\\\ 5 & 3 & 0\\\\ 6 & 7 & 4\\\\ \\end{vmatrix}$. $\\begin {vmatrix} 1 & 0 & 0\\\\ 5 & 3 & 0\\\\ 6 & 7 & 4\\\\ \\end{vmatrix} = 1 \\times 3 \\times 4 = 12$ Let’s check the value of the determinant by expansion. $\\begin {vmatrix} 1 & 0 & 0\\\\ 5 & 3 & 0\\\\ 6 & 7 & 4\\\\ \\end{vmatrix} = 1 \\times \\begin {vmatrix} 3 & 0\\\\ 7 & 4\\\\ \\end{vmatrix} – 0 \\times \\begin {vmatrix} 5 & 0\\\\ 6 & 4\\\\ \\end{vmatrix} + 0 \\times \\begin {vmatrix} 5 & 3\\\\ 6 & 7\\\\ \\end{vmatrix}$ $= 1 \\times (3 \\times 4 – 0 \\times 7) – 0 + 0 = 1 \\times (12 – 0) – 0 + 0 = 12$ ## Examples of Using Properties of Determinants Example 1: Find the value of"],"string":"[\n \"0\\\\end{vmatrix}} \\\\vec{j} + {\\\\begin{vmatrix}-8 & 0\\\\\\\\ 0 & -9\\\\end{vmatrix}} \\\\vec{k} \\\\right\\\\}$ $= <-3 , 0 , 7> \\\\cdot <0 , 0 , 72> = 0 + 0 + 7\\\\cdot72 = 504$ The volume is $V = |504| = 504$ unit3 Updated: 20 April 2017 (A Dendane)\",\n \"could someone help me out please $\\\\begin{vmatrix}7 & 11 & 13 & 17\\\\\\\\ 11 & 13 & 17 & 13\\\\\\\\ 13 & 17 & 13 & 11\\\\\\\\ 17 & 13 & 11 & 7\\\\end{vmatrix}=72^2$\",\n \"$|A| =3\\\\begin{vmatrix} 1 & -2 \\\\\\\\[0.3em] 3 & 1\\\\end{vmatrix}$ $+4\\\\begin{vmatrix} 1 & -2 \\\\\\\\[0.3em] 2 & 1 \\\\end{vmatrix}$ $+5\\\\begin{vmatrix} 1 & 1 \\\\\\\\[0.3em] 2 & 3 \\\\end{vmatrix}$$\\\\\\\\$ $=3(1+6)+4(1+4)+5(3-2) \\\\\\\\ =3(7)+4(5)+5(1)\\\\\\\\ 21+20+5=46$\",\n \"0\\\\end{vmatrix}} \\\\vec{j} + {\\\\begin{vmatrix}-8 & 0\\\\\\\\ 0 & -9\\\\end{vmatrix}} \\\\vec{k} \\\\right\\\\}$ $= <-3 , 0 , 7> \\\\cdot <0 , 0 , 72> = 0 + 0 + 7\\\\cdot72 = 504$ The volume is $V = |504| = 504$ unit3\",\n \"\\\\begin{vmatrix} -1 & -2 & 1\\\\\\\\ -3 & -2 & 1\\\\\\\\ 2& 2 & 1\\\\\\\\ \\\\end{vmatrix}$$ Δ = $$\\\\frac{1}{2}$$[-1(-2 - 2) + 2(-3 - 2) + 1(-6 + 4)] ⇒ Δ = 4 sq. unit\",\n \"\\\\\\\\ 0 & 0 & 1 & -\\\\nicefrac{3}{17} \\\\\\\\ 0 & 0 & 43 & 12 \\\\end{vmatrix*} \\\\begin{matrix*}[l] \\\\\\\\ \\\\\\\\ \\\\leftarrow \\\\nicefrac{1}{17}r_3 \\\\ \\\\text{\\\\scriptsize (factor out 17)} \\\\\\\\ \\\\\\\\ \\\\end{matrix*} \\\\\\\\ &= -17\\\\begin{vmatrix*}[r] 1 & -3 & 5 & -12 \\\\\\\\ 0 & 1 & -12 & -3 \\\\\\\\ 0 & 0 & 1 & -\\\\nicefrac{3}{17} \\\\\\\\ 0 & 0 & 0 & \\\\nicefrac{333}{17} \\\\end{vmatrix*}\",\n \"(n+3)^2 & (n+4)^2 \\\\\\\\ \\\\end{vmatrix} =-4(-1)(-1)(2) = -8$\",\n \"Browse Questions # If $\\\\begin{vmatrix} 3x & 7 \\\\\\\\ 2 & 4 \\\\end{vmatrix}$= 10, find x. Toolbox: • The determinant value of a $2\\\\times 2$ matrix is $|A|=a_{11}\\\\times a_{22}-a_{21}\\\\times a_{12}$ $\\\\begin{vmatrix}3x & 7\\\\\\\\2 & 4\\\\end{vmatrix}$=10 Now let us evaluate the value of the determinant $3x\\\\times 4-7\\\\times 2=10.$ 12x-14=10. 12x=10+14 12x=24 x=$\\\\frac{24}{12}=2$ Hence the value of x=2.\",\n \"4 & 2 \\\\\\\\\\\\\\\\ \\\\end{vmatrix}$$\",\n \"the questions are solved in a way that does not require a calculator. Calculate the determinants of these matrices. For the $3 * 3$ and $4 * 4$ matrices, use at least two methods as applicable. Show all work. (1.) $\\\\begin{vmatrix} -3 & 3 \\\\\\\\[3ex] -1 & -2 \\\\end{vmatrix}$ $\\\\begin{vmatrix} -3 & 3 \\\\\\\\[3ex] -1 & -2 \\\\end{vmatrix} \\\\\\\\[3ex]$ $= (-3)(-2) - (-1)(3) \\\\\\\\[3ex] = 6 - (-3) \\\\\\\\[3ex] = 6 + 3 \\\\\\\\[3ex] = 9$ (2.) $\\\\begin{vmatrix} 7 & \\\\sqrt{7} \\\\\\\\[3ex] \\\\sqrt{7} & 7 \\\\end{vmatrix}$ $\\\\begin{vmatrix} 7 & \\\\sqrt{7} \\\\\\\\[3ex] \\\\sqrt{7} & 7 \\\\end{vmatrix} \\\\\\\\[3ex]$ $= (7)(7) - (\\\\sqrt{7})(\\\\sqrt{7}) \\\\\\\\[3ex] = 49 - 7 \\\\\\\\[3ex] = 42$ (3.) $\\\\begin{vmatrix} 3x^7 & -12 \\\\\\\\[3ex] x^7 & x^5 \\\\end{vmatrix}$ $\\\\begin{vmatrix} 3x^7 & -12 \\\\\\\\[3ex] x^7 & x^5 \\\\end{vmatrix} \\\\\\\\[3ex]$ $= (3x^7)(x^5) - (x^7)(-12) \\\\\\\\[3ex] = 3x^{12} - (-12x^7) \\\\\\\\[3ex] = 3x^{12} + 12x^7$ (4.) $\\\\begin{vmatrix} p & c \\\\\\\\[3ex] -d & -e \\\\end{vmatrix}$ $\\\\begin{vmatrix} p & c \\\\\\\\[3ex] -d & -e \\\\end{vmatrix} \\\\\\\\[3ex]$ $= (p)(-e) - (-d)(c) \\\\\\\\[3ex] = -ep - (-cd) \\\\\\\\[3ex] = -ep + cd \\\\\\\\[3ex] = cd - ep$ (5.) $\\\\begin{vmatrix} 2 & -1 & 6 \\\\\\\\[3ex] 4 & -6 & 0 \\\\\\\\[3ex] 5 & -2 & 6 \\\\end{vmatrix}$ First Method: Formula To solve it faster, we shall use the second row because it contains a\",\n \"\\\\times 7 = \\\\boxed{70} \\\\end{gather*}$$\",\n \"& 3 & 8\\\\\\\\4 & 3 & 2\\\\end {vmatrix}$$ $$det⁡ A=1( 4\\\\begin{vmatrix} 3 & 2 \\\\\\\\9 & 6\\\\end{vmatrix} – 3\\\\begin{vmatrix}4 & 2 \\\\\\\\4 & 6\\\\end{vmatrix}+8\\\\begin{vmatrix}4 & 3 \\\\\\\\4 & 9\\\\end{vmatrix}) -2( 8\\\\begin{vmatrix} 3 & 2 \\\\\\\\9 & 6\\\\end{vmatrix} – 7\\\\begin{vmatrix}4 & 2 \\\\\\\\4 & 6\\\\end{vmatrix}+2\\\\begin{vmatrix}4 & 3 \\\\\\\\4 & 9\\\\end{vmatrix}) +1( 8\\\\begin{vmatrix}3 & 8 \\\\\\\\9 & 6\\\\end{vmatrix} – 7\\\\begin{vmatrix}4 & 8 \\\\\\\\4 & 6\\\\end{vmatrix}+2\\\\begin{vmatrix}4 & 3 \\\\\\\\4 & 9\\\\end{vmatrix}) -1( 8\\\\begin{vmatrix} 3 & 8 \\\\\\\\3 & 2\\\\end{vmatrix} – 7\\\\begin{vmatrix}4 & 8 \\\\\\\\4 & 6\\\\end{vmatrix}+2\\\\begin{vmatrix}4 & 3 \\\\\\\\4 & 3\\\\end{vmatrix})$$ $$det⁡ A = 1[4(18-18)-3(24-8)+ 8(36-12)]-2[ 8(18-18)-7(24-8)+ 2(36-12)]+ 1[ 8(18-72)-7(24-32)+ 2(36-12)] -1[8(6-24)-7(8-32)+ 2(12-12)]$$ $$det⁡ A = 1[4(0)-3(16)+ 8(24)]-2[ 8(0)-7(16)+ 2(24)]+ 1[ 8(-54)-7(-8)+ 2(24)]-1[8(-18)-7(-24)+ 2(0)]$$ $$det⁡ A = 1[0-48+192]-2[0-112+48]+ 1[ -432+56+48]-1[-144+168+0]$$ $$det⁡ A = 1[144]-2[-64]+ 1[-328]-1[24]$$ $$det⁡ A = 144+128-328- 24$$ $$det⁡ A = -80$$ #### Expand Along the Row: For the calculations of matrix A = (aij)4×4 from expansion of row is determined by the following formula: $$det A = \\\\begin{vmatrix} a & b & c & d\\\\\\\\e & f & g &h \\\\\\\\i & j & k & l \\\\\\\\ m & n & o & p \\\\end{vmatrix} \\\\\\\\$$ $$det⁡ A= a\\\\begin{vmatrix} f & g & h\\\\\\\\j & k & l\\\\\\\\n & o & p\\\\end{vmatrix} – b\\\\begin{vmatrix}e & g & h\\\\\\\\i & k & l\\\\\\\\ m & o &\",\n \"-4 & -1 & 0\\\\end{vmatrix}$ but ive no clue how to do this im pretty sure we've never been told could someone help me out please $\\\\begin{vmatrix}7 & 11 & 13 & 17\\\\\\\\ 11 & 13 & 17 & 13\\\\\\\\ 13 & 17 & 13 & 11\\\\\\\\ 17 & 13 & 11 & 7\\\\end{vmatrix}=72^2$\",\n \"26\\\\\\\\ -3 & 5 & -16\\\\\\\\ \\\\end{vmatrix} = \\\\begin {vmatrix} 2 & 4 & 13 – 5 \\\\times 2\\\\\\\\ 5 & 1 & 26 – 5 \\\\times 5\\\\\\\\ -3 & 5 & -16 – 5 \\\\times (-3)\\\\\\\\ \\\\end{vmatrix} = \\\\begin {vmatrix} 2 & 4 & 3\\\\\\\\ 5 & 1 & 1\\\\\\\\ -3 & 5 & -1\\\\\\\\ \\\\end{vmatrix}$ Now expanding the determinant we get $\\\\begin {vmatrix} 2 & 4 & 3\\\\\\\\ 5 & 1 & 1\\\\\\\\ -3 & 5 & -1\\\\\\\\ \\\\end{vmatrix} = 2 \\\\times \\\\begin {vmatrix} 1 & 1\\\\\\\\ 5 & -1\\\\\\\\ \\\\end{vmatrix} – 4 \\\\times \\\\begin {vmatrix} 5 & 1\\\\\\\\ -3 & -1\\\\\\\\ \\\\end{vmatrix} + 3 \\\\times \\\\begin {vmatrix} 5 & 1\\\\\\\\ -3 & 5\\\\\\\\ \\\\end{vmatrix}$ $= 2 \\\\times (1 \\\\times (-1) – 1 \\\\times 5) – 4 \\\\times (5 \\\\times (-1) – 1 \\\\times (-3)) + 3 \\\\times (5 \\\\times 5 – 1 \\\\times (-3))$ $= 2 \\\\times (-1 – 5) – 4 \\\\times (-5 + 3) + 3 \\\\times (25 + 3)$ $= 2 \\\\times (-6) – 4 \\\\times (-2) + 3 \\\\times 28$ $= -12 + 8 + 84 = 80$ ### 7. Triangular Property of Determinant The triangular property of the determinant states that if the elements above or below the main diagonal are equal to zero, then the value of the determinant\",\n \"and 9x - 5y + 8 =0 We have $\\\\begin{vmatrix} 2 & -3 & 5\\\\\\\\ 3 & 4 & -7\\\\\\\\ 9 & -5 & 8\\\\end{vmatrix}$ = 2(32 - 35) - (-3)(24 + 63) + 5(-15 - 36) = 2(-3) + 3(87) + 5(-51) = - 6 + 261 -255 = 0 Therefore, the given three straight lines are concurrent. The Straight Line\",\n \"n & n+1 \\\\ 2 & 3 & … & n & n & n+1 \\\\ … & … & … & … & … & … \\\\ 2 & n & … & n-1 & n & n+1 \\\\ n & 3 & … & n-1 & n & n+1 end{vmatrix}$$ The closest one to mine from StackExchange was this one. But I didn’t manage to link these two determinants. No matter how I transform it, nothing worked for me. Any ideas?\",\n \"0 & 2 \\\\end{vmatrix} + 3 \\\\begin{vmatrix} x & 1 \\\\\\\\ 0 & 3 \\\\end{vmatrix} = 132$$ $$2(2 - 12) + 5(2x) + 3(3x) = 132$$ $$-20 + 10x + 9x = 132$$ $$19x = 152$$ $$x = 8$$\",\n \"\\\\\\\\ { – 7 } & { – 2 } \\\\end{vmatrix}$ $D_{ y } = \\\\begin{vmatrix} 2 & { 7 } \\\\\\\\ { 3 } & { – 7 } \\\\end{vmatrix}$ Recall the formula to evaluate a $2 \\\\times 2$ determinant: For a $2 \\\\timess 2$ matrix — $A = \\\\begin{bmatrix} a & b \\\\\\\\ c & d \\\\end{bmatrix}$ The determinant is calculated as — $det( A ) = | A | = \\\\begin{vmatrix} a & b \\\\\\\\ c & d \\\\end{vmatrix} = ad – bc$ Let’s calculate the determinants: $D = \\\\begin{vmatrix} 2 & 1 \\\\\\\\ { 3 } & { – 2 } \\\\end{vmatrix} = ( 2 )( – 2 ) – ( 1 )( 3 ) = – 4 – 3 = – 7$ $D_{ x } = \\\\begin{vmatrix} 7 & 1 \\\\\\\\ { – 7 } & { – 2 } \\\\end{vmatrix} = ( 7 )( – 2 ) – ( 1 )( – 7 ) = – 14 – ( – 7 ) = – 14 + 7 = – 7$ $D_{ y } = \\\\begin{vmatrix} 2 & { 7 } \\\\\\\\ { 3 } & { – 7 } \\\\end{vmatrix} = ( 2 )( – 7 ) – ( 7 )( 3 ) = -14 – 21 = – 35$ Now, we\",\n \"Note that $$\\\\det \\\\begin{vmatrix} 1 & 1 & 1 \\\\\\\\ 1 & -1 & 1 \\\\\\\\ -1 & 3 & -1 \\\\\\\\ \\\\end{vmatrix} =0$$ so the three $v_i$ are linearly dependent and therefor cannot span the original subspace\",\n \"\\\\times \\\\begin {vmatrix} 0 & 5\\\\\\\\ 0 & 0\\\\\\\\ \\\\end{vmatrix}$ $= 3 \\\\times (5 \\\\times 4 – 7 \\\\times 0) – 0 – 2 \\\\times (0 \\\\times 0 – 5 \\\\times 0)$ $= 3 \\\\times (20 – 0) – 0 – 2 \\\\times (0 – 0) = 3 \\\\times 20 – 0 – 2 \\\\times 0 = 3 \\\\times 20 – 0 – 0 = 60$. Example 2: Find the value of $\\\\begin {vmatrix} 1 & 0 & 0\\\\\\\\ 5 & 3 & 0\\\\\\\\ 6 & 7 & 4\\\\\\\\ \\\\end{vmatrix}$. $\\\\begin {vmatrix} 1 & 0 & 0\\\\\\\\ 5 & 3 & 0\\\\\\\\ 6 & 7 & 4\\\\\\\\ \\\\end{vmatrix} = 1 \\\\times 3 \\\\times 4 = 12$ Let’s check the value of the determinant by expansion. $\\\\begin {vmatrix} 1 & 0 & 0\\\\\\\\ 5 & 3 & 0\\\\\\\\ 6 & 7 & 4\\\\\\\\ \\\\end{vmatrix} = 1 \\\\times \\\\begin {vmatrix} 3 & 0\\\\\\\\ 7 & 4\\\\\\\\ \\\\end{vmatrix} – 0 \\\\times \\\\begin {vmatrix} 5 & 0\\\\\\\\ 6 & 4\\\\\\\\ \\\\end{vmatrix} + 0 \\\\times \\\\begin {vmatrix} 5 & 3\\\\\\\\ 6 & 7\\\\\\\\ \\\\end{vmatrix}$ $= 1 \\\\times (3 \\\\times 4 – 0 \\\\times 7) – 0 + 0 = 1 \\\\times (12 – 0) – 0 + 0 = 12$ ## Examples of Using Properties of Determinants Example 1: Find the value of\"\n]"},"ps_retrievals":{"kind":"list like","value":["2 & 5 \\end{vmatrix} \\begin{vmatrix} 11 \\\\ 15 \\end{vmatrix} = \\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix}$ and continue encryption as follows: $\\begin{vmatrix} 7 \\\\ 8 \\end{vmatrix}, \\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix} \\to \\begin{vmatrix} H \\\\ I \\end{vmatrix}, \\begin{vmatrix} A \\\\ T \\end{vmatrix}$ The matrix K is invertible, hence K − 1 exists such that KK − 1 = K − 1K = I2. To implement decrypting, we compute $K^{-1} = 9^{-1} \\begin{vmatrix} 5 & 23 \\\\ 24 & 3 \\end{vmatrix} = 3 \\begin{vmatrix} 5 & 23 \\\\ 24 & 3 \\end{vmatrix} = \\begin{vmatrix} 15 & 17 \\\\ 20 & 9 \\end{vmatrix}$ $HIAT \\to \\begin{vmatrix} H \\\\ I \\end{vmatrix}, \\begin{vmatrix} A \\\\ T \\end{vmatrix} \\to \\begin{vmatrix} 7 \\\\ 8 \\end{vmatrix}, \\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix}$ Then we compute $\\begin{vmatrix} 15 & 17 \\\\ 20 & 9 \\end{vmatrix}\\begin{vmatrix} 7 \\\\ 8 \\end{vmatrix} = \\begin{vmatrix} 7 \\\\ 4 \\end{vmatrix},$ $\\begin{vmatrix} 15 & 17 \\\\ 20 & 9 \\end{vmatrix}\\begin{vmatrix} 0 \\\\ 19 \\end{vmatrix} = \\begin{vmatrix} 11 \\\\ 15 \\end{vmatrix}$ Therefore $\\begin{vmatrix} 7 \\\\ 4 \\end{vmatrix}, \\begin{vmatrix} 11 \\\\ 15 \\end{vmatrix} \\to \\begin{vmatrix} H \\\\ E \\end{vmatrix}, \\begin{vmatrix} L \\\\ P \\end{vmatrix} \\to HELP$. ## Security Unfortunately, the basic Hill cipher is vulnerable to a known-plaintext attack because it is completely linear. An opponent who intercepts n2 plaintext/ciphertext character pairs can set up a","the questions are solved in a way that does not require a calculator. Calculate the determinants of these matrices. For the $3 * 3$ and $4 * 4$ matrices, use at least two methods as applicable. Show all work. (1.) $\\begin{vmatrix} -3 & 3 \\\\[3ex] -1 & -2 \\end{vmatrix}$ $\\begin{vmatrix} -3 & 3 \\\\[3ex] -1 & -2 \\end{vmatrix} \\\\[3ex]$ $= (-3)(-2) - (-1)(3) \\\\[3ex] = 6 - (-3) \\\\[3ex] = 6 + 3 \\\\[3ex] = 9$ (2.) $\\begin{vmatrix} 7 & \\sqrt{7} \\\\[3ex] \\sqrt{7} & 7 \\end{vmatrix}$ $\\begin{vmatrix} 7 & \\sqrt{7} \\\\[3ex] \\sqrt{7} & 7 \\end{vmatrix} \\\\[3ex]$ $= (7)(7) - (\\sqrt{7})(\\sqrt{7}) \\\\[3ex] = 49 - 7 \\\\[3ex] = 42$ (3.) $\\begin{vmatrix} 3x^7 & -12 \\\\[3ex] x^7 & x^5 \\end{vmatrix}$ $\\begin{vmatrix} 3x^7 & -12 \\\\[3ex] x^7 & x^5 \\end{vmatrix} \\\\[3ex]$ $= (3x^7)(x^5) - (x^7)(-12) \\\\[3ex] = 3x^{12} - (-12x^7) \\\\[3ex] = 3x^{12} + 12x^7$ (4.) $\\begin{vmatrix} p & c \\\\[3ex] -d & -e \\end{vmatrix}$ $\\begin{vmatrix} p & c \\\\[3ex] -d & -e \\end{vmatrix} \\\\[3ex]$ $= (p)(-e) - (-d)(c) \\\\[3ex] = -ep - (-cd) \\\\[3ex] = -ep + cd \\\\[3ex] = cd - ep$ (5.) $\\begin{vmatrix} 2 & -1 & 6 \\\\[3ex] 4 & -6 & 0 \\\\[3ex] 5 & -2 & 6 \\end{vmatrix}$ First Method: Formula To solve it faster, we shall use the second row because it contains a","$|A| =3\\begin{vmatrix} 1 & -2 \\\\[0.3em] 3 & 1\\end{vmatrix}$ $+4\\begin{vmatrix} 1 & -2 \\\\[0.3em] 2 & 1 \\end{vmatrix}$ $+5\\begin{vmatrix} 1 & 1 \\\\[0.3em] 2 & 3 \\end{vmatrix}$$\\\\$ $=3(1+6)+4(1+4)+5(3-2) \\\\ =3(7)+4(5)+5(1)\\\\ 21+20+5=46$","& 4 & 7\\end{vmatrix} = 2\\begin{vmatrix}5 & 2 & 3 \\\\ 1 & 0 & 0 \\\\ 3 & 4 & 7\\end{vmatrix}$$ Where we have first factored out a $2$ from row $2$ and then subtracted row $1$ from row $2$. Now we expand along row $2$ to get $$2\\begin{vmatrix}5 & 2 & 3 \\\\ 1 &0 & 0 \\\\ 3 & 4 & 7\\end{vmatrix} = 2(-1)^{2+1}\\begin{vmatrix} 2 & 3 \\\\ 4 & 7\\end{vmatrix} = -2\\begin{vmatrix} 2 & 3 \\\\ 0 & 1\\end{vmatrix}$$ where in the last step we subtract twice row $1$ from row $2$. Now we simply multiply the diagonal entries to get determinant equal to $-4$.","Q # Using the property of determinants and without expanding, prove that determinant 2 7 65 3 8 75 5 9 86 = 0 Ex 4.2 Q 3 Q : 3 Using the property of determinants and without expanding, prove that $\\dpi{100} \\begin{vmatrix}2 & 7 &65 \\\\3 &8 &75 \\\\5 &9 &86 \\end{vmatrix}=0$ Views Given determinant $\\dpi{100} \\begin{vmatrix}2 & 7 &65 \\\\3 &8 &75 \\\\5 &9 &86 \\end{vmatrix}$ So, we can split it in two addition determinants: $\\dpi{100} \\begin{vmatrix}2 & 7 &65 \\\\3 &8 &75 \\\\5 &9 &86 \\end{vmatrix} = \\begin{vmatrix} 2 &7 &63+2 \\\\ 3& 8 &72+3 \\\\ 5& 9 & 81+5 \\end{vmatrix}$ $\\dpi{100} \\begin{vmatrix} 2 &7 &63+2 \\\\ 3& 8 &72+3 \\\\ 5& 9 & 81+5 \\end{vmatrix} = \\begin{vmatrix} 2 & 7 &2 \\\\ 3& 8& 3\\\\ 5 & 9 & 5 \\end{vmatrix} + \\begin{vmatrix} 2 & 7 &63 \\\\ 3& 8 &72 \\\\ 5 & 9 & 81 \\end{vmatrix}$ $\\dpi{100} \\begin{vmatrix} 2 & 7 &2 \\\\ 3& 8& 3\\\\ 5 & 9 & 5 \\end{vmatrix} = 0$ [$\\dpi{100} \\because$ Here two columns are identical ] and $\\dpi{100} \\begin{vmatrix} 2 & 7 &63 \\\\ 3& 8 &72 \\\\ 5 & 9 & 81 \\end{vmatrix} = \\begin{vmatrix} 2 & 7 &9(7) \\\\ 3& 8 &9(8) \\\\ 5 &9 & 9(9) \\end{vmatrix} = 9 \\begin{vmatrix} 2 & 7","& -8 & 0 & 3 \\\\ 0 & 5 & 0 & -6 \\\\ \\end{vmatrix} = 2 \\cdot \\begin{vmatrix} 0 & 3 & -4 \\\\ -5 & -8 & 3 \\\\ 0 & 5 & -6 \\\\ \\end{vmatrix} = 2 \\cdot -(-5) \\cdot \\begin{vmatrix} 3 & -4 \\\\ 5 & -6 \\\\ \\end{vmatrix} = 2 \\cdot -(-5) \\cdot (-18 -(-20)) = 20.$$ I expanded along the third column in the first case and then along the first column, because these are the ones with the most zeros, so it saves a lot of computations. I am assuming that's what's in your book. If you want to expand along other columns, just keep track of the appropriate minors without making any computation mistakes (I guess that's the hard part ; the only trick is to make it slowly and be careful). So here we go : $$\\det A = \\begin{vmatrix} 5 & -7 & 2 & 2 \\\\ 0 & 3 & 0 & -4 \\\\ -5 & -8 & 0 & 3 \\\\ 0 & 5 & 0 & -6 \\\\ \\end{vmatrix} = 5 \\begin{vmatrix} 3 & 0 & -4 \\\\ -8 & 0 & 3 \\\\ 5 & 0 & -6 \\\\ \\end{vmatrix} -(-7) \\begin{vmatrix} 0 & 0 & -4 \\\\ -5 & 0 & 3","0\\end{vmatrix}} \\vec{j} + {\\begin{vmatrix}-8 & 0\\\\ 0 & -9\\end{vmatrix}} \\vec{k} \\right\\}$ $= <-3 , 0 , 7> \\cdot <0 , 0 , 72> = 0 + 0 + 7\\cdot72 = 504$ The volume is $V = |504| = 504$ unit3","& 3 & 8\\\\4 & 3 & 2\\end {vmatrix}$$ $$det⁡ A=1( 4\\begin{vmatrix} 3 & 2 \\\\9 & 6\\end{vmatrix} – 3\\begin{vmatrix}4 & 2 \\\\4 & 6\\end{vmatrix}+8\\begin{vmatrix}4 & 3 \\\\4 & 9\\end{vmatrix}) -2( 8\\begin{vmatrix} 3 & 2 \\\\9 & 6\\end{vmatrix} – 7\\begin{vmatrix}4 & 2 \\\\4 & 6\\end{vmatrix}+2\\begin{vmatrix}4 & 3 \\\\4 & 9\\end{vmatrix}) +1( 8\\begin{vmatrix}3 & 8 \\\\9 & 6\\end{vmatrix} – 7\\begin{vmatrix}4 & 8 \\\\4 & 6\\end{vmatrix}+2\\begin{vmatrix}4 & 3 \\\\4 & 9\\end{vmatrix}) -1( 8\\begin{vmatrix} 3 & 8 \\\\3 & 2\\end{vmatrix} – 7\\begin{vmatrix}4 & 8 \\\\4 & 6\\end{vmatrix}+2\\begin{vmatrix}4 & 3 \\\\4 & 3\\end{vmatrix})$$ $$det⁡ A = 1[4(18-18)-3(24-8)+ 8(36-12)]-2[ 8(18-18)-7(24-8)+ 2(36-12)]+ 1[ 8(18-72)-7(24-32)+ 2(36-12)] -1[8(6-24)-7(8-32)+ 2(12-12)]$$ $$det⁡ A = 1[4(0)-3(16)+ 8(24)]-2[ 8(0)-7(16)+ 2(24)]+ 1[ 8(-54)-7(-8)+ 2(24)]-1[8(-18)-7(-24)+ 2(0)]$$ $$det⁡ A = 1[0-48+192]-2[0-112+48]+ 1[ -432+56+48]-1[-144+168+0]$$ $$det⁡ A = 1[144]-2[-64]+ 1[-328]-1[24]$$ $$det⁡ A = 144+128-328- 24$$ $$det⁡ A = -80$$ #### Expand Along the Row: For the calculations of matrix A = (aij)4×4 from expansion of row is determined by the following formula: $$det A = \\begin{vmatrix} a & b & c & d\\\\e & f & g &h \\\\i & j & k & l \\\\ m & n & o & p \\end{vmatrix} \\\\$$ $$det⁡ A= a\\begin{vmatrix} f & g & h\\\\j & k & l\\\\n & o & p\\end{vmatrix} – b\\begin{vmatrix}e & g & h\\\\i & k & l\\\\ m & o &","0\\end{vmatrix}} \\vec{j} + {\\begin{vmatrix}-8 & 0\\\\ 0 & -9\\end{vmatrix}} \\vec{k} \\right\\}$ $= <-3 , 0 , 7> \\cdot <0 , 0 , 72> = 0 + 0 + 7\\cdot72 = 504$ The volume is $V = |504| = 504$ unit3 Updated: 20 April 2017 (A Dendane)","# Evaluate the determinants: $\\begin{vmatrix} 0&1&2 \\\\ -1&0&3 \\\\-2&3&0 \\end{vmatrix}$ $\\begin{array}{1 1} 12 \\\\ -12 \\\\ 0 \\\\ \\text{none of the above }\\end{array}$ ## 1 Answer Toolbox: • To evaluate a matrix of order $3\\times 3$ • $\\mid A\\mid=\\begin{vmatrix}a_{11} & a_{12} & a_{13}\\\\a_{21} & a_{22} & a_{23}\\\\a_{31} & a_{32} & a_{33}\\end{vmatrix}$ • Therefore $\\mid A\\mid=a_{11}(a_{22}\\times a_{33}-a_{23}\\times a_{32})-a_{12}(a_{21}\\times a_{32}-a_{23}\\times a_{31})+a_{13}(a_{21}\\times a_{32}-a_{22}\\times a_{31})$ Given:(iii) $Evaluate:\\begin{vmatrix}0 & 1 & 2\\\\-1& 0 &-3\\\\-2 & 3 & 0\\end{vmatrix}$ We know to evaluate the value of the determinant of order $3\\times 3$ Therefore $\\mid A\\mid=a_{11}(a_{22}\\times a_{33}-a_{23}\\times a_{32})-a_{12}(a_{21}\\times a_{32}-a_{23}\\times a_{31})+a_{13}(a_{21}\\times a_{32}- a_{22}\\times a_{31})$ $Hence \\mid A\\mid=0[(0\\times 0-(-3\\times 3)]-1[(-1\\times 0)-(-3\\times -2]+2[(-1\\times 3)-(0\\times -2)]$ $\\qquad=0-1(6)+2(-3)$ $\\qquad=-6-6$ $\\qquad=-12$ answered Mar 6, 2013 edited Apr 30, 2014 1 answer 1 answer 1 answer 1 answer 1 answer 1 answer 1 answer","26\\\\ -3 & 5 & -16\\\\ \\end{vmatrix} = \\begin {vmatrix} 2 & 4 & 13 – 5 \\times 2\\\\ 5 & 1 & 26 – 5 \\times 5\\\\ -3 & 5 & -16 – 5 \\times (-3)\\\\ \\end{vmatrix} = \\begin {vmatrix} 2 & 4 & 3\\\\ 5 & 1 & 1\\\\ -3 & 5 & -1\\\\ \\end{vmatrix}$ Now expanding the determinant we get $\\begin {vmatrix} 2 & 4 & 3\\\\ 5 & 1 & 1\\\\ -3 & 5 & -1\\\\ \\end{vmatrix} = 2 \\times \\begin {vmatrix} 1 & 1\\\\ 5 & -1\\\\ \\end{vmatrix} – 4 \\times \\begin {vmatrix} 5 & 1\\\\ -3 & -1\\\\ \\end{vmatrix} + 3 \\times \\begin {vmatrix} 5 & 1\\\\ -3 & 5\\\\ \\end{vmatrix}$ $= 2 \\times (1 \\times (-1) – 1 \\times 5) – 4 \\times (5 \\times (-1) – 1 \\times (-3)) + 3 \\times (5 \\times 5 – 1 \\times (-3))$ $= 2 \\times (-1 – 5) – 4 \\times (-5 + 3) + 3 \\times (25 + 3)$ $= 2 \\times (-6) – 4 \\times (-2) + 3 \\times 28$ $= -12 + 8 + 84 = 80$ ### 7. Triangular Property of Determinant The triangular property of the determinant states that if the elements above or below the main diagonal are equal to zero, then the value of the determinant","= $$\\begin{vmatrix} 2 & -1 \\\\ -4 & 3 \\end{vmatrix}$$ = 2 $$C_{12}$$ = $$(-1)^{1+2}$$$$M_{12}$$ = -$$M_{12}$$ = -$$\\begin{vmatrix} -3 & -1 \\\\ 2 & 3 \\end{vmatrix}$$ = 7 $$C_{13}$$ = $$(-1)^{1+3}$$$$M_{13}$$ = $$M_{13}$$ = $$\\begin{vmatrix} -3 & 2 \\\\ 2 & -4 \\end{vmatrix}$$ = 8 $$C_{23}$$ = $$(-1)^{2+3}$$$$M_{23}$$ = -$$M_{23}$$ = -$$\\begin{vmatrix} 1 & 2 \\\\ 2 & -4 \\end{vmatrix}$$ = 8 etc.","\\\\ { – 7 } & { – 2 } \\end{vmatrix}$ $D_{ y } = \\begin{vmatrix} 2 & { 7 } \\\\ { 3 } & { – 7 } \\end{vmatrix}$ Recall the formula to evaluate a $2 \\times 2$ determinant: For a $2 \\timess 2$ matrix — $A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$ The determinant is calculated as — $det( A ) = | A | = \\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} = ad – bc$ Let’s calculate the determinants: $D = \\begin{vmatrix} 2 & 1 \\\\ { 3 } & { – 2 } \\end{vmatrix} = ( 2 )( – 2 ) – ( 1 )( 3 ) = – 4 – 3 = – 7$ $D_{ x } = \\begin{vmatrix} 7 & 1 \\\\ { – 7 } & { – 2 } \\end{vmatrix} = ( 7 )( – 2 ) – ( 1 )( – 7 ) = – 14 – ( – 7 ) = – 14 + 7 = – 7$ $D_{ y } = \\begin{vmatrix} 2 & { 7 } \\\\ { 3 } & { – 7 } \\end{vmatrix} = ( 2 )( – 7 ) – ( 7 )( 3 ) = -14 – 21 = – 35$ Now, we","\\begin{vmatrix} -1 & -2 & 1\\\\ -3 & -2 & 1\\\\ 2& 2 & 1\\\\ \\end{vmatrix}$$ Δ = $$\\frac{1}{2}$$[-1(-2 - 2) + 2(-3 - 2) + 1(-6 + 4)] ⇒ Δ = 4 sq. unit","Then you get $\\begin{vmatrix}-6-x & 22 & 11 \\\\ -2 & -7-x & -2 \\\\ -1 & 26 & 10-x \\end{vmatrix} = \\begin{vmatrix}-6-x & 0 & 11 \\\\ -2 & -3-x & -2 \\\\ -1 & 6+2x & 10-x \\end{vmatrix}$. When you see that, you know that you have struck lucky, because there is a factor 3+x going down the middle column. Take this factor out, and you are left with $(3+x)\\begin{vmatrix}-6-x & 0 & 11 \\\\ -2 & -1 & -2 \\\\ -1 & 2 & 10-x \\end{vmatrix}$. Now add twice the middle row to the bottom row, and expand down the middle column, to get $(3+x)\\begin{vmatrix}-6-x & 11 \\\\ -5 & 6-x \\end{vmatrix} = -(3+x)(x^2 - 36 + 55) = -(3+x)(x^2 + 19)$. There's the eigenvalue equation, neatly factorised, without having to do any heavy calculations at all!","be used as a formula for finding the determinant of any square matrix of the order $3 \\times 3$. Now, let’s learn how to derive the determinant of a $3$ by $3$ matrix formula in mathematics. ### Example Find ${\\begin{vmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 9 \\\\ \\end{vmatrix}}$ Let’s use the above steps for calculating the determinant of any square matrix of the order $3 \\times 3$. $\\implies$ ${\\begin{vmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 9 \\\\ \\end{vmatrix}}$ $\\,=\\,$ $1 \\times \\begin{vmatrix} 5 & 6 \\\\ 8 & 9 \\\\ \\end{vmatrix}$ $-$ $2 \\times \\begin{vmatrix} 4 & 6 \\\\ 7 & 9 \\\\ \\end{vmatrix}$ $+$ $3 \\times \\begin{vmatrix} 4 & 5 \\\\ 7 & 8 \\\\ \\end{vmatrix}$ $\\implies$ ${\\begin{vmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 9 \\\\ \\end{vmatrix}}$ $\\,=\\,$ $1 \\times (5 \\times 9-6 \\times 8)$ $-$ $2 \\times (4 \\times 9-6 \\times 7)$ $+$ $3 \\times (4 \\times 8-5 \\times 7)$ $\\implies$ ${\\begin{vmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 9 \\\\ \\end{vmatrix}}$ $\\,=\\,$ $1 \\times (45-48)$ $-$ $2 \\times (36-42)$ $+$ $3 \\times (32-35)$ $\\implies$ ${\\begin{vmatrix}","\\begin{vmatrix} 2^4 & 2^1 & 2^2 & 2^3 \\\\ 3^4 & 3^1 & 3^2 & 3^3 \\\\ (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\\\ (-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\\\ \\end{vmatrix}\\cdot (-3)+\\begin{vmatrix}2^2 & 2^1 & 2^2 & 2^3 \\\\ 3^2 & 3^1 & 3^2 & 3^3 \\\\ (-1)^2 & (-1)^1 &(-1)^2 & (-1)^3 \\\\ (-6)^2 & (-6)^1 & (-6)^2 & (-6)^3 \\\\ \\end{vmatrix} +\\begin{vmatrix}1 & 2^1 & 2^2 & 2^3 \\\\ 1 & 3^1 & 3^2 & 3^3 \\\\ 1 & (-1)^1 &(-1)^2 & (-1)^3 \\\\ 1 & (-6)^1 & (-6)^2 & (-6)^3 \\\\ \\end{vmatrix}$$ The second determinant is zero, and simplify the first one. $$\\begin{vmatrix}5- 3 \\cdot 2^4 & 2^1 & 2^2 & 2^3 \\\\ 10 - 3 \\cdot 3^4 & 3^1 & 3^2 & 3^3 \\\\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\\\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\\\ \\end{vmatrix} = -3 \\cdot 2 \\cdot -1 \\cdot -6 \\begin{vmatrix} 2^3 & 1 & 2^1 & 2^2 \\\\ 3^3 & 1& 3^1 & 3^2 \\\\ (-1)^3 & 1 &(-1)^1 & (-1)^2 \\\\ (-6)^3 & 1 & (-6)^1 & (-6)^2 \\\\ \\end{vmatrix}+\\begin{vmatrix}1 & 2^1 & 2^2 & 2^3 \\\\ 1 & 3^1 & 3^2 & 3^3 \\\\ 1 & (-1)^1 &(-1)^2 & (-1)^3 \\\\ 1 & (-6)^1 & (-6)^2 & (-6)^3 \\\\","\\times \\begin {vmatrix} 0 & 5\\\\ 0 & 0\\\\ \\end{vmatrix}$ $= 3 \\times (5 \\times 4 – 7 \\times 0) – 0 – 2 \\times (0 \\times 0 – 5 \\times 0)$ $= 3 \\times (20 – 0) – 0 – 2 \\times (0 – 0) = 3 \\times 20 – 0 – 2 \\times 0 = 3 \\times 20 – 0 – 0 = 60$. Example 2: Find the value of $\\begin {vmatrix} 1 & 0 & 0\\\\ 5 & 3 & 0\\\\ 6 & 7 & 4\\\\ \\end{vmatrix}$. $\\begin {vmatrix} 1 & 0 & 0\\\\ 5 & 3 & 0\\\\ 6 & 7 & 4\\\\ \\end{vmatrix} = 1 \\times 3 \\times 4 = 12$ Let’s check the value of the determinant by expansion. $\\begin {vmatrix} 1 & 0 & 0\\\\ 5 & 3 & 0\\\\ 6 & 7 & 4\\\\ \\end{vmatrix} = 1 \\times \\begin {vmatrix} 3 & 0\\\\ 7 & 4\\\\ \\end{vmatrix} – 0 \\times \\begin {vmatrix} 5 & 0\\\\ 6 & 4\\\\ \\end{vmatrix} + 0 \\times \\begin {vmatrix} 5 & 3\\\\ 6 & 7\\\\ \\end{vmatrix}$ $= 1 \\times (3 \\times 4 – 0 \\times 7) – 0 + 0 = 1 \\times (12 – 0) – 0 + 0 = 12$ ## Examples of Using Properties of Determinants Example 1: Find the value of","and 9x - 5y + 8 =0 We have $\\begin{vmatrix} 2 & -3 & 5\\\\ 3 & 4 & -7\\\\ 9 & -5 & 8\\end{vmatrix}$ = 2(32 - 35) - (-3)(24 + 63) + 5(-15 - 36) = 2(-3) + 3(87) + 5(-51) = - 6 + 261 -255 = 0 Therefore, the given three straight lines are concurrent. The Straight Line","could someone help me out please $\\begin{vmatrix}7 & 11 & 13 & 17\\\\ 11 & 13 & 17 & 13\\\\ 13 & 17 & 13 & 11\\\\ 17 & 13 & 11 & 7\\end{vmatrix}=72^2$"],"string":"[\n \"2 & 5 \\\\end{vmatrix} \\\\begin{vmatrix} 11 \\\\\\\\ 15 \\\\end{vmatrix} = \\\\begin{vmatrix} 0 \\\\\\\\ 19 \\\\end{vmatrix}$ and continue encryption as follows: $\\\\begin{vmatrix} 7 \\\\\\\\ 8 \\\\end{vmatrix}, \\\\begin{vmatrix} 0 \\\\\\\\ 19 \\\\end{vmatrix} \\\\to \\\\begin{vmatrix} H \\\\\\\\ I \\\\end{vmatrix}, \\\\begin{vmatrix} A \\\\\\\\ T \\\\end{vmatrix}$ The matrix K is invertible, hence K − 1 exists such that KK − 1 = K − 1K = I2. To implement decrypting, we compute $K^{-1} = 9^{-1} \\\\begin{vmatrix} 5 & 23 \\\\\\\\ 24 & 3 \\\\end{vmatrix} = 3 \\\\begin{vmatrix} 5 & 23 \\\\\\\\ 24 & 3 \\\\end{vmatrix} = \\\\begin{vmatrix} 15 & 17 \\\\\\\\ 20 & 9 \\\\end{vmatrix}$ $HIAT \\\\to \\\\begin{vmatrix} H \\\\\\\\ I \\\\end{vmatrix}, \\\\begin{vmatrix} A \\\\\\\\ T \\\\end{vmatrix} \\\\to \\\\begin{vmatrix} 7 \\\\\\\\ 8 \\\\end{vmatrix}, \\\\begin{vmatrix} 0 \\\\\\\\ 19 \\\\end{vmatrix}$ Then we compute $\\\\begin{vmatrix} 15 & 17 \\\\\\\\ 20 & 9 \\\\end{vmatrix}\\\\begin{vmatrix} 7 \\\\\\\\ 8 \\\\end{vmatrix} = \\\\begin{vmatrix} 7 \\\\\\\\ 4 \\\\end{vmatrix},$ $\\\\begin{vmatrix} 15 & 17 \\\\\\\\ 20 & 9 \\\\end{vmatrix}\\\\begin{vmatrix} 0 \\\\\\\\ 19 \\\\end{vmatrix} = \\\\begin{vmatrix} 11 \\\\\\\\ 15 \\\\end{vmatrix}$ Therefore $\\\\begin{vmatrix} 7 \\\\\\\\ 4 \\\\end{vmatrix}, \\\\begin{vmatrix} 11 \\\\\\\\ 15 \\\\end{vmatrix} \\\\to \\\\begin{vmatrix} H \\\\\\\\ E \\\\end{vmatrix}, \\\\begin{vmatrix} L \\\\\\\\ P \\\\end{vmatrix} \\\\to HELP$. ## Security Unfortunately, the basic Hill cipher is vulnerable to a known-plaintext attack because it is completely linear. An opponent who intercepts n2 plaintext/ciphertext character pairs can set up a\",\n \"the questions are solved in a way that does not require a calculator. Calculate the determinants of these matrices. For the $3 * 3$ and $4 * 4$ matrices, use at least two methods as applicable. Show all work. (1.) $\\\\begin{vmatrix} -3 & 3 \\\\\\\\[3ex] -1 & -2 \\\\end{vmatrix}$ $\\\\begin{vmatrix} -3 & 3 \\\\\\\\[3ex] -1 & -2 \\\\end{vmatrix} \\\\\\\\[3ex]$ $= (-3)(-2) - (-1)(3) \\\\\\\\[3ex] = 6 - (-3) \\\\\\\\[3ex] = 6 + 3 \\\\\\\\[3ex] = 9$ (2.) $\\\\begin{vmatrix} 7 & \\\\sqrt{7} \\\\\\\\[3ex] \\\\sqrt{7} & 7 \\\\end{vmatrix}$ $\\\\begin{vmatrix} 7 & \\\\sqrt{7} \\\\\\\\[3ex] \\\\sqrt{7} & 7 \\\\end{vmatrix} \\\\\\\\[3ex]$ $= (7)(7) - (\\\\sqrt{7})(\\\\sqrt{7}) \\\\\\\\[3ex] = 49 - 7 \\\\\\\\[3ex] = 42$ (3.) $\\\\begin{vmatrix} 3x^7 & -12 \\\\\\\\[3ex] x^7 & x^5 \\\\end{vmatrix}$ $\\\\begin{vmatrix} 3x^7 & -12 \\\\\\\\[3ex] x^7 & x^5 \\\\end{vmatrix} \\\\\\\\[3ex]$ $= (3x^7)(x^5) - (x^7)(-12) \\\\\\\\[3ex] = 3x^{12} - (-12x^7) \\\\\\\\[3ex] = 3x^{12} + 12x^7$ (4.) $\\\\begin{vmatrix} p & c \\\\\\\\[3ex] -d & -e \\\\end{vmatrix}$ $\\\\begin{vmatrix} p & c \\\\\\\\[3ex] -d & -e \\\\end{vmatrix} \\\\\\\\[3ex]$ $= (p)(-e) - (-d)(c) \\\\\\\\[3ex] = -ep - (-cd) \\\\\\\\[3ex] = -ep + cd \\\\\\\\[3ex] = cd - ep$ (5.) $\\\\begin{vmatrix} 2 & -1 & 6 \\\\\\\\[3ex] 4 & -6 & 0 \\\\\\\\[3ex] 5 & -2 & 6 \\\\end{vmatrix}$ First Method: Formula To solve it faster, we shall use the second row because it contains a\",\n \"$|A| =3\\\\begin{vmatrix} 1 & -2 \\\\\\\\[0.3em] 3 & 1\\\\end{vmatrix}$ $+4\\\\begin{vmatrix} 1 & -2 \\\\\\\\[0.3em] 2 & 1 \\\\end{vmatrix}$ $+5\\\\begin{vmatrix} 1 & 1 \\\\\\\\[0.3em] 2 & 3 \\\\end{vmatrix}$$\\\\\\\\$ $=3(1+6)+4(1+4)+5(3-2) \\\\\\\\ =3(7)+4(5)+5(1)\\\\\\\\ 21+20+5=46$\",\n \"& 4 & 7\\\\end{vmatrix} = 2\\\\begin{vmatrix}5 & 2 & 3 \\\\\\\\ 1 & 0 & 0 \\\\\\\\ 3 & 4 & 7\\\\end{vmatrix}$$ Where we have first factored out a $2$ from row $2$ and then subtracted row $1$ from row $2$. Now we expand along row $2$ to get $$2\\\\begin{vmatrix}5 & 2 & 3 \\\\\\\\ 1 &0 & 0 \\\\\\\\ 3 & 4 & 7\\\\end{vmatrix} = 2(-1)^{2+1}\\\\begin{vmatrix} 2 & 3 \\\\\\\\ 4 & 7\\\\end{vmatrix} = -2\\\\begin{vmatrix} 2 & 3 \\\\\\\\ 0 & 1\\\\end{vmatrix}$$ where in the last step we subtract twice row $1$ from row $2$. Now we simply multiply the diagonal entries to get determinant equal to $-4$.\",\n \"Q # Using the property of determinants and without expanding, prove that determinant 2 7 65 3 8 75 5 9 86 = 0 Ex 4.2 Q 3 Q : 3 Using the property of determinants and without expanding, prove that $\\\\dpi{100} \\\\begin{vmatrix}2 & 7 &65 \\\\\\\\3 &8 &75 \\\\\\\\5 &9 &86 \\\\end{vmatrix}=0$ Views Given determinant $\\\\dpi{100} \\\\begin{vmatrix}2 & 7 &65 \\\\\\\\3 &8 &75 \\\\\\\\5 &9 &86 \\\\end{vmatrix}$ So, we can split it in two addition determinants: $\\\\dpi{100} \\\\begin{vmatrix}2 & 7 &65 \\\\\\\\3 &8 &75 \\\\\\\\5 &9 &86 \\\\end{vmatrix} = \\\\begin{vmatrix} 2 &7 &63+2 \\\\\\\\ 3& 8 &72+3 \\\\\\\\ 5& 9 & 81+5 \\\\end{vmatrix}$ $\\\\dpi{100} \\\\begin{vmatrix} 2 &7 &63+2 \\\\\\\\ 3& 8 &72+3 \\\\\\\\ 5& 9 & 81+5 \\\\end{vmatrix} = \\\\begin{vmatrix} 2 & 7 &2 \\\\\\\\ 3& 8& 3\\\\\\\\ 5 & 9 & 5 \\\\end{vmatrix} + \\\\begin{vmatrix} 2 & 7 &63 \\\\\\\\ 3& 8 &72 \\\\\\\\ 5 & 9 & 81 \\\\end{vmatrix}$ $\\\\dpi{100} \\\\begin{vmatrix} 2 & 7 &2 \\\\\\\\ 3& 8& 3\\\\\\\\ 5 & 9 & 5 \\\\end{vmatrix} = 0$ [$\\\\dpi{100} \\\\because$ Here two columns are identical ] and $\\\\dpi{100} \\\\begin{vmatrix} 2 & 7 &63 \\\\\\\\ 3& 8 &72 \\\\\\\\ 5 & 9 & 81 \\\\end{vmatrix} = \\\\begin{vmatrix} 2 & 7 &9(7) \\\\\\\\ 3& 8 &9(8) \\\\\\\\ 5 &9 & 9(9) \\\\end{vmatrix} = 9 \\\\begin{vmatrix} 2 & 7\",\n \"& -8 & 0 & 3 \\\\\\\\ 0 & 5 & 0 & -6 \\\\\\\\ \\\\end{vmatrix} = 2 \\\\cdot \\\\begin{vmatrix} 0 & 3 & -4 \\\\\\\\ -5 & -8 & 3 \\\\\\\\ 0 & 5 & -6 \\\\\\\\ \\\\end{vmatrix} = 2 \\\\cdot -(-5) \\\\cdot \\\\begin{vmatrix} 3 & -4 \\\\\\\\ 5 & -6 \\\\\\\\ \\\\end{vmatrix} = 2 \\\\cdot -(-5) \\\\cdot (-18 -(-20)) = 20.$$ I expanded along the third column in the first case and then along the first column, because these are the ones with the most zeros, so it saves a lot of computations. I am assuming that's what's in your book. If you want to expand along other columns, just keep track of the appropriate minors without making any computation mistakes (I guess that's the hard part ; the only trick is to make it slowly and be careful). So here we go : $$\\\\det A = \\\\begin{vmatrix} 5 & -7 & 2 & 2 \\\\\\\\ 0 & 3 & 0 & -4 \\\\\\\\ -5 & -8 & 0 & 3 \\\\\\\\ 0 & 5 & 0 & -6 \\\\\\\\ \\\\end{vmatrix} = 5 \\\\begin{vmatrix} 3 & 0 & -4 \\\\\\\\ -8 & 0 & 3 \\\\\\\\ 5 & 0 & -6 \\\\\\\\ \\\\end{vmatrix} -(-7) \\\\begin{vmatrix} 0 & 0 & -4 \\\\\\\\ -5 & 0 & 3\",\n \"0\\\\end{vmatrix}} \\\\vec{j} + {\\\\begin{vmatrix}-8 & 0\\\\\\\\ 0 & -9\\\\end{vmatrix}} \\\\vec{k} \\\\right\\\\}$ $= <-3 , 0 , 7> \\\\cdot <0 , 0 , 72> = 0 + 0 + 7\\\\cdot72 = 504$ The volume is $V = |504| = 504$ unit3\",\n \"& 3 & 8\\\\\\\\4 & 3 & 2\\\\end {vmatrix}$$ $$det⁡ A=1( 4\\\\begin{vmatrix} 3 & 2 \\\\\\\\9 & 6\\\\end{vmatrix} – 3\\\\begin{vmatrix}4 & 2 \\\\\\\\4 & 6\\\\end{vmatrix}+8\\\\begin{vmatrix}4 & 3 \\\\\\\\4 & 9\\\\end{vmatrix}) -2( 8\\\\begin{vmatrix} 3 & 2 \\\\\\\\9 & 6\\\\end{vmatrix} – 7\\\\begin{vmatrix}4 & 2 \\\\\\\\4 & 6\\\\end{vmatrix}+2\\\\begin{vmatrix}4 & 3 \\\\\\\\4 & 9\\\\end{vmatrix}) +1( 8\\\\begin{vmatrix}3 & 8 \\\\\\\\9 & 6\\\\end{vmatrix} – 7\\\\begin{vmatrix}4 & 8 \\\\\\\\4 & 6\\\\end{vmatrix}+2\\\\begin{vmatrix}4 & 3 \\\\\\\\4 & 9\\\\end{vmatrix}) -1( 8\\\\begin{vmatrix} 3 & 8 \\\\\\\\3 & 2\\\\end{vmatrix} – 7\\\\begin{vmatrix}4 & 8 \\\\\\\\4 & 6\\\\end{vmatrix}+2\\\\begin{vmatrix}4 & 3 \\\\\\\\4 & 3\\\\end{vmatrix})$$ $$det⁡ A = 1[4(18-18)-3(24-8)+ 8(36-12)]-2[ 8(18-18)-7(24-8)+ 2(36-12)]+ 1[ 8(18-72)-7(24-32)+ 2(36-12)] -1[8(6-24)-7(8-32)+ 2(12-12)]$$ $$det⁡ A = 1[4(0)-3(16)+ 8(24)]-2[ 8(0)-7(16)+ 2(24)]+ 1[ 8(-54)-7(-8)+ 2(24)]-1[8(-18)-7(-24)+ 2(0)]$$ $$det⁡ A = 1[0-48+192]-2[0-112+48]+ 1[ -432+56+48]-1[-144+168+0]$$ $$det⁡ A = 1[144]-2[-64]+ 1[-328]-1[24]$$ $$det⁡ A = 144+128-328- 24$$ $$det⁡ A = -80$$ #### Expand Along the Row: For the calculations of matrix A = (aij)4×4 from expansion of row is determined by the following formula: $$det A = \\\\begin{vmatrix} a & b & c & d\\\\\\\\e & f & g &h \\\\\\\\i & j & k & l \\\\\\\\ m & n & o & p \\\\end{vmatrix} \\\\\\\\$$ $$det⁡ A= a\\\\begin{vmatrix} f & g & h\\\\\\\\j & k & l\\\\\\\\n & o & p\\\\end{vmatrix} – b\\\\begin{vmatrix}e & g & h\\\\\\\\i & k & l\\\\\\\\ m & o &\",\n \"0\\\\end{vmatrix}} \\\\vec{j} + {\\\\begin{vmatrix}-8 & 0\\\\\\\\ 0 & -9\\\\end{vmatrix}} \\\\vec{k} \\\\right\\\\}$ $= <-3 , 0 , 7> \\\\cdot <0 , 0 , 72> = 0 + 0 + 7\\\\cdot72 = 504$ The volume is $V = |504| = 504$ unit3 Updated: 20 April 2017 (A Dendane)\",\n \"# Evaluate the determinants: $\\\\begin{vmatrix} 0&1&2 \\\\\\\\ -1&0&3 \\\\\\\\-2&3&0 \\\\end{vmatrix}$ $\\\\begin{array}{1 1} 12 \\\\\\\\ -12 \\\\\\\\ 0 \\\\\\\\ \\\\text{none of the above }\\\\end{array}$ ## 1 Answer Toolbox: • To evaluate a matrix of order $3\\\\times 3$ • $\\\\mid A\\\\mid=\\\\begin{vmatrix}a_{11} & a_{12} & a_{13}\\\\\\\\a_{21} & a_{22} & a_{23}\\\\\\\\a_{31} & a_{32} & a_{33}\\\\end{vmatrix}$ • Therefore $\\\\mid A\\\\mid=a_{11}(a_{22}\\\\times a_{33}-a_{23}\\\\times a_{32})-a_{12}(a_{21}\\\\times a_{32}-a_{23}\\\\times a_{31})+a_{13}(a_{21}\\\\times a_{32}-a_{22}\\\\times a_{31})$ Given:(iii) $Evaluate:\\\\begin{vmatrix}0 & 1 & 2\\\\\\\\-1& 0 &-3\\\\\\\\-2 & 3 & 0\\\\end{vmatrix}$ We know to evaluate the value of the determinant of order $3\\\\times 3$ Therefore $\\\\mid A\\\\mid=a_{11}(a_{22}\\\\times a_{33}-a_{23}\\\\times a_{32})-a_{12}(a_{21}\\\\times a_{32}-a_{23}\\\\times a_{31})+a_{13}(a_{21}\\\\times a_{32}- a_{22}\\\\times a_{31})$ $Hence \\\\mid A\\\\mid=0[(0\\\\times 0-(-3\\\\times 3)]-1[(-1\\\\times 0)-(-3\\\\times -2]+2[(-1\\\\times 3)-(0\\\\times -2)]$ $\\\\qquad=0-1(6)+2(-3)$ $\\\\qquad=-6-6$ $\\\\qquad=-12$ answered Mar 6, 2013 edited Apr 30, 2014 1 answer 1 answer 1 answer 1 answer 1 answer 1 answer 1 answer\",\n \"26\\\\\\\\ -3 & 5 & -16\\\\\\\\ \\\\end{vmatrix} = \\\\begin {vmatrix} 2 & 4 & 13 – 5 \\\\times 2\\\\\\\\ 5 & 1 & 26 – 5 \\\\times 5\\\\\\\\ -3 & 5 & -16 – 5 \\\\times (-3)\\\\\\\\ \\\\end{vmatrix} = \\\\begin {vmatrix} 2 & 4 & 3\\\\\\\\ 5 & 1 & 1\\\\\\\\ -3 & 5 & -1\\\\\\\\ \\\\end{vmatrix}$ Now expanding the determinant we get $\\\\begin {vmatrix} 2 & 4 & 3\\\\\\\\ 5 & 1 & 1\\\\\\\\ -3 & 5 & -1\\\\\\\\ \\\\end{vmatrix} = 2 \\\\times \\\\begin {vmatrix} 1 & 1\\\\\\\\ 5 & -1\\\\\\\\ \\\\end{vmatrix} – 4 \\\\times \\\\begin {vmatrix} 5 & 1\\\\\\\\ -3 & -1\\\\\\\\ \\\\end{vmatrix} + 3 \\\\times \\\\begin {vmatrix} 5 & 1\\\\\\\\ -3 & 5\\\\\\\\ \\\\end{vmatrix}$ $= 2 \\\\times (1 \\\\times (-1) – 1 \\\\times 5) – 4 \\\\times (5 \\\\times (-1) – 1 \\\\times (-3)) + 3 \\\\times (5 \\\\times 5 – 1 \\\\times (-3))$ $= 2 \\\\times (-1 – 5) – 4 \\\\times (-5 + 3) + 3 \\\\times (25 + 3)$ $= 2 \\\\times (-6) – 4 \\\\times (-2) + 3 \\\\times 28$ $= -12 + 8 + 84 = 80$ ### 7. Triangular Property of Determinant The triangular property of the determinant states that if the elements above or below the main diagonal are equal to zero, then the value of the determinant\",\n \"= $$\\\\begin{vmatrix} 2 & -1 \\\\\\\\ -4 & 3 \\\\end{vmatrix}$$ = 2 $$C_{12}$$ = $$(-1)^{1+2}$$$$M_{12}$$ = -$$M_{12}$$ = -$$\\\\begin{vmatrix} -3 & -1 \\\\\\\\ 2 & 3 \\\\end{vmatrix}$$ = 7 $$C_{13}$$ = $$(-1)^{1+3}$$$$M_{13}$$ = $$M_{13}$$ = $$\\\\begin{vmatrix} -3 & 2 \\\\\\\\ 2 & -4 \\\\end{vmatrix}$$ = 8 $$C_{23}$$ = $$(-1)^{2+3}$$$$M_{23}$$ = -$$M_{23}$$ = -$$\\\\begin{vmatrix} 1 & 2 \\\\\\\\ 2 & -4 \\\\end{vmatrix}$$ = 8 etc.\",\n \"\\\\\\\\ { – 7 } & { – 2 } \\\\end{vmatrix}$ $D_{ y } = \\\\begin{vmatrix} 2 & { 7 } \\\\\\\\ { 3 } & { – 7 } \\\\end{vmatrix}$ Recall the formula to evaluate a $2 \\\\times 2$ determinant: For a $2 \\\\timess 2$ matrix — $A = \\\\begin{bmatrix} a & b \\\\\\\\ c & d \\\\end{bmatrix}$ The determinant is calculated as — $det( A ) = | A | = \\\\begin{vmatrix} a & b \\\\\\\\ c & d \\\\end{vmatrix} = ad – bc$ Let’s calculate the determinants: $D = \\\\begin{vmatrix} 2 & 1 \\\\\\\\ { 3 } & { – 2 } \\\\end{vmatrix} = ( 2 )( – 2 ) – ( 1 )( 3 ) = – 4 – 3 = – 7$ $D_{ x } = \\\\begin{vmatrix} 7 & 1 \\\\\\\\ { – 7 } & { – 2 } \\\\end{vmatrix} = ( 7 )( – 2 ) – ( 1 )( – 7 ) = – 14 – ( – 7 ) = – 14 + 7 = – 7$ $D_{ y } = \\\\begin{vmatrix} 2 & { 7 } \\\\\\\\ { 3 } & { – 7 } \\\\end{vmatrix} = ( 2 )( – 7 ) – ( 7 )( 3 ) = -14 – 21 = – 35$ Now, we\",\n \"\\\\begin{vmatrix} -1 & -2 & 1\\\\\\\\ -3 & -2 & 1\\\\\\\\ 2& 2 & 1\\\\\\\\ \\\\end{vmatrix}$$ Δ = $$\\\\frac{1}{2}$$[-1(-2 - 2) + 2(-3 - 2) + 1(-6 + 4)] ⇒ Δ = 4 sq. unit\",\n \"Then you get $\\\\begin{vmatrix}-6-x & 22 & 11 \\\\\\\\ -2 & -7-x & -2 \\\\\\\\ -1 & 26 & 10-x \\\\end{vmatrix} = \\\\begin{vmatrix}-6-x & 0 & 11 \\\\\\\\ -2 & -3-x & -2 \\\\\\\\ -1 & 6+2x & 10-x \\\\end{vmatrix}$. When you see that, you know that you have struck lucky, because there is a factor 3+x going down the middle column. Take this factor out, and you are left with $(3+x)\\\\begin{vmatrix}-6-x & 0 & 11 \\\\\\\\ -2 & -1 & -2 \\\\\\\\ -1 & 2 & 10-x \\\\end{vmatrix}$. Now add twice the middle row to the bottom row, and expand down the middle column, to get $(3+x)\\\\begin{vmatrix}-6-x & 11 \\\\\\\\ -5 & 6-x \\\\end{vmatrix} = -(3+x)(x^2 - 36 + 55) = -(3+x)(x^2 + 19)$. There's the eigenvalue equation, neatly factorised, without having to do any heavy calculations at all!\",\n \"be used as a formula for finding the determinant of any square matrix of the order $3 \\\\times 3$. Now, let’s learn how to derive the determinant of a $3$ by $3$ matrix formula in mathematics. ### Example Find ${\\\\begin{vmatrix} 1 & 2 & 3 \\\\\\\\ 4 & 5 & 6 \\\\\\\\ 7 & 8 & 9 \\\\\\\\ \\\\end{vmatrix}}$ Let’s use the above steps for calculating the determinant of any square matrix of the order $3 \\\\times 3$. $\\\\implies$ ${\\\\begin{vmatrix} 1 & 2 & 3 \\\\\\\\ 4 & 5 & 6 \\\\\\\\ 7 & 8 & 9 \\\\\\\\ \\\\end{vmatrix}}$ $\\\\,=\\\\,$ $1 \\\\times \\\\begin{vmatrix} 5 & 6 \\\\\\\\ 8 & 9 \\\\\\\\ \\\\end{vmatrix}$ $-$ $2 \\\\times \\\\begin{vmatrix} 4 & 6 \\\\\\\\ 7 & 9 \\\\\\\\ \\\\end{vmatrix}$ $+$ $3 \\\\times \\\\begin{vmatrix} 4 & 5 \\\\\\\\ 7 & 8 \\\\\\\\ \\\\end{vmatrix}$ $\\\\implies$ ${\\\\begin{vmatrix} 1 & 2 & 3 \\\\\\\\ 4 & 5 & 6 \\\\\\\\ 7 & 8 & 9 \\\\\\\\ \\\\end{vmatrix}}$ $\\\\,=\\\\,$ $1 \\\\times (5 \\\\times 9-6 \\\\times 8)$ $-$ $2 \\\\times (4 \\\\times 9-6 \\\\times 7)$ $+$ $3 \\\\times (4 \\\\times 8-5 \\\\times 7)$ $\\\\implies$ ${\\\\begin{vmatrix} 1 & 2 & 3 \\\\\\\\ 4 & 5 & 6 \\\\\\\\ 7 & 8 & 9 \\\\\\\\ \\\\end{vmatrix}}$ $\\\\,=\\\\,$ $1 \\\\times (45-48)$ $-$ $2 \\\\times (36-42)$ $+$ $3 \\\\times (32-35)$ $\\\\implies$ ${\\\\begin{vmatrix}\",\n \"\\\\begin{vmatrix} 2^4 & 2^1 & 2^2 & 2^3 \\\\\\\\ 3^4 & 3^1 & 3^2 & 3^3 \\\\\\\\ (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\\\\\\\ (-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\\\\\\\ \\\\end{vmatrix}\\\\cdot (-3)+\\\\begin{vmatrix}2^2 & 2^1 & 2^2 & 2^3 \\\\\\\\ 3^2 & 3^1 & 3^2 & 3^3 \\\\\\\\ (-1)^2 & (-1)^1 &(-1)^2 & (-1)^3 \\\\\\\\ (-6)^2 & (-6)^1 & (-6)^2 & (-6)^3 \\\\\\\\ \\\\end{vmatrix} +\\\\begin{vmatrix}1 & 2^1 & 2^2 & 2^3 \\\\\\\\ 1 & 3^1 & 3^2 & 3^3 \\\\\\\\ 1 & (-1)^1 &(-1)^2 & (-1)^3 \\\\\\\\ 1 & (-6)^1 & (-6)^2 & (-6)^3 \\\\\\\\ \\\\end{vmatrix}$$ The second determinant is zero, and simplify the first one. $$\\\\begin{vmatrix}5- 3 \\\\cdot 2^4 & 2^1 & 2^2 & 2^3 \\\\\\\\ 10 - 3 \\\\cdot 3^4 & 3^1 & 3^2 & 3^3 \\\\\\\\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\\\\\\\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\\\\\\\ \\\\end{vmatrix} = -3 \\\\cdot 2 \\\\cdot -1 \\\\cdot -6 \\\\begin{vmatrix} 2^3 & 1 & 2^1 & 2^2 \\\\\\\\ 3^3 & 1& 3^1 & 3^2 \\\\\\\\ (-1)^3 & 1 &(-1)^1 & (-1)^2 \\\\\\\\ (-6)^3 & 1 & (-6)^1 & (-6)^2 \\\\\\\\ \\\\end{vmatrix}+\\\\begin{vmatrix}1 & 2^1 & 2^2 & 2^3 \\\\\\\\ 1 & 3^1 & 3^2 & 3^3 \\\\\\\\ 1 & (-1)^1 &(-1)^2 & (-1)^3 \\\\\\\\ 1 & (-6)^1 & (-6)^2 & (-6)^3 \\\\\\\\\",\n \"\\\\times \\\\begin {vmatrix} 0 & 5\\\\\\\\ 0 & 0\\\\\\\\ \\\\end{vmatrix}$ $= 3 \\\\times (5 \\\\times 4 – 7 \\\\times 0) – 0 – 2 \\\\times (0 \\\\times 0 – 5 \\\\times 0)$ $= 3 \\\\times (20 – 0) – 0 – 2 \\\\times (0 – 0) = 3 \\\\times 20 – 0 – 2 \\\\times 0 = 3 \\\\times 20 – 0 – 0 = 60$. Example 2: Find the value of $\\\\begin {vmatrix} 1 & 0 & 0\\\\\\\\ 5 & 3 & 0\\\\\\\\ 6 & 7 & 4\\\\\\\\ \\\\end{vmatrix}$. $\\\\begin {vmatrix} 1 & 0 & 0\\\\\\\\ 5 & 3 & 0\\\\\\\\ 6 & 7 & 4\\\\\\\\ \\\\end{vmatrix} = 1 \\\\times 3 \\\\times 4 = 12$ Let’s check the value of the determinant by expansion. $\\\\begin {vmatrix} 1 & 0 & 0\\\\\\\\ 5 & 3 & 0\\\\\\\\ 6 & 7 & 4\\\\\\\\ \\\\end{vmatrix} = 1 \\\\times \\\\begin {vmatrix} 3 & 0\\\\\\\\ 7 & 4\\\\\\\\ \\\\end{vmatrix} – 0 \\\\times \\\\begin {vmatrix} 5 & 0\\\\\\\\ 6 & 4\\\\\\\\ \\\\end{vmatrix} + 0 \\\\times \\\\begin {vmatrix} 5 & 3\\\\\\\\ 6 & 7\\\\\\\\ \\\\end{vmatrix}$ $= 1 \\\\times (3 \\\\times 4 – 0 \\\\times 7) – 0 + 0 = 1 \\\\times (12 – 0) – 0 + 0 = 12$ ## Examples of Using Properties of Determinants Example 1: Find the value of\",\n \"and 9x - 5y + 8 =0 We have $\\\\begin{vmatrix} 2 & -3 & 5\\\\\\\\ 3 & 4 & -7\\\\\\\\ 9 & -5 & 8\\\\end{vmatrix}$ = 2(32 - 35) - (-3)(24 + 63) + 5(-15 - 36) = 2(-3) + 3(87) + 5(-51) = - 6 + 261 -255 = 0 Therefore, the given three straight lines are concurrent. The Straight Line\",\n \"could someone help me out please $\\\\begin{vmatrix}7 & 11 & 13 & 17\\\\\\\\ 11 & 13 & 17 & 13\\\\\\\\ 13 & 17 & 13 & 11\\\\\\\\ 17 & 13 & 11 & 7\\\\end{vmatrix}=72^2$\"\n]"}}},{"rowIdx":6936,"cells":{"problem":{"kind":"string","value":"Find the range of the function\n\\[f(x) = \\frac{\\sin^3 x + 6 \\sin^2 x + \\sin x + 2 \\cos^2 x - 8}{\\sin x - 1},\\]as $x$ ranges over all real numbers such that $\\sin x \\neq 1.$ Enter your answer using interval notation."},"level":{"kind":"string","value":"Level 5"},"type":{"kind":"string","value":"Precalculus"},"solution":{"kind":"string","value":"Since $\\cos^2 x = 1 - \\sin^2 x,$ we can write\n\\begin{align*}\nf(x) &= \\frac{\\sin^3 x + 6 \\sin^2 x + \\sin x + 2(1 - \\sin^2 x) - 8}{\\sin x - 1} \\\\\n&= \\frac{\\sin^3 x + 4 \\sin^2 x + \\sin x - 6}{\\sin x - 1} \\\\\n&= \\frac{(\\sin x - 1)(\\sin x + 2)(\\sin x + 3)}{\\sin x - 1} \\\\\n&= (\\sin x + 2)(\\sin x + 3) \\\\\n&= \\sin^2 x + 5 \\sin x + 6.\n\\end{align*}Let $y = \\sin x.$ Then\n\\[\\sin^2 x + 5 \\sin x + 6 = y^2 + 5y + 6 = \\left( y + \\frac{5}{2} \\right)^2 - \\frac{1}{4}\\]Note that $y = \\sin x$ satisfies $-1 \\le y \\le 1,$ and $\\left( y + \\frac{5}{2} \\right)^2 - \\frac{1}{4}$ is increasing on this interval. Therefore,\n\\[2 \\le (\\sin x + 2)(\\sin x + 3) \\le 12.\\]However, in the original function $f(x),$ $\\sin x$ cannot take on the value of 1, so the range of $f(x)$ is $\\boxed{[2,12)}.$"},"answer":{"kind":"string","value":"[2,12)"},"p_retrievals":{"kind":"list like","value":["+0 help 0 67 2 For what real value of $$N$$ does the range of the function $$y = f(x) = \\frac{4x^2 + Nx + N}{x + 1},$$, where $$x$$ is real (and $$x \\neq -1$$) consist of all real numbers except for a single interval of the form $$-L < y < L$$? Apr 6, 2019 #2 +5172 +1 $$\\dfrac{df}{dx} = 0 \\Rightarrow x = -2,~0\\\\ \\text{we want } f(0) = -f(-2) = L\\\\ f(0) = N\\\\ f(-2) = N-16\\\\ N = 16-N\\\\ N=8$$ $$f(x) \\in \\mathbb{R} - (-8,8)$$ . Apr 6, 2019","# Recent questions tagged range If $f(x) = \\dfrac{\\sqrt{3} \\sin x}{2+\\cos x}$, then the range of $f(x)$ is the interval $[-1 , \\sqrt{3}{/2}]$ the interval $[-\\sqrt{3}{/2}, 1]$ the interval $[-1, 1]$ none of these If $f(x) = \\dfrac{\\sqrt{3}\\sin x}{2+\\cos x}$, then the range of $f(x)$ is the interval $[-1, \\sqrt{3}/2]$ the interval $[- \\sqrt{3}/2, 1]$ the interval $[-1, 1]$ none of the above","+0 # Help pls -1 124 1 +82 Find the range of the function $f(x) = \\frac{x^2 + 14x + 9}{x^2 + 2x + 3},$as $x$ varies over all real numbers. Jul 3, 2020","Browse Questions # What is the range of the real function, $f(x) = \\large\\frac{1}{x^2-1}$ ? Can you answer this question? R − (−1, 0] hence (B) is the correct answer. answered Jun 28, 2014","+0 # Help pls -1 61 1 +82 Find the range of the function $f(x) = \\frac{x^2 + 14x + 9}{x^2 + 2x + 3},$as $x$ varies over all real numbers. Jul 3, 2020","are in the range of the function $f(x) = \\frac{4x^2+75}{2x^2+3}$?","# How do you calculate f '(x) and use calculus to find the maximum value of sin (ln x) on the interval [1, 10]? The answer is: $1$. This function is in the range $\\left[- 1 , 1\\right]$ and oscillates continually between these two values.","\\sin x - 2a &= 0 \\\\ 4y^3 +\\frac{1}{2}\\sin 2y +a &=0 \\end{array} \\right.$$ Compute the value of $\\cos(x+2y)$ Let $f$ be a real-valued function such that $f(x) + 2f(\\frac{2002}{x}) = 3x$ for all $x > 0$. Find $f(2)$.","Browse Questions # What is the range of the real function, $f(x) = 2x^2 + 3x − 4$? $\\bigg[ \\large\\frac{-41}{8} \\infty \\bigg)$","or ${c^2}$ -c < 0 Conclusion: So for 1>c>0 y = ${\\frac{x^2+2x+c}{x^2+4x+3c}}$ Range over all real number when x varies over all real number.","+0 # I need help ASAP! 0 220 2 +10 Find the set of values $$a$$ for which the range of the function $$f(x) = \\frac{x^2 + a}{x + 1}$$ is the set of all real numbers. Aug 24, 2020 edited by Guest Aug 24, 2020 #1 0 The set of possible values of a is [-2,2]. Aug 24, 2020 #2 0 that's incorrect. Guest Aug 24, 2020","The range of sin(x) is [-1, 1]. It's confusing to say \"the range of ##\\sin(x)## in \". The range of ##10^{\\sin(x)}## is [0.1, 10]. Why do you think that 1 is not in the range of ##10^{\\sin x}##? ##f(x) = 10^{\\sin x} + 10^{\\csc x} = e^{\\sin x \\ln 10} + e^{\\frac{1}{\\sin x} \\ln 10}##. Since the function has a period of ##2\\pi##, I think you should study it when ##x \\in (0,2\\pi)##, at the period's bounds, and near ##\\pi##. And maybe you should see when its derivative becomes zero to figure out the lower bound of the range, since, from the definition of ##e^x##, this function can't go below ##0##. Using desmos, the range seems to be ##[0.2, \\infty)##. Last edited: You are confusing the ranges of the subfunctions with the range of the sum of these function. Keep in mind that the combined range is **not** the intersection of the two ranges. For example, consider ##g(x) = \\sqrt x## and ##h(x) = \\left(\\frac x 2\\right)^2## on the interval [0, 1]. On this restricted domain, ##Ran(g) = [0, 1]## and ##Ran(h) = [0, \\frac 1 4]##. The range of their sum, ##\\sqrt x + \\left(\\frac x 2\\right)^2## is [0, 1.25]. Also, in post #1 you have a mistake: The range of sin(x) is [-1, 1]. It's confusing","Browse Questions Let f : A $\\rightarrow$ R be a real function. A $\\subseteq$ R such that f(x) $\\sqrt{x-2}$ . Then find Range of ‘f’ Range of f = $R^+$","Browse Questions # What is the range of the function $f(x) = \\large\\frac{5x-1}{2x+7}$ defined on its maximum domain? Can you answer this question? Range of f = R - $\\bigg\\{ \\large\\frac{ 5}{2} \\bigg\\}$","What is the range of the real function, $f(x) = 2x^2 + 3x − 4$? $\\bigg[ \\large\\frac{-41}{8} \\infty \\bigg)$ Hence (B) is the correct answer.","# Write the Range of the Function F(X) = Sin X, Where − π 4 ≤ X ≤ π 4 . - Mathematics Sum Write the range of the function f(x) = sin [x], where $\\frac{- \\pi}{4} \\leq x \\leq \\frac{\\pi}{4}$ . #### Solution Given : f(x) = sin [x], where $\\frac{- \\pi}{4} \\leq x \\leq \\frac{\\pi}{4}$ $- \\frac{\\pi}{4} \\leq x \\leq \\frac{\\pi}{4}$ $\\Rightarrow - 0 .785 \\leq x \\leq 0 . 785$ $\\therefore x \\in \\left[ - 0.785, 0.785 \\right]$ $\\text{ Or } x = {0, 1}$ $\\text{ Thus, range of } f(x) = \\sin [x] \\text{ is }$ { sin (-1), sin(0) , sin (1) } . Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 11 Mathematics Textbook Chapter 3 Functions Exercise 3.5 | Q 3 | Page 41","the real function:$$f(x)=\\operatorname{arccsc} \\left(\\sqrt{2x-x^2}\\right) Find the range of that function. According to my workbook, the official answer to this ... Assume that $x$ is a rational multiple of $π$ such is $\\cos(x)$ is also rational. Then the number of values of $\\cos(x)$ under the conditions is? I did read niven's proof of this, but it uses complex ... $\\tan a =\\frac{5}{12}$, $\\pi 0$. Find $f(2)$.\",\n \"Browse Questions # What is the range of the real function, $f(x) = 2x^2 + 3x − 4$? $\\\\bigg[ \\\\large\\\\frac{-41}{8} \\\\infty \\\\bigg)$\",\n \"or ${c^2}$ -c < 0 Conclusion: So for 1>c>0 y = ${\\\\frac{x^2+2x+c}{x^2+4x+3c}}$ Range over all real number when x varies over all real number.\",\n \"+0 # I need help ASAP! 0 220 2 +10 Find the set of values $$a$$ for which the range of the function $$f(x) = \\\\frac{x^2 + a}{x + 1}$$ is the set of all real numbers. Aug 24, 2020 edited by Guest Aug 24, 2020 #1 0 The set of possible values of a is [-2,2]. Aug 24, 2020 #2 0 that's incorrect. Guest Aug 24, 2020\",\n \"The range of sin(x) is [-1, 1]. It's confusing to say \\\"the range of ##\\\\sin(x)## in \\\". The range of ##10^{\\\\sin(x)}## is [0.1, 10]. Why do you think that 1 is not in the range of ##10^{\\\\sin x}##? ##f(x) = 10^{\\\\sin x} + 10^{\\\\csc x} = e^{\\\\sin x \\\\ln 10} + e^{\\\\frac{1}{\\\\sin x} \\\\ln 10}##. Since the function has a period of ##2\\\\pi##, I think you should study it when ##x \\\\in (0,2\\\\pi)##, at the period's bounds, and near ##\\\\pi##. And maybe you should see when its derivative becomes zero to figure out the lower bound of the range, since, from the definition of ##e^x##, this function can't go below ##0##. Using desmos, the range seems to be ##[0.2, \\\\infty)##. Last edited: You are confusing the ranges of the subfunctions with the range of the sum of these function. Keep in mind that the combined range is **not** the intersection of the two ranges. For example, consider ##g(x) = \\\\sqrt x## and ##h(x) = \\\\left(\\\\frac x 2\\\\right)^2## on the interval [0, 1]. On this restricted domain, ##Ran(g) = [0, 1]## and ##Ran(h) = [0, \\\\frac 1 4]##. The range of their sum, ##\\\\sqrt x + \\\\left(\\\\frac x 2\\\\right)^2## is [0, 1.25]. Also, in post #1 you have a mistake: The range of sin(x) is [-1, 1]. It's confusing\",\n \"Browse Questions Let f : A $\\\\rightarrow$ R be a real function. A $\\\\subseteq$ R such that f(x) $\\\\sqrt{x-2}$ . Then find Range of ‘f’ Range of f = $R^+$\",\n \"Browse Questions # What is the range of the function $f(x) = \\\\large\\\\frac{5x-1}{2x+7}$ defined on its maximum domain? Can you answer this question? Range of f = R - $\\\\bigg\\\\{ \\\\large\\\\frac{ 5}{2} \\\\bigg\\\\}$\",\n \"What is the range of the real function, $f(x) = 2x^2 + 3x − 4$? $\\\\bigg[ \\\\large\\\\frac{-41}{8} \\\\infty \\\\bigg)$ Hence (B) is the correct answer.\",\n \"# Write the Range of the Function F(X) = Sin X, Where − π 4 ≤ X ≤ π 4 . - Mathematics Sum Write the range of the function f(x) = sin [x], where $\\\\frac{- \\\\pi}{4} \\\\leq x \\\\leq \\\\frac{\\\\pi}{4}$ . #### Solution Given : f(x) = sin [x], where $\\\\frac{- \\\\pi}{4} \\\\leq x \\\\leq \\\\frac{\\\\pi}{4}$ $- \\\\frac{\\\\pi}{4} \\\\leq x \\\\leq \\\\frac{\\\\pi}{4}$ $\\\\Rightarrow - 0 .785 \\\\leq x \\\\leq 0 . 785$ $\\\\therefore x \\\\in \\\\left[ - 0.785, 0.785 \\\\right]$ $\\\\text{ Or } x = {0, 1}$ $\\\\text{ Thus, range of } f(x) = \\\\sin [x] \\\\text{ is }$ { sin (-1), sin(0) , sin (1) } . Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 11 Mathematics Textbook Chapter 3 Functions Exercise 3.5 | Q 3 | Page 41\",\n \"the real function:$$f(x)=\\\\operatorname{arccsc} \\\\left(\\\\sqrt{2x-x^2}\\\\right) Find the range of that function. According to my workbook, the official answer to this ... Assume that $x$ is a rational multiple of $π$ such is $\\\\cos(x)$ is also rational. Then the number of values of $\\\\cos(x)$ under the conditions is? I did read niven's proof of this, but it uses complex ... $\\\\tan a =\\\\frac{5}{12}$, $\\\\pi \\frac 1 {\\sqrt{1+x^2}}$ unless we have $\\tan c < x. \\qquad$ Oct 18, 2020 at 17:02 • @MichaelHardy I don't think we have an assumption of $\\tan c < x$. But I have attached the snapshot of the question from the book here: imgur.com/a/UHtGZGb The question number is 12(b) and I hope I haven't missed anything. – Sibi Oct 18, 2020 at 17:10 • \\begin{align} & \\frac{\\sin x} x = \\cos c \\\\ {} \\\\ > {} & \\cos x \\text{ since the cosine function decreases on this interval} \\\\ {} \\\\ = {} & \\sqrt{1-\\sin^2 x} \\\\ {} \\\\ > {} & \\sqrt{1-x^2} \\text{ as shown in part (a) of the same exercise} \\\\{} \\\\ \\not> {} & \\frac 1 {\\sqrt{1+x^2}}\\quad \\text{So this is not there yet.} \\end{align} Oct 18, 2020 at 17:30 You are on the right path: square both sides of $$\\sin(x)>x (1-\\sin^2(x))^{1/2}$$ and get $$\\sin^2(x)>x^2(1-\\sin^2(x))=x^2-x^2\\sin^2(x)$$, then add $$x^2\\sin^2(x)$$ to both sides to obtain $$\\sin^2(x)(1+x^2)>x^2$$. Now divide both sides by $$(1+x^2)$$ and take the square root. • Can you expand on how squaring has given you $\\sin^2(x)(1+x^2)>x$ ? • @Sibi, I think it's supposed to be an $x^2$ on the right hand side of the $\\gt$. Oct 18, 2020 at 23:14 • @BarryCipra Thanks! If that's the case, then do you where","function instead. In my opinion, these are major differences. Well, we have to use something. The product expansion also immediately gives the result $$\\sin(x) = x \\cdot \\prod_{k=1}^\\infty \\left( 1 - \\frac{x^2}{k^2 \\pi^2} \\right)$$ But if you do not allow analytical methods, then I'll have to ask you: What is $\\sin(x)\\,$? And if the answer is a geometric one, I'll ask, which linear distance is $x$ then, to compare its length with the sine value? 5. Sep 20, 2017 ### mathman Why is it obvious? 6. Sep 20, 2017 ### I like Serena Alternatively, we have: $$\\sin'(0)=\\lim_{x\\to 0} \\frac{\\sin(x) - \\sin(0)}{x-0}=\\lim_{x\\to 0} \\frac{\\sin(x)}{x}$$ So: $$\\lim_{x\\to 0} \\frac{\\sin(x)}{x} = \\sin'(0) = \\cos(0) = 1$$ Or if we want to stick to a Taylor expansion, with Lagrange's remainder theorem we have: $$\\sin(x) = \\sin(0) + \\frac{x}{1!}\\sin'(0) + \\frac{x^2}{2!}\\sin''(0) + \\frac{x^3}{3!}\\sin'''(\\xi)$$ where $\\xi$ is a value between $x$ and 0. It follows that: $$\\sin(x)=x + \\frac{x^3}{3!}\\cdot -\\cos(\\xi)$$ And since cosine is limited to [-1,1], we get: $$x-\\left|\\frac{x^3}{3!}\\right| \\le \\sin(x) \\le x+ \\left|\\frac{x^3}{3!}\\right|$$ Now we can apply the squeeze theorem - and there is no cheating involved! Last edited: Sep 20, 2017 7. Sep 20, 2017 ### Staff: Mentor Don't you mean $\\sin(x) = \\sin(0) + \\frac{x}{1!}\\sin'(0) + \\frac{x^2}{2!}\\sin''(0) + \\frac{x^3}{3!}\\sin'''(\\xi)$? Easier to read as $\\sin(x)=x - \\frac{x^3}{3!} cos(\\xi)$ 8. Sep 20,","- {\\cot ^{ - 1}}x$ We have seen that $${f^{ - 1}}(f(x)) = f({f^{ - 1}}(x)) = x$$. For example, \\begin{align}&\\sin \\left( {{{\\sin }^{ - 1}}\\left( {\\frac{1}{2}} \\right)} \\right) = \\sin \\left( {\\frac{\\pi }{6}} \\right) = \\frac{1}{2} \\\\ &{\\sin ^{ - 1}}\\left( {\\sin \\left( {\\frac{\\pi }{6}} \\right)} \\right) = {\\sin ^{ - 1}}\\left( {\\frac{1}{2}} \\right) = \\frac{\\pi }{6}\\quad etc. \\\\ \\end{align} But this relation must be used with caution, since it is valid only if x lies in the appropriate domain. For example, ${\\sin ^{ - 1}}\\left( {\\sin \\left( {\\frac{{5\\pi }}{6}} \\right)} \\right) = {\\sin ^{ - 1}}\\left( {\\frac{1}{2}} \\right) = \\frac{\\pi }{6}\\;\\;{\\text{and}}\\;{\\mathbf{not}}\\;\\frac{{5\\pi }}{6}$ Therefore $${\\sin ^{ - 1}}(\\sin \\theta )$$, will not always be $$\\theta$$, since the output of $${\\sin ^{ - 1}}$$ will necessarily lie in the range \\begin{align}\\left[ {\\frac{{ - \\pi }}{2},\\;\\frac{\\pi }{2}} \\right]\\end{align}. To adjust for this fact, note the following: \\begin{align}&{\\sin ^{ - 1}}(\\sin \\theta ) = \\left\\{ \\begin{gathered}- \\pi - \\theta ,\\;\\;{\\text{if}}\\;\\theta \\in \\;\\left[ {\\frac{{ - 3\\pi }}{2},\\;\\frac{{ - \\pi }}{2}} \\right] \\\\ \\;\\;\\;\\;\\;\\theta ,\\;\\;\\;\\;{\\text{if}}\\;\\theta \\in \\;\\left[ {\\frac{{ - \\pi }}{2},\\;\\frac{\\pi }{2}} \\right] \\\\ \\pi - \\theta ,\\;\\;\\;\\;{\\text{if}}\\;\\theta \\in \\;\\left[ {\\frac{\\pi }{2},\\;\\frac{{3\\pi }}{2}} \\right] \\\\ - 2\\pi + \\theta ,\\mathop {{\\text{if}}}\\limits_ \\vdots \\;\\theta \\in \\;3\\pi /2,\\;5\\pi /2 \\\\ \\end{gathered} \\right\\} \\qquad\\left({{\\text{Verify!}}} \\right) \\\\ &\\qquad\\qquad\\qquad\\qquad\\qquad{\\text{and}}\\;{\\text{so}}\\;{\\text{on}} \\\\ \\end{align} That is, we have adjusted the","We have \\begin{align} \\text{LHS} &= (\\cos x)(1+\\cos x+\\sin x)\\tag{definition}\\\\[0.5em] &= \\cos x +\\cos^2(x)+\\cos x\\sin x\\tag{expand} \\end{align} and \\begin{align} \\text{RHS} &= (1+\\sin x)(1+\\cos x-\\sin x)\\tag{definition}\\\\[0.5em] &= 1+\\cos x-\\sin x+\\sin x+\\sin x\\cos x-\\sin^2(x)\\tag{expand}\\\\[0.5em] &= 1+\\cos x+\\sin x\\cos x-\\sin^2(x).\\tag{simplify} \\end{align} Now compare the LHS and RHS. We clearly have that $$\\cos^2(x)=1-\\sin^2(x)\\Longleftrightarrow \\cos^2(x)+\\sin^2(x)=1,$$ and we know this famous result to be true (or easily established otherwise). Hence, we have proved $(1)$. Rearrange and Cross multiply it is required to prove that $$\\dfrac{\\cos x +1 +\\sin x}{1-\\sin x +\\cos x} = \\dfrac{1+\\sin x}{\\cos x}$$ or $$\\cos^2 x + \\cos x (1 +\\sin x ) = (1-\\sin^2 x) + \\cos x (1 +\\sin x )$$ or $$\\cos x (1 +\\sin x ) = \\cos x (1 +\\sin x ).$$ • That was cute. +1 – Daniel W. Farlow Mar 11 '15 at 9:57 Hint: $$= \\frac{(1 + \\cos x + \\sin x)(1 - \\cos x + \\sin x)}{(1 + \\cos x - \\sin x)(1 - \\cos x + \\sin x)}$$","decrease to the left). Taking a simpler example, try f(x) = 2x. This function approaches 12 as x approaches 6 from the left-hand side (it also approaches 12 as x approaches 6 from the right-hand side, so we simply say that f(x) approaches 12 as x approaches 6). So if you take values for x of 5, 5.5, 5.9, 5.99, 5.999, 5.999999, etc., then your values for f(x) will get closer and closer to 12. lurflurf Homework Helper This is a common limit. in fact it arises in finding the derivative of sine $$\\frac{d}{dx}\\sin(x)=\\lim_{h\\rightarrow 0} \\frac{\\sin(x)}{x}\\cos(x+\\frac{h}{2})$$ the limit can be shown several ways including showing that cos(x)0$$ $$270 Again, $$-90 Finally $$\\sin2y=\\sin(x+y+(x-y))=?$$ Try to solve the first equation for $$x$$ and plug this in the second equation . I got","one can replace $\\sin x$ with $x$ and in the numerator the third order formula to get $$\\frac{x-(x-\\frac16 x^3+O(x^5))}{-4x^3+O(x^5)} = -\\frac{1-O(x^2)}{24}$$ $$\\lim_{x\\to0}\\frac{\\sin(x)}x=1$$ doesn't allow you to write $$\\frac{\\sin(x)}x=1\\text{ nor } \\sin(x)-x=0\\ !$$ There are circumstances where $\\sin(x)-x$ cannot be neglected because it is amplified. For the sake of illustration, here is a plot of $$\\frac{\\sin(2x)-2x}{\\sin(x)-x}.$$ The problem you encountered was that both $\\sin(x)-\\tan(x)=O\\!\\left(x^3\\right)$ and $x-\\sin(x)=O\\!\\left(x^3\\right)$, this means that substituting $x$ for $\\sin(x)$ may substantively change the limit. To be precise, \\begin{align} \\sin(x)-\\tan(x) &=\\left(x-\\frac{x^3}6+O\\!\\left(x^5\\right)\\right)-\\left(x+\\frac{x^3}3+O\\!\\left(x^5\\right)\\right)\\\\ &=-\\frac{x^3}2+O\\!\\left(x^5\\right) \\end{align} whereas \\begin{align} x-\\tan(x) &=x-\\left(x+\\frac{x^3}3+O\\!\\left(x^5\\right)\\right)\\\\ &=-\\frac{x^3}3+O\\!\\left(x^5\\right) \\end{align} This explains why the limits you got differ by a factor of $\\frac32$. I think the limit notation has been made so confusing thanks to various educators (including book authors) who try to simplify it too much. When you say $$\\lim_{x \\to 0}\\frac{\\sin x}{x} = 1 \\tag{1}$$ you don't mean $(\\sin x)/x = 1$ when $x \\to 0$. Hence it is just not possible to replace $\\sin x$ by $x$. However whenever you see the expression $\\lim_{x \\to 0}\\dfrac{\\sin x}{x}$ you can replace it by $1$ (because of the equation $(1)$ above). You can always replace a thing called $A$ by a thing called $B$ when you know that $A$ and $B$ are equal but if $A \\neq B$ then you can't replace $A$ by $B$. Unfortunately","O(x^6)-x^2}{\\left(\\arcsin \\:x\\right)^2-x^2} \\end{align} (The above is true as $\\ln(1+x) \\approx x-x^2/2 +O(x^3)$ when $x$ is small) \\begin{align} \\lim _{x\\to \\:0}\\frac{\\sin \\left(x^2\\right)-\\frac{\\sin^2(x^2)}{2} + O(x^6)-x^2}{\\left(\\arcsin \\:x\\right)^2-x^2} &= \\lim _{x\\to \\:0}\\frac{x^2 -x^4/2 + O(x^{6})-x^2}{\\left(x + \\frac{x^3}{6}+O(x^5)\\right)^2-x^2} \\\\ &=\\lim _{x\\to \\:0}\\frac{\\frac{-x^4}{2}}{\\frac{2x^4}{6}}\\\\ &= -\\frac{3}{2} \\end{align} (I used Taylor expansions) • Unfortunately the answer has errors. For example, $\\sin x^2\\ne \\sin^2 x$. $\\sin x^2=x^2-\\frac16x^6+O(x^{10})$. $\\sin ^2x=(x-\\frac16 x^3+O(x^5))^2=x^2-\\frac13x^4+O(x^6)$. And please explain the justification to arrive at the second to last equality. – Mark Viola Sep 23 '15 at 20:59 • the answer is $-\\dfrac{3}{2}$ so please would you provide more details – Educ Sep 23 '15 at 21:09 • Thanks. I made the changes and edited the answer. Hope it helps! – chandu1729 Sep 23 '15 at 21:15","at 15:15 \\begin{align}\\tanh(x+2y)+\\tanh(x-2y)&=\\frac{\\sinh(x+2y)}{\\cosh(x+2y)}+\\frac{\\sinh(x-2y)}{\\cosh(x-2y)}\\\\&=\\frac{\\sinh(x+2y)\\cosh(x-2y)+\\sinh(x-2y)\\cosh(x+2y)}{\\cosh(x+2y)\\cosh(x-2y)}\\\\&=\\frac{\\sinh(2x)}{\\cosh(x+2y)\\cosh(x-2y)}\\\\&=\\frac{\\frac{e^{2x}-e^{-2x}}2}{\\frac{(e^{x+2y}+e^{-(x+2y)})(e^{x-2y}+e^{-(x-2y)})}4}\\\\&=2\\times\\frac{1-e^{-6x}}{(e^{2y}+e^{-2x-2y})(e^{-2y}+e^{-2x-2y})}\\end{align} Then note that as $x\\to\\infty$ and $y$ is held constant, then it goes to $2$. Likewise, as $y\\to\\infty$ and $x$ is held constant, then it goes to $0$. Thus, the limit doesn't exist. • But if $y$ is held constant ($< \\infty$) isn't $\\lim_{x \\to \\infty} \\left[ \\tanh(x+2y) + \\tanh(x-2y) \\right] = 1 + 1 = 2$? And not infinity? – 655321 May 26 '17 at 15:28 • Yes but $\\sinh(x+2y) \\cosh(x-2y) + \\sinh(x-2y) \\cosh(x+2y) = \\sinh(2x) \\neq \\sinh(4x)$ and the last term in the denominator should be $e^{-(2x-2y)} = e^{-2x+2y}$ so $$\\tanh(x+2y) - \\tanh(x-2y) = 2 \\cdot \\frac{1-e^{-4x}}{(e^{2y} + e^{-2x-2y})(e^{-2y}+e^{-2x+2y})} \\mathrm{.}$$ I find it hard to believe that a sum on $\\tanh$-functions would tend to infinity as $\\tanh(x) \\in (-1, 1)$. Anyway I think that this limit really doesn't exist. – 655321 May 26 '17 at 17:08"],"string":"[\n \"## Precalculus (6th Edition) Blitzer The values of x are $\\\\frac{\\\\pi }{6},\\\\frac{5\\\\pi }{6},\\\\,\\\\text{ and }\\\\,\\\\frac{3\\\\pi }{2}$. $\\\\cos 2x-\\\\sin x=0,\\\\,0\\\\le x\\\\,<2\\\\pi$ …… (1) Substituting the value of $\\\\cos 2x$ as $\\\\cos 2x=1-2{{\\\\sin }^{2}}x$ in equation (1) we get, \\\\begin{align} & 1-2{{\\\\sin }^{2}}x-\\\\sin x=0 \\\\\\\\ & -2{{\\\\sin }^{2}}x-\\\\sin x+1=0 \\\\end{align} Multiplying the above equation by $-1$ we get \\\\begin{align} & -2{{\\\\sin }^{2}}x-\\\\sin x+1=0 \\\\\\\\ & 2{{\\\\sin }^{2}}x+\\\\sin x-1=0 \\\\end{align} We can also express the equation as below: \\\\begin{align} & 2{{\\\\sin }^{2}}x+\\\\sin x-1=0 \\\\\\\\ & 2{{\\\\sin }^{2}}x+2\\\\sin x-\\\\sin x-1=0 \\\\end{align} Now, factorize the equation. \\\\begin{align} & 2\\\\sin x\\\\left( \\\\sin x+1 \\\\right)-1\\\\left( \\\\sin x+1 \\\\right)=0 \\\\\\\\ & \\\\left( 2\\\\sin x-1 \\\\right)\\\\left( \\\\sin x+1 \\\\right)=0 \\\\end{align} Solve for x. $2\\\\sin x-1=0$ or $\\\\sin x+1=0$ $2\\\\sin x=1$ $\\\\sin x=-1$ $\\\\sin x=\\\\frac{1}{2}$ $x=\\\\frac{3\\\\pi }{2}$ $x=\\\\frac{\\\\pi }{6}$ In the range $0\\\\le x\\\\,<2\\\\pi$, the values of x for $2\\\\sin x-1=0$ are $x=\\\\pi -\\\\frac{\\\\pi }{6}=\\\\frac{5\\\\pi }{6}$.\",\n \"so \\\\begin{align} \\\\cos^4x\\\\sin^4x &=\\\\frac{1}{2^8}(e^{2ix}-e^{-2ix})^4\\\\\\\\[6px] &=\\\\frac{1}{2^8}(e^{8ix}-4e^{4ix}+6-4e^{-4ix}+e^{8ix})\\\\\\\\[6px] &=\\\\frac{1}{128}\\\\cos8x-\\\\frac{1}{32}\\\\cos4x+\\\\frac{3}{128} \\\\end{align} we now that $$\\\\sin 2x=2\\\\cos x\\\\sin x$$ $$\\\\cos x\\\\sin x=\\\\frac{\\\\sin 2x}{2}$$ so... $$\\\\cos^4 x\\\\sin^4 x=\\\\frac{\\\\sin^4 2x}{16}$$ $$\\\\frac{\\\\sin^4 2x}{16}=\\\\frac{(\\\\frac{1-\\\\cos 4x}{2})^2}{16}=\\\\frac{1-2\\\\cos 4x+\\\\cos^24x}{64}=\\\\frac{1-2\\\\cos 4x+\\\\frac{1+\\\\cos 8x}{2}}{64}$$\",\n \"the origin. Let's try to evaluate $\\\\lim_{(x, y) \\\\to (0, 0)} \\\\frac{\\\\sin x \\\\sin^3 y}{1 - \\\\cos (x^2 + y^2)} = L$. Along the line $y = 0$ we have that: (1) \\\\begin{align} \\\\quad \\\\lim_{(x, 0) \\\\to (0, 0)} \\\\frac{\\\\sin x \\\\sin^3 y}{1 - \\\\cos (x^2 + y^2)} = \\\\lim_{(x \\\\to 0} 0 = 0 \\\\end{align} However, along the line $x = y$ we have that: (2) \\\\begin{align} \\\\quad \\\\lim_{(y, y) \\\\to (0, 0)} \\\\frac{\\\\sin x \\\\sin^3 y}{1 - \\\\cos (x^2 + y^2)} = \\\\lim_{y \\\\to 0} \\\\frac{\\\\sin^4 y}{1 - \\\\cos (2y^2)} \\\\end{align} We now use the trigonometric identity that $\\\\sin^2 u = \\\\frac{1 - \\\\cos (2u)}{2}$. We then have that $2\\\\sin^2 (y^2) = 1 - \\\\cos (2y^2)$. Plugging this into the limit above and: (3) \\\\begin{align} \\\\quad = \\\\lim_{y \\\\to 0} \\\\frac{\\\\sin^4 y}{2 \\\\sin^2 (y^2)} \\\\end{align} Using L'Hospital's rule multiple times and we see that this limit is equal to $\\\\frac{1}{2}$. Therefore $L$ does not exist and $\\\\hat{f}$ cannot be a redefinition of $f$ and be continuous at the origin. The graph of $f$ is given below:\",\n \"that $r\\\\sin(-a)$ is positive. This happens when $r=5$ and $a=-36.87^{\\\\circ}$; or when $r=-5$ and $a=143.13$. The two choices work, the other two possibilities do not. \\\\begin{align*} f(x) &= 3 \\\\sin(x) + 4 \\\\cos(x) \\\\\\\\ f(x) &= 5\\\\left(\\\\frac{3}{5} \\\\sin(x) + \\\\frac{4}{5} \\\\cos(x) \\\\right) \\\\end{align*} Let $\\\\sin(y) = 4/5$ and $\\\\cos(y) =3/5$, for some real number $y$. \\\\begin{align*} f(x) &= 5\\\\left(\\\\cos(y)\\\\sin(x) + \\\\sin(y)\\\\cos(x)\\\\right)\\\\\\\\ f(x) &= 5\\\\sin(x+y)&\\\\text{(where }y = \\\\arcsin(4/5)\\\\text{)} \\\\end{align*} Using trigonometric tables, $y = 36.87$. $$f(x) = 5\\\\sin(x + 36.87^{\\\\circ})$$ which can also be written as $$f(x) = 5 \\\\sin (180 - x - 36.87) = -5\\\\sin(x - 144.13).$$ This justifies the 2 values of $a$ and $r$ in $r\\\\sin(x−a)$.\",\n \"really desperate i need to this for tommorow and I don´t know how to do this a really need some help - as I said I have already asked the question in I answered that i would tried to do the problem but I can´t do it... – David Hernandez Nov 30 '13 at 5:13 Why not use commas and periods to separate sentences and paragraghs? – awllower Nov 30 '13 at 5:15 I am sorry :) I am not very good at writing in mathjax – David Hernandez Nov 30 '13 at 5:24 Helpful identity: note that $$1 - \\\\cos(x) = 2\\\\frac{1 - \\\\cos(2\\\\frac{x}{2})}{2} = 2 \\\\sin^2\\\\left(\\\\frac{x}{2}\\\\right)$$ so that our function is simply $$f(x)= \\\\begin{cases} 2 {\\\\sin^2(\\\\frac x2)\\\\over x^2}, & \\\\text{if x\\\\neq 0} \\\\\\\\ 1/2, & \\\\text{if x= 0} \\\\\\\\ \\\\end{cases}$$ With that in mind, we can find (assuming $x,y \\\\neq 0$) \\\\begin{align} \\\\left|f(x)-f(y)\\\\right| &= 2\\\\left| \\\\left(\\\\frac{\\\\sin(\\\\frac y2)}{y}\\\\right)^2 - \\\\left(\\\\frac{\\\\sin(\\\\frac x2)}{x}\\\\right)^2 \\\\right|\\\\\\\\ &=2\\\\left| \\\\left(\\\\frac{\\\\sin(\\\\frac y2)}{y}\\\\right) + \\\\left(\\\\frac{\\\\sin(\\\\frac x2)}{x}\\\\right) \\\\right| \\\\left| \\\\left(\\\\frac{\\\\sin(\\\\frac y2)}{y}\\\\right) - \\\\left(\\\\frac{\\\\sin(\\\\frac x2)}{x}\\\\right) \\\\right| \\\\end{align} Now, we just have to work with $\\\\left| \\\\left(\\\\frac{\\\\sin(\\\\frac y2)}{y}\\\\right) - \\\\left(\\\\frac{\\\\sin(\\\\frac x2)}{x}\\\\right) \\\\right|$ to get out a $|x-y|$. thank you!! and we also have that ${|xsin(y)-ysin(x)|\\\\over |xy|}\\\\le{|x-y|\\\\over |xy|}$ is this correct? – David Hernandez Nov 30 '13 at 5:50 I think there is a typo. $\\\\frac{2\\\\sin^2(x)}{x^2}$ should be $\\\\frac{2\\\\sin^2(x/2)}{x^2}$\",\n \"& =0.9980 \\\\end{align} At $x=-0.0001$ Substitute, $x=-0.0001$ in the provided limit notation, $\\\\underset{x\\\\to 0}{\\\\mathop{\\\\lim }}\\\\,\\\\frac{2{{x}^{2}}+x}{\\\\sin x}$. $\\\\underset{x\\\\to -0.0001}{\\\\mathop{\\\\lim }}\\\\,\\\\frac{2{{x}^{2}}+x}{\\\\sin x}=\\\\frac{2{{\\\\left( 0.0001 \\\\right)}^{2}}+\\\\left( -0.0001 \\\\right)}{\\\\sin \\\\left( -0.0001 \\\\right)}$ Apply the trigonometric property, $\\\\sin \\\\left( -\\\\theta \\\\right)=-\\\\sin \\\\left( \\\\theta \\\\right)$ \\\\begin{align} & \\\\underset{x\\\\to -0.0001}{\\\\mathop{\\\\lim }}\\\\,\\\\frac{2{{x}^{2}}+x}{\\\\sin x}=\\\\frac{2\\\\left( 1\\\\times {{10}^{-8}} \\\\right)-0.0001}{-\\\\sin \\\\left( 0.0001 \\\\right)} \\\\\\\\ & =\\\\frac{\\\\left( 2\\\\times {{10}^{-8}} \\\\right)-0.0001}{-\\\\left( 9.99\\\\times {{10}^{-5}} \\\\right)} \\\\\\\\ & =\\\\frac{-9.998\\\\times {{10}^{-5}}}{-\\\\left( 9.99\\\\times {{10}^{-5}} \\\\right)} \\\\\\\\ & =0.9998 \\\\end{align} Now, as x approaches $0$ from the right, arbitrarily start with $x=0.01$. Then select two additional values of x that are closer to 0, but still greater than 0; we choose $0.001$ and $0.0001$. At $x=0.01$ Substitute, $x=0.01$ in the provided limit notation, $\\\\underset{x\\\\to 0}{\\\\mathop{\\\\lim }}\\\\,\\\\frac{2{{x}^{2}}+x}{\\\\sin x}$. Therefore, $\\\\underset{x\\\\to 0.01}{\\\\mathop{\\\\lim }}\\\\,\\\\frac{2{{x}^{2}}+x}{\\\\sin x}=\\\\frac{2{{\\\\left( 0.01 \\\\right)}^{2}}+\\\\left( 0.01 \\\\right)}{\\\\sin \\\\left( 0.01 \\\\right)}$ \\\\begin{align} & \\\\underset{x\\\\to 0.01}{\\\\mathop{\\\\lim }}\\\\,\\\\frac{2{{x}^{2}}+x}{\\\\sin x}=\\\\frac{2\\\\left( 0.0001 \\\\right)+0.01}{\\\\sin \\\\left( 0.01 \\\\right)} \\\\\\\\ & =\\\\frac{0.0002+0.01}{0.0099} \\\\\\\\ & =\\\\frac{0.0102}{0.0099} \\\\\\\\ & =1.0200 \\\\end{align} Now, at $x=0.001$ Substitute, $x=0.001$ in the provided limit notation, $\\\\underset{x\\\\to 0}{\\\\mathop{\\\\lim }}\\\\,\\\\frac{2{{x}^{2}}+x}{\\\\sin x}$ \\\\begin{align} & \\\\underset{x\\\\to 0.001}{\\\\mathop{\\\\lim }}\\\\,\\\\frac{2{{x}^{2}}+x}{\\\\sin x}=\\\\frac{2{{\\\\left( 0.001 \\\\right)}^{2}}+\\\\left( 0.001 \\\\right)}{\\\\sin \\\\left( 0.001 \\\\right)} \\\\\\\\ & =\\\\frac{2\\\\left( 1\\\\times {{10}^{-6}} \\\\right)+0.001}{9.99\\\\times {{10}^{-4}}} \\\\\\\\ & =\\\\frac{\\\\left( 2\\\\times {{10}^{-6}} \\\\right)+0.001}{9.99\\\\times {{10}^{-4}}} \\\\\\\\ & =\\\\frac{1.002\\\\times {{10}^{-3}}}{9.99\\\\times {{10}^{-4}}} \\\\end{align} Further solve, $\\\\underset{x\\\\to 0.001}{\\\\mathop{\\\\lim }}\\\\,\\\\frac{2{{x}^{2}}+x}{\\\\sin x}=1.0020$ At $x=0.0001$ Substitute, $x=0.0001$ in the provided limit notation, $\\\\underset{x\\\\to 0}{\\\\mathop{\\\\lim }}\\\\,\\\\frac{2{{x}^{2}}+x}{\\\\sin x}$. Therefore, \\\\begin{align}\",\n \"## Precalculus (6th Edition) Blitzer The solutions to the given equation are $\\\\frac{7\\\\pi }{6},\\\\,and\\\\,\\\\frac{11\\\\pi }{6}$. We have to solve the equation on the interval $[0,2\\\\pi )$: $5\\\\text{sin}\\\\ x=2\\\\text{co}{{\\\\text{s}}^{2}}x-4$ And use the Pythagorean identity of trigonometry ${{\\\\sin }^{2}}x+{{\\\\cos }^{2}}x=1$. And the expression ${{\\\\cos }^{2}}x$ can be written as $1-{{\\\\sin }^{2}}x$: \\\\begin{align} & 5\\\\text{sin}\\\\ x=2\\\\left( 1-\\\\text{si}{{\\\\text{n}}^{2}}x \\\\right)-4 \\\\\\\\ & =2-2\\\\text{si}{{\\\\text{n}}^{2}}x-4 \\\\\\\\ & =-2{{\\\\sin }^{2}}x-2 \\\\end{align} After that, simplify it as: \\\\begin{align} & 2{{\\\\sin }^{2}}x+5\\\\sin x+2=0 \\\\\\\\ & 2{{\\\\sin }^{2}}x+4\\\\sin x+\\\\sin x+2=0 \\\\end{align} And factorize the equation: \\\\begin{align} & 2\\\\sin x\\\\left( \\\\sin x+2 \\\\right)+1\\\\left( \\\\sin x+2 \\\\right)=0 \\\\\\\\ & \\\\left( 2\\\\sin x+1 \\\\right)\\\\left( \\\\sin x+2 \\\\right)=0 \\\\end{align} And each factor needs to be calculated as: \\\\begin{align} & 2\\\\text{sin}\\\\ x+1=0 \\\\\\\\ & 2\\\\sin x=0-1 \\\\\\\\ & \\\\text{sin}\\\\ x=-\\\\frac{1}{2} \\\\end{align} Or, \\\\begin{align} & \\\\sin x+2=0 \\\\\\\\ & \\\\sin x=0-2 \\\\\\\\ & =-2 \\\\end{align} So, in the quadrant graph, the value of sine is $-\\\\frac{1}{2}$ at $\\\\frac{7\\\\pi }{6}$, and $\\\\frac{11\\\\pi }{6}$. This implies, \\\\begin{align} & sinx=sin\\\\frac{7\\\\pi }{6} \\\\\\\\ & x=\\\\frac{7\\\\pi }{6} \\\\end{align} Then, take the value of $x$ as $\\\\frac{11\\\\pi }{6}$: \\\\begin{align} & sinx=sin\\\\frac{11\\\\pi }{6} \\\\\\\\ & x=\\\\frac{11\\\\pi }{6} \\\\end{align} And, So, in the quadrant graph, the value of sine is $-2$. This value of sine lies between the values of $-1$ and $1$. Therefore, the value of sine $-2$ will not be counted\",\n \"sin2(x))2 gives: $$= \\\\int sin^4(x) [1 - 2 sin^2(x) + sin^4(x)] cos(x) dx$$ and multiplying through by sin4(x) gives us three separate terms which can be integrated separately. \\\\begin{align} = \\\\int sin^4(&x) cos(x) \\\\, dx \\\\\\\\ &- 2 \\\\int sin^6(x) cos(x) \\\\, dx \\\\\\\\ &+ \\\\int sin^8(x) cos(x) \\\\, dx \\\\end{align} But now each of our terms contains cos(x), whcih we can make use of in a u-substitution: let u = sin(x), then du = cos(x) dx Notice that cos(x) dx is in each of our integrals, so the substitution gives integrals that are easy to evaluate: \\\\begin{align} &\\\\int u^4 \\\\, du - 2 \\\\int u^6 \\\\, du + \\\\int u^8 \\\\, du \\\\\\\\ \\\\\\\\ &= \\\\frac{1}{5} u^5 = \\\\frac{2}{7} u^7 + \\\\frac{1}{9} u^9 + C \\\\end{align} Finally, we can substitute sin(x) back in for u to get the result: \\\\begin{align} \\\\int &sin^4(x) cos^5(x) \\\\, dx \\\\\\\\ &= \\\\frac{1}{5} sin^5(x) - \\\\frac{2}{7} sin^7(x) + \\\\frac{1}{9} sin^9(x) + C \\\\end{align} Just so we don't leave the tangent function out, we'll do another example below with a mixed tangent-secant integrand. ### Example 5 #### A mixed tan-sec integrand $\\\\int tan^3(x) sec^2(x) \\\\, dx$ Solution: By now I hope you're getting the hang of these. Focus on the odd power on the tangent by replacing tan2(x) with sec2(x) - 1. $$\\\\int\",\n \"say$\\\\displaystyle {} > \\\\frac 1 {\\\\sqrt{1+x^2}}$ unless we have $\\\\tan c < x. \\\\qquad$ Oct 18, 2020 at 17:02 • @MichaelHardy I don't think we have an assumption of $\\\\tan c < x$. But I have attached the snapshot of the question from the book here: imgur.com/a/UHtGZGb The question number is 12(b) and I hope I haven't missed anything. – Sibi Oct 18, 2020 at 17:10 • \\\\begin{align} & \\\\frac{\\\\sin x} x = \\\\cos c \\\\\\\\ {} \\\\\\\\ > {} & \\\\cos x \\\\text{ since the cosine function decreases on this interval} \\\\\\\\ {} \\\\\\\\ = {} & \\\\sqrt{1-\\\\sin^2 x} \\\\\\\\ {} \\\\\\\\ > {} & \\\\sqrt{1-x^2} \\\\text{ as shown in part (a) of the same exercise} \\\\\\\\{} \\\\\\\\ \\\\not> {} & \\\\frac 1 {\\\\sqrt{1+x^2}}\\\\quad \\\\text{So this is not there yet.} \\\\end{align} Oct 18, 2020 at 17:30 You are on the right path: square both sides of $$\\\\sin(x)>x (1-\\\\sin^2(x))^{1/2}$$ and get $$\\\\sin^2(x)>x^2(1-\\\\sin^2(x))=x^2-x^2\\\\sin^2(x)$$, then add $$x^2\\\\sin^2(x)$$ to both sides to obtain $$\\\\sin^2(x)(1+x^2)>x^2$$. Now divide both sides by $$(1+x^2)$$ and take the square root. • Can you expand on how squaring has given you $\\\\sin^2(x)(1+x^2)>x$ ? • @Sibi, I think it's supposed to be an $x^2$ on the right hand side of the $\\\\gt$. Oct 18, 2020 at 23:14 • @BarryCipra Thanks! If that's the case, then do you where\",\n \"function instead. In my opinion, these are major differences. Well, we have to use something. The product expansion also immediately gives the result $$\\\\sin(x) = x \\\\cdot \\\\prod_{k=1}^\\\\infty \\\\left( 1 - \\\\frac{x^2}{k^2 \\\\pi^2} \\\\right)$$ But if you do not allow analytical methods, then I'll have to ask you: What is $\\\\sin(x)\\\\,$? And if the answer is a geometric one, I'll ask, which linear distance is $x$ then, to compare its length with the sine value? 5. Sep 20, 2017 ### mathman Why is it obvious? 6. Sep 20, 2017 ### I like Serena Alternatively, we have: $$\\\\sin'(0)=\\\\lim_{x\\\\to 0} \\\\frac{\\\\sin(x) - \\\\sin(0)}{x-0}=\\\\lim_{x\\\\to 0} \\\\frac{\\\\sin(x)}{x}$$ So: $$\\\\lim_{x\\\\to 0} \\\\frac{\\\\sin(x)}{x} = \\\\sin'(0) = \\\\cos(0) = 1$$ Or if we want to stick to a Taylor expansion, with Lagrange's remainder theorem we have: $$\\\\sin(x) = \\\\sin(0) + \\\\frac{x}{1!}\\\\sin'(0) + \\\\frac{x^2}{2!}\\\\sin''(0) + \\\\frac{x^3}{3!}\\\\sin'''(\\\\xi)$$ where $\\\\xi$ is a value between $x$ and 0. It follows that: $$\\\\sin(x)=x + \\\\frac{x^3}{3!}\\\\cdot -\\\\cos(\\\\xi)$$ And since cosine is limited to [-1,1], we get: $$x-\\\\left|\\\\frac{x^3}{3!}\\\\right| \\\\le \\\\sin(x) \\\\le x+ \\\\left|\\\\frac{x^3}{3!}\\\\right|$$ Now we can apply the squeeze theorem - and there is no cheating involved! Last edited: Sep 20, 2017 7. Sep 20, 2017 ### Staff: Mentor Don't you mean $\\\\sin(x) = \\\\sin(0) + \\\\frac{x}{1!}\\\\sin'(0) + \\\\frac{x^2}{2!}\\\\sin''(0) + \\\\frac{x^3}{3!}\\\\sin'''(\\\\xi)$? Easier to read as $\\\\sin(x)=x - \\\\frac{x^3}{3!} cos(\\\\xi)$ 8. Sep 20,\",\n \"- {\\\\cot ^{ - 1}}x$ We have seen that $${f^{ - 1}}(f(x)) = f({f^{ - 1}}(x)) = x$$. For example, \\\\begin{align}&\\\\sin \\\\left( {{{\\\\sin }^{ - 1}}\\\\left( {\\\\frac{1}{2}} \\\\right)} \\\\right) = \\\\sin \\\\left( {\\\\frac{\\\\pi }{6}} \\\\right) = \\\\frac{1}{2} \\\\\\\\ &{\\\\sin ^{ - 1}}\\\\left( {\\\\sin \\\\left( {\\\\frac{\\\\pi }{6}} \\\\right)} \\\\right) = {\\\\sin ^{ - 1}}\\\\left( {\\\\frac{1}{2}} \\\\right) = \\\\frac{\\\\pi }{6}\\\\quad etc. \\\\\\\\ \\\\end{align} But this relation must be used with caution, since it is valid only if x lies in the appropriate domain. For example, ${\\\\sin ^{ - 1}}\\\\left( {\\\\sin \\\\left( {\\\\frac{{5\\\\pi }}{6}} \\\\right)} \\\\right) = {\\\\sin ^{ - 1}}\\\\left( {\\\\frac{1}{2}} \\\\right) = \\\\frac{\\\\pi }{6}\\\\;\\\\;{\\\\text{and}}\\\\;{\\\\mathbf{not}}\\\\;\\\\frac{{5\\\\pi }}{6}$ Therefore $${\\\\sin ^{ - 1}}(\\\\sin \\\\theta )$$, will not always be $$\\\\theta$$, since the output of $${\\\\sin ^{ - 1}}$$ will necessarily lie in the range \\\\begin{align}\\\\left[ {\\\\frac{{ - \\\\pi }}{2},\\\\;\\\\frac{\\\\pi }{2}} \\\\right]\\\\end{align}. To adjust for this fact, note the following: \\\\begin{align}&{\\\\sin ^{ - 1}}(\\\\sin \\\\theta ) = \\\\left\\\\{ \\\\begin{gathered}- \\\\pi - \\\\theta ,\\\\;\\\\;{\\\\text{if}}\\\\;\\\\theta \\\\in \\\\;\\\\left[ {\\\\frac{{ - 3\\\\pi }}{2},\\\\;\\\\frac{{ - \\\\pi }}{2}} \\\\right] \\\\\\\\ \\\\;\\\\;\\\\;\\\\;\\\\;\\\\theta ,\\\\;\\\\;\\\\;\\\\;{\\\\text{if}}\\\\;\\\\theta \\\\in \\\\;\\\\left[ {\\\\frac{{ - \\\\pi }}{2},\\\\;\\\\frac{\\\\pi }{2}} \\\\right] \\\\\\\\ \\\\pi - \\\\theta ,\\\\;\\\\;\\\\;\\\\;{\\\\text{if}}\\\\;\\\\theta \\\\in \\\\;\\\\left[ {\\\\frac{\\\\pi }{2},\\\\;\\\\frac{{3\\\\pi }}{2}} \\\\right] \\\\\\\\ - 2\\\\pi + \\\\theta ,\\\\mathop {{\\\\text{if}}}\\\\limits_ \\\\vdots \\\\;\\\\theta \\\\in \\\\;3\\\\pi /2,\\\\;5\\\\pi /2 \\\\\\\\ \\\\end{gathered} \\\\right\\\\} \\\\qquad\\\\left({{\\\\text{Verify!}}} \\\\right) \\\\\\\\ &\\\\qquad\\\\qquad\\\\qquad\\\\qquad\\\\qquad{\\\\text{and}}\\\\;{\\\\text{so}}\\\\;{\\\\text{on}} \\\\\\\\ \\\\end{align} That is, we have adjusted the\",\n \"We have \\\\begin{align} \\\\text{LHS} &= (\\\\cos x)(1+\\\\cos x+\\\\sin x)\\\\tag{definition}\\\\\\\\[0.5em] &= \\\\cos x +\\\\cos^2(x)+\\\\cos x\\\\sin x\\\\tag{expand} \\\\end{align} and \\\\begin{align} \\\\text{RHS} &= (1+\\\\sin x)(1+\\\\cos x-\\\\sin x)\\\\tag{definition}\\\\\\\\[0.5em] &= 1+\\\\cos x-\\\\sin x+\\\\sin x+\\\\sin x\\\\cos x-\\\\sin^2(x)\\\\tag{expand}\\\\\\\\[0.5em] &= 1+\\\\cos x+\\\\sin x\\\\cos x-\\\\sin^2(x).\\\\tag{simplify} \\\\end{align} Now compare the LHS and RHS. We clearly have that $$\\\\cos^2(x)=1-\\\\sin^2(x)\\\\Longleftrightarrow \\\\cos^2(x)+\\\\sin^2(x)=1,$$ and we know this famous result to be true (or easily established otherwise). Hence, we have proved $(1)$. Rearrange and Cross multiply it is required to prove that $$\\\\dfrac{\\\\cos x +1 +\\\\sin x}{1-\\\\sin x +\\\\cos x} = \\\\dfrac{1+\\\\sin x}{\\\\cos x}$$ or $$\\\\cos^2 x + \\\\cos x (1 +\\\\sin x ) = (1-\\\\sin^2 x) + \\\\cos x (1 +\\\\sin x )$$ or $$\\\\cos x (1 +\\\\sin x ) = \\\\cos x (1 +\\\\sin x ).$$ • That was cute. +1 – Daniel W. Farlow Mar 11 '15 at 9:57 Hint: $$= \\\\frac{(1 + \\\\cos x + \\\\sin x)(1 - \\\\cos x + \\\\sin x)}{(1 + \\\\cos x - \\\\sin x)(1 - \\\\cos x + \\\\sin x)}$$\",\n \"decrease to the left). Taking a simpler example, try f(x) = 2x. This function approaches 12 as x approaches 6 from the left-hand side (it also approaches 12 as x approaches 6 from the right-hand side, so we simply say that f(x) approaches 12 as x approaches 6). So if you take values for x of 5, 5.5, 5.9, 5.99, 5.999, 5.999999, etc., then your values for f(x) will get closer and closer to 12. lurflurf Homework Helper This is a common limit. in fact it arises in finding the derivative of sine $$\\\\frac{d}{dx}\\\\sin(x)=\\\\lim_{h\\\\rightarrow 0} \\\\frac{\\\\sin(x)}{x}\\\\cos(x+\\\\frac{h}{2})$$ the limit can be shown several ways including showing that cos(x)0$$ $$270 Again, $$-90 Finally $$\\\\sin2y=\\\\sin(x+y+(x-y))=?$$ Try to solve the first equation for $$x$$ and plug this in the second equation . I got\",\n \"one can replace $\\\\sin x$ with $x$ and in the numerator the third order formula to get $$\\\\frac{x-(x-\\\\frac16 x^3+O(x^5))}{-4x^3+O(x^5)} = -\\\\frac{1-O(x^2)}{24}$$ $$\\\\lim_{x\\\\to0}\\\\frac{\\\\sin(x)}x=1$$ doesn't allow you to write $$\\\\frac{\\\\sin(x)}x=1\\\\text{ nor } \\\\sin(x)-x=0\\\\ !$$ There are circumstances where $\\\\sin(x)-x$ cannot be neglected because it is amplified. For the sake of illustration, here is a plot of $$\\\\frac{\\\\sin(2x)-2x}{\\\\sin(x)-x}.$$ The problem you encountered was that both $\\\\sin(x)-\\\\tan(x)=O\\\\!\\\\left(x^3\\\\right)$ and $x-\\\\sin(x)=O\\\\!\\\\left(x^3\\\\right)$, this means that substituting $x$ for $\\\\sin(x)$ may substantively change the limit. To be precise, \\\\begin{align} \\\\sin(x)-\\\\tan(x) &=\\\\left(x-\\\\frac{x^3}6+O\\\\!\\\\left(x^5\\\\right)\\\\right)-\\\\left(x+\\\\frac{x^3}3+O\\\\!\\\\left(x^5\\\\right)\\\\right)\\\\\\\\ &=-\\\\frac{x^3}2+O\\\\!\\\\left(x^5\\\\right) \\\\end{align} whereas \\\\begin{align} x-\\\\tan(x) &=x-\\\\left(x+\\\\frac{x^3}3+O\\\\!\\\\left(x^5\\\\right)\\\\right)\\\\\\\\ &=-\\\\frac{x^3}3+O\\\\!\\\\left(x^5\\\\right) \\\\end{align} This explains why the limits you got differ by a factor of $\\\\frac32$. I think the limit notation has been made so confusing thanks to various educators (including book authors) who try to simplify it too much. When you say $$\\\\lim_{x \\\\to 0}\\\\frac{\\\\sin x}{x} = 1 \\\\tag{1}$$ you don't mean $(\\\\sin x)/x = 1$ when $x \\\\to 0$. Hence it is just not possible to replace $\\\\sin x$ by $x$. However whenever you see the expression $\\\\lim_{x \\\\to 0}\\\\dfrac{\\\\sin x}{x}$ you can replace it by $1$ (because of the equation $(1)$ above). You can always replace a thing called $A$ by a thing called $B$ when you know that $A$ and $B$ are equal but if $A \\\\neq B$ then you can't replace $A$ by $B$. Unfortunately\",\n \"O(x^6)-x^2}{\\\\left(\\\\arcsin \\\\:x\\\\right)^2-x^2} \\\\end{align} (The above is true as $\\\\ln(1+x) \\\\approx x-x^2/2 +O(x^3)$ when $x$ is small) \\\\begin{align} \\\\lim _{x\\\\to \\\\:0}\\\\frac{\\\\sin \\\\left(x^2\\\\right)-\\\\frac{\\\\sin^2(x^2)}{2} + O(x^6)-x^2}{\\\\left(\\\\arcsin \\\\:x\\\\right)^2-x^2} &= \\\\lim _{x\\\\to \\\\:0}\\\\frac{x^2 -x^4/2 + O(x^{6})-x^2}{\\\\left(x + \\\\frac{x^3}{6}+O(x^5)\\\\right)^2-x^2} \\\\\\\\ &=\\\\lim _{x\\\\to \\\\:0}\\\\frac{\\\\frac{-x^4}{2}}{\\\\frac{2x^4}{6}}\\\\\\\\ &= -\\\\frac{3}{2} \\\\end{align} (I used Taylor expansions) • Unfortunately the answer has errors. For example, $\\\\sin x^2\\\\ne \\\\sin^2 x$. $\\\\sin x^2=x^2-\\\\frac16x^6+O(x^{10})$. $\\\\sin ^2x=(x-\\\\frac16 x^3+O(x^5))^2=x^2-\\\\frac13x^4+O(x^6)$. And please explain the justification to arrive at the second to last equality. – Mark Viola Sep 23 '15 at 20:59 • the answer is $-\\\\dfrac{3}{2}$ so please would you provide more details – Educ Sep 23 '15 at 21:09 • Thanks. I made the changes and edited the answer. Hope it helps! – chandu1729 Sep 23 '15 at 21:15\",\n \"at 15:15 \\\\begin{align}\\\\tanh(x+2y)+\\\\tanh(x-2y)&=\\\\frac{\\\\sinh(x+2y)}{\\\\cosh(x+2y)}+\\\\frac{\\\\sinh(x-2y)}{\\\\cosh(x-2y)}\\\\\\\\&=\\\\frac{\\\\sinh(x+2y)\\\\cosh(x-2y)+\\\\sinh(x-2y)\\\\cosh(x+2y)}{\\\\cosh(x+2y)\\\\cosh(x-2y)}\\\\\\\\&=\\\\frac{\\\\sinh(2x)}{\\\\cosh(x+2y)\\\\cosh(x-2y)}\\\\\\\\&=\\\\frac{\\\\frac{e^{2x}-e^{-2x}}2}{\\\\frac{(e^{x+2y}+e^{-(x+2y)})(e^{x-2y}+e^{-(x-2y)})}4}\\\\\\\\&=2\\\\times\\\\frac{1-e^{-6x}}{(e^{2y}+e^{-2x-2y})(e^{-2y}+e^{-2x-2y})}\\\\end{align} Then note that as $x\\\\to\\\\infty$ and $y$ is held constant, then it goes to $2$. Likewise, as $y\\\\to\\\\infty$ and $x$ is held constant, then it goes to $0$. Thus, the limit doesn't exist. • But if $y$ is held constant ($< \\\\infty$) isn't $\\\\lim_{x \\\\to \\\\infty} \\\\left[ \\\\tanh(x+2y) + \\\\tanh(x-2y) \\\\right] = 1 + 1 = 2$? And not infinity? – 655321 May 26 '17 at 15:28 • Yes but $\\\\sinh(x+2y) \\\\cosh(x-2y) + \\\\sinh(x-2y) \\\\cosh(x+2y) = \\\\sinh(2x) \\\\neq \\\\sinh(4x)$ and the last term in the denominator should be $e^{-(2x-2y)} = e^{-2x+2y}$ so $$\\\\tanh(x+2y) - \\\\tanh(x-2y) = 2 \\\\cdot \\\\frac{1-e^{-4x}}{(e^{2y} + e^{-2x-2y})(e^{-2y}+e^{-2x+2y})} \\\\mathrm{.}$$ I find it hard to believe that a sum on $\\\\tanh$-functions would tend to infinity as $\\\\tanh(x) \\\\in (-1, 1)$. Anyway I think that this limit really doesn't exist. – 655321 May 26 '17 at 17:08\"\n]"},"ps_retrievals":{"kind":"list like","value":["# Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I'm trying to solve this problem. Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I have simplified this problem to $$f(x)= 8\\cos^2x+6\\sin x\\cos x+8$$ and tried working with $g(x)= 8\\cos^2x+6\\sin x\\cos x$. I factored out the $2\\cos x$ and rewrote the other factor as a linear combination of cosine. It reduces down to $$g(x)=10\\cos x\\cos\\left(x-\\tan^{-1}\\frac{3}{4}\\right)$$ • Have you tried finding the maximum and minimum valueof $f$? Jan 6 '18 at 15:44 Rewrite as following \\begin{align} f(x) & = 8 + 6 \\sin(x ) \\cos(x ) + 8\\cos^2(x) - 4 + 4 \\\\ &= 12 + 3 \\sin(2x) + 4 \\cos(2x) \\\\ &= 12 + 5 \\left(\\frac{3}{5} \\sin(2x) + \\frac{4}{5} \\cos(2x)\\right) \\\\ &= 12 + 5 \\sin(2x + \\arctan{\\tfrac{4}{3}}) \\end{align} Now its easy since $\\sin(...)$ always lies in $[-1,1]$, max/min values are $12 \\pm 5$. So maximum value is $17$ and minimum is $7$ write your function as $$f(x)=(\\sin(x)+3\\cos(x))^2+7$$ • This is good for minimum, but then for maximum? But answer is good! +1! Jan 6 '18 at 15:59 • You should write $3\\cos x + \\sin x$ in the form $R\\cos(x-\\alpha)$, where $R>0$ and $0 < \\alpha < \\frac{1}{2}\\pi$. Jan 6 '18 at 16:12 • you Can write your function as $$f(t)=4\\,{\\frac {4\\,{t}^{4}-3\\,{t}^{3}+3\\,t+4}{ \\left( {t}^{2}+1 \\right) ^{2 }}}$$ with $$\\tan(\\frac{x}{2})=t$$ and","Que.7. The maximum value for the function sin 2y can be attained at what point in the interval (0,2n). Ans. f(y) = sin 2y ${f}'$(y) = 2cos 2y ${f}'$(y) = 0 cos 2y = 0 2y = $\\frac{\\pi }{2}$, $\\frac{3\\pi }{2}$, $\\frac{5\\pi }{2}$, $\\frac{7\\pi }{2}$ y = $\\frac{\\pi }{4}$, $\\frac{3\\pi }{4}$, $\\frac{5\\pi }{4}$, $\\frac{7\\pi }{4}$ f($\\frac{\\pi }{4}$) = sin $\\frac{\\pi }{2}$ = 1 f($\\frac{3\\pi }{4}$) = sin $\\frac{3\\pi }{2}$ = -1 f($\\frac{5\\pi }{4}$) = sin $\\frac{5\\pi }{2}$ = 1 f($\\frac{7\\pi }{2}$) = sin $\\frac{7\\pi }{2}$ = -1 f(0) = sin 0 = 0, f(2$\\pi$) = sin 2$\\pi$ = 0 Thus, the absolute maximum is occurring at y = $\\frac{\\pi }{4}$ and y = ($\\frac{5\\pi }{4}$). Que.8. In a function sin y + cos y, find the maximum value. f(y) = sin y + cos y Ans. ${f}'$(y) = cos y – sin y ${f}'$(y) = 0 sin y = cos y = – (sin y + cos y) $\\Rightarrow$ tan y = 1 y = $\\frac{\\pi }{4}$, $\\frac{5\\pi }{4}$ ${f}”$(y) = – sin y – cos y = – (sin y + cos y) When (sin y + cos y) is positive, ${f}”$(y) will be negative. ${f}”$(y) is negative when y $\\epsilon$ (0, $\\frac{\\pi }{2}$) Thus, y = $\\frac{\\pi }{4}$ ${f}”$( $\\frac{\\pi }{4}$) = – (sin $\\frac{\\pi }{4}$","49 views If $f(x) = \\dfrac{\\sqrt{3} \\sin x}{2+\\cos x}$, then the range of $f(x)$ is 1. the interval $[-1 , \\sqrt{3}{/2}]$ 2. the interval $[-\\sqrt{3}{/2}, 1]$ 3. the interval $[-1, 1]$ 4. none of these in Calculus recategorized | 49 views Answer: $\\mathbf C$ $f^2(x) =\\dfrac{3-3\\cos^2x}{\\cos^2x+4\\cos x+4}$ Now, Let t = $\\cos x$ then, $f(t) = \\dfrac{3-3t^2}{t^2+4t+4}$ $f_{max}.f'(t) = \\frac{-6(2t+1)}{t+2} = 0$, when$t = \\frac{-1}{2}$ $\\therefore f(-1) =0, f(\\frac{-1}{2}) = 1, \\;and\\; f(1) = 0$ So, $f^2\\text{max} = 1 \\implies f^2(x) \\leq 1 \\implies -1\\leq f(x) \\leq 1$ So, the range of $f$ is $[-1,1]$ $\\therefore \\mathbf C$ is the right option. by Boss (18.9k points) edited by To find the range of $f(x)$, actually we need to find the maximum and minimum value. Now \\begin{align}f(x)&=\\frac{\\sqrt{3}\\sin x}{2+\\cos x}\\\\ \\Rightarrow f'(x)&=\\sqrt{3}\\cos x \\cdot \\frac{1}{2+\\cos x}+\\sqrt{3}\\sin x \\cdot \\frac{-\\sin x}{(2+\\cos x)^2}\\\\ &=\\sqrt{3}\\cdot \\frac{2\\cos x+\\cos^2 x+\\sin^2 x}{(2+\\cos x)^2} \\end{align} At the point of which $f(x)$ is maximum or minimum, the slope is $0$. \\begin{align}\\therefore f'(x)&=0\\\\ \\Rightarrow \\sqrt{3}\\cdot \\frac{2\\cos x+\\cos^2 x+\\sin^2 x}{(2+\\cos x)^2} &=0\\\\ \\Rightarrow 2\\cos x +1&=0;~[\\because \\sin^2 x+\\cos^2 x=1] \\\\ \\Rightarrow \\cos x &= -\\frac{1}{2}\\\\ \\Rightarrow x &= \\frac{2\\pi}{3}, \\frac{4\\pi}{3}\\end{align} Now putting those values of $x$ to $f(x)$, we obtain $f(\\frac{2\\pi}{3})=\\frac{\\sqrt{3}(\\frac{\\sqrt{3}}{2})}{2-\\frac{1}{2}}=1$ and $f(\\frac{4\\pi}{3})=\\frac{\\sqrt{3}(-\\frac{\\sqrt{3}}{2})}{2-\\frac{1}{2}}=-1$. It means $f(x)$ has minimum value $-1$ and maximum $1$. $$\\therefore f(x) \\in [-1,1]$$ So the correct","decrease to the left). Taking a simpler example, try f(x) = 2x. This function approaches 12 as x approaches 6 from the left-hand side (it also approaches 12 as x approaches 6 from the right-hand side, so we simply say that f(x) approaches 12 as x approaches 6). So if you take values for x of 5, 5.5, 5.9, 5.99, 5.999, 5.999999, etc., then your values for f(x) will get closer and closer to 12. lurflurf Homework Helper This is a common limit. in fact it arises in finding the derivative of sine $$\\frac{d}{dx}\\sin(x)=\\lim_{h\\rightarrow 0} \\frac{\\sin(x)}{x}\\cos(x+\\frac{h}{2})$$ the limit can be shown several ways including showing that cos(x) 0$ for all $x \\setminus \\in \\mathbb{R}$, implying that $f \\left(x\\right)$ is strictly increasing over $\\mathbb{R}$, and also over $\\left[- 1 , 2\\right]$. Therefore, the minimum value of $f$ over $\\left[- 1 , 2\\right]$ is $f \\left(- 1\\right) = 2 \\sinh \\left(- 1\\right) = {e}^{- 1} - {e}^{1} = \\frac{1 - {e}^{2}}{e} \\approx - 2.3504$ and the maximum value of $f$ over $\\left[- 1 , 2\\right]$ is $f \\left(2\\right) = 2 \\sinh \\left(2\\right) = {e}^{2} - {e}^{- 2} = \\frac{{e}^{4} - 1}{{e}^{2}} \\approx 7.2537$. This implies that the range is the closed interval $= \\left[2 \\sinh \\left(- 1\\right) , 2 \\sinh \\left(2\\right)\\right] = \\left[\\frac{1 - {e}^{2}}{e} , \\frac{{e}^{4} - 1}{{e}^{2}}\\right] \\approx \\left[- 2.3504 , 7.2537\\right]$.","# Range of $(\\sin x)^6+ (\\cos x)^6$ I need to find the range of the function $y = (\\sin x)^6+ (\\cos x)^6$ I did find the answer but working in a crude way rather than a methodical step by step approach. I give below the steps I used , please help with a methodical approach to such problems. 1) To find the max value of the function I noticed that in the range where x is $[0,2\\pi]$ , when $\\cos x$ hits $+1$ then $\\sin x$ is $0$ , when $\\cos x$ is $0$ then $\\sin x$ is $+1$ ..etc so the max value at any of these points could be either $1^6+0^6$ or $(-1)^6+0^6$ so the max value is 1. to find the minimum I differentiated the function $f' (x) = 6(\\sin x)^5\\cos x- 6(\\cos x^5)\\sin x =0$ , equating this to zero we have $(\\sin x)^4 = (\\cos x)^4 => x = \\pi/4 =>$ min value of function is $(\\sin(\\pi/4))^6 + (\\cos(\\pi/4))^6 = 1/4$ . so the range is $(\\frac{1}{4},1)$. Please can someone help with how can this type of problems be methodically approached ? - Thanks. • Well, you can do some basic trig manipulation to view it as $\\frac18(3\\cos(4x)+5)$. This gives you a maximum of $\\frac18(3+5)=1$ and a minimum of $\\frac18(-3+5)=\\frac14$. So the range","\\le x < \\frac{3 + \\sqrt{621}}{2}$. Now, we find the length of the working intervals and divide it by the length of the total interval, $15 - 5 = 10$: \\begin{align*} \\frac{\\left( \\frac{3 + \\sqrt{61}}{2} - 5 \\right) + \\left( \\frac{3 + \\sqrt{109}}{2} - 6 \\right) + \\left( \\frac{3 + \\sqrt{621}}{2} - 13 \\right)}{10} \\\\ &= \\frac{\\sqrt{61} + \\sqrt{109} + \\sqrt{621} - 39}{20} \\end{align*} Thus, the answer is $61 + 109 + 621 + 39 + 20 = \\fbox{850}$.","# Examples on Domains and Ranges of Functions Set 7 Go back to 'Functions' Example- 23 Find the range for (a) f(x) = \\begin{align}\\frac{1}{{2 + \\sin 3x + \\cos 3x}}\\end{align} (b) $$f(x) = [{x^2}] - {[x]^2}$$ (c) $$f(x) = \\sqrt {a - x} \\,\\,\\, + \\,\\,\\,\\sqrt {x - b}$$ $$\\quad a > b > 0$$ (d) $$f(x) = {x^3} + 3{x^2} + 4x + 5$$ Solution: In case of linear, quadratic and other simple functions, we can express x in terms of f(x) and find the values of f(x) for which x is defined (These values form the range) We did so in the last two questions. However, this expression is not always easily possible. So we have to find other ways that could yield the answer more easily: (a) f (x) = \\begin{align} \\frac{1}{{2 + \\sin 3x + \\cos 3x}} = \\frac{1}{{2 + \\sqrt 2 \\sin (3x + \\pi /4)}}\\end{align} The denominator can vary from $$2 - \\sqrt 2$$ to $$2 + \\sqrt 2$$ because $$-{\\rm{ }}1 \\le sin\\theta \\le 1$$ (The denominator is never 0 and hence D = $$\\mathbb{R}$$ ) Therefore, $\\frac{1}{{2 + \\sqrt 2 }} \\le f(x){\\rm{ }} \\le \\frac{1}{{2 - \\sqrt 2 }}$ (b) Let x be expressed as I + f where I is the integral part and f the fractional part","# Thread: Maximum Value of a function on a closed interval 1. ## Maximum Value of a function on a closed interval On the closed interval [0,2pi], the maximum value of the function f(x)=4sinx-3cosx is: A.)3 B.)4 C.)24/5 D.)5 E.)None of these Explanations appreciated. 2. Originally Posted by Manabu On the closed interval [0,2pi], the maximum value of the function f(x)=4sinx-3cosx is: A.)3 B.)4 C.)24/5 D.)5 E.)None of these Explanations appreciated. $f(x)=4\\sin x-3\\cos x$ $f'(x)=4\\cos x+3\\sin x$ for max let $f'(x)=0$ $0=4\\cos x+3\\sin x$ $4\\cos x=-3\\sin x$ $\\frac{-4}{3}=\\frac{\\sin x}{\\cos x}$ $\\tan x =\\frac{-4}{3}$ Can you finish this? 3. Originally Posted by Manabu On the closed interval [0,2pi], the maximum value of the function f(x)=4sinx-3cosx is: A.)3 B.)4 C.)24/5 D.)5 E.)None of these Explanations appreciated. A couple of suggestions: Have you learnt to write that function in the form sin (a+b) or cos(a+b)? If so, that'd be the easiest way. 2. Calculus?? 4. Originally Posted by pickslides $f(x)=4\\sin x-3\\cos x$ $f'(x)=4\\cos x+3\\sin x$ for max let $f'(x)=0$ $0=4\\cos x+3\\sin x$ $4\\cos x=-3\\sin x$ $\\frac{-4}{3}=\\frac{\\sin x}{\\cos x}$ $\\tan x =\\frac{-4}{3}$ Can you finish this? Not really. I understand how you got to that point, but I'm not 100% sure how to calculate maximum values. 5. Originally Posted by Manabu On the closed interval [0,2pi], the maximum value of the function","\\times 28}}{5} + 32 = \\frac{{252 + 160}}{5} = \\frac{{412}}{5} = 82.4 \\end{align} (iii) \\begin{align}t\\left( { - 10} \\right) = \\frac{{9 \\times ( - 10)}}{5} + 32 = 9 \\times ( - 2) + 32 = - 18 + 32 = 14 \\end{align} (iv) It is given that \\begin{align}\\;\\;\\;\\;t&\\left( C \\right) = 212\\\\ \\Rightarrow \\;&\\frac{{9C}}{5} + 32 = 212\\\\ \\Rightarrow \\;&\\frac{{9C}}{5} = 212 - 32\\\\ \\Rightarrow \\;&C = \\frac{{180 \\times 5}}{9}\\\\&\\quad= 100\\end{align} Thus, the value of $$'t'$$, when $$t\\left( C \\right) = 212$$ is $$100.$$ ## Chapter 2 Ex.2.3 Question 5 Find the range of each of the following functions. (i) $$f\\left( x \\right)~=2-3x,\\,x\\in R,x>0.$$ (ii) $$f\\left( x \\right)~={{x}^{2}}+2,\\,x$$ is a real number. (iii) $$f\\left( x \\right)~=x,\\,~x$$ is a real number. ### Solution (i) $$f\\left( x \\right)~=2-3x,x\\in \\mathbf{R},x>0.$$ The values of $$f\\left( x \\right)$$ for various values of real numbers $$x > 0$$ can be written in the tabular form as $$x$$ $$0.01$$ $$0.1$$ $$0.9$$ $$1$$ $$2$$ $$2.5$$ $$4$$ $$5$$ $$\\ldots$$ $$f\\left( x \\right)$$ $$1.97$$ $$1.7$$ $$-0.7$$ $$-1$$ $$-4$$ $$-5.5$$ $$-10$$ $$-13$$ $$\\ldots$$ Thus, it can be clearly observed that the range of $$f$$ is the set of all real numbers less than $$2.$$ i.e., range of $$f = \\left( {-\\infty ,2} \\right)$$ Alternative Method: Let $$x > 0$$ $\\begin{array}{l} \\Rightarrow 3x > 0\\\\ \\Rightarrow 2 - 3x","# Find the range of function $f(x) =\\cos\\sin\\ln\\left(\\frac{x^2+e}{x^2+1} \\right)+\\sin\\cos\\ln\\left(\\frac{x^2+e}{x^2+1} \\right)$ Problem : Find the range of function $$f\\left( x \\right) =\\cos \\left( \\sin \\left( \\ln \\left( \\frac{x^2+e}{x^2+1} \\right) \\right) \\right) +\\sin \\left( \\cos \\left( \\ln \\left( \\frac{x^2+e}{x^2+1} \\right) \\right) \\right)$$ My approach : maximum value of the function is when denominator term is minimum i.e. $$x^2+1$$ is minimum. It is minimum when $$x^2 =0$$ therefore, maximum value of function $$\\cos(\\sin(\\ln e))+\\sin(\\cos(\\ln e))$$ $$\\cos(\\sin(1))+\\sin(\\cos(1))$$ now how to find the minimum value of the function please suggest . Thanks... • \"maximum value of the function is when denominator term is minimum \" of which function? Oct 24, 2014 at 18:01 • Maybe you are just supposed to graph it? Oct 24, 2014 at 18:12 Well, first you need to see the range of $\\frac{x^2+e}{x^2+1}$, which you can easily verify to be $(1,e]$. Then, the range of $\\log$ in this domain is $(0,1]$. As pointed out in a comment: Since $\\sin$ is increasing and $\\cos$ decreasing in $(0,1]$, then $\\sin(0,1] =(0,\\sin(1)]$ and $\\cos(0,1] =[\\cos(1),1)$. Finally, since in $(0,\\sin(1)]$ the cosine is decreasing and in $[\\cos(1),1)$ the sine is increasing, we get that the range of each part of the sum is $[\\cos(\\sin(1)),1)$ and $[\\sin(\\cos(1)),1)$. So the range of the complete function is $$[\\cos(\\sin(1))+\\sin(\\cos(1)), 2)$$ Another way could be using the fact","# What is the range of f(x) = ln(sin^(-1)(x^2+x+3/4)) ? Jun 15, 2017 $\\left[\\ln \\left(\\frac{\\pi}{6}\\right) , \\ln \\left(\\frac{\\pi}{2}\\right)\\right] \\approx \\left[- 0.6470 , 0.4516\\right]$ #### Explanation: Given: $f \\left(x\\right) = \\ln \\left({\\sin}^{- 1} \\left({x}^{2} + x + \\frac{3}{4}\\right)\\right)$ First note that: ${x}^{2} + x + \\frac{3}{4} = {x}^{2} + x + \\frac{1}{4} + \\frac{1}{2} = {\\left(x + \\frac{1}{2}\\right)}^{2} + \\frac{1}{2}$ which can take any value in the range $\\left[\\frac{1}{2} , \\infty\\right)$ The domain of ${\\sin}^{- 1}$ as a real valued function of real arguments is $\\left[- 1 , 1\\right]$. So the possible valid arguments to it in $f \\left(x\\right)$ are all in: $\\left[\\frac{1}{2} , \\infty\\right) \\cap \\left[- 1 , 1\\right] = \\left[\\frac{1}{2} , 1\\right]$ ${\\sin}^{- 1} \\left(\\frac{1}{2}\\right) = \\frac{\\pi}{6}$ ${\\sin}^{- 1} \\left(1\\right) = \\frac{\\pi}{2}$ and $\\sin$ is monotonically increasing between these two endpoints. So the range of ${\\sin}^{- 1} \\left({x}^{2} + x + \\frac{3}{4}\\right)$ is $\\left[\\frac{\\pi}{6} , \\frac{\\pi}{2}\\right]$ Hence the range of $f \\left(x\\right)$ is: $\\left[\\ln \\left(\\frac{\\pi}{6}\\right) , \\ln \\left(\\frac{\\pi}{2}\\right)\\right]$","+ 3$, and $n = N$, $$|S_m{(x)} - S_n{(x)}| \\geq \\frac{2\\sqrt{2}}5$$ And then ultimately show the original sequence doesn't uniformly converge on $\\mathbb{R}$ Part C I think I'm on the right track and some tips should suffice, but for part d I'm relatively lost. Any help's greatly appreciated - thank you Part 2 Let $N + 1 \\le k \\le 4N + 3$. Then $\\frac{N+1}{k}$ satisfies the inequalities $\\frac{N+1}{4N+3}\\le\\frac{N+1}{k}\\le1$. Now, for $x$ in the interval $\\left(\\frac{N+1}{4N+3},1\\right)$, $\\sin x$ is monotonically increasing and we have $$1\\ge \\sin(1)\\ge \\sin\\left(\\frac{N+1}{k}\\right)\\ge\\sin\\left(\\frac{N+1}{4N+3}\\right)$$ and thus \\begin{align} \\sin\\left(\\frac{N+1}{k}\\right)&\\ge\\sin\\left(\\frac{N+1}{4N+3}\\right)\\\\\\\\ &\\ge\\sin(1/4)\\\\\\\\ &\\ge \\frac14-\\frac{1}{3!}(\\frac14)^3\\\\\\\\ &=\\frac14\\left(1-\\frac{1}{96}\\right)\\\\\\\\ &>\\frac15 \\end{align} which was to be shown. Part 3 For this part, we will use the result from the following integral \\begin{align} \\int_{N+1}^{4N+4}\\frac{dx}{\\sqrt{x}}&=2\\left(\\sqrt{4N+4}-\\sqrt{N+1}\\right)\\\\ &=2\\sqrt{N+1}(2-1)\\\\ &\\ge 2\\sqrt{2} \\end{align} for $N\\ge1$. Now, we note that this integral can be represented by the following summation: \\begin{align} \\int_{N+1}^{4N+4}\\frac{dx}{\\sqrt{x}}&=\\sum_{k=N+1}^{4N+3}\\int_{k}^{k+1} \\frac{dx}{\\sqrt{x}}\\\\ &\\le\\sum_{k=N+1}^{4N+3} \\frac{1}{\\sqrt{k}} &\\ge 2\\sqrt{2} \\end{align} and we have the desired inequality! Part 4 From the Part 2, we showed that if $N + 1 \\le k \\le 4N + 3$, then $$1 \\ge \\sin\\left(\\frac{N+1}{k}\\right) \\ge \\frac15$$ From the Part 3, we showed that for all $N$ $$\\sum_{k=N+1}^{4N+3}{\\frac{1}{\\sqrt{k}}} \\ge 2\\sqrt{2}$$ Then, putting these together we have \\begin{align} |S_m(x)-S_n(x)|&=\\left|\\sum_{k=N+1}^{4N+3} \\frac{\\sin((N+1)/k)}{\\sqrt{k}}\\right|\\\\ &\\ge\\left|\\sum_{k=N+1}^{4N+3} \\frac{\\frac15}{\\sqrt{k}}\\right|\\\\\\\\ &\\ge\\frac15\\left|\\sum_{k=N+1}^{4N+3} \\frac{1}{\\sqrt{k}}\\right|\\\\\\\\ &\\ge\\frac{2\\sqrt{2}}{5} \\end{align}","= \\frac 7 2$$. This becomes our next interval. Step 6 Find the third approximation and its associated error. The midpoint of the interval $$\\left[3, \\frac 7 2\\right]$$ is at $$x = \\frac{13} 4$$, as shown on the graph below. This is the next approximation. Max Error: The interval has a length of $$1/2$$, so the maximum possible error is $$\\pm1/4$$ of a unit. The table below summarizes the approximations we found and their associated errors. $$\\begin{array}{ccc} \\mbox{Approximation} & x\\mbox{-value} & \\mbox{Possible Error}\\\\ \\hline 1^{st} & x = 3 & \\pm1\\\\[6pt] 2^{nd} & x = \\frac 7 2 & \\pm\\frac 1 2\\\\[6pt] 3^{rd} & x = \\frac{13} 4 & \\pm\\frac 1 4 \\end{array}$$ ##### Example 2 Use the bisection method to approximate the solution to the equation below to within less than 0.1 of its real value. Assume $$x$$ is in radians. $$\\sin x = 6 - x$$ Step 1 Rewrite the equation so it is equal to 0. $$x - 6 + \\sin x = 0$$ The function we'll work with is $$f(x) = x - 6 + \\sin x$$. Notice that the function is continuous everywhere. Step 2 Find an initial interval to work with. Setting up a table of values, we see the following. $$\\begin{array}{cl} x & {f(x)}\\\\ \\hline 0 & f(0) = -6\\\\ 1","really desperate i need to this for tommorow and I don´t know how to do this a really need some help - as I said I have already asked the question in I answered that i would tried to do the problem but I can´t do it... – David Hernandez Nov 30 '13 at 5:13 Why not use commas and periods to separate sentences and paragraghs? – awllower Nov 30 '13 at 5:15 I am sorry :) I am not very good at writing in mathjax – David Hernandez Nov 30 '13 at 5:24 Helpful identity: note that $$1 - \\cos(x) = 2\\frac{1 - \\cos(2\\frac{x}{2})}{2} = 2 \\sin^2\\left(\\frac{x}{2}\\right)$$ so that our function is simply $$f(x)= \\begin{cases} 2 {\\sin^2(\\frac x2)\\over x^2}, & \\text{if x\\neq 0} \\\\ 1/2, & \\text{if x= 0} \\\\ \\end{cases}$$ With that in mind, we can find (assuming $x,y \\neq 0$) \\begin{align} \\left|f(x)-f(y)\\right| &= 2\\left| \\left(\\frac{\\sin(\\frac y2)}{y}\\right)^2 - \\left(\\frac{\\sin(\\frac x2)}{x}\\right)^2 \\right|\\\\ &=2\\left| \\left(\\frac{\\sin(\\frac y2)}{y}\\right) + \\left(\\frac{\\sin(\\frac x2)}{x}\\right) \\right| \\left| \\left(\\frac{\\sin(\\frac y2)}{y}\\right) - \\left(\\frac{\\sin(\\frac x2)}{x}\\right) \\right| \\end{align} Now, we just have to work with $\\left| \\left(\\frac{\\sin(\\frac y2)}{y}\\right) - \\left(\\frac{\\sin(\\frac x2)}{x}\\right) \\right|$ to get out a $|x-y|$. thank you!! and we also have that ${|xsin(y)-ysin(x)|\\over |xy|}\\le{|x-y|\\over |xy|}$ is this correct? – David Hernandez Nov 30 '13 at 5:50 I think there is a typo. $\\frac{2\\sin^2(x)}{x^2}$ should be $\\frac{2\\sin^2(x/2)}{x^2}$","Share Books Shortlist # Solution for If the Function F(X) = Cos |X| − 2ax + B Increases Along the Entire Number Scale, Then (A) a = B (B) a = 1 2 B (C) a ≤ − 1 2 (D) a > − 3 2 - CBSE (Science) Class 12 - Mathematics ConceptIncreasing and Decreasing Functions #### Question If the function f(x) = cos |x| − 2ax + b increases along the entire number scale, then (a) a = b (b) $a = \\frac{1}{2}b$ (c) $a \\leq - \\frac{1}{2}$ (d) $a > - \\frac{3}{2}$ #### Solution (c) $a \\leq - \\frac{1}{2}$ $Given: f\\left( x \\right) = \\cos \\left| x \\right| - 2ax + b$ $\\text { Now}, \\left| x \\right| =\\begin{cases} x ,& x \\geq 0 \\\\ - x, & x < 0 \\end{cases}$ $\\text { and } \\cos \\left| x \\right| = \\begin{cases} \\cos\\left( x \\right) , & x \\geq 0 \\\\cos\\left( - x \\right) = cos\\left( x \\right), & x < 0\\end{cases}$ $\\therefore \\cos\\left| x \\right| = \\cos x , \\forall x \\in R$ $\\therefore f\\left( x \\right) = \\cos x - 2ax + b$ $\\Rightarrow f'\\left( x \\right) = - \\sin x - 2a$ $\\text { It is given that f(x) is increasing } .$ $\\Rightarrow f'\\left( x \\right) \\geq 0$ $\\Rightarrow - \\sin","$$0 < c < \\frac{\\pi}{2}$$ and $$g(c) = 0$$. $$h(\\theta) = 1 + \\theta - 3\\tan{\\theta}$$ Here, $$h(0) = 1$$ and $$h\\left(\\frac{\\pi}{4}\\right) \\approx -\\frac{5}{4}$$. This means that there is a $$c$$ in $$\\left[0, \\frac{\\pi}{4}\\right]$$ where $$h(c) = 0$$. #### 3.2.3 Usage: Finding a value inside a set interval Given a function $$f$$, a closed interval $$[a, b]$$, and a $$k$$, prove that there is a $$c$$ such that $$f(c) = k$$ and $$a < c < b$$. $$f(x) = x^{2} + x - 1$$, $$[0, 5]$$, $$f(c) = 11$$ Substituting $$c$$ and setting equal to $$11$$: \\begin{align} & c^{2} + c - 1 &= 11 & \\\\ & c^{2} + c - 12 &= 0 & \\\\ & c &= \\frac{-1 \\pm \\sqrt{1^{2} - 4(1)(-12)}}{2} & \\\\ & c &= \\frac{-1 \\pm \\sqrt{49}}{2} & \\\\ & c &= \\frac{-1 \\pm 7}{2} & \\\\ c &= \\frac{-8}{2} & c &= \\frac{6}{2} \\\\ c &= -4 & c &= 3 \\end{align} $$-4$$ is not in the domain, so the solution is $$c = 3$$. $$f(x) = x^{2} - 6x + 8$$, $$[0, 3]$$, $$f(c) = 0$$ Again, substitute $$c$$ and set equal to $$0$$. \\begin{align} & c^{2} - 6c + 8 &= 0 & \\\\ & c &= \\frac{6 \\pm \\sqrt{(-6)^{2} - 4(1)(8)}}{2} & \\\\ & c &= \\frac{6 \\pm","\\begin{align} \\text{Range}(g) &= \\left[ \\frac{-1 - \\sin \\left( \\frac{1}{2} \\right)}{2 \\sin \\left( \\frac{1}{2} \\right)},\\frac{1 - \\sin \\left( \\frac{1}{2} \\right)}{2 \\sin \\left( \\frac{1}{2} \\right)} \\right] \\\\ &= [-1.54291482 \\ldots,0.54291482 \\ldots] \\\\ &\\subseteq [-2,2]. \\end{align} Conclusion: $\\displaystyle \\left| \\sum_{k=1}^{n} \\sin(k) \\right| < 2$ and $\\displaystyle \\left| \\sum_{k=1}^{n} \\cos(k) \\right| < 2$ for all $n \\in \\mathbb{N}$. - You have the formula $\\frac{1-x^{n+1}}{1-x}=\\sum_{k=1}^{n}x^k$, which is easy enough to see by algebra. So you can replace the LHS with $\\frac{1-e^{i(n+1)\\theta}}{1-e^{i\\theta}}$. – minimalrho Feb 1 '13 at 22:03 +1 Very nice hint. This is not a series, @Iuli, but a finite sum and, in fact, a rather easy one: a geometric sequence sum. Above you can read what you get from it, which with absolute values gives at once the wanted upper bound $\\,2\\,$ – DonAntonio Feb 1 '13 at 22:15 @CalvinLin , yes...so what? The absolute value of the expression hinted in the answer above is less than or equal what you wrote (in fact, it is always less). Besides this, in the given formula you must take the imaginary part as that the part that corresponds to the sum of sines... – DonAntonio Feb 1 '13 at 23:05 @HaskellCurry Yes, with the trig version I could see how to do it. The exp version was less obvious, since you still had"],"string":"[\n \"# Find the range of $f(x)=11\\\\cos^2x+3\\\\sin^2x+6\\\\sin x\\\\cos x+5$ I'm trying to solve this problem. Find the range of $f(x)=11\\\\cos^2x+3\\\\sin^2x+6\\\\sin x\\\\cos x+5$ I have simplified this problem to $$f(x)= 8\\\\cos^2x+6\\\\sin x\\\\cos x+8$$ and tried working with $g(x)= 8\\\\cos^2x+6\\\\sin x\\\\cos x$. I factored out the $2\\\\cos x$ and rewrote the other factor as a linear combination of cosine. It reduces down to $$g(x)=10\\\\cos x\\\\cos\\\\left(x-\\\\tan^{-1}\\\\frac{3}{4}\\\\right)$$ • Have you tried finding the maximum and minimum valueof $f$? Jan 6 '18 at 15:44 Rewrite as following \\\\begin{align} f(x) & = 8 + 6 \\\\sin(x ) \\\\cos(x ) + 8\\\\cos^2(x) - 4 + 4 \\\\\\\\ &= 12 + 3 \\\\sin(2x) + 4 \\\\cos(2x) \\\\\\\\ &= 12 + 5 \\\\left(\\\\frac{3}{5} \\\\sin(2x) + \\\\frac{4}{5} \\\\cos(2x)\\\\right) \\\\\\\\ &= 12 + 5 \\\\sin(2x + \\\\arctan{\\\\tfrac{4}{3}}) \\\\end{align} Now its easy since $\\\\sin(...)$ always lies in $[-1,1]$, max/min values are $12 \\\\pm 5$. So maximum value is $17$ and minimum is $7$ write your function as $$f(x)=(\\\\sin(x)+3\\\\cos(x))^2+7$$ • This is good for minimum, but then for maximum? But answer is good! +1! Jan 6 '18 at 15:59 • You should write $3\\\\cos x + \\\\sin x$ in the form $R\\\\cos(x-\\\\alpha)$, where $R>0$ and $0 < \\\\alpha < \\\\frac{1}{2}\\\\pi$. Jan 6 '18 at 16:12 • you Can write your function as $$f(t)=4\\\\,{\\\\frac {4\\\\,{t}^{4}-3\\\\,{t}^{3}+3\\\\,t+4}{ \\\\left( {t}^{2}+1 \\\\right) ^{2 }}}$$ with $$\\\\tan(\\\\frac{x}{2})=t$$ and\",\n \"Que.7. The maximum value for the function sin 2y can be attained at what point in the interval (0,2n). Ans. f(y) = sin 2y ${f}'$(y) = 2cos 2y ${f}'$(y) = 0 cos 2y = 0 2y = $\\\\frac{\\\\pi }{2}$, $\\\\frac{3\\\\pi }{2}$, $\\\\frac{5\\\\pi }{2}$, $\\\\frac{7\\\\pi }{2}$ y = $\\\\frac{\\\\pi }{4}$, $\\\\frac{3\\\\pi }{4}$, $\\\\frac{5\\\\pi }{4}$, $\\\\frac{7\\\\pi }{4}$ f($\\\\frac{\\\\pi }{4}$) = sin $\\\\frac{\\\\pi }{2}$ = 1 f($\\\\frac{3\\\\pi }{4}$) = sin $\\\\frac{3\\\\pi }{2}$ = -1 f($\\\\frac{5\\\\pi }{4}$) = sin $\\\\frac{5\\\\pi }{2}$ = 1 f($\\\\frac{7\\\\pi }{2}$) = sin $\\\\frac{7\\\\pi }{2}$ = -1 f(0) = sin 0 = 0, f(2$\\\\pi$) = sin 2$\\\\pi$ = 0 Thus, the absolute maximum is occurring at y = $\\\\frac{\\\\pi }{4}$ and y = ($\\\\frac{5\\\\pi }{4}$). Que.8. In a function sin y + cos y, find the maximum value. f(y) = sin y + cos y Ans. ${f}'$(y) = cos y – sin y ${f}'$(y) = 0 sin y = cos y = – (sin y + cos y) $\\\\Rightarrow$ tan y = 1 y = $\\\\frac{\\\\pi }{4}$, $\\\\frac{5\\\\pi }{4}$ ${f}”$(y) = – sin y – cos y = – (sin y + cos y) When (sin y + cos y) is positive, ${f}”$(y) will be negative. ${f}”$(y) is negative when y $\\\\epsilon$ (0, $\\\\frac{\\\\pi }{2}$) Thus, y = $\\\\frac{\\\\pi }{4}$ ${f}”$( $\\\\frac{\\\\pi }{4}$) = – (sin $\\\\frac{\\\\pi }{4}$\",\n \"49 views If $f(x) = \\\\dfrac{\\\\sqrt{3} \\\\sin x}{2+\\\\cos x}$, then the range of $f(x)$ is 1. the interval $[-1 , \\\\sqrt{3}{/2}]$ 2. the interval $[-\\\\sqrt{3}{/2}, 1]$ 3. the interval $[-1, 1]$ 4. none of these in Calculus recategorized | 49 views Answer: $\\\\mathbf C$ $f^2(x) =\\\\dfrac{3-3\\\\cos^2x}{\\\\cos^2x+4\\\\cos x+4}$ Now, Let t = $\\\\cos x$ then, $f(t) = \\\\dfrac{3-3t^2}{t^2+4t+4}$ $f_{max}.f'(t) = \\\\frac{-6(2t+1)}{t+2} = 0$, when$t = \\\\frac{-1}{2}$ $\\\\therefore f(-1) =0, f(\\\\frac{-1}{2}) = 1, \\\\;and\\\\; f(1) = 0$ So, $f^2\\\\text{max} = 1 \\\\implies f^2(x) \\\\leq 1 \\\\implies -1\\\\leq f(x) \\\\leq 1$ So, the range of $f$ is $[-1,1]$ $\\\\therefore \\\\mathbf C$ is the right option. by Boss (18.9k points) edited by To find the range of $f(x)$, actually we need to find the maximum and minimum value. Now \\\\begin{align}f(x)&=\\\\frac{\\\\sqrt{3}\\\\sin x}{2+\\\\cos x}\\\\\\\\ \\\\Rightarrow f'(x)&=\\\\sqrt{3}\\\\cos x \\\\cdot \\\\frac{1}{2+\\\\cos x}+\\\\sqrt{3}\\\\sin x \\\\cdot \\\\frac{-\\\\sin x}{(2+\\\\cos x)^2}\\\\\\\\ &=\\\\sqrt{3}\\\\cdot \\\\frac{2\\\\cos x+\\\\cos^2 x+\\\\sin^2 x}{(2+\\\\cos x)^2} \\\\end{align} At the point of which $f(x)$ is maximum or minimum, the slope is $0$. \\\\begin{align}\\\\therefore f'(x)&=0\\\\\\\\ \\\\Rightarrow \\\\sqrt{3}\\\\cdot \\\\frac{2\\\\cos x+\\\\cos^2 x+\\\\sin^2 x}{(2+\\\\cos x)^2} &=0\\\\\\\\ \\\\Rightarrow 2\\\\cos x +1&=0;~[\\\\because \\\\sin^2 x+\\\\cos^2 x=1] \\\\\\\\ \\\\Rightarrow \\\\cos x &= -\\\\frac{1}{2}\\\\\\\\ \\\\Rightarrow x &= \\\\frac{2\\\\pi}{3}, \\\\frac{4\\\\pi}{3}\\\\end{align} Now putting those values of $x$ to $f(x)$, we obtain $f(\\\\frac{2\\\\pi}{3})=\\\\frac{\\\\sqrt{3}(\\\\frac{\\\\sqrt{3}}{2})}{2-\\\\frac{1}{2}}=1$ and $f(\\\\frac{4\\\\pi}{3})=\\\\frac{\\\\sqrt{3}(-\\\\frac{\\\\sqrt{3}}{2})}{2-\\\\frac{1}{2}}=-1$. It means $f(x)$ has minimum value $-1$ and maximum $1$. $$\\\\therefore f(x) \\\\in [-1,1]$$ So the correct\",\n \"decrease to the left). Taking a simpler example, try f(x) = 2x. This function approaches 12 as x approaches 6 from the left-hand side (it also approaches 12 as x approaches 6 from the right-hand side, so we simply say that f(x) approaches 12 as x approaches 6). So if you take values for x of 5, 5.5, 5.9, 5.99, 5.999, 5.999999, etc., then your values for f(x) will get closer and closer to 12. lurflurf Homework Helper This is a common limit. in fact it arises in finding the derivative of sine $$\\\\frac{d}{dx}\\\\sin(x)=\\\\lim_{h\\\\rightarrow 0} \\\\frac{\\\\sin(x)}{x}\\\\cos(x+\\\\frac{h}{2})$$ the limit can be shown several ways including showing that cos(x) 0$ for all $x \\\\setminus \\\\in \\\\mathbb{R}$, implying that $f \\\\left(x\\\\right)$ is strictly increasing over $\\\\mathbb{R}$, and also over $\\\\left[- 1 , 2\\\\right]$. Therefore, the minimum value of $f$ over $\\\\left[- 1 , 2\\\\right]$ is $f \\\\left(- 1\\\\right) = 2 \\\\sinh \\\\left(- 1\\\\right) = {e}^{- 1} - {e}^{1} = \\\\frac{1 - {e}^{2}}{e} \\\\approx - 2.3504$ and the maximum value of $f$ over $\\\\left[- 1 , 2\\\\right]$ is $f \\\\left(2\\\\right) = 2 \\\\sinh \\\\left(2\\\\right) = {e}^{2} - {e}^{- 2} = \\\\frac{{e}^{4} - 1}{{e}^{2}} \\\\approx 7.2537$. This implies that the range is the closed interval $= \\\\left[2 \\\\sinh \\\\left(- 1\\\\right) , 2 \\\\sinh \\\\left(2\\\\right)\\\\right] = \\\\left[\\\\frac{1 - {e}^{2}}{e} , \\\\frac{{e}^{4} - 1}{{e}^{2}}\\\\right] \\\\approx \\\\left[- 2.3504 , 7.2537\\\\right]$.\",\n \"# Range of $(\\\\sin x)^6+ (\\\\cos x)^6$ I need to find the range of the function $y = (\\\\sin x)^6+ (\\\\cos x)^6$ I did find the answer but working in a crude way rather than a methodical step by step approach. I give below the steps I used , please help with a methodical approach to such problems. 1) To find the max value of the function I noticed that in the range where x is $[0,2\\\\pi]$ , when $\\\\cos x$ hits $+1$ then $\\\\sin x$ is $0$ , when $\\\\cos x$ is $0$ then $\\\\sin x$ is $+1$ ..etc so the max value at any of these points could be either $1^6+0^6$ or $(-1)^6+0^6$ so the max value is 1. to find the minimum I differentiated the function $f' (x) = 6(\\\\sin x)^5\\\\cos x- 6(\\\\cos x^5)\\\\sin x =0$ , equating this to zero we have $(\\\\sin x)^4 = (\\\\cos x)^4 => x = \\\\pi/4 =>$ min value of function is $(\\\\sin(\\\\pi/4))^6 + (\\\\cos(\\\\pi/4))^6 = 1/4$ . so the range is $(\\\\frac{1}{4},1)$. Please can someone help with how can this type of problems be methodically approached ? - Thanks. • Well, you can do some basic trig manipulation to view it as $\\\\frac18(3\\\\cos(4x)+5)$. This gives you a maximum of $\\\\frac18(3+5)=1$ and a minimum of $\\\\frac18(-3+5)=\\\\frac14$. So the range\",\n \"\\\\le x < \\\\frac{3 + \\\\sqrt{621}}{2}$. Now, we find the length of the working intervals and divide it by the length of the total interval, $15 - 5 = 10$: \\\\begin{align*} \\\\frac{\\\\left( \\\\frac{3 + \\\\sqrt{61}}{2} - 5 \\\\right) + \\\\left( \\\\frac{3 + \\\\sqrt{109}}{2} - 6 \\\\right) + \\\\left( \\\\frac{3 + \\\\sqrt{621}}{2} - 13 \\\\right)}{10} \\\\\\\\ &= \\\\frac{\\\\sqrt{61} + \\\\sqrt{109} + \\\\sqrt{621} - 39}{20} \\\\end{align*} Thus, the answer is $61 + 109 + 621 + 39 + 20 = \\\\fbox{850}$.\",\n \"# Examples on Domains and Ranges of Functions Set 7 Go back to 'Functions' Example- 23 Find the range for (a) f(x) = \\\\begin{align}\\\\frac{1}{{2 + \\\\sin 3x + \\\\cos 3x}}\\\\end{align} (b) $$f(x) = [{x^2}] - {[x]^2}$$ (c) $$f(x) = \\\\sqrt {a - x} \\\\,\\\\,\\\\, + \\\\,\\\\,\\\\,\\\\sqrt {x - b}$$ $$\\\\quad a > b > 0$$ (d) $$f(x) = {x^3} + 3{x^2} + 4x + 5$$ Solution: In case of linear, quadratic and other simple functions, we can express x in terms of f(x) and find the values of f(x) for which x is defined (These values form the range) We did so in the last two questions. However, this expression is not always easily possible. So we have to find other ways that could yield the answer more easily: (a) f (x) = \\\\begin{align} \\\\frac{1}{{2 + \\\\sin 3x + \\\\cos 3x}} = \\\\frac{1}{{2 + \\\\sqrt 2 \\\\sin (3x + \\\\pi /4)}}\\\\end{align} The denominator can vary from $$2 - \\\\sqrt 2$$ to $$2 + \\\\sqrt 2$$ because $$-{\\\\rm{ }}1 \\\\le sin\\\\theta \\\\le 1$$ (The denominator is never 0 and hence D = $$\\\\mathbb{R}$$ ) Therefore, $\\\\frac{1}{{2 + \\\\sqrt 2 }} \\\\le f(x){\\\\rm{ }} \\\\le \\\\frac{1}{{2 - \\\\sqrt 2 }}$ (b) Let x be expressed as I + f where I is the integral part and f the fractional part\",\n \"# Thread: Maximum Value of a function on a closed interval 1. ## Maximum Value of a function on a closed interval On the closed interval [0,2pi], the maximum value of the function f(x)=4sinx-3cosx is: A.)3 B.)4 C.)24/5 D.)5 E.)None of these Explanations appreciated. 2. Originally Posted by Manabu On the closed interval [0,2pi], the maximum value of the function f(x)=4sinx-3cosx is: A.)3 B.)4 C.)24/5 D.)5 E.)None of these Explanations appreciated. $f(x)=4\\\\sin x-3\\\\cos x$ $f'(x)=4\\\\cos x+3\\\\sin x$ for max let $f'(x)=0$ $0=4\\\\cos x+3\\\\sin x$ $4\\\\cos x=-3\\\\sin x$ $\\\\frac{-4}{3}=\\\\frac{\\\\sin x}{\\\\cos x}$ $\\\\tan x =\\\\frac{-4}{3}$ Can you finish this? 3. Originally Posted by Manabu On the closed interval [0,2pi], the maximum value of the function f(x)=4sinx-3cosx is: A.)3 B.)4 C.)24/5 D.)5 E.)None of these Explanations appreciated. A couple of suggestions: Have you learnt to write that function in the form sin (a+b) or cos(a+b)? If so, that'd be the easiest way. 2. Calculus?? 4. Originally Posted by pickslides $f(x)=4\\\\sin x-3\\\\cos x$ $f'(x)=4\\\\cos x+3\\\\sin x$ for max let $f'(x)=0$ $0=4\\\\cos x+3\\\\sin x$ $4\\\\cos x=-3\\\\sin x$ $\\\\frac{-4}{3}=\\\\frac{\\\\sin x}{\\\\cos x}$ $\\\\tan x =\\\\frac{-4}{3}$ Can you finish this? Not really. I understand how you got to that point, but I'm not 100% sure how to calculate maximum values. 5. Originally Posted by Manabu On the closed interval [0,2pi], the maximum value of the function\",\n \"\\\\times 28}}{5} + 32 = \\\\frac{{252 + 160}}{5} = \\\\frac{{412}}{5} = 82.4 \\\\end{align} (iii) \\\\begin{align}t\\\\left( { - 10} \\\\right) = \\\\frac{{9 \\\\times ( - 10)}}{5} + 32 = 9 \\\\times ( - 2) + 32 = - 18 + 32 = 14 \\\\end{align} (iv) It is given that \\\\begin{align}\\\\;\\\\;\\\\;\\\\;t&\\\\left( C \\\\right) = 212\\\\\\\\ \\\\Rightarrow \\\\;&\\\\frac{{9C}}{5} + 32 = 212\\\\\\\\ \\\\Rightarrow \\\\;&\\\\frac{{9C}}{5} = 212 - 32\\\\\\\\ \\\\Rightarrow \\\\;&C = \\\\frac{{180 \\\\times 5}}{9}\\\\\\\\&\\\\quad= 100\\\\end{align} Thus, the value of $$'t'$$, when $$t\\\\left( C \\\\right) = 212$$ is $$100.$$ ## Chapter 2 Ex.2.3 Question 5 Find the range of each of the following functions. (i) $$f\\\\left( x \\\\right)~=2-3x,\\\\,x\\\\in R,x>0.$$ (ii) $$f\\\\left( x \\\\right)~={{x}^{2}}+2,\\\\,x$$ is a real number. (iii) $$f\\\\left( x \\\\right)~=x,\\\\,~x$$ is a real number. ### Solution (i) $$f\\\\left( x \\\\right)~=2-3x,x\\\\in \\\\mathbf{R},x>0.$$ The values of $$f\\\\left( x \\\\right)$$ for various values of real numbers $$x > 0$$ can be written in the tabular form as $$x$$ $$0.01$$ $$0.1$$ $$0.9$$ $$1$$ $$2$$ $$2.5$$ $$4$$ $$5$$ $$\\\\ldots$$ $$f\\\\left( x \\\\right)$$ $$1.97$$ $$1.7$$ $$-0.7$$ $$-1$$ $$-4$$ $$-5.5$$ $$-10$$ $$-13$$ $$\\\\ldots$$ Thus, it can be clearly observed that the range of $$f$$ is the set of all real numbers less than $$2.$$ i.e., range of $$f = \\\\left( {-\\\\infty ,2} \\\\right)$$ Alternative Method: Let $$x > 0$$ $\\\\begin{array}{l} \\\\Rightarrow 3x > 0\\\\\\\\ \\\\Rightarrow 2 - 3x\",\n \"# Find the range of function $f(x) =\\\\cos\\\\sin\\\\ln\\\\left(\\\\frac{x^2+e}{x^2+1} \\\\right)+\\\\sin\\\\cos\\\\ln\\\\left(\\\\frac{x^2+e}{x^2+1} \\\\right)$ Problem : Find the range of function $$f\\\\left( x \\\\right) =\\\\cos \\\\left( \\\\sin \\\\left( \\\\ln \\\\left( \\\\frac{x^2+e}{x^2+1} \\\\right) \\\\right) \\\\right) +\\\\sin \\\\left( \\\\cos \\\\left( \\\\ln \\\\left( \\\\frac{x^2+e}{x^2+1} \\\\right) \\\\right) \\\\right)$$ My approach : maximum value of the function is when denominator term is minimum i.e. $$x^2+1$$ is minimum. It is minimum when $$x^2 =0$$ therefore, maximum value of function $$\\\\cos(\\\\sin(\\\\ln e))+\\\\sin(\\\\cos(\\\\ln e))$$ $$\\\\cos(\\\\sin(1))+\\\\sin(\\\\cos(1))$$ now how to find the minimum value of the function please suggest . Thanks... • \\\"maximum value of the function is when denominator term is minimum \\\" of which function? Oct 24, 2014 at 18:01 • Maybe you are just supposed to graph it? Oct 24, 2014 at 18:12 Well, first you need to see the range of $\\\\frac{x^2+e}{x^2+1}$, which you can easily verify to be $(1,e]$. Then, the range of $\\\\log$ in this domain is $(0,1]$. As pointed out in a comment: Since $\\\\sin$ is increasing and $\\\\cos$ decreasing in $(0,1]$, then $\\\\sin(0,1] =(0,\\\\sin(1)]$ and $\\\\cos(0,1] =[\\\\cos(1),1)$. Finally, since in $(0,\\\\sin(1)]$ the cosine is decreasing and in $[\\\\cos(1),1)$ the sine is increasing, we get that the range of each part of the sum is $[\\\\cos(\\\\sin(1)),1)$ and $[\\\\sin(\\\\cos(1)),1)$. So the range of the complete function is $$[\\\\cos(\\\\sin(1))+\\\\sin(\\\\cos(1)), 2)$$ Another way could be using the fact\",\n \"# What is the range of f(x) = ln(sin^(-1)(x^2+x+3/4)) ? Jun 15, 2017 $\\\\left[\\\\ln \\\\left(\\\\frac{\\\\pi}{6}\\\\right) , \\\\ln \\\\left(\\\\frac{\\\\pi}{2}\\\\right)\\\\right] \\\\approx \\\\left[- 0.6470 , 0.4516\\\\right]$ #### Explanation: Given: $f \\\\left(x\\\\right) = \\\\ln \\\\left({\\\\sin}^{- 1} \\\\left({x}^{2} + x + \\\\frac{3}{4}\\\\right)\\\\right)$ First note that: ${x}^{2} + x + \\\\frac{3}{4} = {x}^{2} + x + \\\\frac{1}{4} + \\\\frac{1}{2} = {\\\\left(x + \\\\frac{1}{2}\\\\right)}^{2} + \\\\frac{1}{2}$ which can take any value in the range $\\\\left[\\\\frac{1}{2} , \\\\infty\\\\right)$ The domain of ${\\\\sin}^{- 1}$ as a real valued function of real arguments is $\\\\left[- 1 , 1\\\\right]$. So the possible valid arguments to it in $f \\\\left(x\\\\right)$ are all in: $\\\\left[\\\\frac{1}{2} , \\\\infty\\\\right) \\\\cap \\\\left[- 1 , 1\\\\right] = \\\\left[\\\\frac{1}{2} , 1\\\\right]$ ${\\\\sin}^{- 1} \\\\left(\\\\frac{1}{2}\\\\right) = \\\\frac{\\\\pi}{6}$ ${\\\\sin}^{- 1} \\\\left(1\\\\right) = \\\\frac{\\\\pi}{2}$ and $\\\\sin$ is monotonically increasing between these two endpoints. So the range of ${\\\\sin}^{- 1} \\\\left({x}^{2} + x + \\\\frac{3}{4}\\\\right)$ is $\\\\left[\\\\frac{\\\\pi}{6} , \\\\frac{\\\\pi}{2}\\\\right]$ Hence the range of $f \\\\left(x\\\\right)$ is: $\\\\left[\\\\ln \\\\left(\\\\frac{\\\\pi}{6}\\\\right) , \\\\ln \\\\left(\\\\frac{\\\\pi}{2}\\\\right)\\\\right]$\",\n \"+ 3$, and $n = N$, $$|S_m{(x)} - S_n{(x)}| \\\\geq \\\\frac{2\\\\sqrt{2}}5$$ And then ultimately show the original sequence doesn't uniformly converge on $\\\\mathbb{R}$ Part C I think I'm on the right track and some tips should suffice, but for part d I'm relatively lost. Any help's greatly appreciated - thank you Part 2 Let $N + 1 \\\\le k \\\\le 4N + 3$. Then $\\\\frac{N+1}{k}$ satisfies the inequalities $\\\\frac{N+1}{4N+3}\\\\le\\\\frac{N+1}{k}\\\\le1$. Now, for $x$ in the interval $\\\\left(\\\\frac{N+1}{4N+3},1\\\\right)$, $\\\\sin x$ is monotonically increasing and we have $$1\\\\ge \\\\sin(1)\\\\ge \\\\sin\\\\left(\\\\frac{N+1}{k}\\\\right)\\\\ge\\\\sin\\\\left(\\\\frac{N+1}{4N+3}\\\\right)$$ and thus \\\\begin{align} \\\\sin\\\\left(\\\\frac{N+1}{k}\\\\right)&\\\\ge\\\\sin\\\\left(\\\\frac{N+1}{4N+3}\\\\right)\\\\\\\\\\\\\\\\ &\\\\ge\\\\sin(1/4)\\\\\\\\\\\\\\\\ &\\\\ge \\\\frac14-\\\\frac{1}{3!}(\\\\frac14)^3\\\\\\\\\\\\\\\\ &=\\\\frac14\\\\left(1-\\\\frac{1}{96}\\\\right)\\\\\\\\\\\\\\\\ &>\\\\frac15 \\\\end{align} which was to be shown. Part 3 For this part, we will use the result from the following integral \\\\begin{align} \\\\int_{N+1}^{4N+4}\\\\frac{dx}{\\\\sqrt{x}}&=2\\\\left(\\\\sqrt{4N+4}-\\\\sqrt{N+1}\\\\right)\\\\\\\\ &=2\\\\sqrt{N+1}(2-1)\\\\\\\\ &\\\\ge 2\\\\sqrt{2} \\\\end{align} for $N\\\\ge1$. Now, we note that this integral can be represented by the following summation: \\\\begin{align} \\\\int_{N+1}^{4N+4}\\\\frac{dx}{\\\\sqrt{x}}&=\\\\sum_{k=N+1}^{4N+3}\\\\int_{k}^{k+1} \\\\frac{dx}{\\\\sqrt{x}}\\\\\\\\ &\\\\le\\\\sum_{k=N+1}^{4N+3} \\\\frac{1}{\\\\sqrt{k}} &\\\\ge 2\\\\sqrt{2} \\\\end{align} and we have the desired inequality! Part 4 From the Part 2, we showed that if $N + 1 \\\\le k \\\\le 4N + 3$, then $$1 \\\\ge \\\\sin\\\\left(\\\\frac{N+1}{k}\\\\right) \\\\ge \\\\frac15$$ From the Part 3, we showed that for all $N$ $$\\\\sum_{k=N+1}^{4N+3}{\\\\frac{1}{\\\\sqrt{k}}} \\\\ge 2\\\\sqrt{2}$$ Then, putting these together we have \\\\begin{align} |S_m(x)-S_n(x)|&=\\\\left|\\\\sum_{k=N+1}^{4N+3} \\\\frac{\\\\sin((N+1)/k)}{\\\\sqrt{k}}\\\\right|\\\\\\\\ &\\\\ge\\\\left|\\\\sum_{k=N+1}^{4N+3} \\\\frac{\\\\frac15}{\\\\sqrt{k}}\\\\right|\\\\\\\\\\\\\\\\ &\\\\ge\\\\frac15\\\\left|\\\\sum_{k=N+1}^{4N+3} \\\\frac{1}{\\\\sqrt{k}}\\\\right|\\\\\\\\\\\\\\\\ &\\\\ge\\\\frac{2\\\\sqrt{2}}{5} \\\\end{align}\",\n \"= \\\\frac 7 2$$. This becomes our next interval. Step 6 Find the third approximation and its associated error. The midpoint of the interval $$\\\\left[3, \\\\frac 7 2\\\\right]$$ is at $$x = \\\\frac{13} 4$$, as shown on the graph below. This is the next approximation. Max Error: The interval has a length of $$1/2$$, so the maximum possible error is $$\\\\pm1/4$$ of a unit. The table below summarizes the approximations we found and their associated errors. $$\\\\begin{array}{ccc} \\\\mbox{Approximation} & x\\\\mbox{-value} & \\\\mbox{Possible Error}\\\\\\\\ \\\\hline 1^{st} & x = 3 & \\\\pm1\\\\\\\\[6pt] 2^{nd} & x = \\\\frac 7 2 & \\\\pm\\\\frac 1 2\\\\\\\\[6pt] 3^{rd} & x = \\\\frac{13} 4 & \\\\pm\\\\frac 1 4 \\\\end{array}$$ ##### Example 2 Use the bisection method to approximate the solution to the equation below to within less than 0.1 of its real value. Assume $$x$$ is in radians. $$\\\\sin x = 6 - x$$ Step 1 Rewrite the equation so it is equal to 0. $$x - 6 + \\\\sin x = 0$$ The function we'll work with is $$f(x) = x - 6 + \\\\sin x$$. Notice that the function is continuous everywhere. Step 2 Find an initial interval to work with. Setting up a table of values, we see the following. $$\\\\begin{array}{cl} x & {f(x)}\\\\\\\\ \\\\hline 0 & f(0) = -6\\\\\\\\ 1\",\n \"really desperate i need to this for tommorow and I don´t know how to do this a really need some help - as I said I have already asked the question in I answered that i would tried to do the problem but I can´t do it... – David Hernandez Nov 30 '13 at 5:13 Why not use commas and periods to separate sentences and paragraghs? – awllower Nov 30 '13 at 5:15 I am sorry :) I am not very good at writing in mathjax – David Hernandez Nov 30 '13 at 5:24 Helpful identity: note that $$1 - \\\\cos(x) = 2\\\\frac{1 - \\\\cos(2\\\\frac{x}{2})}{2} = 2 \\\\sin^2\\\\left(\\\\frac{x}{2}\\\\right)$$ so that our function is simply $$f(x)= \\\\begin{cases} 2 {\\\\sin^2(\\\\frac x2)\\\\over x^2}, & \\\\text{if x\\\\neq 0} \\\\\\\\ 1/2, & \\\\text{if x= 0} \\\\\\\\ \\\\end{cases}$$ With that in mind, we can find (assuming $x,y \\\\neq 0$) \\\\begin{align} \\\\left|f(x)-f(y)\\\\right| &= 2\\\\left| \\\\left(\\\\frac{\\\\sin(\\\\frac y2)}{y}\\\\right)^2 - \\\\left(\\\\frac{\\\\sin(\\\\frac x2)}{x}\\\\right)^2 \\\\right|\\\\\\\\ &=2\\\\left| \\\\left(\\\\frac{\\\\sin(\\\\frac y2)}{y}\\\\right) + \\\\left(\\\\frac{\\\\sin(\\\\frac x2)}{x}\\\\right) \\\\right| \\\\left| \\\\left(\\\\frac{\\\\sin(\\\\frac y2)}{y}\\\\right) - \\\\left(\\\\frac{\\\\sin(\\\\frac x2)}{x}\\\\right) \\\\right| \\\\end{align} Now, we just have to work with $\\\\left| \\\\left(\\\\frac{\\\\sin(\\\\frac y2)}{y}\\\\right) - \\\\left(\\\\frac{\\\\sin(\\\\frac x2)}{x}\\\\right) \\\\right|$ to get out a $|x-y|$. thank you!! and we also have that ${|xsin(y)-ysin(x)|\\\\over |xy|}\\\\le{|x-y|\\\\over |xy|}$ is this correct? – David Hernandez Nov 30 '13 at 5:50 I think there is a typo. $\\\\frac{2\\\\sin^2(x)}{x^2}$ should be $\\\\frac{2\\\\sin^2(x/2)}{x^2}$\",\n \"Share Books Shortlist # Solution for If the Function F(X) = Cos |X| − 2ax + B Increases Along the Entire Number Scale, Then (A) a = B (B) a = 1 2 B (C) a ≤ − 1 2 (D) a > − 3 2 - CBSE (Science) Class 12 - Mathematics ConceptIncreasing and Decreasing Functions #### Question If the function f(x) = cos |x| − 2ax + b increases along the entire number scale, then (a) a = b (b) $a = \\\\frac{1}{2}b$ (c) $a \\\\leq - \\\\frac{1}{2}$ (d) $a > - \\\\frac{3}{2}$ #### Solution (c) $a \\\\leq - \\\\frac{1}{2}$ $Given: f\\\\left( x \\\\right) = \\\\cos \\\\left| x \\\\right| - 2ax + b$ $\\\\text { Now}, \\\\left| x \\\\right| =\\\\begin{cases} x ,& x \\\\geq 0 \\\\\\\\ - x, & x < 0 \\\\end{cases}$ $\\\\text { and } \\\\cos \\\\left| x \\\\right| = \\\\begin{cases} \\\\cos\\\\left( x \\\\right) , & x \\\\geq 0 \\\\\\\\cos\\\\left( - x \\\\right) = cos\\\\left( x \\\\right), & x < 0\\\\end{cases}$ $\\\\therefore \\\\cos\\\\left| x \\\\right| = \\\\cos x , \\\\forall x \\\\in R$ $\\\\therefore f\\\\left( x \\\\right) = \\\\cos x - 2ax + b$ $\\\\Rightarrow f'\\\\left( x \\\\right) = - \\\\sin x - 2a$ $\\\\text { It is given that f(x) is increasing } .$ $\\\\Rightarrow f'\\\\left( x \\\\right) \\\\geq 0$ $\\\\Rightarrow - \\\\sin\",\n \"$$0 < c < \\\\frac{\\\\pi}{2}$$ and $$g(c) = 0$$. $$h(\\\\theta) = 1 + \\\\theta - 3\\\\tan{\\\\theta}$$ Here, $$h(0) = 1$$ and $$h\\\\left(\\\\frac{\\\\pi}{4}\\\\right) \\\\approx -\\\\frac{5}{4}$$. This means that there is a $$c$$ in $$\\\\left[0, \\\\frac{\\\\pi}{4}\\\\right]$$ where $$h(c) = 0$$. #### 3.2.3 Usage: Finding a value inside a set interval Given a function $$f$$, a closed interval $$[a, b]$$, and a $$k$$, prove that there is a $$c$$ such that $$f(c) = k$$ and $$a < c < b$$. $$f(x) = x^{2} + x - 1$$, $$[0, 5]$$, $$f(c) = 11$$ Substituting $$c$$ and setting equal to $$11$$: \\\\begin{align} & c^{2} + c - 1 &= 11 & \\\\\\\\ & c^{2} + c - 12 &= 0 & \\\\\\\\ & c &= \\\\frac{-1 \\\\pm \\\\sqrt{1^{2} - 4(1)(-12)}}{2} & \\\\\\\\ & c &= \\\\frac{-1 \\\\pm \\\\sqrt{49}}{2} & \\\\\\\\ & c &= \\\\frac{-1 \\\\pm 7}{2} & \\\\\\\\ c &= \\\\frac{-8}{2} & c &= \\\\frac{6}{2} \\\\\\\\ c &= -4 & c &= 3 \\\\end{align} $$-4$$ is not in the domain, so the solution is $$c = 3$$. $$f(x) = x^{2} - 6x + 8$$, $$[0, 3]$$, $$f(c) = 0$$ Again, substitute $$c$$ and set equal to $$0$$. \\\\begin{align} & c^{2} - 6c + 8 &= 0 & \\\\\\\\ & c &= \\\\frac{6 \\\\pm \\\\sqrt{(-6)^{2} - 4(1)(8)}}{2} & \\\\\\\\ & c &= \\\\frac{6 \\\\pm\",\n \"\\\\begin{align} \\\\text{Range}(g) &= \\\\left[ \\\\frac{-1 - \\\\sin \\\\left( \\\\frac{1}{2} \\\\right)}{2 \\\\sin \\\\left( \\\\frac{1}{2} \\\\right)},\\\\frac{1 - \\\\sin \\\\left( \\\\frac{1}{2} \\\\right)}{2 \\\\sin \\\\left( \\\\frac{1}{2} \\\\right)} \\\\right] \\\\\\\\ &= [-1.54291482 \\\\ldots,0.54291482 \\\\ldots] \\\\\\\\ &\\\\subseteq [-2,2]. \\\\end{align} Conclusion: $\\\\displaystyle \\\\left| \\\\sum_{k=1}^{n} \\\\sin(k) \\\\right| < 2$ and $\\\\displaystyle \\\\left| \\\\sum_{k=1}^{n} \\\\cos(k) \\\\right| < 2$ for all $n \\\\in \\\\mathbb{N}$. - You have the formula $\\\\frac{1-x^{n+1}}{1-x}=\\\\sum_{k=1}^{n}x^k$, which is easy enough to see by algebra. So you can replace the LHS with $\\\\frac{1-e^{i(n+1)\\\\theta}}{1-e^{i\\\\theta}}$. – minimalrho Feb 1 '13 at 22:03 +1 Very nice hint. This is not a series, @Iuli, but a finite sum and, in fact, a rather easy one: a geometric sequence sum. Above you can read what you get from it, which with absolute values gives at once the wanted upper bound $\\\\,2\\\\,$ – DonAntonio Feb 1 '13 at 22:15 @CalvinLin , yes...so what? The absolute value of the expression hinted in the answer above is less than or equal what you wrote (in fact, it is always less). Besides this, in the given formula you must take the imaginary part as that the part that corresponds to the sum of sines... – DonAntonio Feb 1 '13 at 23:05 @HaskellCurry Yes, with the trig version I could see how to do it. The exp version was less obvious, since you still had\"\n]"}}},{"rowIdx":6937,"cells":{"problem":{"kind":"string","value":"The roots of\n\\[z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}\\]are $\\text{cis } \\theta_1$, $\\text{cis } \\theta_2$, $\\dots$, $\\text{cis } \\theta_7$, where $0^\\circ \\le \\theta_k < 360^\\circ$ for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees."},"level":{"kind":"string","value":"Level 4"},"type":{"kind":"string","value":"Precalculus"},"solution":{"kind":"string","value":"Let $z = \\text{cis } \\theta$. Then\n\\[z^7 = \\text{cis } 7 \\theta.\\]Using cis notation,\n\\[-\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}} = \\text{cis } 225^\\circ,\\]so we want\n\\[\\text{cis } 7 \\theta = \\text{cis } 225^\\circ.\\]This equation holds if and only if\n\\[7 \\theta = 225^\\circ + 360^\\circ k\\]for some integer $k$, or\n\\[\\theta = \\frac{225^\\circ + 360^\\circ k}{7}.\\][asy]\nunitsize(2 cm);\n\ndraw((-1.2,0)--(1.2,0));\ndraw((0,-1.2)--(0,1.2));\ndraw(Circle((0,0),1));\n\ndot(\"cis $\\frac{225^\\circ}{7}$\", dir(225/7), dir(225/7));\n\nfor(int i = 1; i <= 6; ++i) {\n dot(dir(225/7 + 360*i/7));\n}\n\nlabel(\"Re\", (1.2,0), NE);\nlabel(\"Im\", (0,1.2), NE);\n[/asy]\n\nThe angles of this form that are between $0^\\circ$ and $360^\\circ$ are\n\\[\\frac{225^\\circ}{7}, \\quad \\frac{225^\\circ + 360^\\circ}{7}, \\quad \\frac{225^\\circ + 2 \\cdot 360^\\circ}{7}, \\quad \\dots, \\quad \\frac{225^\\circ + 6 \\cdot 360^\\circ}{7}.\\]By the formula for an arithmetic series, the sum of these angles is\n\\[\\frac{1}{2} \\cdot \\left( \\frac{225^\\circ}{7} + \\frac{225^\\circ + 6 \\cdot 360^\\circ}{7} \\right) \\cdot 7 = \\boxed{1305^\\circ}.\\]"},"answer":{"kind":"string","value":"1305^\\circ"},"p_retrievals":{"kind":"list like","value":["# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Rod Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? - The roots are $e^{2k\\pi i/7}$ ($k=0,\\ldots,6$). Write them as $r^k$ where $r = e^{2\\pi i/7}$. So the sum is (using the simple formula for summing a finite geometric series) $$\\sum_{k=0}^6 r^k = \\frac{1-r^7}{1-r} = \\frac{1 - e^{2\\pi i}}{1-e^{2\\pi i/7}} = 0$$. - This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not","# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Mr.Fry Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? If I am understanding correctly...for $k=1$, we take $$\\operatorname{cis}(7\\theta_1)=\\operatorname{cis} 225^\\circ$$ and then somehow determine $\\theta_1$, and repeat for $k=2,3,4,5,6,7$? – Mathy Person Jul 18 '15 at 21:59 No, that's getting further from the answer. Rather, think about the solutions set for the equation $$\\operatorname{cis}\\theta=1. This will let you get rid of the \\operatorname{cis}, and find all solutions to the original","equation. Then you can find the seven solutions in the desired interval. – Cameron Buie Jul 18 '15 at 22:10 I'm not sure exactly what you mean. Can you elaborate? – Mathy Person Jul 18 '15 at 22:18 Well, what are the solutions to$$\\operatorname{cis}\\theta=1,$$or, more usefully put,$$\\cos\\theta+i\\sin\\theta=1+0i$$– Cameron Buie Jul 18 '15 at 22:24 \\cos \\theta = 1 and \\sin \\theta = 0, so \\theta = 360 degrees/2\\pi. – Mathy Person Jul 18 '15 at 22:30 The roots are e^{2k\\pi i/7} (k=0,\\ldots,6). Write them as r^k where r = e^{2\\pi i/7}. So the sum is (using the simple formula for summing a finite geometric series)$$\\sum_{k=0}^6 r^k = \\frac{1-r^7}{1-r} = \\frac{1 - e^{2\\pi i}}{1-e^{2\\pi i/7}} = 0. This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not summing up the roots, but rather the arguments of said roots lying in $[0^\\circ,360^\\circ).$ – Cameron Buie Jan 24 '14 at 17:15 $\\def\\cis{\\operatorname{cis}}$Let's consider a more general problem. We have $a=r\\cis\\alpha$ and we write its $n$th roots as \\begin{gather} \\sqrt[n]{r}\\cis\\frac{\\beta}{n}\\\\ \\sqrt[n]{r}\\cis\\left(\\frac{\\beta}{n}+\\frac{2\\pi}{n}\\right)\\\\ \\sqrt[n]{r}\\cis\\left(\\frac{\\beta}{n}+2\\frac{2\\pi}{n}\\right)\\\\ \\dots\\\\ \\sqrt[n]{r}\\cis\\left(\\frac{\\beta}{n}+(n-1)\\frac{2\\pi}{n}\\right) \\end{gather} Can you sum up all those angles? Use $360$ instead of $2\\pi$ if you so prefer or are forced to.","to get the imaginary … 5. ### precalc express the roots of unity in standard form a+bi. 1.) cube roots of unity 2.) fourth roots of unity 3.) sixth roots of unity 4.) square roots of unity 6. ### precalculus express the roots of unity in standard form a+bi. 1.) cube roots of unity 2.) fourth roots of unity 3.) sixth roots of unity 4.) square roots of unity 7. ### PRECALCULUS find the cube roots of the complex number. 27((cos(11pi/6)) + i(sin(11pi/6)) 8. ### PRECALCULUS 1.) determine two pairs of polar coordinates for the point (-4sqrt3, 4sqrt3)with 0degrees less than or equal to theta less than or equal to 360degrees 2.) Use DeMoivre's Theorem to find (1-i)^10. write your answer in the form a+bi. … The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$are $\\text{cis } \\theta_1$, $\\text{cis } \\theta_2$, $\\dots$, $\\text{cis } \\theta_7$, where $0^\\circ \\le \\theta_k < 360^\\circ$ for all $1 \\le k \\le 7$. Find $\\theta_1 … 10. ### Precalculus The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$are$\\text{cis } \\theta_1$,$\\text{cis } \\theta_2$,$\\dots$,$\\text{cis } \\theta_7$, where$0^\\circ \\le \\theta_k < 360^\\circ$for all$1 \\le k \\le 7$. Find$\\theta_1 … More Similar Questions","These correspond to the complex numbers $$w_{\\text{\\tiny \\(0$$}} = 2 \\text{cis}(0)\\), $$w_{\\text{\\tiny \\(1$$}} = 2 \\text{cis}\\left(\\frac{2\\pi}{3}\\right)\\) and $$w_{\\text{\\tiny \\(2$$}} = 2 \\text{cis}\\left(\\frac{4\\pi}{3}\\right)\\), respectively. Writing these out in rectangular form yields $$w_{\\text{\\tiny \\(0$$}} = 2\\), $$w_{\\text{\\tiny \\(1$$}} = -1 + i\\sqrt{3}\\) and $$w_{\\text{\\tiny \\(2$$}} = -1-i\\sqrt{3}\\). While this process seems a tad more involved than our previous factoring approach, this procedure can be generalized to find, for example, all of the fifth roots of $$32$$. (Try using Chapter \\ref{Polynomials} techniques on that!) If we start with a generic complex number in polar form $$z = |z| \\text{cis}(\\theta)$$ and solve $$w^{n} = z$$ in the same manner as above, we arrive at the following theorem. Note: The $$n^{th}$$ roots of a Complex Number Let $$z \\neq 0$$ be a complex number with polar form $$z = r\\text{cis}(\\theta)$$. For each natural number $$n$$, $$z$$ has $$n$$ distinct $$n^{\\text{th}}$$ roots, which we denote by $$w_{\\text{\\tiny$$0\\)}}\\), $$w_{\\text{\\tiny$$1\\)}}\\), \\ldots, $$w_{\\text{\\tiny$$n-1\\)}}\\), and they are given by the formula \\index{\\)n^{\\textrm{th}}\\) root ! of a complex number} \\index{complex number ! $$n^{\\textrm{th}}$$ root} $w_{\\text{k}} = \\sqrt[n]{r}\\text{cis}\\left(\\frac{\\theta}{n} + \\frac{2\\pi}{n} k \\right)$ The proof of Theorem \\ref{nthrootscomplexthm} breaks into to two parts: first, showing that each $$w_{\\text{k}}$$ is an $$n^{\\text{th}}$$ root, and second, showing that the set $$\\left\\{ w_{\\text{k}} \\, | \\, k = 0, 1, \\ldots, (n-1)\\right\\}$$ consists of","+0 # help +1 50 1 The number $$\\text{cis } 75^\\circ + \\text{cis } 83^\\circ + \\text{cis } 91^\\circ + \\dots + \\text{cis } 147^\\circ$$ is expressed in the form $$r \\, \\text{cis } \\theta$$, where $$0 \\le \\theta < 360^\\circ$$. Find $$\\theta$$ in degrees. Dec 20, 2019","For L cuts C at two distinct points, there should be two distinct roots in (*), hence $\\Delta > 0$. Thus $(-6\\cos\\theta)^2 - 4(6) > 0$ $\\cos^2\\theta > \\frac{2}{3}$ $-\\cos^{-1}(\\sqrt{\\frac{2}{3}}) < \\theta < \\cos^{-1}(\\sqrt{\\frac{2}{3}})$ It may not be easy to yield the last inequality above, just sketch a graph to see it. F.5 students, don’t turn a blind eye to this problem, it IS IN THE SYLLABUS.","summing up the roots, but rather the arguments of said roots lying in $[0^\\circ,360^\\circ).$ – Cameron Buie Jan 24 '14 at 17:15 $\\def\\cis{\\operatorname{cis}}$Let's consider a more general problem. We have $a=r\\cis\\alpha$ and we write its $n$th roots as \\begin{gather} \\sqrt[n]{r}\\cis\\frac{\\beta}{n}\\\\ \\sqrt[n]{r}\\cis\\left(\\frac{\\beta}{n}+\\frac{2\\pi}{n}\\right)\\\\ \\sqrt[n]{r}\\cis\\left(\\frac{\\beta}{n}+2\\frac{2\\pi}{n}\\right)\\\\ \\dots\\\\ \\sqrt[n]{r}\\cis\\left(\\frac{\\beta}{n}+(n-1)\\frac{2\\pi}{n}\\right) \\end{gather} Can you sum up all those angles? Use $360$ instead of $2\\pi$ if you so prefer or are forced to. -","are: $\\set {\\sqrt [6] 2 \\cis 20 \\degrees, \\sqrt [6] 2 \\cis 80 \\degrees, \\sqrt [6] 2 \\cis 140 \\degrees, \\sqrt [6] 2 \\cis 200 \\degrees, \\sqrt [6] 2 \\cis 260 \\degrees, \\sqrt [6] 2 \\cis 320 \\degrees}$ ### Roots of $z^8 + 1 = 0$ The roots of the polynomial: $z^8 + 1 = 0$ are: $\\set {\\cos \\dfrac {\\paren {2 k + 1} \\pi} 8 + i \\sin \\dfrac {\\paren {2 k + 1} \\pi} 8: k \\in \\set {0, 1, \\ldots, 7} }$","# roots of unity questions (1 Viewer) #### poptarts12345 ##### Member 1. Let w be a seventh root of unity. Find the equation of the quadratic polynomial with roots w+w^2+w^4 and w^3+w^5+w^6. ANS : x^2+x+2 2. Let w be the principal nth root of unity. Prove that w conjugate = w^(n-1) #### idkkdi ##### Well-Known Member 1. Let w be a seventh root of unity. Find the equation of the quadratic polynomial with roots w+w^2+w^4 and w^3+w^5+w^6. ANS : x^2+x+2 2. Let w be the principal nth root of unity. Prove that w conjugate = w^(n-1) w^7-1 = 0. (w-1)(w^6+w^5+...+1 ) = 0. For w =/ 1, (w^6+w^5+...+1) = 0. clearly w+w^2+w^4 + w^3+w^5+w^6 = -1. (w+w^2+w^4)(w^3+w^5+w^6) = w^4 + w^6 + w^7 + w^5 + w^7 + w^8 + w^7 + w^9 + w^10 = w^4 + w^5 + w^6 + w^7 + w^7 + w + w^2 + w^3 + w^7 = -1 + 3 = 2. (note that w is assumed to be the principal root of unity) let w = cis theta w cong = cis -theta w^n-1 = cis (n-1) theta = cis n theta/cis theta = cis -theta #### Qeru ##### Member 1. Let w be a seventh root of unity. Find the equation of the quadratic polynomial with roots w+w^2+w^4 and","# How to find the roots of $x³-2$? I'm trying to find the roots of $x^3 -2$, I know that one of the roots are $\\sqrt[3] 2$ and $\\sqrt[3] {2}e^{\\frac{2\\pi}{3}i}$ but I don't why. The first one is easy to find, but the another two roots? I need help Thank you If $\\omega^3 = 1$ and $x^3 = 2$ then $(\\omega x)^3 = \\omega^3 x^3 = 2$. Possible values of $\\omega$ are $e^{\\frac{1}{3}2 i \\pi}$, $e^{\\frac{2}{3}2 i \\pi}$ and $e^{\\frac{3}{3}2 i \\pi}$. This is because $1 = e^{2 i \\pi} = (e^{\\frac{1}{k} 2 i \\pi})^k$. So the solutions of $x^3 - 2 = 0$ are $e^{\\frac{1}{3}2 i \\pi} \\sqrt[3]{2}$, $e^{\\frac{2}{3}2 i \\pi} \\sqrt[3]{2}$ and $\\sqrt[3]{2}$. Hint: $x^3-2=(x-2^{\\frac{1}{3}})(x^2+2^{\\frac{1}{3}}x+2^{\\frac{2}{3}})$. Let $x=z\\sqrt[3]2$. Then $0=x^3-2=2z^3-2$ and so $z^3=1$. Write $z=e^{i\\theta}$ and solve for $\\theta$.","told to do so, we leave this as a good, but messy, exercise. 4. To find the five fifth roots of $$1$$, we write $$1 = 1 \\text{cis}(0)$$. We have $$r = 1$$, $$\\theta = 0$$ and $$n = 5$$. Since $$\\sqrt[5]{1} = 1$$, the roots are $$w_{\\text{\\tiny$$0\\)}} = \\text{cis}(0) = 1\\), $$w_{\\text{\\tiny$$1\\)}} = \\text{cis}\\left(\\frac{2\\pi}{5}\\right)\\), $$w_{\\text{\\tiny$$2\\)}} = \\text{cis}\\left(\\frac{4\\pi}{5}\\right)\\), $$w_{\\text{\\tiny$$3\\)}} = \\text{cis}\\left(\\frac{6\\pi}{5}\\right)\\) and $$w_{\\text{\\tiny$$4\\)}} = \\text{cis}\\left(\\frac{8\\pi}{5}\\right)\\). The situation here is even graver than in the previous example, since we have not developed any identities to help us determine the cosine or sine of $$\\frac{2\\pi}{5}$$. At this stage, we could approximate our answers using a calculator, and we leave this as an exercise. Now that we have done some computations using Theorem \\ref{nthrootscomplexthm}, we take a step back to look at things geometrically. Essentially, Theorem \\ref{nthrootscomplexthm} says that to find the $$n^{\\text{th}}$$ roots of a complex number, we first take the $$n^{\\text{th}}$$ root of the modulus and divide the argument by $$n$$. This gives the first root $$w_{\\text{\\tiny$$0\\)}}\\). Each succeessive root is found by adding $$\\frac{2\\pi}{n}$$ to the argument, which amounts to rotating $$w_{\\text{\\tiny$$0\\)}}\\) by $$\\frac{2\\pi}{n}$$ radians. This results in $$n$$ roots, spaced equally around the complex plane. As an example of this, we plot our answers to number \\ref{fourthrootsneg16} in Example \\ref{nthrootscomplexex} below. We have only glimpsed at the","# Difference between revisions of \"1996 AIME Problems/Problem 11\" ## Problem Let $\\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\\mathrm {P}=r(\\cos{\\theta^{\\circ}}+i\\sin{\\theta^{\\circ}})$, where $0 and $0\\leq \\theta <360$. Find $\\theta$. ## Solution 1 $\\begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \\frac{z^5-1}{z-1}\\\\ 0 &=& \\frac{(z^5 - 1)(z(z-1)+1)}{z-1} = \\frac{(z^2-z+1)(z^5-1)}{z-1} \\end{eqnarray*}$ Thus $z^5 = 1, z \\neq 1 \\Longrightarrow z = \\mathrm{cis}\\ 72, 144, 216, 288$, or $z^2 - z + 1 = 0 \\Longrightarrow z = \\frac{1 \\pm \\sqrt{-3}}{2} = \\mathrm{cis}\\ 60, 300$ (see cis). Discarding the roots with negative imaginary parts (leaving us with $\\mathrm{cis} \\theta,\\ 0 < \\theta < 180$), we are left with $\\mathrm{cis}\\ 60, 72, 144$; their product is $P = \\mathrm{cis} (60 + 72 + 144) = \\mathrm{cis} \\boxed{276}$. ## Solution 2 Let $w =$ the fifth roots of unity, except for $1$. Then $w^6 + w^4 + w^3 + w^2 + 1 = w^4 + w^3 + w^2 + w + 1 = 0$, and since both sides have the fifth roots of unity as roots, we have $z^4 + z^3 + z^2 + z + 1 | z^6 + z^4 + z^3 + z^2 + 1$. Long","{th }}$$ root of unity. Then all the $$n^{\\text {th }}$$ roots of $$z$$ are $$\\left\\{w_{0}, \\omega w_{0}, \\omega^{2} w_{0}, \\ldots, \\omega^{n-1} w_{0}\\right\\}$$. ## Tartaglia-Cardano Revisited Let us consider again Example 9.9. We wanted to find the cube roots of $\\zeta_{\\pm}=\\frac{-1 \\pm \\sqrt{-3}}{2} .$ If we take the $$+$$ sign, we get $\\zeta_{+}=\\operatorname{Cis}(2 \\pi / 3),$ and if we take the - sign, we get $\\zeta_{-}=\\operatorname{Cis}(4 \\pi / 3) .$ So $$\\zeta_{+}$$has 3 roots, namely $\\left\\{\\operatorname{Cis}\\left(\\frac{2 \\pi}{9}+\\frac{2 k \\pi}{3}\\right): k=0,1,2\\right\\} \\text {, }$ and $$\\zeta_{-}$$has 3 roots, namely $\\left\\{\\operatorname{Cis}\\left(\\frac{4 \\pi}{9}+\\frac{2 k \\pi}{3}\\right): k=0,1,2\\right\\} \\text {, }$ Knowing $$w$$, we want to find $$x$$, which for Example $$9.9$$ is given by $$w+1 / w$$. For any number $$w$$ that can be written as $$\\operatorname{Cis}(\\theta)$$ (i.e. any complex number of modulus 1), we have \\begin{aligned} w+\\frac{1}{w} &=\\cos \\theta+i \\sin \\theta+\\cos (-\\theta)+i \\sin (-\\theta) \\\\ &=2 \\cos \\theta . \\end{aligned} Therefore the roots of the polynomial given in (9.10) are $\\left\\{2 \\cos \\frac{2 \\pi}{9}, 2 \\cos \\frac{8 \\pi}{9}, 2 \\cos \\frac{14 \\pi}{9}, 2 \\cos \\frac{4 \\pi}{9}, 2 \\cos \\frac{10 \\pi}{9}, 2 \\cos \\frac{16 \\pi}{9}\\right\\} .$ Are these 6 different roots? Theorem $$4.10$$ says that $$p$$ can have at most 3 different roots. As $$\\cos (\\theta)=\\cos (2 \\pi-\\theta)$$, we see our set (9.27) may be written as $\\left\\{2 \\cos \\frac{2 \\pi}{9}, 2","has got n different n-th roots, $$\\sqrt [n] {1} = \\cos(\\frac{2\\pi}{n}k) + i \\sin (\\frac{2\\pi}{n}k)$$ k=0, ...n-1. 8. Feb 17, 2005 Thank you for all the help! I think I finally get what you're saying... $$z=i\\, ^{\\frac{4}{3}}=\\left( i^4 \\right)^{1/3} = \\sqrt[3]{1}$$ $$h= 1 = \\mbox{cis } 0$$ $$z=h_w ^3 = \\mbox{ cis } \\left( \\frac{0+2k\\pi}{3} \\right) = \\cos \\left( \\frac{2k\\pi}{3} \\right) +\\mbox{ } i\\mbox{ } \\sin \\left( \\frac{2k\\pi}{3} \\right) = -\\frac{1}{2} +\\mbox{ } i \\mbox{ } \\frac{\\sqrt{3}}{2}$$","1. Cube roots of -8 I am totally stuck on this problem. I need to find the cube roots of -8 in polar coordinates. I know that -8 lies on the real axis, so the angle of reference is $\\pi$. I also know that $cos \\pi=-1$. I know that the formula I need to use is cube root of $|z|= cis \\frac{\\theta}{3}$ and I know that $cos \\frac{\\theta}{3}=Sqrt(\\frac{1+cos\\theta}{3})$, $sin\\theta=Sqrt(\\frac{1-cos\\theta}{3})$ and the cube root $=2(cos\\frac{\\theta}{3}+i sin\\frac{\\theta}3}$. I must be tired and doing something wrong with the signs or something. I keep getting a zero for the first term and I am pretty sure the answer is $1+i \\sqrt{3}$. Please help! Thanks! 2. Re: Complex Analysis - Cube roots of -8 First of all, you'll need a 2, somewhere. |z| = cis(x/3) really doesn't make quite enough sense. It is as if you ALMOST know where you are going. Start with $z^{3} = -8 = 8\\cdot cis(\\pi)$ Just apply the theorem from DeMoivre. $z = 8^{\\frac{1}{3}}\\cdot cis\\left(\\frac{\\pi +2k\\pi}{3}\\right)$ for k = 0, 1, 2 k = 0 leads to $2\\cdot cis(\\pi/3)$ k = 1 leads to $2\\cdot cis(\\pi)$ k = 2 leads to $2\\cdot cis(5\\pi/3)$ 3. Re: Complex Analysis - Cube roots of -8 I think you're right; I ALMOST know where I am going! OK... I began with $2","still a little confused about how to approach this particular question. Once I have the i term, where do i go from there? Is there any particular formula that i need to use? Thanks, Function • August 21st 2009, 02:30 AM mr fantastic Quote: Originally Posted by function Hey Mr. Fantastic, Thanks for your help, although I'm still a little confused about how to approach this particular question. Once I have the i term, where do i go from there? Is there any particular formula that i need to use? Thanks, Function You have $z^5 = r^5 \\text{cis} (5 \\theta) = \\text{cis} \\left( \\frac{\\pi}{2} + 2n \\pi\\right)$ where $n$ is an integer. Therefore $z = r \\text{cis} (\\theta) = \\text{cis} \\left( \\frac{\\pi}{10} + \\frac{2n \\pi}{5} \\right)$. Now substitute five consecutive values of n to get the five distinct roots. • August 22nd 2009, 01:59 AM pacman Roots in the complex plane, z1 = i, z2 = (-1)^(9/10), z3 = -(-1)^(3/10), z4 = -(-1)^(7/10), z5 = (-1)(1/10)","root, the negative one leads to t<0. thus $$t = \\frac{1}{\\sqrt{2c}} \\int_d^0 \\sqrt{ \\frac{ dz}{ d - z}} \\. d z$$ The substitution z = d cos2(theta) makes this doable. 4. Nov 10, 2005 Euclid Yeah, that's what I indicated I tried above. It appeared to fail the first time I did it, but the second time it worked out. Thanks!","\\frac{2^k\\pi i}{9}}, k=1,2,3$ $\\text{By letting }y=z+\\frac{1}{z}\\text{ deduce that }\\cos \\frac{2\\pi}{9}+\\cos \\frac{4\\pi}{9}+\\cos \\frac{8\\pi}{9}=0$ $\\text{I'm not sure what to do after }y^3-y+1=0$ Originally Posted by leehuan Sigh... $\\text{Proven already: The roots of }z^6+z^3+1=0\\text{ are }e^{\\pm \\frac{2^k\\pi i}{9}}, k=1,2,3$ $\\text{By letting }y=z+\\frac{1}{z}\\text{ deduce that }\\cos \\frac{2\\pi}{9}+\\cos \\frac{4\\pi}{9}+\\cos \\frac{8\\pi}{9}=0$ $\\text{I'm not sure what to do after }y^3-y+1=0$ $\\noindent As you have already proven, the roots of p(z):= z^{6} + z^{3} + 1 = 0 are z_1 = e^{i \\frac{2\\pi}{9}}, z_2 = e^{i \\frac{4\\pi}{9}}, z_{3} = e^{i \\frac{8\\pi}{9}}, and the conjugates of these, \\overline{z_j}, j=1,2,3.$ $\\noindent Note that if w = e^{i \\theta}, then w + \\overline{w} = 2\\cos \\theta. Since the sum of roots of p is 0 (as there is no quintic term), we have \\sum _{j=1}^{3}\\left(z_j + \\overline{z_j}\\right) = 0, where z_j = e^{i \\frac{2^{j}\\pi}{9}} (the L.H.S. is just the sum of the roots). Since the L.H.S. equals \\sum _{j=1}^{3} 2\\cos \\left(\\frac{2^{j}\\pi}{9}\\right), dividing the last equation through by 2 yields the result.$ $\\noindent Realised I didn't do it by doing the y = z + z^{-1} method. Using this method, you got it down to y^{3} -y + 1=0, where z is a root of the equation z^6 + z^3 + 1 = 0. Since the roots of this equation produce exactly three values of z+ z^{-1}, namely 2\\cos \\left(\\frac{2^j","you : what are the complex cube roots of $$8i$$? Thankfully, these can be answered via a use of De Moivre's theorem, which will tell you that the roots are precisely those of the form $$\\sqrt[3]{8} e^{\\frac {i\\pi}{6} + \\frac{2in \\pi}{3}}$$ where $$n=0,1,2$$. Now, this can be simplified: $$\\sqrt[3]{8} e^{\\frac \\pi{6} + \\frac{2n \\pi}{3}} = 2 \\operatorname{cis} \\frac \\pi {6}, 2 \\operatorname{cis} \\frac {5\\pi} {6}, 2 \\operatorname{cis} \\frac {9\\pi} {6}$$ Expand using the fact that $$\\operatorname{cis} \\theta = \\cos \\theta + i \\sin \\theta$$, and don't forget to subtract $$2i - \\sqrt 3$$ from each of the cube roots. This gives an answer that involves only sine and cosine evaluation. Alternately, since you figured out a root, you could also have done polynomial division, but for that you would have to expand the cube root, which is laborious."],"string":"[\n \"# Complex Numbers and exponential form and roots The roots of $z^7 = -\\\\frac{1}{\\\\sqrt{2}} - \\\\frac{i}{\\\\sqrt{2}}$ are $\\\\text{cis } \\\\theta_1, \\\\text{cis } \\\\theta_2, \\\\dots, \\\\text{cis } \\\\theta_7,$ where $0^\\\\circ \\\\le \\\\theta_k < 360^\\\\circ$for all $1 \\\\le k \\\\le 7$. Find $\\\\theta_1 + \\\\theta_2 + \\\\dots + \\\\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\\\left(5 \\\\pi i/4 \\\\right)$. How should I simplify? Thanks - What is $\\\\text{cis}$? Do you by chance mean $\\\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\\\theta)+isin(\\\\theta)$ – Rod Jan 24 '14 at 16:27 @AlexR: \\\"cis\\\" is a commonly used abbreviation for \\\"cosine plus $i$ times the sine.\\\" In other words, $$\\\\operatorname{cis}\\\\theta=\\\\cos\\\\theta+i\\\\sin\\\\theta=e^{i\\\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis}(7\\\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis} 225^\\\\circ,$$ as you've already determined, and so we need $$\\\\operatorname{cis}(7\\\\theta_k)=\\\\operatorname{cis} 225^\\\\circ\\\\\\\\\\\\frac{\\\\operatorname{cis}(7\\\\theta_k)}{\\\\operatorname{cis} 225^\\\\circ}=1\\\\\\\\\\\\operatorname{cis}(7\\\\theta_k-225^\\\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? - The roots are $e^{2k\\\\pi i/7}$ ($k=0,\\\\ldots,6$). Write them as $r^k$ where $r = e^{2\\\\pi i/7}$. So the sum is (using the simple formula for summing a finite geometric series) $$\\\\sum_{k=0}^6 r^k = \\\\frac{1-r^7}{1-r} = \\\\frac{1 - e^{2\\\\pi i}}{1-e^{2\\\\pi i/7}} = 0$$. - This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not\",\n \"# Complex Numbers and exponential form and roots The roots of $z^7 = -\\\\frac{1}{\\\\sqrt{2}} - \\\\frac{i}{\\\\sqrt{2}}$ are $\\\\text{cis } \\\\theta_1, \\\\text{cis } \\\\theta_2, \\\\dots, \\\\text{cis } \\\\theta_7,$ where $0^\\\\circ \\\\le \\\\theta_k < 360^\\\\circ$for all $1 \\\\le k \\\\le 7$. Find $\\\\theta_1 + \\\\theta_2 + \\\\dots + \\\\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\\\left(5 \\\\pi i/4 \\\\right)$. How should I simplify? Thanks - What is $\\\\text{cis}$? Do you by chance mean $\\\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\\\theta)+isin(\\\\theta)$ – Mr.Fry Jan 24 '14 at 16:27 @AlexR: \\\"cis\\\" is a commonly used abbreviation for \\\"cosine plus $i$ times the sine.\\\" In other words, $$\\\\operatorname{cis}\\\\theta=\\\\cos\\\\theta+i\\\\sin\\\\theta=e^{i\\\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis}(7\\\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis} 225^\\\\circ,$$ as you've already determined, and so we need $$\\\\operatorname{cis}(7\\\\theta_k)=\\\\operatorname{cis} 225^\\\\circ\\\\\\\\\\\\frac{\\\\operatorname{cis}(7\\\\theta_k)}{\\\\operatorname{cis} 225^\\\\circ}=1\\\\\\\\\\\\operatorname{cis}(7\\\\theta_k-225^\\\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? If I am understanding correctly...for $k=1$, we take $$\\\\operatorname{cis}(7\\\\theta_1)=\\\\operatorname{cis} 225^\\\\circ$$ and then somehow determine $\\\\theta_1$, and repeat for $k=2,3,4,5,6,7$? – Mathy Person Jul 18 '15 at 21:59 No, that's getting further from the answer. Rather, think about the solutions set for the equation $$\\\\operatorname{cis}\\\\theta=1. This will let you get rid of the \\\\operatorname{cis}, and find all solutions to the original\",\n \"equation. Then you can find the seven solutions in the desired interval. – Cameron Buie Jul 18 '15 at 22:10 I'm not sure exactly what you mean. Can you elaborate? – Mathy Person Jul 18 '15 at 22:18 Well, what are the solutions to$$\\\\operatorname{cis}\\\\theta=1,$$or, more usefully put,$$\\\\cos\\\\theta+i\\\\sin\\\\theta=1+0i$$– Cameron Buie Jul 18 '15 at 22:24 \\\\cos \\\\theta = 1 and \\\\sin \\\\theta = 0, so \\\\theta = 360 degrees/2\\\\pi. – Mathy Person Jul 18 '15 at 22:30 The roots are e^{2k\\\\pi i/7} (k=0,\\\\ldots,6). Write them as r^k where r = e^{2\\\\pi i/7}. So the sum is (using the simple formula for summing a finite geometric series)$$\\\\sum_{k=0}^6 r^k = \\\\frac{1-r^7}{1-r} = \\\\frac{1 - e^{2\\\\pi i}}{1-e^{2\\\\pi i/7}} = 0. This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not summing up the roots, but rather the arguments of said roots lying in $[0^\\\\circ,360^\\\\circ).$ – Cameron Buie Jan 24 '14 at 17:15 $\\\\def\\\\cis{\\\\operatorname{cis}}$Let's consider a more general problem. We have $a=r\\\\cis\\\\alpha$ and we write its $n$th roots as \\\\begin{gather} \\\\sqrt[n]{r}\\\\cis\\\\frac{\\\\beta}{n}\\\\\\\\ \\\\sqrt[n]{r}\\\\cis\\\\left(\\\\frac{\\\\beta}{n}+\\\\frac{2\\\\pi}{n}\\\\right)\\\\\\\\ \\\\sqrt[n]{r}\\\\cis\\\\left(\\\\frac{\\\\beta}{n}+2\\\\frac{2\\\\pi}{n}\\\\right)\\\\\\\\ \\\\dots\\\\\\\\ \\\\sqrt[n]{r}\\\\cis\\\\left(\\\\frac{\\\\beta}{n}+(n-1)\\\\frac{2\\\\pi}{n}\\\\right) \\\\end{gather} Can you sum up all those angles? Use $360$ instead of $2\\\\pi$ if you so prefer or are forced to.\",\n \"to get the imaginary … 5. ### precalc express the roots of unity in standard form a+bi. 1.) cube roots of unity 2.) fourth roots of unity 3.) sixth roots of unity 4.) square roots of unity 6. ### precalculus express the roots of unity in standard form a+bi. 1.) cube roots of unity 2.) fourth roots of unity 3.) sixth roots of unity 4.) square roots of unity 7. ### PRECALCULUS find the cube roots of the complex number. 27((cos(11pi/6)) + i(sin(11pi/6)) 8. ### PRECALCULUS 1.) determine two pairs of polar coordinates for the point (-4sqrt3, 4sqrt3)with 0degrees less than or equal to theta less than or equal to 360degrees 2.) Use DeMoivre's Theorem to find (1-i)^10. write your answer in the form a+bi. … The roots of $z^7 = -\\\\frac{1}{\\\\sqrt{2}} - \\\\frac{i}{\\\\sqrt{2}}$are $\\\\text{cis } \\\\theta_1$, $\\\\text{cis } \\\\theta_2$, $\\\\dots$, $\\\\text{cis } \\\\theta_7$, where $0^\\\\circ \\\\le \\\\theta_k < 360^\\\\circ$ for all $1 \\\\le k \\\\le 7$. Find $\\\\theta_1 … 10. ### Precalculus The roots of $z^7 = -\\\\frac{1}{\\\\sqrt{2}} - \\\\frac{i}{\\\\sqrt{2}}$are$\\\\text{cis } \\\\theta_1$,$\\\\text{cis } \\\\theta_2$,$\\\\dots$,$\\\\text{cis } \\\\theta_7$, where$0^\\\\circ \\\\le \\\\theta_k < 360^\\\\circ$for all$1 \\\\le k \\\\le 7$. Find$\\\\theta_1 … More Similar Questions\",\n \"These correspond to the complex numbers $$w_{\\\\text{\\\\tiny \\\\(0$$}} = 2 \\\\text{cis}(0)\\\\), $$w_{\\\\text{\\\\tiny \\\\(1$$}} = 2 \\\\text{cis}\\\\left(\\\\frac{2\\\\pi}{3}\\\\right)\\\\) and $$w_{\\\\text{\\\\tiny \\\\(2$$}} = 2 \\\\text{cis}\\\\left(\\\\frac{4\\\\pi}{3}\\\\right)\\\\), respectively. Writing these out in rectangular form yields $$w_{\\\\text{\\\\tiny \\\\(0$$}} = 2\\\\), $$w_{\\\\text{\\\\tiny \\\\(1$$}} = -1 + i\\\\sqrt{3}\\\\) and $$w_{\\\\text{\\\\tiny \\\\(2$$}} = -1-i\\\\sqrt{3}\\\\). While this process seems a tad more involved than our previous factoring approach, this procedure can be generalized to find, for example, all of the fifth roots of $$32$$. (Try using Chapter \\\\ref{Polynomials} techniques on that!) If we start with a generic complex number in polar form $$z = |z| \\\\text{cis}(\\\\theta)$$ and solve $$w^{n} = z$$ in the same manner as above, we arrive at the following theorem. Note: The $$n^{th}$$ roots of a Complex Number Let $$z \\\\neq 0$$ be a complex number with polar form $$z = r\\\\text{cis}(\\\\theta)$$. For each natural number $$n$$, $$z$$ has $$n$$ distinct $$n^{\\\\text{th}}$$ roots, which we denote by $$w_{\\\\text{\\\\tiny$$0\\\\)}}\\\\), $$w_{\\\\text{\\\\tiny$$1\\\\)}}\\\\), \\\\ldots, $$w_{\\\\text{\\\\tiny$$n-1\\\\)}}\\\\), and they are given by the formula \\\\index{\\\\)n^{\\\\textrm{th}}\\\\) root ! of a complex number} \\\\index{complex number ! $$n^{\\\\textrm{th}}$$ root} $w_{\\\\text{k}} = \\\\sqrt[n]{r}\\\\text{cis}\\\\left(\\\\frac{\\\\theta}{n} + \\\\frac{2\\\\pi}{n} k \\\\right)$ The proof of Theorem \\\\ref{nthrootscomplexthm} breaks into to two parts: first, showing that each $$w_{\\\\text{k}}$$ is an $$n^{\\\\text{th}}$$ root, and second, showing that the set $$\\\\left\\\\{ w_{\\\\text{k}} \\\\, | \\\\, k = 0, 1, \\\\ldots, (n-1)\\\\right\\\\}$$ consists of\",\n \"+0 # help +1 50 1 The number $$\\\\text{cis } 75^\\\\circ + \\\\text{cis } 83^\\\\circ + \\\\text{cis } 91^\\\\circ + \\\\dots + \\\\text{cis } 147^\\\\circ$$ is expressed in the form $$r \\\\, \\\\text{cis } \\\\theta$$, where $$0 \\\\le \\\\theta < 360^\\\\circ$$. Find $$\\\\theta$$ in degrees. Dec 20, 2019\",\n \"For L cuts C at two distinct points, there should be two distinct roots in (*), hence $\\\\Delta > 0$. Thus $(-6\\\\cos\\\\theta)^2 - 4(6) > 0$ $\\\\cos^2\\\\theta > \\\\frac{2}{3}$ $-\\\\cos^{-1}(\\\\sqrt{\\\\frac{2}{3}}) < \\\\theta < \\\\cos^{-1}(\\\\sqrt{\\\\frac{2}{3}})$ It may not be easy to yield the last inequality above, just sketch a graph to see it. F.5 students, don’t turn a blind eye to this problem, it IS IN THE SYLLABUS.\",\n \"summing up the roots, but rather the arguments of said roots lying in $[0^\\\\circ,360^\\\\circ).$ – Cameron Buie Jan 24 '14 at 17:15 $\\\\def\\\\cis{\\\\operatorname{cis}}$Let's consider a more general problem. We have $a=r\\\\cis\\\\alpha$ and we write its $n$th roots as \\\\begin{gather} \\\\sqrt[n]{r}\\\\cis\\\\frac{\\\\beta}{n}\\\\\\\\ \\\\sqrt[n]{r}\\\\cis\\\\left(\\\\frac{\\\\beta}{n}+\\\\frac{2\\\\pi}{n}\\\\right)\\\\\\\\ \\\\sqrt[n]{r}\\\\cis\\\\left(\\\\frac{\\\\beta}{n}+2\\\\frac{2\\\\pi}{n}\\\\right)\\\\\\\\ \\\\dots\\\\\\\\ \\\\sqrt[n]{r}\\\\cis\\\\left(\\\\frac{\\\\beta}{n}+(n-1)\\\\frac{2\\\\pi}{n}\\\\right) \\\\end{gather} Can you sum up all those angles? Use $360$ instead of $2\\\\pi$ if you so prefer or are forced to. -\",\n \"are: $\\\\set {\\\\sqrt [6] 2 \\\\cis 20 \\\\degrees, \\\\sqrt [6] 2 \\\\cis 80 \\\\degrees, \\\\sqrt [6] 2 \\\\cis 140 \\\\degrees, \\\\sqrt [6] 2 \\\\cis 200 \\\\degrees, \\\\sqrt [6] 2 \\\\cis 260 \\\\degrees, \\\\sqrt [6] 2 \\\\cis 320 \\\\degrees}$ ### Roots of $z^8 + 1 = 0$ The roots of the polynomial: $z^8 + 1 = 0$ are: $\\\\set {\\\\cos \\\\dfrac {\\\\paren {2 k + 1} \\\\pi} 8 + i \\\\sin \\\\dfrac {\\\\paren {2 k + 1} \\\\pi} 8: k \\\\in \\\\set {0, 1, \\\\ldots, 7} }$\",\n \"# roots of unity questions (1 Viewer) #### poptarts12345 ##### Member 1. Let w be a seventh root of unity. Find the equation of the quadratic polynomial with roots w+w^2+w^4 and w^3+w^5+w^6. ANS : x^2+x+2 2. Let w be the principal nth root of unity. Prove that w conjugate = w^(n-1) #### idkkdi ##### Well-Known Member 1. Let w be a seventh root of unity. Find the equation of the quadratic polynomial with roots w+w^2+w^4 and w^3+w^5+w^6. ANS : x^2+x+2 2. Let w be the principal nth root of unity. Prove that w conjugate = w^(n-1) w^7-1 = 0. (w-1)(w^6+w^5+...+1 ) = 0. For w =/ 1, (w^6+w^5+...+1) = 0. clearly w+w^2+w^4 + w^3+w^5+w^6 = -1. (w+w^2+w^4)(w^3+w^5+w^6) = w^4 + w^6 + w^7 + w^5 + w^7 + w^8 + w^7 + w^9 + w^10 = w^4 + w^5 + w^6 + w^7 + w^7 + w + w^2 + w^3 + w^7 = -1 + 3 = 2. (note that w is assumed to be the principal root of unity) let w = cis theta w cong = cis -theta w^n-1 = cis (n-1) theta = cis n theta/cis theta = cis -theta #### Qeru ##### Member 1. Let w be a seventh root of unity. Find the equation of the quadratic polynomial with roots w+w^2+w^4 and\",\n \"# How to find the roots of $x³-2$? I'm trying to find the roots of $x^3 -2$, I know that one of the roots are $\\\\sqrt[3] 2$ and $\\\\sqrt[3] {2}e^{\\\\frac{2\\\\pi}{3}i}$ but I don't why. The first one is easy to find, but the another two roots? I need help Thank you If $\\\\omega^3 = 1$ and $x^3 = 2$ then $(\\\\omega x)^3 = \\\\omega^3 x^3 = 2$. Possible values of $\\\\omega$ are $e^{\\\\frac{1}{3}2 i \\\\pi}$, $e^{\\\\frac{2}{3}2 i \\\\pi}$ and $e^{\\\\frac{3}{3}2 i \\\\pi}$. This is because $1 = e^{2 i \\\\pi} = (e^{\\\\frac{1}{k} 2 i \\\\pi})^k$. So the solutions of $x^3 - 2 = 0$ are $e^{\\\\frac{1}{3}2 i \\\\pi} \\\\sqrt[3]{2}$, $e^{\\\\frac{2}{3}2 i \\\\pi} \\\\sqrt[3]{2}$ and $\\\\sqrt[3]{2}$. Hint: $x^3-2=(x-2^{\\\\frac{1}{3}})(x^2+2^{\\\\frac{1}{3}}x+2^{\\\\frac{2}{3}})$. Let $x=z\\\\sqrt[3]2$. Then $0=x^3-2=2z^3-2$ and so $z^3=1$. Write $z=e^{i\\\\theta}$ and solve for $\\\\theta$.\",\n \"told to do so, we leave this as a good, but messy, exercise. 4. To find the five fifth roots of $$1$$, we write $$1 = 1 \\\\text{cis}(0)$$. We have $$r = 1$$, $$\\\\theta = 0$$ and $$n = 5$$. Since $$\\\\sqrt[5]{1} = 1$$, the roots are $$w_{\\\\text{\\\\tiny$$0\\\\)}} = \\\\text{cis}(0) = 1\\\\), $$w_{\\\\text{\\\\tiny$$1\\\\)}} = \\\\text{cis}\\\\left(\\\\frac{2\\\\pi}{5}\\\\right)\\\\), $$w_{\\\\text{\\\\tiny$$2\\\\)}} = \\\\text{cis}\\\\left(\\\\frac{4\\\\pi}{5}\\\\right)\\\\), $$w_{\\\\text{\\\\tiny$$3\\\\)}} = \\\\text{cis}\\\\left(\\\\frac{6\\\\pi}{5}\\\\right)\\\\) and $$w_{\\\\text{\\\\tiny$$4\\\\)}} = \\\\text{cis}\\\\left(\\\\frac{8\\\\pi}{5}\\\\right)\\\\). The situation here is even graver than in the previous example, since we have not developed any identities to help us determine the cosine or sine of $$\\\\frac{2\\\\pi}{5}$$. At this stage, we could approximate our answers using a calculator, and we leave this as an exercise. Now that we have done some computations using Theorem \\\\ref{nthrootscomplexthm}, we take a step back to look at things geometrically. Essentially, Theorem \\\\ref{nthrootscomplexthm} says that to find the $$n^{\\\\text{th}}$$ roots of a complex number, we first take the $$n^{\\\\text{th}}$$ root of the modulus and divide the argument by $$n$$. This gives the first root $$w_{\\\\text{\\\\tiny$$0\\\\)}}\\\\). Each succeessive root is found by adding $$\\\\frac{2\\\\pi}{n}$$ to the argument, which amounts to rotating $$w_{\\\\text{\\\\tiny$$0\\\\)}}\\\\) by $$\\\\frac{2\\\\pi}{n}$$ radians. This results in $$n$$ roots, spaced equally around the complex plane. As an example of this, we plot our answers to number \\\\ref{fourthrootsneg16} in Example \\\\ref{nthrootscomplexex} below. We have only glimpsed at the\",\n \"# Difference between revisions of \\\"1996 AIME Problems/Problem 11\\\" ## Problem Let $\\\\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\\\\mathrm {P}=r(\\\\cos{\\\\theta^{\\\\circ}}+i\\\\sin{\\\\theta^{\\\\circ}})$, where $0 and $0\\\\leq \\\\theta <360$. Find $\\\\theta$. ## Solution 1 $\\\\begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \\\\frac{z^5-1}{z-1}\\\\\\\\ 0 &=& \\\\frac{(z^5 - 1)(z(z-1)+1)}{z-1} = \\\\frac{(z^2-z+1)(z^5-1)}{z-1} \\\\end{eqnarray*}$ Thus $z^5 = 1, z \\\\neq 1 \\\\Longrightarrow z = \\\\mathrm{cis}\\\\ 72, 144, 216, 288$, or $z^2 - z + 1 = 0 \\\\Longrightarrow z = \\\\frac{1 \\\\pm \\\\sqrt{-3}}{2} = \\\\mathrm{cis}\\\\ 60, 300$ (see cis). Discarding the roots with negative imaginary parts (leaving us with $\\\\mathrm{cis} \\\\theta,\\\\ 0 < \\\\theta < 180$), we are left with $\\\\mathrm{cis}\\\\ 60, 72, 144$; their product is $P = \\\\mathrm{cis} (60 + 72 + 144) = \\\\mathrm{cis} \\\\boxed{276}$. ## Solution 2 Let $w =$ the fifth roots of unity, except for $1$. Then $w^6 + w^4 + w^3 + w^2 + 1 = w^4 + w^3 + w^2 + w + 1 = 0$, and since both sides have the fifth roots of unity as roots, we have $z^4 + z^3 + z^2 + z + 1 | z^6 + z^4 + z^3 + z^2 + 1$. Long\",\n \"{th }}$$ root of unity. Then all the $$n^{\\\\text {th }}$$ roots of $$z$$ are $$\\\\left\\\\{w_{0}, \\\\omega w_{0}, \\\\omega^{2} w_{0}, \\\\ldots, \\\\omega^{n-1} w_{0}\\\\right\\\\}$$. ## Tartaglia-Cardano Revisited Let us consider again Example 9.9. We wanted to find the cube roots of $\\\\zeta_{\\\\pm}=\\\\frac{-1 \\\\pm \\\\sqrt{-3}}{2} .$ If we take the $$+$$ sign, we get $\\\\zeta_{+}=\\\\operatorname{Cis}(2 \\\\pi / 3),$ and if we take the - sign, we get $\\\\zeta_{-}=\\\\operatorname{Cis}(4 \\\\pi / 3) .$ So $$\\\\zeta_{+}$$has 3 roots, namely $\\\\left\\\\{\\\\operatorname{Cis}\\\\left(\\\\frac{2 \\\\pi}{9}+\\\\frac{2 k \\\\pi}{3}\\\\right): k=0,1,2\\\\right\\\\} \\\\text {, }$ and $$\\\\zeta_{-}$$has 3 roots, namely $\\\\left\\\\{\\\\operatorname{Cis}\\\\left(\\\\frac{4 \\\\pi}{9}+\\\\frac{2 k \\\\pi}{3}\\\\right): k=0,1,2\\\\right\\\\} \\\\text {, }$ Knowing $$w$$, we want to find $$x$$, which for Example $$9.9$$ is given by $$w+1 / w$$. For any number $$w$$ that can be written as $$\\\\operatorname{Cis}(\\\\theta)$$ (i.e. any complex number of modulus 1), we have \\\\begin{aligned} w+\\\\frac{1}{w} &=\\\\cos \\\\theta+i \\\\sin \\\\theta+\\\\cos (-\\\\theta)+i \\\\sin (-\\\\theta) \\\\\\\\ &=2 \\\\cos \\\\theta . \\\\end{aligned} Therefore the roots of the polynomial given in (9.10) are $\\\\left\\\\{2 \\\\cos \\\\frac{2 \\\\pi}{9}, 2 \\\\cos \\\\frac{8 \\\\pi}{9}, 2 \\\\cos \\\\frac{14 \\\\pi}{9}, 2 \\\\cos \\\\frac{4 \\\\pi}{9}, 2 \\\\cos \\\\frac{10 \\\\pi}{9}, 2 \\\\cos \\\\frac{16 \\\\pi}{9}\\\\right\\\\} .$ Are these 6 different roots? Theorem $$4.10$$ says that $$p$$ can have at most 3 different roots. As $$\\\\cos (\\\\theta)=\\\\cos (2 \\\\pi-\\\\theta)$$, we see our set (9.27) may be written as $\\\\left\\\\{2 \\\\cos \\\\frac{2 \\\\pi}{9}, 2\",\n \"has got n different n-th roots, $$\\\\sqrt [n] {1} = \\\\cos(\\\\frac{2\\\\pi}{n}k) + i \\\\sin (\\\\frac{2\\\\pi}{n}k)$$ k=0, ...n-1. 8. Feb 17, 2005 Thank you for all the help! I think I finally get what you're saying... $$z=i\\\\, ^{\\\\frac{4}{3}}=\\\\left( i^4 \\\\right)^{1/3} = \\\\sqrt[3]{1}$$ $$h= 1 = \\\\mbox{cis } 0$$ $$z=h_w ^3 = \\\\mbox{ cis } \\\\left( \\\\frac{0+2k\\\\pi}{3} \\\\right) = \\\\cos \\\\left( \\\\frac{2k\\\\pi}{3} \\\\right) +\\\\mbox{ } i\\\\mbox{ } \\\\sin \\\\left( \\\\frac{2k\\\\pi}{3} \\\\right) = -\\\\frac{1}{2} +\\\\mbox{ } i \\\\mbox{ } \\\\frac{\\\\sqrt{3}}{2}$$\",\n \"1. Cube roots of -8 I am totally stuck on this problem. I need to find the cube roots of -8 in polar coordinates. I know that -8 lies on the real axis, so the angle of reference is $\\\\pi$. I also know that $cos \\\\pi=-1$. I know that the formula I need to use is cube root of $|z|= cis \\\\frac{\\\\theta}{3}$ and I know that $cos \\\\frac{\\\\theta}{3}=Sqrt(\\\\frac{1+cos\\\\theta}{3})$, $sin\\\\theta=Sqrt(\\\\frac{1-cos\\\\theta}{3})$ and the cube root $=2(cos\\\\frac{\\\\theta}{3}+i sin\\\\frac{\\\\theta}3}$. I must be tired and doing something wrong with the signs or something. I keep getting a zero for the first term and I am pretty sure the answer is $1+i \\\\sqrt{3}$. Please help! Thanks! 2. Re: Complex Analysis - Cube roots of -8 First of all, you'll need a 2, somewhere. |z| = cis(x/3) really doesn't make quite enough sense. It is as if you ALMOST know where you are going. Start with $z^{3} = -8 = 8\\\\cdot cis(\\\\pi)$ Just apply the theorem from DeMoivre. $z = 8^{\\\\frac{1}{3}}\\\\cdot cis\\\\left(\\\\frac{\\\\pi +2k\\\\pi}{3}\\\\right)$ for k = 0, 1, 2 k = 0 leads to $2\\\\cdot cis(\\\\pi/3)$ k = 1 leads to $2\\\\cdot cis(\\\\pi)$ k = 2 leads to $2\\\\cdot cis(5\\\\pi/3)$ 3. Re: Complex Analysis - Cube roots of -8 I think you're right; I ALMOST know where I am going! OK... I began with $2\",\n \"still a little confused about how to approach this particular question. Once I have the i term, where do i go from there? Is there any particular formula that i need to use? Thanks, Function • August 21st 2009, 02:30 AM mr fantastic Quote: Originally Posted by function Hey Mr. Fantastic, Thanks for your help, although I'm still a little confused about how to approach this particular question. Once I have the i term, where do i go from there? Is there any particular formula that i need to use? Thanks, Function You have $z^5 = r^5 \\\\text{cis} (5 \\\\theta) = \\\\text{cis} \\\\left( \\\\frac{\\\\pi}{2} + 2n \\\\pi\\\\right)$ where $n$ is an integer. Therefore $z = r \\\\text{cis} (\\\\theta) = \\\\text{cis} \\\\left( \\\\frac{\\\\pi}{10} + \\\\frac{2n \\\\pi}{5} \\\\right)$. Now substitute five consecutive values of n to get the five distinct roots. • August 22nd 2009, 01:59 AM pacman Roots in the complex plane, z1 = i, z2 = (-1)^(9/10), z3 = -(-1)^(3/10), z4 = -(-1)^(7/10), z5 = (-1)(1/10)\",\n \"root, the negative one leads to t<0. thus $$t = \\\\frac{1}{\\\\sqrt{2c}} \\\\int_d^0 \\\\sqrt{ \\\\frac{ dz}{ d - z}} \\\\. d z$$ The substitution z = d cos2(theta) makes this doable. 4. Nov 10, 2005 Euclid Yeah, that's what I indicated I tried above. It appeared to fail the first time I did it, but the second time it worked out. Thanks!\",\n \"\\\\frac{2^k\\\\pi i}{9}}, k=1,2,3$ $\\\\text{By letting }y=z+\\\\frac{1}{z}\\\\text{ deduce that }\\\\cos \\\\frac{2\\\\pi}{9}+\\\\cos \\\\frac{4\\\\pi}{9}+\\\\cos \\\\frac{8\\\\pi}{9}=0$ $\\\\text{I'm not sure what to do after }y^3-y+1=0$ Originally Posted by leehuan Sigh... $\\\\text{Proven already: The roots of }z^6+z^3+1=0\\\\text{ are }e^{\\\\pm \\\\frac{2^k\\\\pi i}{9}}, k=1,2,3$ $\\\\text{By letting }y=z+\\\\frac{1}{z}\\\\text{ deduce that }\\\\cos \\\\frac{2\\\\pi}{9}+\\\\cos \\\\frac{4\\\\pi}{9}+\\\\cos \\\\frac{8\\\\pi}{9}=0$ $\\\\text{I'm not sure what to do after }y^3-y+1=0$ $\\\\noindent As you have already proven, the roots of p(z):= z^{6} + z^{3} + 1 = 0 are z_1 = e^{i \\\\frac{2\\\\pi}{9}}, z_2 = e^{i \\\\frac{4\\\\pi}{9}}, z_{3} = e^{i \\\\frac{8\\\\pi}{9}}, and the conjugates of these, \\\\overline{z_j}, j=1,2,3.$ $\\\\noindent Note that if w = e^{i \\\\theta}, then w + \\\\overline{w} = 2\\\\cos \\\\theta. Since the sum of roots of p is 0 (as there is no quintic term), we have \\\\sum _{j=1}^{3}\\\\left(z_j + \\\\overline{z_j}\\\\right) = 0, where z_j = e^{i \\\\frac{2^{j}\\\\pi}{9}} (the L.H.S. is just the sum of the roots). Since the L.H.S. equals \\\\sum _{j=1}^{3} 2\\\\cos \\\\left(\\\\frac{2^{j}\\\\pi}{9}\\\\right), dividing the last equation through by 2 yields the result.$ $\\\\noindent Realised I didn't do it by doing the y = z + z^{-1} method. Using this method, you got it down to y^{3} -y + 1=0, where z is a root of the equation z^6 + z^3 + 1 = 0. Since the roots of this equation produce exactly three values of z+ z^{-1}, namely 2\\\\cos \\\\left(\\\\frac{2^j\",\n \"you : what are the complex cube roots of $$8i$$? Thankfully, these can be answered via a use of De Moivre's theorem, which will tell you that the roots are precisely those of the form $$\\\\sqrt[3]{8} e^{\\\\frac {i\\\\pi}{6} + \\\\frac{2in \\\\pi}{3}}$$ where $$n=0,1,2$$. Now, this can be simplified: $$\\\\sqrt[3]{8} e^{\\\\frac \\\\pi{6} + \\\\frac{2n \\\\pi}{3}} = 2 \\\\operatorname{cis} \\\\frac \\\\pi {6}, 2 \\\\operatorname{cis} \\\\frac {5\\\\pi} {6}, 2 \\\\operatorname{cis} \\\\frac {9\\\\pi} {6}$$ Expand using the fact that $$\\\\operatorname{cis} \\\\theta = \\\\cos \\\\theta + i \\\\sin \\\\theta$$, and don't forget to subtract $$2i - \\\\sqrt 3$$ from each of the cube roots. This gives an answer that involves only sine and cosine evaluation. Alternately, since you figured out a root, you could also have done polynomial division, but for that you would have to expand the cube root, which is laborious.\"\n]"},"s_retrievals":{"kind":"list like","value":["# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Rod Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? - The roots are $e^{2k\\pi i/7}$ ($k=0,\\ldots,6$). Write them as $r^k$ where $r = e^{2\\pi i/7}$. So the sum is (using the simple formula for summing a finite geometric series) $$\\sum_{k=0}^6 r^k = \\frac{1-r^7}{1-r} = \\frac{1 - e^{2\\pi i}}{1-e^{2\\pi i/7}} = 0$$. - This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not","# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Mr.Fry Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? If I am understanding correctly...for $k=1$, we take $$\\operatorname{cis}(7\\theta_1)=\\operatorname{cis} 225^\\circ$$ and then somehow determine $\\theta_1$, and repeat for $k=2,3,4,5,6,7$? – Mathy Person Jul 18 '15 at 21:59 No, that's getting further from the answer. Rather, think about the solutions set for the equation $$\\operatorname{cis}\\theta=1. This will let you get rid of the \\operatorname{cis}, and find all solutions to the original","34^\\circ \\cdot 5 \\ \\text{cis} \\ 16^\\circ \\cdot \\frac{1}{2} \\ \\text{cis} \\ 100^\\circ\\\\& =4 \\cdot 5 \\cdot \\frac{1}{2} \\cdot \\text{cis} \\ (34^\\circ+16^\\circ+100^\\circ)=10 \\ \\text{cis} \\ 150$ Note how much easier it is to do products and quotients in trigonometric polar form. #### Practice Translate the following complex numbers from trigonometric polar form to rectangular form. 1. $5 \\ \\text{cis} \\ 270^\\circ$ 2. $2 \\ \\text{cis} \\ 30^\\circ$ 3. $-4 \\ \\text{cis} \\ \\frac{\\pi}{4}$ 4. $6 \\ \\text{cis} \\ \\frac{\\pi}{3}$ 5. $2 \\ \\text{cis} \\ \\frac{5 \\pi}{2}$ Translate the following complex numbers from rectangular form into trigonometric polar form. 6. $2-i$ 7. $5+12i$ 8. $6i+8$ 9. $i$ Complete the following calculations and simplify. 10. $2 \\ \\text{cis} \\ 22^\\circ \\cdot \\frac{1}{5} \\ \\text{cis} \\ 15^\\circ \\cdot 3 \\ \\text{cis} \\ 95^\\circ$ 11. $9 \\ \\text{cis} \\ 98^\\circ \\div 3 \\ \\text{cis} \\ 12^\\circ$ 12. $15 \\ \\text{cis} \\ \\frac{\\pi}{4} \\cdot 2 \\ \\text{cis} \\ \\frac{\\pi}{6}$ 13. $-2 \\ \\text{cis} \\ \\frac{2 \\pi}{3} \\div 15 \\ \\text{cis} \\ \\frac{7 \\pi}{6}$ Let $z_1=r_1 \\cdot \\text{cis} \\ \\theta_1$ and $z_2=r_2 \\cdot \\text{cis} \\ \\theta_2$ with $r_2 \\neq 0$ . 14. Use the trigonometric sum and difference identities to prove that $z_1 \\cdot z_2=r_1 \\cdot r_2 \\cdot \\text{cis} \\ (\\theta_1+\\theta_2)$ . 15. Use the trigonometric sum and difference identities to prove that $z_1","\\pi}{4}\\right)\\right)=4 \\left(-\\frac{\\sqrt{2}}{2}+\\frac{\\sqrt{2}}{2}i\\right)=-2 \\sqrt{2}+2 \\sqrt{2}i$ Example C Divide the following complex numbers. $\\frac{4 \\ \\text{cis} \\ 32^\\circ}{2 \\ \\text{cis} \\ 2^\\circ}$ Solution: $\\frac{4 \\ \\text{cis} \\ 32^\\circ}{2 \\ \\text{cis} \\ 2^\\circ}=\\frac{4}{2} \\ \\text{cis} \\ (32^\\circ-2^\\circ)=2 \\ \\text{cis} \\ (30^\\circ)$ Concept Problem Revisited In rectangular coordinates: $\\left(1+\\sqrt{3}i\\right)\\left(\\sqrt{2}-\\sqrt{2}i\\right)=\\sqrt{2}-\\sqrt{2}i+\\sqrt{6}i+\\sqrt{6}$ In trigonometric polar coordinates, $1+\\sqrt{3}i=2 \\ \\text{cis} \\ 60^\\circ$ and $\\sqrt{2}-\\sqrt{2}i=2 \\ \\text{cis}-45^\\circ$ . Therefore: $\\left(1+\\sqrt{3}i\\right) \\left(\\sqrt{2}-\\sqrt{2}i \\right)=2 \\ \\text{cis} \\ 60^\\circ \\cdot 2 \\ \\text{cis} - 45^\\circ=4 \\ \\text{cis} \\ 105^\\circ$ #### Vocabulary Trigonometric polar form of a complex number describes the location of a point on the complex plane using the angle and the radius of the point. The abbreviation $r \\cdot \\text{cis} \\ \\theta$ stands for $r \\cdot (\\cos \\theta+i \\cdot \\sin \\theta)$ and is how trigonometric polar form is typically observed. #### Guided Practice 1. Translate the following complex number from trigonometric polar form to rectangular form. $5 \\ \\text{cis} \\ 270^\\circ$ 2. Translate the following complex number from rectangular form into trigonometric polar form. 8 3. Multiply the following complex numbers in trigonometric polar form. $4 \\ \\text{cis} \\ 34^\\circ \\cdot 5 \\ \\text{cis} \\ 16^\\circ \\cdot \\frac{1}{2} \\ \\text{cis} \\ 100^\\circ$ 1. $5 \\ \\text{cis} \\ 270^\\circ =5 (\\cos 270^\\circ+i \\cdot \\sin 270^\\circ)=5 (0-i)=-5 i$ 2. $8=8 \\ \\text{cis} \\ 0^\\circ$ 3. $& 4 \\ \\text{cis} \\","and third quadrants. $\\scriptsize x={{360}^\\circ}-{{138.59}^\\circ}={{221.41}^\\circ}$ General solution: $\\scriptsize x={{138.59}^\\circ}+k{{.360}^\\circ}\\text{ or }x={{221.49}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ For the interval $\\scriptsize \\left[ {{{0}^\\circ},{{{180}}^\\circ}} \\right]$: $\\scriptsize x={{90}^\\circ}\\text{ or }x={{138.59}^\\circ}$ 4. . \\scriptsize \\begin{align*}6\\cos \\theta -5 & =\\displaystyle \\frac{4}{{\\cos \\theta }}\\quad \\quad \\cos \\theta \\ne 0\\therefore \\theta \\ne {{90}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore 6{{\\cos }^{2}}\\theta -5\\cos \\theta -4 & =0\\\\\\therefore (3\\cos \\theta -4)(2\\cos \\theta +1) & =0\\\\\\therefore \\cos \\theta =\\displaystyle \\frac{4}{3}\\text{ } & \\text{or }\\cos \\theta =-\\displaystyle \\frac{1}{2}\\end{align*} $\\scriptsize \\cos \\theta =\\displaystyle \\frac{4}{3}$ – No solution $\\scriptsize \\cos \\theta =-\\displaystyle \\frac{1}{2}$: Ref angle: $\\scriptsize \\theta ={{120}^\\circ}$ Cosine is negative in the second and third quadrants. $\\scriptsize \\theta ={{360}^\\circ}-({{120}^\\circ})={{240}^\\circ}$ General solution: $\\scriptsize \\theta ={{120}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{240}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ 5. . \\scriptsize \\begin{align*}\\cos 2x-\\cos x+1 & =0\\\\\\therefore 2{{\\cos }^{2}}x-1-\\cos x+1 & =0\\\\\\therefore 2{{\\cos }^{2}}x-\\cos x & =0\\\\\\therefore \\cos x(2\\cos x-1) & =0\\\\\\therefore \\cos x=0\\text{ } & \\text{or }\\cos x=\\displaystyle \\frac{1}{2}\\end{align*} $\\scriptsize \\cos x=0$: Ref angle: $\\scriptsize x={{90}^\\circ}$ General solution: $\\scriptsize x={{90}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}$ $\\scriptsize \\cos x=\\displaystyle \\frac{1}{2}$ Ref angle: $\\scriptsize \\theta ={{60}^\\circ}$ Cosine is positive in the first and fourth quadrants. $\\scriptsize \\theta ={{360}^\\circ}-{{60}^\\circ}={{300}^\\circ}$ General solution: $\\scriptsize \\theta ={{60}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{300}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ For the interval $\\scriptsize \\left[ {{{0}^\\circ},{{{360}}^\\circ}} \\right]$: $\\scriptsize x=\\text{6}{{\\text{0}}^\\circ}\\text{ or }x={{90}^\\circ}\\text{ or }x={{180}^\\circ}\\text{ or }x=\\text{30}{{\\text{0}}^\\circ}\\text{ or }x=\\text{36}{{\\text{0}}^\\circ}$ 2. $\\scriptsize \\cos 2\\theta =2{{\\cos }^{2}}\\theta -1$ 1. . \\scriptsize \\begin{align*}\\text{LHS}=\\cos 2\\theta +3\\cos \\theta -1\\\\=2{{\\cos }^{2}}\\theta -1+3\\cos \\theta -1\\\\=2{{\\cos }^{2}}\\theta +3\\cos","solution is correct. ### Example 2.2 Determine the general solution for $\\scriptsize 7\\cos 2\\theta +4=0$. Solution \\scriptsize \\displaystyle \\begin{align*}7\\cos 2\\theta +4&=0\\\\\\therefore \\cos 2\\theta &=-\\displaystyle \\frac{4}{7}\\end{align*} Step 1: Use a calculator to determine the reference angle \\scriptsize \\displaystyle \\begin{align*}\\cos 2\\theta &=-\\displaystyle \\frac{4}{7}\\\\\\therefore 2\\theta &={{124.8}^\\circ}\\end{align*} Note: We keep working with the reference angle of $\\scriptsize 2\\theta$ until we generate the general solution. Step 2: Use the CAST diagram to determine any other possible solutions Our equation is $\\scriptsize \\displaystyle \\cos 2\\theta =-\\displaystyle \\frac{4}{7}$. $\\scriptsize \\cos 2\\theta \\lt 0$. Cosine is negative in the second and third quadrants. Our reference angle is in the second quadrant. Second quadrant: $\\scriptsize 2\\theta ={{124.8}^\\circ}$ Third quadrant: $\\scriptsize \\cos ({{360}^\\circ}-\\theta )=\\cos \\theta$ $\\scriptsize 2\\theta ={{360}^\\circ}-{{124.8}^\\circ}={{235.2}^\\circ}$. Step 3: Generate the general solution \\scriptsize \\begin{align*}2\\theta & ={{124.8}^\\circ}+k{{.360}^\\circ}\\text{ or 2}\\theta ={{235.2}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\text{ }\\\\\\therefore \\theta & ={{62.4}^\\circ}+k{{.180}^\\circ}\\text{ or }\\theta ={{117.6}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\text{ }\\end{align*} Step 4: Check your general solution $\\scriptsize k=2$: $\\scriptsize \\therefore \\theta ={{62.4}^\\circ}+2\\times {{180}^\\circ}={{422.4}^\\circ}\\text{ or }\\theta ={{117.6}^\\circ}+2\\times {{180}^\\circ}={{477.6}^\\circ}$ $\\scriptsize \\displaystyle \\cos (2\\times {{422.4}^\\circ})=-\\displaystyle \\frac{4}{7}$ $\\scriptsize \\displaystyle \\cos (2\\times {{477.6}^\\circ})=-\\displaystyle \\frac{4}{7}$ ### Take note! The general solutions for equations involving the three basic trigonometric ratios can be written as follows: If $\\scriptsize \\sin \\theta =x$ then: $\\scriptsize \\theta =\\left( {{{{\\sin }}^{{-1}}}x+k{{{.360}}^\\circ}} \\right)\\text{ or }\\theta =\\left( {\\left( {{{{180}}^\\circ}-{{{\\sin }}^{{-1}}}x} \\right)+k{{{.360}}^\\circ}} \\right),k\\in \\mathbb{Z}$ If $\\scriptsize \\cos \\theta =x$ then: $\\scriptsize \\theta","\\theta & =-0.7\\end{align*} Ref angle: $\\scriptsize \\theta ={{44.4}^\\circ}$ $\\scriptsize \\sin \\theta \\lt 0$: $\\scriptsize \\theta ={{180}^\\circ}+{{44.4}^\\circ}={{224.4}^\\circ}\\text{ or }\\theta ={{360}^\\circ}-{{44.4}^\\circ}={{315.6}^\\circ}$ General solution: $\\scriptsize \\theta ={{224.4}^\\circ}+k\\text{.36}{{\\text{0}}^\\circ}\\text{ or }\\theta ={{315.6}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\text{ }$ 2. . \\scriptsize \\begin{align*}2\\tan\\theta&=0.279\\\\\\therefore \\tan\\theta&=0.1395\\end{align*} Ref angle: $\\scriptsize \\theta ={{8.0}^\\circ}$ $\\scriptsize \\tan \\theta \\gt 0:$ $\\scriptsize \\theta ={{8.0}^\\circ}\\text{ or }\\theta ={{180}^\\circ}+{{8.0}^\\circ}={{188}^\\circ}$ General solution: $\\scriptsize \\theta ={{8.0}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\text{ }$ Specific solution for $\\scriptsize \\theta \\in [{{0}^\\circ},{{360}^\\circ}]$: $\\scriptsize \\theta ={{8}^\\circ}+0\\times {{180}^\\circ}={{8}^\\circ}$ or $\\scriptsize \\theta ={{8.0}^\\circ}+1\\times {{180}^\\circ}={{188.0}^\\circ}$ 3. . \\scriptsize \\begin{align*}5\\cos2\\theta&=-2.7\\\\\\therefore \\cos2\\theta&=-0.54\\end{align*} Ref angle: $\\scriptsize 2\\theta ={{57.3}^\\circ}$ $\\scriptsize \\cos 2\\theta \\lt 0$: $\\scriptsize 2\\theta ={{180}^\\circ}-{{57.3}^\\circ}={{122.7}^\\circ}\\text{ or }2\\theta ={{180}^\\circ}+{{57.3}^\\circ}={{237.3}^\\circ}$ General solution: \\scriptsize \\begin{align*}2\\theta ={{122.7}^\\circ}+k{{.360}^\\circ}\\text{ or }2\\theta ={{237.3}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\text{ }\\\\\\therefore \\theta ={{61.35}^\\circ}+k{{.180}^\\circ}\\text{ or }\\theta ={{118.65}^\\circ}+{{180}^\\circ},k\\in \\mathbb{Z}\\text{ }\\end{align*} Specific solution for $\\scriptsize \\theta \\in [{{0}^\\circ},{{360}^\\circ}]$: $\\scriptsize \\theta ={{61.35}^\\circ}+0\\times {{180}^\\circ}={{61.35}^\\circ}$ or $\\scriptsize \\theta ={{61.35}^\\circ}+1\\times {{180}^\\circ}={{241.35}^\\circ}$ or $\\scriptsize \\theta ={{118.65}^\\circ}+0\\times {{180}^\\circ}={{118.65}^\\circ}$ or $\\scriptsize \\theta ={{118.65}^\\circ}+1\\times {{180}^\\circ}={{298.65}^\\circ}$ 4. $\\scriptsize \\tan (3\\theta -{{48}^\\circ})=3.2$ Ref angle: $\\scriptsize 3\\theta -{{48}^\\circ}={{72.6}^\\circ}$ $\\scriptsize 3\\theta -{{48}^\\circ}={{72.6}^\\circ}\\text{ or }3\\theta -{{48}^\\circ}={{180}^\\circ}+{{72.6}^\\circ}={{252.6}^\\circ}$: General solution: \\scriptsize \\begin{align*}3\\theta -{{48}^\\circ} & ={{72.6}^\\circ}+k{{.180}^\\circ}\\text{ or }3\\theta -{{48}^\\circ}={{252.6}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\text{ }\\\\\\therefore 3\\theta & ={{120.6}^\\circ}+k{{.180}^\\circ}\\text{ or }3\\theta ={{300.6}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\text{ }\\\\\\therefore \\theta & ={{40.2}^\\circ}+k{{.60}^\\circ}\\text{ or }\\theta ={{100.2}^\\circ}+k{{.60}^\\circ},k\\in \\mathbb{Z}\\text{ }\\end{align*} But $\\scriptsize {{40.2}^\\circ}+1\\times {{60}^\\circ}={{100.2}^\\circ}$ Therefore, simplest general solution is $\\scriptsize \\theta ={{40.2}^\\circ}+k{{.60}^\\circ},k\\in \\mathbb{Z}\\text{ }$ Specific solution for $\\scriptsize \\theta \\in [{{0}^\\circ},{{180}^\\circ}]$: $\\scriptsize \\theta ={{40.2}^\\circ}+0\\times {{60}^\\circ}={{40.2}^\\circ}$ or $\\scriptsize \\theta ={{40.2}^\\circ}+1\\times {{60}^\\circ}={{100.2}^\\circ}$ or $\\scriptsize \\theta ={{40.2}^\\circ}+2\\times {{60}^\\circ}={{160.2}^\\circ}$ 5. .","or }x={{61.475}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\end{align*} 2. Solutions for the interval $\\scriptsize \\left[ {{{0}^\\circ},{{{90}}^\\circ}} \\right]$: $\\scriptsize x ={{28.525}^\\circ}\\text{ or }x={{61.475}^\\circ}$ Back to Exercise 2.2 ### Unit 2: Assessment 1. . 1. . \\scriptsize \\begin{align*}2\\cos \\theta -1 & =0\\\\\\therefore \\cos \\theta & =\\displaystyle \\frac{1}{2}\\end{align*} Ref angle: $\\scriptsize \\theta ={{60}^\\circ}$ Cosine is positive in the first and fourth quadrants. $\\scriptsize \\theta ={{360}^\\circ}-{{60}^\\circ}={{300}^\\circ}$ General solution: $\\scriptsize \\theta ={{60}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{300}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ 2. . \\scriptsize \\begin{align*}2{{\\sin }^{2}}\\theta +\\sin \\theta -1 & =0\\\\\\therefore (2\\sin \\theta -1)(\\sin \\theta +1) & =0\\\\\\therefore \\sin \\theta =\\displaystyle \\frac{1}{2}\\text{ } & \\text{or }\\sin \\theta =-1\\end{align*} $\\scriptsize \\sin \\theta =\\displaystyle \\frac{1}{2}$: Ref angle: $\\scriptsize \\theta ={{30}^\\circ}$ Sine is positive in the first and second quadrants. $\\scriptsize \\theta ={{180}^\\circ}-{{30}^\\circ}={{150}^\\circ}$ General solution: $\\scriptsize \\theta ={{30}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{150}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ $\\scriptsize \\sin \\theta =1$: Ref angle: $\\scriptsize \\theta ={{90}^\\circ}$ General solution: $\\scriptsize \\theta ={{90}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}$ 3. . \\scriptsize \\displaystyle \\begin{align*}4\\sin x+3\\tan x & =\\displaystyle \\frac{3}{{\\cos x}}+4\\quad \\quad \\cos x\\ne 0\\therefore x\\ne {{90}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\\\\\therefore 4\\sin x+3\\displaystyle \\frac{{\\sin x}}{{\\cos x}}-\\displaystyle \\frac{3}{{\\cos x}}-4 & =0\\\\\\therefore 4\\sin x\\cos x+3\\sin x-3-4\\cos x & =0\\\\\\therefore 4\\cos x(\\sin x-1)+3(\\sin x-1) & =0\\\\\\therefore (\\sin x-1)(4\\cos x+3) & =0\\\\\\therefore \\sin x=1\\text{ } & \\text{or }\\cos x=-\\displaystyle \\frac{3}{4}\\end{align*} $\\scriptsize \\sin x=1$: Ref angle: $\\scriptsize x={{90}^\\circ}$ General solution: $\\scriptsize x={{90}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}$ $\\scriptsize \\cos x=-\\displaystyle \\frac{3}{4}$: Ref angle: $\\scriptsize x={{138.59}^\\circ}$ Cosine is negative in the second","form: $\\scriptsize z=13\\text{cis}{{292.62}^\\circ}$ 2. $\\scriptsize z=-7+2i$ Find $\\scriptsize r$: \\scriptsize \\begin{align*}r & =\\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\\\ & =\\sqrt{{{{{(-7)}}^{2}}+{{2}^{2}}}}\\\\& =\\sqrt{{49+4}}\\\\ & =\\sqrt{{53}}\\end{align*} Find $\\scriptsize \\theta$ ($\\scriptsize \\theta$ is in the second quadrant): $\\scriptsize \\cos \\theta =\\displaystyle \\frac{x}{r}=\\displaystyle \\frac{{-7}}{{\\sqrt{{53}}}}$ Work with $\\scriptsize \\cos \\theta =\\displaystyle \\frac{7}{{\\sqrt{{53}}}}$ to find a referance angle. \\scriptsize \\begin{align*}\\therefore \\text{reference angle} & ={{15.95}^\\circ}\\\\\\therefore \\theta & ={{180}^\\circ}-{{15.95}^\\circ}={{164.05}^\\circ}\\end{align*} Polar form: $\\scriptsize z=\\sqrt{{53}}\\text{cis164}\\text{.0}{{\\text{5}}^\\circ}$ Back to Exercise 3.1 ### Exercise 3.2 1. $\\scriptsize z=6\\text{cis}{{45}^\\circ}$ $\\scriptsize {{45}^\\circ}$ is a special angle and the complex number is in the first quadrant. $\\scriptsize \\cos {{45}^\\circ}=\\displaystyle \\frac{1}{{\\sqrt{2}}}$ $\\scriptsize \\sin {{45}^\\circ}=\\displaystyle \\frac{1}{{\\sqrt{2}}}$ \\scriptsize \\displaystyle \\begin{align*}z & =6(\\cos {{45}^\\circ}+i\\sin {{45}^\\circ})\\\\ & =6\\left( {\\displaystyle \\frac{1}{{\\sqrt{2}}}+\\displaystyle \\frac{1}{{\\sqrt{2}}}i} \\right)\\\\ & =\\displaystyle \\frac{6}{{\\sqrt{2}}}+\\displaystyle \\frac{6}{{\\sqrt{2}}}i\\end{align*} The complex number in standard form is $\\scriptsize \\displaystyle z=\\displaystyle \\frac{6}{{\\sqrt{2}}}+\\displaystyle \\frac{6}{{\\sqrt{2}}}i$. 2. $\\scriptsize z=\\sqrt{5}\\text{cis21}{{\\text{0}}^\\circ}$ $\\scriptsize {{180}^\\circ}+{{30}^\\circ}={{210}^\\circ}$. Therefore, we use the special angle of $\\scriptsize {{30}^\\circ}$ in the third quadrant. $\\scriptsize \\cos {{210}^\\circ}=\\cos ({{180}^\\circ}+{{30}^\\circ})=-\\cos {{30}^\\circ}=-\\displaystyle \\frac{{\\sqrt{3}}}{2}$ $\\scriptsize \\sin {{210}^\\circ}=\\sin ({{180}^\\circ}+{{30}^\\circ})=-\\sin {{30}^\\circ}=-\\displaystyle \\frac{1}{2}$ \\scriptsize \\displaystyle \\begin{align*}z & =\\sqrt{5}(\\cos {{210}^\\circ}+i\\sin {{210}^\\circ})\\\\ & =\\sqrt{5}\\left( {-\\displaystyle \\frac{{\\sqrt{3}}}{2}-\\displaystyle \\frac{1}{2}i} \\right)\\quad \\text{Both cosine and sine are negative in the third quadrant}\\\\ & =-\\displaystyle \\frac{{\\sqrt{{15}}}}{2}-\\displaystyle \\frac{{\\sqrt{5}}}{2}i\\end{align*} The complex number in standard form is $\\scriptsize \\displaystyle z=-\\displaystyle \\frac{{\\sqrt{{15}}}}{2}-\\displaystyle \\frac{{\\sqrt{5}}}{2}i$. 3. $\\scriptsize z=2\\text{cis}{{40}^\\circ}$ $\\scriptsize {{40}^\\circ}$ is not a special angle. The complex number is in the first quadrant. $\\scriptsize \\cos {{40}^\\circ}=0.7660$ $\\scriptsize \\sin {{40}^\\circ}=0.6428$ \\scriptsize \\displaystyle \\begin{align*}z &","multiply the numerator and denominator by the conjugate of the number in the denominator. $$\\frac{\\textrm{cis} 75^o - \\textrm{cis} 155^o}{ 1 - \\cos{8^o} -i \\sin{8^o}} \\frac{1-\\cos{8^o} + i \\sin{8^o}}{1-\\cos{8^o} + i \\sin{8^o}}$$ This will give us a completely real denominator. And from there, we just have a lot of algebra to grind out to get the remaining expression down to a form of $r \\textrm{cis}{\\theta}$. I hope you find the algebra to be doable. The problem is basically a vector addition problem. $cis\\theta = cos\\theta + i sin\\theta$. So, $cis 75 + cis 83 + ... = cos 75 + cos 83 + ... + i(sin 75 + sin 83 + ...)$ Actually doing the math would be difficult, but note that you're adding a bunch of vectors of equal length, each 8 degrees apart. They form an arc and what you want is the average direction of the arc since that is also the net direction of the vectors. Hope that gets you started. I can comment further if necessary. this is a geometric series $$cis(75^\\circ) +cis(75^\\circ + 8^\\circ)+\\cdots + cis(75^\\circ+9\\times 8^\\circ)=cis(75^\\circ)\\frac{1-cis(80^\\circ)}{1-cis(8^\\circ)}$$ i hope you can take it from here.","form. 1. $$5 \\; cis \\; 270^{\\circ}$$ 2. $$2 \\; cis \\; 30^{\\circ}$$ 3. $$−4 \\; cis \\; \\dfrac{\\pi}{4}$$ 4. $$6 \\; cis \\; \\dfrac{\\pi}{3}$$ 5. $$2 \\; cis \\; \\dfrac{5\\pi}{2}$$ Translate the following complex numbers from rectangular form into trigonometric polar form. 6. $$2−i$$ 7. $$5+12i$$ 8. $$6i+8$$ 9. $$i$$ Complete the following calculations and simplify. 10. $$2 \\; cis \\; 22^{\\circ} \\cdot \\dfrac{1}{5} \\; cis \\; 15^{\\circ} \\cdot 3 \\; cis \\; 95^{\\circ}$$ 11. $$9 \\; cis \\; 98^{\\circ} \\div 3 \\; cis \\; 12^{\\circ}$$ 12. $$15 \\; cis \\; \\dfrac{\\pi}{4}\\cdot 2 \\; cis \\; \\dfrac{\\pi}{6}$$ 13. $$−2 \\; cis \\; \\dfrac{2\\pi}{3}\\div 15 \\; cis \\; \\dfrac{7 \\pi}{6}$$ Let $$z_1 =r_1 \\cdot \\; cis \\; \\theta_1$$ and $$z_2 =r_2 \\cdot \\; cis \\; \\theta_2$$ with r_2 \\neq 0\\). 14. Use the trigonometric sum and difference identities to prove that $$z_1 \\cdot z_2 =r_1 \\cdot r_2 \\cdot \\; cis \\; (\\theta_1 +\\theta_2)$$. 15. Use the trigonometric sum and difference identities to prove that $$z_1 \\div z_2 =r_1 r_2 \\cdot \\; cis \\; (\\theta_1 −\\theta_2)$$. To see the Review answers, open this PDF file and look for section 11.3. ## Vocabulary Term Definition $$r\\; cis \\;\\theta$$ $$r\\; cis \\;\\theta$$ is shorthand for the expression $$r\\cos \\theta +ri\\sin \\theta$$. complex plane The complex plane is the graphical representation of","and suppose that P=r(costheta+isintheta) where $0 \\lt r$ and $0 \\leq \\theta \\lt 360$ find $\\theta$ • is 107 • is 276 • is 840 • cannot be determined from the given information ### Key Concepts Equations Complex Numbers Integers AIME, 1996, Question 11 Complex Numbers from A to Z by Titu Andreescue ## Try with Hints First hint here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\\frac{(z^{5}-1)}{(z-1)}$=$\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0 Second Hint gives $z^{5}=1 for z\\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \\neq 1$ gives z=$\\frac{1+-(-3)^\\frac{1}{2}}{2}$=$cis60,300$ where cis$\\theta$=cos$\\theta$+isin$\\theta$ Final Step taking $0 \\lt theta \\lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276. . ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed.","P be the product of the roots of $z^{6}+z^{4}+z^{2}+1=0$ that have a positive imaginary part and suppose that P=r(costheta+isintheta) where $0 \\lt r$ and $0 \\leq \\theta \\lt 360$ find $\\theta$ • is 107 • is 276 • is 840 • cannot be determined from the given information Equations Complex Numbers Integers ## Check the Answer Answer: is 276. AIME, 1996, Question 11 Complex Numbers from A to Z by Titu Andreescue ## Try with Hints here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\\frac{(z^{5}-1)}{(z-1)}$=$\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0 gives $z^{5}=1 for z\\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \\neq 1$ gives z=$\\frac{1+-(-3)^\\frac{1}{2}}{2}$=$cis60,300$ where cis$\\theta$=cos$\\theta$+isin$\\theta$ taking $0 \\lt theta \\lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276. . Categories ## Tetrahedron Problem | AIME I, 1992 | Question 6 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron. ## Tetrahedron Problem – AIME I, 1992 Faces ABC and BCD of tetrahedron ABCD meet at an angle of 30,The area of face ABC=120, the area of face BCD is 80, BC=10. Find volume of tetrahedron. • is 107 • is 320 • is 840 • cannot be determined from the given information Area Volume Tetrahedron ## Check the Answer Answer: is 320. AIME I, 1992, Question 6 Coordinate Geometry by Loney ## Try","# Geometric sequence of complex numbers in polar fom The number $[\\text{cis } 75^\\circ + \\text{cis } 83^\\circ + \\text{cis } 91^\\circ + \\dots + \\text{cis } 147^\\circ]$ is expressed in the form $r \\, \\text{cis } \\theta$, where $0 \\le \\theta < 360^\\circ$. Find $\\theta$ in degrees. How would I start this problem? I am sort of stuck on it. Thanks Firstly, you can begin by using Euler's formula to rewrite the series. $$\\left[ e^{i 75^o} + e^{i 83^o} + \\cdots + e^{i 147^o} \\right] = e^{i 75^o} \\left[ 1 +e^{i 8^o} + e^{i 16^o} + \\cdots + e^{i 72^o} \\right]$$ Or $$e^{i 75^o} \\left[ 1 +e^{i 8^o} + (e^{i 8^o})^2 + \\cdots + (e^{i 8^o})^9 \\right] = \\Sigma_{n=0}^9 e^{i 75^o} (e^{i 8^o})^n$$ Which is a geometric series with a common ratio of $e^{i 8^o}$ and a first term of $a= e^{i 75^o}$, which can be computed as $$\\Sigma_{n=0}^{n-1} ar^n = a \\frac{1-r^n}{1-r}$$ So we can plug in our values for $a$, $r$, and $n$. $$S = e^{i 75^o} \\frac{1-(e^{i 8^o})^{10}}{1- e^{i 8^o}} = e^{i 75^o} \\frac{1-e^{i 80^o}}{1- e^{i 8^o}}$$ Now is when we have to get a little more clever to return to $r \\textrm{cis}{\\theta}$ form, so we'll rewrite the series as $$\\frac{\\textrm{cis} 75^o - \\textrm{cis} 155^o}{ 1 - \\cos{8^o} -i \\sin{8^o}}$$ Next, we have to","\\right )\\prod_{n=0}^{\\infty}\\cos\\left ( \\frac{x}{n} \\right )~\\mathrm dx\\approx \\frac{\\pi}{8}-7.41\\times 10^{-43}$$ One can easily be fooled into thinking that it is exactly $\\dfrac{\\pi}{8}$. References: - Let $\\sigma(n)$ denote the sum of the divisors of $n$. If $$p=1+\\sigma(k),$$ then $$p^a=1+\\sigma(kp^{a-1})$$ where $a,k$ are positive integers and $p$ is a prime such that $p\\not\\mid k$. - $$27\\cdot56=2\\cdot756,$$ $$277\\cdot756=27\\cdot7756,$$ $$2777\\cdot7756=277\\cdot77756,$$ and so on. - \\begin{align}\\frac{64}{16}&=\\frac{6\\!\\!/\\,4}{16\\!\\!/}\\\\&=\\frac41\\\\&=4\\end{align} For more examples of these weird fractions, see \"How Weird Are Weird Fractions?\", Ryan Stuffelbeam, The College Mathematics Journal, Vol. 44, No. 3 (May 2013), pp. 202-209. - show 1 more comment $$\\sin \\theta \\cdot \\sin \\bigl(60^\\circ - \\theta \\bigr) \\cdot \\sin \\bigl(60^\\circ + \\theta \\bigr) = \\frac{1}{4} \\sin 3\\theta$$ $$\\cos \\theta \\cdot \\cos \\bigl(60^\\circ - \\theta \\bigr) \\cdot \\cos \\bigl(60^\\circ + \\theta \\bigr) = \\frac{1}{4} \\cos 3\\theta$$ $$\\tan \\theta \\cdot \\tan \\bigl(60^\\circ - \\theta \\bigr) \\cdot \\tan \\bigl(60^\\circ + \\theta \\bigr) = \\tan 3\\theta$$ - I just wanted to mention that your first identity is equivalent to the case $n=3$ of the formula for $\\sin nx$ given there. (Just replace $\\sin(60^{\\circ}-\\theta)$ by $\\sin(\\theta+120^{\\circ})$.) – Hans Lundmark Nov 4 '10 at 9:56 considering your first two identities the thirth should be $$\\tan \\theta \\cdot \\tan \\bigl(60 - \\theta \\bigr) \\cdot \\tan \\bigl(60 + \\theta \\bigr) = \\tan 3\\theta$$ – Neves Mar 6 '11 at 16:08 $\\textbf{Claim:}\\quad$$\\frac{\\sin x}{n}=6$$ for","\\frac{1}{cos( 270 + \\Theta )}$ But $cos( 270 - \\Theta )= -sin \\Theta$ ∴ sec( $270 - \\Theta )= -\\frac{1}{sin \\Theta}$ ∴ $\\underline {sec( 270 - \\Theta )=-csc \\Theta}$ 6) $cot( 270 - \\Theta ) = \\frac{1}{sin( 270 - \\Theta )}$ But $tan( 270 - \\Theta )= cot \\Theta$ ∴ csc( $270 - \\Theta )= \\frac{1}{-cos \\Theta}$ ∴ $\\underline {cot( 270 - \\Theta )= tan \\Theta}$ ## Examples on Trigonometric ratios of 270 degrees minus theta(270 - θ) 1) Find the value of $cos 270^{0}$ Solution : $cos 270^{0}$ 270 = 270 - 0 As we know that $cos (270 - \\Theta) = -sin\\Theta$ ∴ $cos 270^{0}$ = cos (270 -0) = - sin0 ⇒ $cos 270^{0}$ = 0 2) Find the value of $csc 225^{0}$ Solution : $csc 225^{0}$ 225 = 270 - 45 As we know that $csc (270 - \\Theta) = -sec\\Theta$ ∴ $csc 225^{0}$ = csc (270 -45) = - sec 45 ⇒ $csc 225 = -{\\sqrt{2}}$","multiply the argument by n. This makes sense geometrically because if you do repeated multiplication, you will have to dilate the radius n times and you will have to rotate the argument n times. On the unit circle where $$r=1$$, this can be easily visualized. On the complex unit circle, if you raise some number with an argument of 17.1 degrees ($$z=cis(17.1^o)$$) to the power of 3.54 , the resulting number $$z^{3.54}$$ would be $$cis(17.1^o \\cdot 3.54)=cis(60.534^o)\\approx 0.4919 + 0.8706 i$$ Image by Jennifer Sun ### Complex Numbers Taking roots is similar, except that instead of multiplying the argument by n, we divide the argument by n: $z^{\\frac{1}{n}} = (re^{i\\theta})^{\\frac{1}{n}} = r^{\\frac{1}{n}}e^{i({\\frac{\\theta}{n}})}= r^{\\frac{1}{n}}(cis(\\frac{\\theta}{n}))$ This is not the final formula however. If the power of $$z$$ is greater than 1, like in the previous example where the power was 3.54, there is only one principal solution to the problem. However, when dealing with an argument of $$\\frac{\\theta}{n}$$ where there are powers of $$\\frac{1}{2}$$,$$\\frac{1}{3}$$,$$\\frac{1}{4}…$$ , there are several principal solutions. You can add $$2{\\pi}k$$ to the argument several times and it would still be a unique principal complex number. $$z^{\\frac{1}{n}}=r^{\\frac{1}{n}}(cis(\\frac{\\theta + 2{\\pi}k}{n}))$$ where $$k = 0,1,2...,n-1$$ The 7th root of a complex number would have 7 unique solutions. The cube root of a complex number would have 3 unique solutions.","-1 ; cot 180^\\circ = text(Adjacent)/( Opposite ) = (-1)/( 0 ) = \\infty ### For \\theta = 270^\\circ A point (x, y) = (0, -1) lies on the terminal side of angle \\theta = 270^\\circ as shown below: In this case, x = 0 & y = -1 => Adjacent = 0 & Opposite = -1 According to pythagorean theorem ### (Adjacent)^2 + (Opposite)^2 = text(Hypotenuse)2 => text(Hypotenuse) = \\sqrt(text(Adjacent)^2 + text(Opposite)^2) => Hypotenuse = \\sqrt((0)^2+ (-1)^2 ) = \\sqrt(0+1 ) = \\sqrt(1 ) = 1 cos 270^\\circ = text(Adjacent)/text( Hypotenuse) = 0/( 1 ) = 0 ; sin 270^\\circ = text(Opposite)/text( Hypotenuse) = (-1)/( 1 ) = -1 tan 270^\\circ = text(Opposite)/( Adjacent ) = (-1)/( 0 ) = \\infty ; cosec 270^\\circ = text( Hypotenuse )/( Opposite) = 1/(-1 ) = -1 sec 270^\\circ = text( Hypotenuse )/( Adjacent) = 1/( 0 ) = \\infty ; cot 270^\\circ = text(Adjacent)/( Opposite ) = ( 0)/(-1 ) = 0 ### Trigonometric Ratios of 0^\\circ, 90^\\circ, 180^\\circ, 270^\\circ The table given below summarizes the trigonometric ratios of angles 0^\\circ, 90^\\circ, 180^\\circ, 270^\\circ \\theta Cos \\theta Sin \\theta Tan \\theta Cosec \\theta Sec \\theta Cot \\theta 0^\\circ 1 0 0 \\infty 1 \\infty 90^\\circ 0 1 \\infty 1 \\infty 0 180^\\circ -1 0 0 \\infty -1 \\infty 270^\\circ","we obtain: $\\theta_1=\\theta_2=\\frac{7\\pi}{5}$ Your $*$ is true by the De Moivre's rule. Your $**$ is true because $\\frac{cis\\theta}{cis\\phi}=cis(\\theta-\\phi)$ and $\\overline{cis\\theta}=cis(-\\theta)$. $\\frac{cis\\theta}{cis\\phi}=cis(\\theta-\\phi)$ because $$\\frac{cis\\theta}{cis\\phi}=\\frac{cis\\theta\\overline{cis\\phi}}{cis\\phi\\overline{cis\\phi}}=cis(\\theta-\\phi)=$$ • Please read the last line in the OP. I want to understand the points $(\\ast)$ and $(\\ast\\ast)$, I know the final answer. – Yos Jul 7 '17 at 7:45 • @Yos I fixed. See now. – Michael Rozenberg Jul 7 '17 at 7:56 • Why is $\\frac{cis\\theta_1}{cis\\theta_2}=cis(\\theta_1-\\theta_2)$? Is there an identity for that? – Yos Jul 7 '17 at 8:01 • @Yos See now please. – Michael Rozenberg Jul 7 '17 at 8:18 • @Yos Yes, but I think it's better to use the rule: $\\frac{cis\\theta}{cis\\phi}=cis(\\theta-\\phi)$ – Michael Rozenberg Jul 7 '17 at 8:28 The line where you have $(*)$ should better be $$\\left|\\frac{z_1}{\\overline{z_2}^{\\,2}}\\right|= \\frac{|z_1|}{\\bigl|\\overline{z_2}^{\\,2}\\bigr|}=\\frac{r_1}{r_2^2}$$ There is no reason for the equality before the one you mark with $(*)$. The equality you mark with $({*}{*})$ is indeed wrong and it should be $$\\frac{r_1(\\cos\\theta_1+i\\sin\\theta_1)}{r_2^2(\\cos2\\theta_2-i\\sin2\\theta_2)}=\\frac{r_1}{r_2^2}(\\cos(\\theta_1+2\\theta_2)+i\\sin(\\theta_1+2\\theta_2))=\\frac{2}{25}(\\cos\\frac{\\pi}{5}+i\\sin\\frac{\\pi}{5})$$ because if $z=r(\\cos\\theta+i\\sin\\theta)$, then $\\bar{z}=r(\\cos\\theta-i\\sin\\theta)$. Next apply the standard rules \\begin{gather} \\cos\\theta-i\\sin\\theta=(\\cos\\theta+i\\sin\\theta)^{-1}\\\\[4px] (\\cos\\alpha+i\\sin\\alpha)(\\cos\\beta+i\\sin\\beta)= \\cos(\\alpha+\\beta)+i\\sin(\\alpha+\\beta) \\end{gather} ### Further notes. The equality $$\\frac{z_1z_2^2}{\\overline{z_2}^{\\,2}z_2^2}=\\frac{r_1r_2^2}{|z_2|^4}$$ is generally false, because there's no reason for $z_1z_2^2$ to be real. When you expand $\\dfrac{z_1}{\\overline{z_2}^{\\,2}}$, you should write $$\\frac{r_1(\\cos\\theta_1+i\\sin\\theta_1)} {\\bigl(\\overline{r_2(\\cos\\theta_2+i\\sin\\theta_2}\\bigr)^2} = \\frac{r_1(\\cos\\theta_1+i\\sin\\theta_1)} {(r_2(\\cos\\theta_2-i\\sin\\theta_2)^2} = \\frac{r_1(\\cos\\theta_1+i\\sin\\theta_1)} {r_2^2(\\cos2\\theta_2-i\\sin2\\theta_2)}$$ • Can you please explain why $\\ast\\ast$ equality is a mistake? As far","The solution $\\scriptsize x={{150}^\\circ}$ from before is now $\\scriptsize x=\\displaystyle \\frac{{{{{150}}^\\circ}}}{2}={{75}^\\circ}$. 3. Solutions to $\\scriptsize \\sin x=0.5$ are separated by integer multiples of $\\scriptsize {{360}^\\circ}$. Solutions to $\\scriptsize \\sin 2x=0.5$ are seperated by interger multiples of $\\scriptsize \\displaystyle \\frac{{{{{360}}^\\circ}}}{2}={{180}^\\circ}$. Therefore, when we solve equations such as $\\scriptsize \\sin 2\\theta =\\displaystyle \\frac{1}{2}$, we need to modify the general solution to take into account the period of the function is now $\\scriptsize {{180}^\\circ}$. So, if the general solution to $\\scriptsize \\sin \\theta =\\displaystyle \\frac{1}{2}$ is $\\scriptsize \\theta ={{30}^\\circ}+k{{.360}^\\circ}\\text{ or }\\theta ={{150}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\text{ }$, then the general solution to $\\scriptsize \\sin 2\\theta =\\displaystyle \\frac{1}{2}$ is $\\scriptsize \\theta ={{15}^\\circ}+k{{.180}^\\circ}\\text{ or }\\theta ={{75}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\text{ }$. The same is true for cosine and tangent equations. ### Example 4.5 Determine the general solution for $\\scriptsize \\theta$ if $\\scriptsize 5\\sin 2\\theta =3$. Solution \\scriptsize \\begin{align*}5\\sin 2\\theta & =3\\\\\\therefore \\sin 2\\theta & =\\displaystyle \\frac{3}{5}\\end{align*} Step 1: Use a calculator to determine the reference angle \\scriptsize \\begin{align*}\\sin 2\\theta & =\\displaystyle \\frac{3}{5}\\\\\\therefore 2\\theta & ={{36.9}^\\circ}\\end{align*} Note: We keep working with the reference angle as $\\scriptsize 2\\theta$ until we generate the general solution. Step 2: Use the CAST diagram to determine any other possible solutions Our equation is $\\scriptsize \\sin 2\\theta =\\displaystyle \\frac{3}{5}$. $\\scriptsize \\sin 2\\theta \\gt 0$. Sine is positive in the first and second quadrants. First quadrant: $\\scriptsize"],"string":"[\n \"# Complex Numbers and exponential form and roots The roots of $z^7 = -\\\\frac{1}{\\\\sqrt{2}} - \\\\frac{i}{\\\\sqrt{2}}$ are $\\\\text{cis } \\\\theta_1, \\\\text{cis } \\\\theta_2, \\\\dots, \\\\text{cis } \\\\theta_7,$ where $0^\\\\circ \\\\le \\\\theta_k < 360^\\\\circ$for all $1 \\\\le k \\\\le 7$. Find $\\\\theta_1 + \\\\theta_2 + \\\\dots + \\\\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\\\left(5 \\\\pi i/4 \\\\right)$. How should I simplify? Thanks - What is $\\\\text{cis}$? Do you by chance mean $\\\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\\\theta)+isin(\\\\theta)$ – Rod Jan 24 '14 at 16:27 @AlexR: \\\"cis\\\" is a commonly used abbreviation for \\\"cosine plus $i$ times the sine.\\\" In other words, $$\\\\operatorname{cis}\\\\theta=\\\\cos\\\\theta+i\\\\sin\\\\theta=e^{i\\\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis}(7\\\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis} 225^\\\\circ,$$ as you've already determined, and so we need $$\\\\operatorname{cis}(7\\\\theta_k)=\\\\operatorname{cis} 225^\\\\circ\\\\\\\\\\\\frac{\\\\operatorname{cis}(7\\\\theta_k)}{\\\\operatorname{cis} 225^\\\\circ}=1\\\\\\\\\\\\operatorname{cis}(7\\\\theta_k-225^\\\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? - The roots are $e^{2k\\\\pi i/7}$ ($k=0,\\\\ldots,6$). Write them as $r^k$ where $r = e^{2\\\\pi i/7}$. So the sum is (using the simple formula for summing a finite geometric series) $$\\\\sum_{k=0}^6 r^k = \\\\frac{1-r^7}{1-r} = \\\\frac{1 - e^{2\\\\pi i}}{1-e^{2\\\\pi i/7}} = 0$$. - This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not\",\n \"# Complex Numbers and exponential form and roots The roots of $z^7 = -\\\\frac{1}{\\\\sqrt{2}} - \\\\frac{i}{\\\\sqrt{2}}$ are $\\\\text{cis } \\\\theta_1, \\\\text{cis } \\\\theta_2, \\\\dots, \\\\text{cis } \\\\theta_7,$ where $0^\\\\circ \\\\le \\\\theta_k < 360^\\\\circ$for all $1 \\\\le k \\\\le 7$. Find $\\\\theta_1 + \\\\theta_2 + \\\\dots + \\\\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\\\left(5 \\\\pi i/4 \\\\right)$. How should I simplify? Thanks - What is $\\\\text{cis}$? Do you by chance mean $\\\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\\\theta)+isin(\\\\theta)$ – Mr.Fry Jan 24 '14 at 16:27 @AlexR: \\\"cis\\\" is a commonly used abbreviation for \\\"cosine plus $i$ times the sine.\\\" In other words, $$\\\\operatorname{cis}\\\\theta=\\\\cos\\\\theta+i\\\\sin\\\\theta=e^{i\\\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis}(7\\\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis} 225^\\\\circ,$$ as you've already determined, and so we need $$\\\\operatorname{cis}(7\\\\theta_k)=\\\\operatorname{cis} 225^\\\\circ\\\\\\\\\\\\frac{\\\\operatorname{cis}(7\\\\theta_k)}{\\\\operatorname{cis} 225^\\\\circ}=1\\\\\\\\\\\\operatorname{cis}(7\\\\theta_k-225^\\\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? If I am understanding correctly...for $k=1$, we take $$\\\\operatorname{cis}(7\\\\theta_1)=\\\\operatorname{cis} 225^\\\\circ$$ and then somehow determine $\\\\theta_1$, and repeat for $k=2,3,4,5,6,7$? – Mathy Person Jul 18 '15 at 21:59 No, that's getting further from the answer. Rather, think about the solutions set for the equation $$\\\\operatorname{cis}\\\\theta=1. This will let you get rid of the \\\\operatorname{cis}, and find all solutions to the original\",\n \"34^\\\\circ \\\\cdot 5 \\\\ \\\\text{cis} \\\\ 16^\\\\circ \\\\cdot \\\\frac{1}{2} \\\\ \\\\text{cis} \\\\ 100^\\\\circ\\\\\\\\& =4 \\\\cdot 5 \\\\cdot \\\\frac{1}{2} \\\\cdot \\\\text{cis} \\\\ (34^\\\\circ+16^\\\\circ+100^\\\\circ)=10 \\\\ \\\\text{cis} \\\\ 150$ Note how much easier it is to do products and quotients in trigonometric polar form. #### Practice Translate the following complex numbers from trigonometric polar form to rectangular form. 1. $5 \\\\ \\\\text{cis} \\\\ 270^\\\\circ$ 2. $2 \\\\ \\\\text{cis} \\\\ 30^\\\\circ$ 3. $-4 \\\\ \\\\text{cis} \\\\ \\\\frac{\\\\pi}{4}$ 4. $6 \\\\ \\\\text{cis} \\\\ \\\\frac{\\\\pi}{3}$ 5. $2 \\\\ \\\\text{cis} \\\\ \\\\frac{5 \\\\pi}{2}$ Translate the following complex numbers from rectangular form into trigonometric polar form. 6. $2-i$ 7. $5+12i$ 8. $6i+8$ 9. $i$ Complete the following calculations and simplify. 10. $2 \\\\ \\\\text{cis} \\\\ 22^\\\\circ \\\\cdot \\\\frac{1}{5} \\\\ \\\\text{cis} \\\\ 15^\\\\circ \\\\cdot 3 \\\\ \\\\text{cis} \\\\ 95^\\\\circ$ 11. $9 \\\\ \\\\text{cis} \\\\ 98^\\\\circ \\\\div 3 \\\\ \\\\text{cis} \\\\ 12^\\\\circ$ 12. $15 \\\\ \\\\text{cis} \\\\ \\\\frac{\\\\pi}{4} \\\\cdot 2 \\\\ \\\\text{cis} \\\\ \\\\frac{\\\\pi}{6}$ 13. $-2 \\\\ \\\\text{cis} \\\\ \\\\frac{2 \\\\pi}{3} \\\\div 15 \\\\ \\\\text{cis} \\\\ \\\\frac{7 \\\\pi}{6}$ Let $z_1=r_1 \\\\cdot \\\\text{cis} \\\\ \\\\theta_1$ and $z_2=r_2 \\\\cdot \\\\text{cis} \\\\ \\\\theta_2$ with $r_2 \\\\neq 0$ . 14. Use the trigonometric sum and difference identities to prove that $z_1 \\\\cdot z_2=r_1 \\\\cdot r_2 \\\\cdot \\\\text{cis} \\\\ (\\\\theta_1+\\\\theta_2)$ . 15. Use the trigonometric sum and difference identities to prove that $z_1\",\n \"\\\\pi}{4}\\\\right)\\\\right)=4 \\\\left(-\\\\frac{\\\\sqrt{2}}{2}+\\\\frac{\\\\sqrt{2}}{2}i\\\\right)=-2 \\\\sqrt{2}+2 \\\\sqrt{2}i$ Example C Divide the following complex numbers. $\\\\frac{4 \\\\ \\\\text{cis} \\\\ 32^\\\\circ}{2 \\\\ \\\\text{cis} \\\\ 2^\\\\circ}$ Solution: $\\\\frac{4 \\\\ \\\\text{cis} \\\\ 32^\\\\circ}{2 \\\\ \\\\text{cis} \\\\ 2^\\\\circ}=\\\\frac{4}{2} \\\\ \\\\text{cis} \\\\ (32^\\\\circ-2^\\\\circ)=2 \\\\ \\\\text{cis} \\\\ (30^\\\\circ)$ Concept Problem Revisited In rectangular coordinates: $\\\\left(1+\\\\sqrt{3}i\\\\right)\\\\left(\\\\sqrt{2}-\\\\sqrt{2}i\\\\right)=\\\\sqrt{2}-\\\\sqrt{2}i+\\\\sqrt{6}i+\\\\sqrt{6}$ In trigonometric polar coordinates, $1+\\\\sqrt{3}i=2 \\\\ \\\\text{cis} \\\\ 60^\\\\circ$ and $\\\\sqrt{2}-\\\\sqrt{2}i=2 \\\\ \\\\text{cis}-45^\\\\circ$ . Therefore: $\\\\left(1+\\\\sqrt{3}i\\\\right) \\\\left(\\\\sqrt{2}-\\\\sqrt{2}i \\\\right)=2 \\\\ \\\\text{cis} \\\\ 60^\\\\circ \\\\cdot 2 \\\\ \\\\text{cis} - 45^\\\\circ=4 \\\\ \\\\text{cis} \\\\ 105^\\\\circ$ #### Vocabulary Trigonometric polar form of a complex number describes the location of a point on the complex plane using the angle and the radius of the point. The abbreviation $r \\\\cdot \\\\text{cis} \\\\ \\\\theta$ stands for $r \\\\cdot (\\\\cos \\\\theta+i \\\\cdot \\\\sin \\\\theta)$ and is how trigonometric polar form is typically observed. #### Guided Practice 1. Translate the following complex number from trigonometric polar form to rectangular form. $5 \\\\ \\\\text{cis} \\\\ 270^\\\\circ$ 2. Translate the following complex number from rectangular form into trigonometric polar form. 8 3. Multiply the following complex numbers in trigonometric polar form. $4 \\\\ \\\\text{cis} \\\\ 34^\\\\circ \\\\cdot 5 \\\\ \\\\text{cis} \\\\ 16^\\\\circ \\\\cdot \\\\frac{1}{2} \\\\ \\\\text{cis} \\\\ 100^\\\\circ$ 1. $5 \\\\ \\\\text{cis} \\\\ 270^\\\\circ =5 (\\\\cos 270^\\\\circ+i \\\\cdot \\\\sin 270^\\\\circ)=5 (0-i)=-5 i$ 2. $8=8 \\\\ \\\\text{cis} \\\\ 0^\\\\circ$ 3. $& 4 \\\\ \\\\text{cis} \\\\\",\n \"and third quadrants. $\\\\scriptsize x={{360}^\\\\circ}-{{138.59}^\\\\circ}={{221.41}^\\\\circ}$ General solution: $\\\\scriptsize x={{138.59}^\\\\circ}+k{{.360}^\\\\circ}\\\\text{ or }x={{221.49}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}$ For the interval $\\\\scriptsize \\\\left[ {{{0}^\\\\circ},{{{180}}^\\\\circ}} \\\\right]$: $\\\\scriptsize x={{90}^\\\\circ}\\\\text{ or }x={{138.59}^\\\\circ}$ 4. . \\\\scriptsize \\\\begin{align*}6\\\\cos \\\\theta -5 & =\\\\displaystyle \\\\frac{4}{{\\\\cos \\\\theta }}\\\\quad \\\\quad \\\\cos \\\\theta \\\\ne 0\\\\therefore \\\\theta \\\\ne {{90}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\\\\\\\\\therefore 6{{\\\\cos }^{2}}\\\\theta -5\\\\cos \\\\theta -4 & =0\\\\\\\\\\\\therefore (3\\\\cos \\\\theta -4)(2\\\\cos \\\\theta +1) & =0\\\\\\\\\\\\therefore \\\\cos \\\\theta =\\\\displaystyle \\\\frac{4}{3}\\\\text{ } & \\\\text{or }\\\\cos \\\\theta =-\\\\displaystyle \\\\frac{1}{2}\\\\end{align*} $\\\\scriptsize \\\\cos \\\\theta =\\\\displaystyle \\\\frac{4}{3}$ – No solution $\\\\scriptsize \\\\cos \\\\theta =-\\\\displaystyle \\\\frac{1}{2}$: Ref angle: $\\\\scriptsize \\\\theta ={{120}^\\\\circ}$ Cosine is negative in the second and third quadrants. $\\\\scriptsize \\\\theta ={{360}^\\\\circ}-({{120}^\\\\circ})={{240}^\\\\circ}$ General solution: $\\\\scriptsize \\\\theta ={{120}^\\\\circ}+k{{.360}^\\\\circ}\\\\text{ or }\\\\theta ={{240}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}$ 5. . \\\\scriptsize \\\\begin{align*}\\\\cos 2x-\\\\cos x+1 & =0\\\\\\\\\\\\therefore 2{{\\\\cos }^{2}}x-1-\\\\cos x+1 & =0\\\\\\\\\\\\therefore 2{{\\\\cos }^{2}}x-\\\\cos x & =0\\\\\\\\\\\\therefore \\\\cos x(2\\\\cos x-1) & =0\\\\\\\\\\\\therefore \\\\cos x=0\\\\text{ } & \\\\text{or }\\\\cos x=\\\\displaystyle \\\\frac{1}{2}\\\\end{align*} $\\\\scriptsize \\\\cos x=0$: Ref angle: $\\\\scriptsize x={{90}^\\\\circ}$ General solution: $\\\\scriptsize x={{90}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}$ $\\\\scriptsize \\\\cos x=\\\\displaystyle \\\\frac{1}{2}$ Ref angle: $\\\\scriptsize \\\\theta ={{60}^\\\\circ}$ Cosine is positive in the first and fourth quadrants. $\\\\scriptsize \\\\theta ={{360}^\\\\circ}-{{60}^\\\\circ}={{300}^\\\\circ}$ General solution: $\\\\scriptsize \\\\theta ={{60}^\\\\circ}+k{{.360}^\\\\circ}\\\\text{ or }\\\\theta ={{300}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}$ For the interval $\\\\scriptsize \\\\left[ {{{0}^\\\\circ},{{{360}}^\\\\circ}} \\\\right]$: $\\\\scriptsize x=\\\\text{6}{{\\\\text{0}}^\\\\circ}\\\\text{ or }x={{90}^\\\\circ}\\\\text{ or }x={{180}^\\\\circ}\\\\text{ or }x=\\\\text{30}{{\\\\text{0}}^\\\\circ}\\\\text{ or }x=\\\\text{36}{{\\\\text{0}}^\\\\circ}$ 2. $\\\\scriptsize \\\\cos 2\\\\theta =2{{\\\\cos }^{2}}\\\\theta -1$ 1. . \\\\scriptsize \\\\begin{align*}\\\\text{LHS}=\\\\cos 2\\\\theta +3\\\\cos \\\\theta -1\\\\\\\\=2{{\\\\cos }^{2}}\\\\theta -1+3\\\\cos \\\\theta -1\\\\\\\\=2{{\\\\cos }^{2}}\\\\theta +3\\\\cos\",\n \"solution is correct. ### Example 2.2 Determine the general solution for $\\\\scriptsize 7\\\\cos 2\\\\theta +4=0$. Solution \\\\scriptsize \\\\displaystyle \\\\begin{align*}7\\\\cos 2\\\\theta +4&=0\\\\\\\\\\\\therefore \\\\cos 2\\\\theta &=-\\\\displaystyle \\\\frac{4}{7}\\\\end{align*} Step 1: Use a calculator to determine the reference angle \\\\scriptsize \\\\displaystyle \\\\begin{align*}\\\\cos 2\\\\theta &=-\\\\displaystyle \\\\frac{4}{7}\\\\\\\\\\\\therefore 2\\\\theta &={{124.8}^\\\\circ}\\\\end{align*} Note: We keep working with the reference angle of $\\\\scriptsize 2\\\\theta$ until we generate the general solution. Step 2: Use the CAST diagram to determine any other possible solutions Our equation is $\\\\scriptsize \\\\displaystyle \\\\cos 2\\\\theta =-\\\\displaystyle \\\\frac{4}{7}$. $\\\\scriptsize \\\\cos 2\\\\theta \\\\lt 0$. Cosine is negative in the second and third quadrants. Our reference angle is in the second quadrant. Second quadrant: $\\\\scriptsize 2\\\\theta ={{124.8}^\\\\circ}$ Third quadrant: $\\\\scriptsize \\\\cos ({{360}^\\\\circ}-\\\\theta )=\\\\cos \\\\theta$ $\\\\scriptsize 2\\\\theta ={{360}^\\\\circ}-{{124.8}^\\\\circ}={{235.2}^\\\\circ}$. Step 3: Generate the general solution \\\\scriptsize \\\\begin{align*}2\\\\theta & ={{124.8}^\\\\circ}+k{{.360}^\\\\circ}\\\\text{ or 2}\\\\theta ={{235.2}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\\\\\\\\\therefore \\\\theta & ={{62.4}^\\\\circ}+k{{.180}^\\\\circ}\\\\text{ or }\\\\theta ={{117.6}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\end{align*} Step 4: Check your general solution $\\\\scriptsize k=2$: $\\\\scriptsize \\\\therefore \\\\theta ={{62.4}^\\\\circ}+2\\\\times {{180}^\\\\circ}={{422.4}^\\\\circ}\\\\text{ or }\\\\theta ={{117.6}^\\\\circ}+2\\\\times {{180}^\\\\circ}={{477.6}^\\\\circ}$ $\\\\scriptsize \\\\displaystyle \\\\cos (2\\\\times {{422.4}^\\\\circ})=-\\\\displaystyle \\\\frac{4}{7}$ $\\\\scriptsize \\\\displaystyle \\\\cos (2\\\\times {{477.6}^\\\\circ})=-\\\\displaystyle \\\\frac{4}{7}$ ### Take note! The general solutions for equations involving the three basic trigonometric ratios can be written as follows: If $\\\\scriptsize \\\\sin \\\\theta =x$ then: $\\\\scriptsize \\\\theta =\\\\left( {{{{\\\\sin }}^{{-1}}}x+k{{{.360}}^\\\\circ}} \\\\right)\\\\text{ or }\\\\theta =\\\\left( {\\\\left( {{{{180}}^\\\\circ}-{{{\\\\sin }}^{{-1}}}x} \\\\right)+k{{{.360}}^\\\\circ}} \\\\right),k\\\\in \\\\mathbb{Z}$ If $\\\\scriptsize \\\\cos \\\\theta =x$ then: $\\\\scriptsize \\\\theta\",\n \"\\\\theta & =-0.7\\\\end{align*} Ref angle: $\\\\scriptsize \\\\theta ={{44.4}^\\\\circ}$ $\\\\scriptsize \\\\sin \\\\theta \\\\lt 0$: $\\\\scriptsize \\\\theta ={{180}^\\\\circ}+{{44.4}^\\\\circ}={{224.4}^\\\\circ}\\\\text{ or }\\\\theta ={{360}^\\\\circ}-{{44.4}^\\\\circ}={{315.6}^\\\\circ}$ General solution: $\\\\scriptsize \\\\theta ={{224.4}^\\\\circ}+k\\\\text{.36}{{\\\\text{0}}^\\\\circ}\\\\text{ or }\\\\theta ={{315.6}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }$ 2. . \\\\scriptsize \\\\begin{align*}2\\\\tan\\\\theta&=0.279\\\\\\\\\\\\therefore \\\\tan\\\\theta&=0.1395\\\\end{align*} Ref angle: $\\\\scriptsize \\\\theta ={{8.0}^\\\\circ}$ $\\\\scriptsize \\\\tan \\\\theta \\\\gt 0:$ $\\\\scriptsize \\\\theta ={{8.0}^\\\\circ}\\\\text{ or }\\\\theta ={{180}^\\\\circ}+{{8.0}^\\\\circ}={{188}^\\\\circ}$ General solution: $\\\\scriptsize \\\\theta ={{8.0}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }$ Specific solution for $\\\\scriptsize \\\\theta \\\\in [{{0}^\\\\circ},{{360}^\\\\circ}]$: $\\\\scriptsize \\\\theta ={{8}^\\\\circ}+0\\\\times {{180}^\\\\circ}={{8}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{8.0}^\\\\circ}+1\\\\times {{180}^\\\\circ}={{188.0}^\\\\circ}$ 3. . \\\\scriptsize \\\\begin{align*}5\\\\cos2\\\\theta&=-2.7\\\\\\\\\\\\therefore \\\\cos2\\\\theta&=-0.54\\\\end{align*} Ref angle: $\\\\scriptsize 2\\\\theta ={{57.3}^\\\\circ}$ $\\\\scriptsize \\\\cos 2\\\\theta \\\\lt 0$: $\\\\scriptsize 2\\\\theta ={{180}^\\\\circ}-{{57.3}^\\\\circ}={{122.7}^\\\\circ}\\\\text{ or }2\\\\theta ={{180}^\\\\circ}+{{57.3}^\\\\circ}={{237.3}^\\\\circ}$ General solution: \\\\scriptsize \\\\begin{align*}2\\\\theta ={{122.7}^\\\\circ}+k{{.360}^\\\\circ}\\\\text{ or }2\\\\theta ={{237.3}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\\\\\\\\\therefore \\\\theta ={{61.35}^\\\\circ}+k{{.180}^\\\\circ}\\\\text{ or }\\\\theta ={{118.65}^\\\\circ}+{{180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\end{align*} Specific solution for $\\\\scriptsize \\\\theta \\\\in [{{0}^\\\\circ},{{360}^\\\\circ}]$: $\\\\scriptsize \\\\theta ={{61.35}^\\\\circ}+0\\\\times {{180}^\\\\circ}={{61.35}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{61.35}^\\\\circ}+1\\\\times {{180}^\\\\circ}={{241.35}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{118.65}^\\\\circ}+0\\\\times {{180}^\\\\circ}={{118.65}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{118.65}^\\\\circ}+1\\\\times {{180}^\\\\circ}={{298.65}^\\\\circ}$ 4. $\\\\scriptsize \\\\tan (3\\\\theta -{{48}^\\\\circ})=3.2$ Ref angle: $\\\\scriptsize 3\\\\theta -{{48}^\\\\circ}={{72.6}^\\\\circ}$ $\\\\scriptsize 3\\\\theta -{{48}^\\\\circ}={{72.6}^\\\\circ}\\\\text{ or }3\\\\theta -{{48}^\\\\circ}={{180}^\\\\circ}+{{72.6}^\\\\circ}={{252.6}^\\\\circ}$: General solution: \\\\scriptsize \\\\begin{align*}3\\\\theta -{{48}^\\\\circ} & ={{72.6}^\\\\circ}+k{{.180}^\\\\circ}\\\\text{ or }3\\\\theta -{{48}^\\\\circ}={{252.6}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\\\\\\\\\therefore 3\\\\theta & ={{120.6}^\\\\circ}+k{{.180}^\\\\circ}\\\\text{ or }3\\\\theta ={{300.6}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\\\\\\\\\therefore \\\\theta & ={{40.2}^\\\\circ}+k{{.60}^\\\\circ}\\\\text{ or }\\\\theta ={{100.2}^\\\\circ}+k{{.60}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\end{align*} But $\\\\scriptsize {{40.2}^\\\\circ}+1\\\\times {{60}^\\\\circ}={{100.2}^\\\\circ}$ Therefore, simplest general solution is $\\\\scriptsize \\\\theta ={{40.2}^\\\\circ}+k{{.60}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }$ Specific solution for $\\\\scriptsize \\\\theta \\\\in [{{0}^\\\\circ},{{180}^\\\\circ}]$: $\\\\scriptsize \\\\theta ={{40.2}^\\\\circ}+0\\\\times {{60}^\\\\circ}={{40.2}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{40.2}^\\\\circ}+1\\\\times {{60}^\\\\circ}={{100.2}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{40.2}^\\\\circ}+2\\\\times {{60}^\\\\circ}={{160.2}^\\\\circ}$ 5. .\",\n \"or }x={{61.475}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\end{align*} 2. Solutions for the interval $\\\\scriptsize \\\\left[ {{{0}^\\\\circ},{{{90}}^\\\\circ}} \\\\right]$: $\\\\scriptsize x ={{28.525}^\\\\circ}\\\\text{ or }x={{61.475}^\\\\circ}$ Back to Exercise 2.2 ### Unit 2: Assessment 1. . 1. . \\\\scriptsize \\\\begin{align*}2\\\\cos \\\\theta -1 & =0\\\\\\\\\\\\therefore \\\\cos \\\\theta & =\\\\displaystyle \\\\frac{1}{2}\\\\end{align*} Ref angle: $\\\\scriptsize \\\\theta ={{60}^\\\\circ}$ Cosine is positive in the first and fourth quadrants. $\\\\scriptsize \\\\theta ={{360}^\\\\circ}-{{60}^\\\\circ}={{300}^\\\\circ}$ General solution: $\\\\scriptsize \\\\theta ={{60}^\\\\circ}+k{{.360}^\\\\circ}\\\\text{ or }\\\\theta ={{300}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}$ 2. . \\\\scriptsize \\\\begin{align*}2{{\\\\sin }^{2}}\\\\theta +\\\\sin \\\\theta -1 & =0\\\\\\\\\\\\therefore (2\\\\sin \\\\theta -1)(\\\\sin \\\\theta +1) & =0\\\\\\\\\\\\therefore \\\\sin \\\\theta =\\\\displaystyle \\\\frac{1}{2}\\\\text{ } & \\\\text{or }\\\\sin \\\\theta =-1\\\\end{align*} $\\\\scriptsize \\\\sin \\\\theta =\\\\displaystyle \\\\frac{1}{2}$: Ref angle: $\\\\scriptsize \\\\theta ={{30}^\\\\circ}$ Sine is positive in the first and second quadrants. $\\\\scriptsize \\\\theta ={{180}^\\\\circ}-{{30}^\\\\circ}={{150}^\\\\circ}$ General solution: $\\\\scriptsize \\\\theta ={{30}^\\\\circ}+k{{.360}^\\\\circ}\\\\text{ or }\\\\theta ={{150}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}$ $\\\\scriptsize \\\\sin \\\\theta =1$: Ref angle: $\\\\scriptsize \\\\theta ={{90}^\\\\circ}$ General solution: $\\\\scriptsize \\\\theta ={{90}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}$ 3. . \\\\scriptsize \\\\displaystyle \\\\begin{align*}4\\\\sin x+3\\\\tan x & =\\\\displaystyle \\\\frac{3}{{\\\\cos x}}+4\\\\quad \\\\quad \\\\cos x\\\\ne 0\\\\therefore x\\\\ne {{90}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\\\\\\\\\therefore 4\\\\sin x+3\\\\displaystyle \\\\frac{{\\\\sin x}}{{\\\\cos x}}-\\\\displaystyle \\\\frac{3}{{\\\\cos x}}-4 & =0\\\\\\\\\\\\therefore 4\\\\sin x\\\\cos x+3\\\\sin x-3-4\\\\cos x & =0\\\\\\\\\\\\therefore 4\\\\cos x(\\\\sin x-1)+3(\\\\sin x-1) & =0\\\\\\\\\\\\therefore (\\\\sin x-1)(4\\\\cos x+3) & =0\\\\\\\\\\\\therefore \\\\sin x=1\\\\text{ } & \\\\text{or }\\\\cos x=-\\\\displaystyle \\\\frac{3}{4}\\\\end{align*} $\\\\scriptsize \\\\sin x=1$: Ref angle: $\\\\scriptsize x={{90}^\\\\circ}$ General solution: $\\\\scriptsize x={{90}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}$ $\\\\scriptsize \\\\cos x=-\\\\displaystyle \\\\frac{3}{4}$: Ref angle: $\\\\scriptsize x={{138.59}^\\\\circ}$ Cosine is negative in the second\",\n \"form: $\\\\scriptsize z=13\\\\text{cis}{{292.62}^\\\\circ}$ 2. $\\\\scriptsize z=-7+2i$ Find $\\\\scriptsize r$: \\\\scriptsize \\\\begin{align*}r & =\\\\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\\\\\\\ & =\\\\sqrt{{{{{(-7)}}^{2}}+{{2}^{2}}}}\\\\\\\\& =\\\\sqrt{{49+4}}\\\\\\\\ & =\\\\sqrt{{53}}\\\\end{align*} Find $\\\\scriptsize \\\\theta$ ($\\\\scriptsize \\\\theta$ is in the second quadrant): $\\\\scriptsize \\\\cos \\\\theta =\\\\displaystyle \\\\frac{x}{r}=\\\\displaystyle \\\\frac{{-7}}{{\\\\sqrt{{53}}}}$ Work with $\\\\scriptsize \\\\cos \\\\theta =\\\\displaystyle \\\\frac{7}{{\\\\sqrt{{53}}}}$ to find a referance angle. \\\\scriptsize \\\\begin{align*}\\\\therefore \\\\text{reference angle} & ={{15.95}^\\\\circ}\\\\\\\\\\\\therefore \\\\theta & ={{180}^\\\\circ}-{{15.95}^\\\\circ}={{164.05}^\\\\circ}\\\\end{align*} Polar form: $\\\\scriptsize z=\\\\sqrt{{53}}\\\\text{cis164}\\\\text{.0}{{\\\\text{5}}^\\\\circ}$ Back to Exercise 3.1 ### Exercise 3.2 1. $\\\\scriptsize z=6\\\\text{cis}{{45}^\\\\circ}$ $\\\\scriptsize {{45}^\\\\circ}$ is a special angle and the complex number is in the first quadrant. $\\\\scriptsize \\\\cos {{45}^\\\\circ}=\\\\displaystyle \\\\frac{1}{{\\\\sqrt{2}}}$ $\\\\scriptsize \\\\sin {{45}^\\\\circ}=\\\\displaystyle \\\\frac{1}{{\\\\sqrt{2}}}$ \\\\scriptsize \\\\displaystyle \\\\begin{align*}z & =6(\\\\cos {{45}^\\\\circ}+i\\\\sin {{45}^\\\\circ})\\\\\\\\ & =6\\\\left( {\\\\displaystyle \\\\frac{1}{{\\\\sqrt{2}}}+\\\\displaystyle \\\\frac{1}{{\\\\sqrt{2}}}i} \\\\right)\\\\\\\\ & =\\\\displaystyle \\\\frac{6}{{\\\\sqrt{2}}}+\\\\displaystyle \\\\frac{6}{{\\\\sqrt{2}}}i\\\\end{align*} The complex number in standard form is $\\\\scriptsize \\\\displaystyle z=\\\\displaystyle \\\\frac{6}{{\\\\sqrt{2}}}+\\\\displaystyle \\\\frac{6}{{\\\\sqrt{2}}}i$. 2. $\\\\scriptsize z=\\\\sqrt{5}\\\\text{cis21}{{\\\\text{0}}^\\\\circ}$ $\\\\scriptsize {{180}^\\\\circ}+{{30}^\\\\circ}={{210}^\\\\circ}$. Therefore, we use the special angle of $\\\\scriptsize {{30}^\\\\circ}$ in the third quadrant. $\\\\scriptsize \\\\cos {{210}^\\\\circ}=\\\\cos ({{180}^\\\\circ}+{{30}^\\\\circ})=-\\\\cos {{30}^\\\\circ}=-\\\\displaystyle \\\\frac{{\\\\sqrt{3}}}{2}$ $\\\\scriptsize \\\\sin {{210}^\\\\circ}=\\\\sin ({{180}^\\\\circ}+{{30}^\\\\circ})=-\\\\sin {{30}^\\\\circ}=-\\\\displaystyle \\\\frac{1}{2}$ \\\\scriptsize \\\\displaystyle \\\\begin{align*}z & =\\\\sqrt{5}(\\\\cos {{210}^\\\\circ}+i\\\\sin {{210}^\\\\circ})\\\\\\\\ & =\\\\sqrt{5}\\\\left( {-\\\\displaystyle \\\\frac{{\\\\sqrt{3}}}{2}-\\\\displaystyle \\\\frac{1}{2}i} \\\\right)\\\\quad \\\\text{Both cosine and sine are negative in the third quadrant}\\\\\\\\ & =-\\\\displaystyle \\\\frac{{\\\\sqrt{{15}}}}{2}-\\\\displaystyle \\\\frac{{\\\\sqrt{5}}}{2}i\\\\end{align*} The complex number in standard form is $\\\\scriptsize \\\\displaystyle z=-\\\\displaystyle \\\\frac{{\\\\sqrt{{15}}}}{2}-\\\\displaystyle \\\\frac{{\\\\sqrt{5}}}{2}i$. 3. $\\\\scriptsize z=2\\\\text{cis}{{40}^\\\\circ}$ $\\\\scriptsize {{40}^\\\\circ}$ is not a special angle. The complex number is in the first quadrant. $\\\\scriptsize \\\\cos {{40}^\\\\circ}=0.7660$ $\\\\scriptsize \\\\sin {{40}^\\\\circ}=0.6428$ \\\\scriptsize \\\\displaystyle \\\\begin{align*}z &\",\n \"multiply the numerator and denominator by the conjugate of the number in the denominator. $$\\\\frac{\\\\textrm{cis} 75^o - \\\\textrm{cis} 155^o}{ 1 - \\\\cos{8^o} -i \\\\sin{8^o}} \\\\frac{1-\\\\cos{8^o} + i \\\\sin{8^o}}{1-\\\\cos{8^o} + i \\\\sin{8^o}}$$ This will give us a completely real denominator. And from there, we just have a lot of algebra to grind out to get the remaining expression down to a form of $r \\\\textrm{cis}{\\\\theta}$. I hope you find the algebra to be doable. The problem is basically a vector addition problem. $cis\\\\theta = cos\\\\theta + i sin\\\\theta$. So, $cis 75 + cis 83 + ... = cos 75 + cos 83 + ... + i(sin 75 + sin 83 + ...)$ Actually doing the math would be difficult, but note that you're adding a bunch of vectors of equal length, each 8 degrees apart. They form an arc and what you want is the average direction of the arc since that is also the net direction of the vectors. Hope that gets you started. I can comment further if necessary. this is a geometric series $$cis(75^\\\\circ) +cis(75^\\\\circ + 8^\\\\circ)+\\\\cdots + cis(75^\\\\circ+9\\\\times 8^\\\\circ)=cis(75^\\\\circ)\\\\frac{1-cis(80^\\\\circ)}{1-cis(8^\\\\circ)}$$ i hope you can take it from here.\",\n \"form. 1. $$5 \\\\; cis \\\\; 270^{\\\\circ}$$ 2. $$2 \\\\; cis \\\\; 30^{\\\\circ}$$ 3. $$−4 \\\\; cis \\\\; \\\\dfrac{\\\\pi}{4}$$ 4. $$6 \\\\; cis \\\\; \\\\dfrac{\\\\pi}{3}$$ 5. $$2 \\\\; cis \\\\; \\\\dfrac{5\\\\pi}{2}$$ Translate the following complex numbers from rectangular form into trigonometric polar form. 6. $$2−i$$ 7. $$5+12i$$ 8. $$6i+8$$ 9. $$i$$ Complete the following calculations and simplify. 10. $$2 \\\\; cis \\\\; 22^{\\\\circ} \\\\cdot \\\\dfrac{1}{5} \\\\; cis \\\\; 15^{\\\\circ} \\\\cdot 3 \\\\; cis \\\\; 95^{\\\\circ}$$ 11. $$9 \\\\; cis \\\\; 98^{\\\\circ} \\\\div 3 \\\\; cis \\\\; 12^{\\\\circ}$$ 12. $$15 \\\\; cis \\\\; \\\\dfrac{\\\\pi}{4}\\\\cdot 2 \\\\; cis \\\\; \\\\dfrac{\\\\pi}{6}$$ 13. $$−2 \\\\; cis \\\\; \\\\dfrac{2\\\\pi}{3}\\\\div 15 \\\\; cis \\\\; \\\\dfrac{7 \\\\pi}{6}$$ Let $$z_1 =r_1 \\\\cdot \\\\; cis \\\\; \\\\theta_1$$ and $$z_2 =r_2 \\\\cdot \\\\; cis \\\\; \\\\theta_2$$ with r_2 \\\\neq 0\\\\). 14. Use the trigonometric sum and difference identities to prove that $$z_1 \\\\cdot z_2 =r_1 \\\\cdot r_2 \\\\cdot \\\\; cis \\\\; (\\\\theta_1 +\\\\theta_2)$$. 15. Use the trigonometric sum and difference identities to prove that $$z_1 \\\\div z_2 =r_1 r_2 \\\\cdot \\\\; cis \\\\; (\\\\theta_1 −\\\\theta_2)$$. To see the Review answers, open this PDF file and look for section 11.3. ## Vocabulary Term Definition $$r\\\\; cis \\\\;\\\\theta$$ $$r\\\\; cis \\\\;\\\\theta$$ is shorthand for the expression $$r\\\\cos \\\\theta +ri\\\\sin \\\\theta$$. complex plane The complex plane is the graphical representation of\",\n \"and suppose that P=r(costheta+isintheta) where $0 \\\\lt r$ and $0 \\\\leq \\\\theta \\\\lt 360$ find $\\\\theta$ • is 107 • is 276 • is 840 • cannot be determined from the given information ### Key Concepts Equations Complex Numbers Integers AIME, 1996, Question 11 Complex Numbers from A to Z by Titu Andreescue ## Try with Hints First hint here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\\\\frac{(z^{5}-1)}{(z-1)}$=$\\\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\\\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0 Second Hint gives $z^{5}=1 for z\\\\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \\\\neq 1$ gives z=$\\\\frac{1+-(-3)^\\\\frac{1}{2}}{2}$=$cis60,300$ where cis$\\\\theta$=cos$\\\\theta$+isin$\\\\theta$ Final Step taking $0 \\\\lt theta \\\\lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276. . ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed.\",\n \"P be the product of the roots of $z^{6}+z^{4}+z^{2}+1=0$ that have a positive imaginary part and suppose that P=r(costheta+isintheta) where $0 \\\\lt r$ and $0 \\\\leq \\\\theta \\\\lt 360$ find $\\\\theta$ • is 107 • is 276 • is 840 • cannot be determined from the given information Equations Complex Numbers Integers ## Check the Answer Answer: is 276. AIME, 1996, Question 11 Complex Numbers from A to Z by Titu Andreescue ## Try with Hints here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\\\\frac{(z^{5}-1)}{(z-1)}$=$\\\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\\\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0 gives $z^{5}=1 for z\\\\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \\\\neq 1$ gives z=$\\\\frac{1+-(-3)^\\\\frac{1}{2}}{2}$=$cis60,300$ where cis$\\\\theta$=cos$\\\\theta$+isin$\\\\theta$ taking $0 \\\\lt theta \\\\lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276. . Categories ## Tetrahedron Problem | AIME I, 1992 | Question 6 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron. ## Tetrahedron Problem – AIME I, 1992 Faces ABC and BCD of tetrahedron ABCD meet at an angle of 30,The area of face ABC=120, the area of face BCD is 80, BC=10. Find volume of tetrahedron. • is 107 • is 320 • is 840 • cannot be determined from the given information Area Volume Tetrahedron ## Check the Answer Answer: is 320. AIME I, 1992, Question 6 Coordinate Geometry by Loney ## Try\",\n \"# Geometric sequence of complex numbers in polar fom The number $[\\\\text{cis } 75^\\\\circ + \\\\text{cis } 83^\\\\circ + \\\\text{cis } 91^\\\\circ + \\\\dots + \\\\text{cis } 147^\\\\circ]$ is expressed in the form $r \\\\, \\\\text{cis } \\\\theta$, where $0 \\\\le \\\\theta < 360^\\\\circ$. Find $\\\\theta$ in degrees. How would I start this problem? I am sort of stuck on it. Thanks Firstly, you can begin by using Euler's formula to rewrite the series. $$\\\\left[ e^{i 75^o} + e^{i 83^o} + \\\\cdots + e^{i 147^o} \\\\right] = e^{i 75^o} \\\\left[ 1 +e^{i 8^o} + e^{i 16^o} + \\\\cdots + e^{i 72^o} \\\\right]$$ Or $$e^{i 75^o} \\\\left[ 1 +e^{i 8^o} + (e^{i 8^o})^2 + \\\\cdots + (e^{i 8^o})^9 \\\\right] = \\\\Sigma_{n=0}^9 e^{i 75^o} (e^{i 8^o})^n$$ Which is a geometric series with a common ratio of $e^{i 8^o}$ and a first term of $a= e^{i 75^o}$, which can be computed as $$\\\\Sigma_{n=0}^{n-1} ar^n = a \\\\frac{1-r^n}{1-r}$$ So we can plug in our values for $a$, $r$, and $n$. $$S = e^{i 75^o} \\\\frac{1-(e^{i 8^o})^{10}}{1- e^{i 8^o}} = e^{i 75^o} \\\\frac{1-e^{i 80^o}}{1- e^{i 8^o}}$$ Now is when we have to get a little more clever to return to $r \\\\textrm{cis}{\\\\theta}$ form, so we'll rewrite the series as $$\\\\frac{\\\\textrm{cis} 75^o - \\\\textrm{cis} 155^o}{ 1 - \\\\cos{8^o} -i \\\\sin{8^o}}$$ Next, we have to\",\n \"\\\\right )\\\\prod_{n=0}^{\\\\infty}\\\\cos\\\\left ( \\\\frac{x}{n} \\\\right )~\\\\mathrm dx\\\\approx \\\\frac{\\\\pi}{8}-7.41\\\\times 10^{-43}$$ One can easily be fooled into thinking that it is exactly $\\\\dfrac{\\\\pi}{8}$. References: - Let $\\\\sigma(n)$ denote the sum of the divisors of $n$. If $$p=1+\\\\sigma(k),$$ then $$p^a=1+\\\\sigma(kp^{a-1})$$ where $a,k$ are positive integers and $p$ is a prime such that $p\\\\not\\\\mid k$. - $$27\\\\cdot56=2\\\\cdot756,$$ $$277\\\\cdot756=27\\\\cdot7756,$$ $$2777\\\\cdot7756=277\\\\cdot77756,$$ and so on. - \\\\begin{align}\\\\frac{64}{16}&=\\\\frac{6\\\\!\\\\!/\\\\,4}{16\\\\!\\\\!/}\\\\\\\\&=\\\\frac41\\\\\\\\&=4\\\\end{align} For more examples of these weird fractions, see \\\"How Weird Are Weird Fractions?\\\", Ryan Stuffelbeam, The College Mathematics Journal, Vol. 44, No. 3 (May 2013), pp. 202-209. - show 1 more comment $$\\\\sin \\\\theta \\\\cdot \\\\sin \\\\bigl(60^\\\\circ - \\\\theta \\\\bigr) \\\\cdot \\\\sin \\\\bigl(60^\\\\circ + \\\\theta \\\\bigr) = \\\\frac{1}{4} \\\\sin 3\\\\theta$$ $$\\\\cos \\\\theta \\\\cdot \\\\cos \\\\bigl(60^\\\\circ - \\\\theta \\\\bigr) \\\\cdot \\\\cos \\\\bigl(60^\\\\circ + \\\\theta \\\\bigr) = \\\\frac{1}{4} \\\\cos 3\\\\theta$$ $$\\\\tan \\\\theta \\\\cdot \\\\tan \\\\bigl(60^\\\\circ - \\\\theta \\\\bigr) \\\\cdot \\\\tan \\\\bigl(60^\\\\circ + \\\\theta \\\\bigr) = \\\\tan 3\\\\theta$$ - I just wanted to mention that your first identity is equivalent to the case $n=3$ of the formula for $\\\\sin nx$ given there. (Just replace $\\\\sin(60^{\\\\circ}-\\\\theta)$ by $\\\\sin(\\\\theta+120^{\\\\circ})$.) – Hans Lundmark Nov 4 '10 at 9:56 considering your first two identities the thirth should be $$\\\\tan \\\\theta \\\\cdot \\\\tan \\\\bigl(60 - \\\\theta \\\\bigr) \\\\cdot \\\\tan \\\\bigl(60 + \\\\theta \\\\bigr) = \\\\tan 3\\\\theta$$ – Neves Mar 6 '11 at 16:08 $\\\\textbf{Claim:}\\\\quad$$\\\\frac{\\\\sin x}{n}=6$$ for\",\n \"\\\\frac{1}{cos( 270 + \\\\Theta )}$ But $cos( 270 - \\\\Theta )= -sin \\\\Theta$ ∴ sec( $270 - \\\\Theta )= -\\\\frac{1}{sin \\\\Theta}$ ∴ $\\\\underline {sec( 270 - \\\\Theta )=-csc \\\\Theta}$ 6) $cot( 270 - \\\\Theta ) = \\\\frac{1}{sin( 270 - \\\\Theta )}$ But $tan( 270 - \\\\Theta )= cot \\\\Theta$ ∴ csc( $270 - \\\\Theta )= \\\\frac{1}{-cos \\\\Theta}$ ∴ $\\\\underline {cot( 270 - \\\\Theta )= tan \\\\Theta}$ ## Examples on Trigonometric ratios of 270 degrees minus theta(270 - θ) 1) Find the value of $cos 270^{0}$ Solution : $cos 270^{0}$ 270 = 270 - 0 As we know that $cos (270 - \\\\Theta) = -sin\\\\Theta$ ∴ $cos 270^{0}$ = cos (270 -0) = - sin0 ⇒ $cos 270^{0}$ = 0 2) Find the value of $csc 225^{0}$ Solution : $csc 225^{0}$ 225 = 270 - 45 As we know that $csc (270 - \\\\Theta) = -sec\\\\Theta$ ∴ $csc 225^{0}$ = csc (270 -45) = - sec 45 ⇒ $csc 225 = -{\\\\sqrt{2}}$\",\n \"multiply the argument by n. This makes sense geometrically because if you do repeated multiplication, you will have to dilate the radius n times and you will have to rotate the argument n times. On the unit circle where $$r=1$$, this can be easily visualized. On the complex unit circle, if you raise some number with an argument of 17.1 degrees ($$z=cis(17.1^o)$$) to the power of 3.54 , the resulting number $$z^{3.54}$$ would be $$cis(17.1^o \\\\cdot 3.54)=cis(60.534^o)\\\\approx 0.4919 + 0.8706 i$$ Image by Jennifer Sun ### Complex Numbers Taking roots is similar, except that instead of multiplying the argument by n, we divide the argument by n: $z^{\\\\frac{1}{n}} = (re^{i\\\\theta})^{\\\\frac{1}{n}} = r^{\\\\frac{1}{n}}e^{i({\\\\frac{\\\\theta}{n}})}= r^{\\\\frac{1}{n}}(cis(\\\\frac{\\\\theta}{n}))$ This is not the final formula however. If the power of $$z$$ is greater than 1, like in the previous example where the power was 3.54, there is only one principal solution to the problem. However, when dealing with an argument of $$\\\\frac{\\\\theta}{n}$$ where there are powers of $$\\\\frac{1}{2}$$,$$\\\\frac{1}{3}$$,$$\\\\frac{1}{4}…$$ , there are several principal solutions. You can add $$2{\\\\pi}k$$ to the argument several times and it would still be a unique principal complex number. $$z^{\\\\frac{1}{n}}=r^{\\\\frac{1}{n}}(cis(\\\\frac{\\\\theta + 2{\\\\pi}k}{n}))$$ where $$k = 0,1,2...,n-1$$ The 7th root of a complex number would have 7 unique solutions. The cube root of a complex number would have 3 unique solutions.\",\n \"-1 ; cot 180^\\\\circ = text(Adjacent)/( Opposite ) = (-1)/( 0 ) = \\\\infty ### For \\\\theta = 270^\\\\circ A point (x, y) = (0, -1) lies on the terminal side of angle \\\\theta = 270^\\\\circ as shown below: In this case, x = 0 & y = -1 => Adjacent = 0 & Opposite = -1 According to pythagorean theorem ### (Adjacent)^2 + (Opposite)^2 = text(Hypotenuse)2 => text(Hypotenuse) = \\\\sqrt(text(Adjacent)^2 + text(Opposite)^2) => Hypotenuse = \\\\sqrt((0)^2+ (-1)^2 ) = \\\\sqrt(0+1 ) = \\\\sqrt(1 ) = 1 cos 270^\\\\circ = text(Adjacent)/text( Hypotenuse) = 0/( 1 ) = 0 ; sin 270^\\\\circ = text(Opposite)/text( Hypotenuse) = (-1)/( 1 ) = -1 tan 270^\\\\circ = text(Opposite)/( Adjacent ) = (-1)/( 0 ) = \\\\infty ; cosec 270^\\\\circ = text( Hypotenuse )/( Opposite) = 1/(-1 ) = -1 sec 270^\\\\circ = text( Hypotenuse )/( Adjacent) = 1/( 0 ) = \\\\infty ; cot 270^\\\\circ = text(Adjacent)/( Opposite ) = ( 0)/(-1 ) = 0 ### Trigonometric Ratios of 0^\\\\circ, 90^\\\\circ, 180^\\\\circ, 270^\\\\circ The table given below summarizes the trigonometric ratios of angles 0^\\\\circ, 90^\\\\circ, 180^\\\\circ, 270^\\\\circ \\\\theta Cos \\\\theta Sin \\\\theta Tan \\\\theta Cosec \\\\theta Sec \\\\theta Cot \\\\theta 0^\\\\circ 1 0 0 \\\\infty 1 \\\\infty 90^\\\\circ 0 1 \\\\infty 1 \\\\infty 0 180^\\\\circ -1 0 0 \\\\infty -1 \\\\infty 270^\\\\circ\",\n \"we obtain: $\\\\theta_1=\\\\theta_2=\\\\frac{7\\\\pi}{5}$ Your $*$ is true by the De Moivre's rule. Your $**$ is true because $\\\\frac{cis\\\\theta}{cis\\\\phi}=cis(\\\\theta-\\\\phi)$ and $\\\\overline{cis\\\\theta}=cis(-\\\\theta)$. $\\\\frac{cis\\\\theta}{cis\\\\phi}=cis(\\\\theta-\\\\phi)$ because $$\\\\frac{cis\\\\theta}{cis\\\\phi}=\\\\frac{cis\\\\theta\\\\overline{cis\\\\phi}}{cis\\\\phi\\\\overline{cis\\\\phi}}=cis(\\\\theta-\\\\phi)=$$ • Please read the last line in the OP. I want to understand the points $(\\\\ast)$ and $(\\\\ast\\\\ast)$, I know the final answer. – Yos Jul 7 '17 at 7:45 • @Yos I fixed. See now. – Michael Rozenberg Jul 7 '17 at 7:56 • Why is $\\\\frac{cis\\\\theta_1}{cis\\\\theta_2}=cis(\\\\theta_1-\\\\theta_2)$? Is there an identity for that? – Yos Jul 7 '17 at 8:01 • @Yos See now please. – Michael Rozenberg Jul 7 '17 at 8:18 • @Yos Yes, but I think it's better to use the rule: $\\\\frac{cis\\\\theta}{cis\\\\phi}=cis(\\\\theta-\\\\phi)$ – Michael Rozenberg Jul 7 '17 at 8:28 The line where you have $(*)$ should better be $$\\\\left|\\\\frac{z_1}{\\\\overline{z_2}^{\\\\,2}}\\\\right|= \\\\frac{|z_1|}{\\\\bigl|\\\\overline{z_2}^{\\\\,2}\\\\bigr|}=\\\\frac{r_1}{r_2^2}$$ There is no reason for the equality before the one you mark with $(*)$. The equality you mark with $({*}{*})$ is indeed wrong and it should be $$\\\\frac{r_1(\\\\cos\\\\theta_1+i\\\\sin\\\\theta_1)}{r_2^2(\\\\cos2\\\\theta_2-i\\\\sin2\\\\theta_2)}=\\\\frac{r_1}{r_2^2}(\\\\cos(\\\\theta_1+2\\\\theta_2)+i\\\\sin(\\\\theta_1+2\\\\theta_2))=\\\\frac{2}{25}(\\\\cos\\\\frac{\\\\pi}{5}+i\\\\sin\\\\frac{\\\\pi}{5})$$ because if $z=r(\\\\cos\\\\theta+i\\\\sin\\\\theta)$, then $\\\\bar{z}=r(\\\\cos\\\\theta-i\\\\sin\\\\theta)$. Next apply the standard rules \\\\begin{gather} \\\\cos\\\\theta-i\\\\sin\\\\theta=(\\\\cos\\\\theta+i\\\\sin\\\\theta)^{-1}\\\\\\\\[4px] (\\\\cos\\\\alpha+i\\\\sin\\\\alpha)(\\\\cos\\\\beta+i\\\\sin\\\\beta)= \\\\cos(\\\\alpha+\\\\beta)+i\\\\sin(\\\\alpha+\\\\beta) \\\\end{gather} ### Further notes. The equality $$\\\\frac{z_1z_2^2}{\\\\overline{z_2}^{\\\\,2}z_2^2}=\\\\frac{r_1r_2^2}{|z_2|^4}$$ is generally false, because there's no reason for $z_1z_2^2$ to be real. When you expand $\\\\dfrac{z_1}{\\\\overline{z_2}^{\\\\,2}}$, you should write $$\\\\frac{r_1(\\\\cos\\\\theta_1+i\\\\sin\\\\theta_1)} {\\\\bigl(\\\\overline{r_2(\\\\cos\\\\theta_2+i\\\\sin\\\\theta_2}\\\\bigr)^2} = \\\\frac{r_1(\\\\cos\\\\theta_1+i\\\\sin\\\\theta_1)} {(r_2(\\\\cos\\\\theta_2-i\\\\sin\\\\theta_2)^2} = \\\\frac{r_1(\\\\cos\\\\theta_1+i\\\\sin\\\\theta_1)} {r_2^2(\\\\cos2\\\\theta_2-i\\\\sin2\\\\theta_2)}$$ • Can you please explain why $\\\\ast\\\\ast$ equality is a mistake? As far\",\n \"The solution $\\\\scriptsize x={{150}^\\\\circ}$ from before is now $\\\\scriptsize x=\\\\displaystyle \\\\frac{{{{{150}}^\\\\circ}}}{2}={{75}^\\\\circ}$. 3. Solutions to $\\\\scriptsize \\\\sin x=0.5$ are separated by integer multiples of $\\\\scriptsize {{360}^\\\\circ}$. Solutions to $\\\\scriptsize \\\\sin 2x=0.5$ are seperated by interger multiples of $\\\\scriptsize \\\\displaystyle \\\\frac{{{{{360}}^\\\\circ}}}{2}={{180}^\\\\circ}$. Therefore, when we solve equations such as $\\\\scriptsize \\\\sin 2\\\\theta =\\\\displaystyle \\\\frac{1}{2}$, we need to modify the general solution to take into account the period of the function is now $\\\\scriptsize {{180}^\\\\circ}$. So, if the general solution to $\\\\scriptsize \\\\sin \\\\theta =\\\\displaystyle \\\\frac{1}{2}$ is $\\\\scriptsize \\\\theta ={{30}^\\\\circ}+k{{.360}^\\\\circ}\\\\text{ or }\\\\theta ={{150}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }$, then the general solution to $\\\\scriptsize \\\\sin 2\\\\theta =\\\\displaystyle \\\\frac{1}{2}$ is $\\\\scriptsize \\\\theta ={{15}^\\\\circ}+k{{.180}^\\\\circ}\\\\text{ or }\\\\theta ={{75}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }$. The same is true for cosine and tangent equations. ### Example 4.5 Determine the general solution for $\\\\scriptsize \\\\theta$ if $\\\\scriptsize 5\\\\sin 2\\\\theta =3$. Solution \\\\scriptsize \\\\begin{align*}5\\\\sin 2\\\\theta & =3\\\\\\\\\\\\therefore \\\\sin 2\\\\theta & =\\\\displaystyle \\\\frac{3}{5}\\\\end{align*} Step 1: Use a calculator to determine the reference angle \\\\scriptsize \\\\begin{align*}\\\\sin 2\\\\theta & =\\\\displaystyle \\\\frac{3}{5}\\\\\\\\\\\\therefore 2\\\\theta & ={{36.9}^\\\\circ}\\\\end{align*} Note: We keep working with the reference angle as $\\\\scriptsize 2\\\\theta$ until we generate the general solution. Step 2: Use the CAST diagram to determine any other possible solutions Our equation is $\\\\scriptsize \\\\sin 2\\\\theta =\\\\displaystyle \\\\frac{3}{5}$. $\\\\scriptsize \\\\sin 2\\\\theta \\\\gt 0$. Sine is positive in the first and second quadrants. First quadrant: $\\\\scriptsize\"\n]"},"ps_retrievals":{"kind":"list like","value":["# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Rod Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? - The roots are $e^{2k\\pi i/7}$ ($k=0,\\ldots,6$). Write them as $r^k$ where $r = e^{2\\pi i/7}$. So the sum is (using the simple formula for summing a finite geometric series) $$\\sum_{k=0}^6 r^k = \\frac{1-r^7}{1-r} = \\frac{1 - e^{2\\pi i}}{1-e^{2\\pi i/7}} = 0$$. - This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not","# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Mr.Fry Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? If I am understanding correctly...for $k=1$, we take $$\\operatorname{cis}(7\\theta_1)=\\operatorname{cis} 225^\\circ$$ and then somehow determine $\\theta_1$, and repeat for $k=2,3,4,5,6,7$? – Mathy Person Jul 18 '15 at 21:59 No, that's getting further from the answer. Rather, think about the solutions set for the equation $$\\operatorname{cis}\\theta=1. This will let you get rid of the \\operatorname{cis}, and find all solutions to the original","equation. Then you can find the seven solutions in the desired interval. – Cameron Buie Jul 18 '15 at 22:10 I'm not sure exactly what you mean. Can you elaborate? – Mathy Person Jul 18 '15 at 22:18 Well, what are the solutions to$$\\operatorname{cis}\\theta=1,$$or, more usefully put,$$\\cos\\theta+i\\sin\\theta=1+0i$$– Cameron Buie Jul 18 '15 at 22:24 \\cos \\theta = 1 and \\sin \\theta = 0, so \\theta = 360 degrees/2\\pi. – Mathy Person Jul 18 '15 at 22:30 The roots are e^{2k\\pi i/7} (k=0,\\ldots,6). Write them as r^k where r = e^{2\\pi i/7}. So the sum is (using the simple formula for summing a finite geometric series)$$\\sum_{k=0}^6 r^k = \\frac{1-r^7}{1-r} = \\frac{1 - e^{2\\pi i}}{1-e^{2\\pi i/7}} = 0. This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not summing up the roots, but rather the arguments of said roots lying in $[0^\\circ,360^\\circ).$ – Cameron Buie Jan 24 '14 at 17:15 $\\def\\cis{\\operatorname{cis}}$Let's consider a more general problem. We have $a=r\\cis\\alpha$ and we write its $n$th roots as \\begin{gather} \\sqrt[n]{r}\\cis\\frac{\\beta}{n}\\\\ \\sqrt[n]{r}\\cis\\left(\\frac{\\beta}{n}+\\frac{2\\pi}{n}\\right)\\\\ \\sqrt[n]{r}\\cis\\left(\\frac{\\beta}{n}+2\\frac{2\\pi}{n}\\right)\\\\ \\dots\\\\ \\sqrt[n]{r}\\cis\\left(\\frac{\\beta}{n}+(n-1)\\frac{2\\pi}{n}\\right) \\end{gather} Can you sum up all those angles? Use $360$ instead of $2\\pi$ if you so prefer or are forced to.","# Difference between revisions of \"1996 AIME Problems/Problem 11\" ## Problem Let $\\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\\mathrm {P}=r(\\cos{\\theta^{\\circ}}+i\\sin{\\theta^{\\circ}})$, where $0 and $0\\leq \\theta <360$. Find $\\theta$. ## Solution 1 $\\begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \\frac{z^5-1}{z-1}\\\\ 0 &=& \\frac{(z^5 - 1)(z(z-1)+1)}{z-1} = \\frac{(z^2-z+1)(z^5-1)}{z-1} \\end{eqnarray*}$ Thus $z^5 = 1, z \\neq 1 \\Longrightarrow z = \\mathrm{cis}\\ 72, 144, 216, 288$, or $z^2 - z + 1 = 0 \\Longrightarrow z = \\frac{1 \\pm \\sqrt{-3}}{2} = \\mathrm{cis}\\ 60, 300$ (see cis). Discarding the roots with negative imaginary parts (leaving us with $\\mathrm{cis} \\theta,\\ 0 < \\theta < 180$), we are left with $\\mathrm{cis}\\ 60, 72, 144$; their product is $P = \\mathrm{cis} (60 + 72 + 144) = \\mathrm{cis} \\boxed{276}$. ## Solution 2 Let $w =$ the fifth roots of unity, except for $1$. Then $w^6 + w^4 + w^3 + w^2 + 1 = w^4 + w^3 + w^2 + w + 1 = 0$, and since both sides have the fifth roots of unity as roots, we have $z^4 + z^3 + z^2 + z + 1 | z^6 + z^4 + z^3 + z^2 + 1$. Long","+ \\frac{2k\\pi}{3}$$ $$z = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18} + \\frac{2k\\pi}{3})$$ $$OR = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18})$$ $$OR = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18} + \\frac{2\\pi}{3})$$ $$OR = (\\sqrt[3]{2})cis(-\\frac{\\pi}{18} + \\frac{4\\pi}{3})$$ Should they be written like this instead, seeing as I've used the exponential form earlier: $$z = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18} + \\frac{2k\\pi}{3})}$$ $$OR = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18})}$$ $$OR = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18} + \\frac{2\\pi}{3})}$$ $$OR = (\\sqrt[3]{2})e^{i(-\\frac{\\pi}{18} + \\frac{4\\pi}{3})}$$ Would that be $$(\\sqrt3 - i)^{\\frac{1}{3}}$$? 5. Feb 28, 2010 ### Staff: Mentor Well, yes, but that doesn't tell your what the three cube roots are. 6. Feb 28, 2010 ### Char. Limit I W-A'd it... and only got 1 answer. 7. Mar 1, 2010 ### HallsofIvy \"$cis(\\theta)$\", which is short for $cos(\\theta)+ i sin(\\theta)= e^{i\\theta}$ is primarily an engineering notation. Which is better to use depends on what kind of course this is. If it is a mathematics course being taught be a mathematician, I would say definitely use $e^{i\\theta}$.","we obtain: $\\theta_1=\\theta_2=\\frac{7\\pi}{5}$ Your $*$ is true by the De Moivre's rule. Your $**$ is true because $\\frac{cis\\theta}{cis\\phi}=cis(\\theta-\\phi)$ and $\\overline{cis\\theta}=cis(-\\theta)$. $\\frac{cis\\theta}{cis\\phi}=cis(\\theta-\\phi)$ because $$\\frac{cis\\theta}{cis\\phi}=\\frac{cis\\theta\\overline{cis\\phi}}{cis\\phi\\overline{cis\\phi}}=cis(\\theta-\\phi)=$$ • Please read the last line in the OP. I want to understand the points $(\\ast)$ and $(\\ast\\ast)$, I know the final answer. – Yos Jul 7 '17 at 7:45 • @Yos I fixed. See now. – Michael Rozenberg Jul 7 '17 at 7:56 • Why is $\\frac{cis\\theta_1}{cis\\theta_2}=cis(\\theta_1-\\theta_2)$? Is there an identity for that? – Yos Jul 7 '17 at 8:01 • @Yos See now please. – Michael Rozenberg Jul 7 '17 at 8:18 • @Yos Yes, but I think it's better to use the rule: $\\frac{cis\\theta}{cis\\phi}=cis(\\theta-\\phi)$ – Michael Rozenberg Jul 7 '17 at 8:28 The line where you have $(*)$ should better be $$\\left|\\frac{z_1}{\\overline{z_2}^{\\,2}}\\right|= \\frac{|z_1|}{\\bigl|\\overline{z_2}^{\\,2}\\bigr|}=\\frac{r_1}{r_2^2}$$ There is no reason for the equality before the one you mark with $(*)$. The equality you mark with $({*}{*})$ is indeed wrong and it should be $$\\frac{r_1(\\cos\\theta_1+i\\sin\\theta_1)}{r_2^2(\\cos2\\theta_2-i\\sin2\\theta_2)}=\\frac{r_1}{r_2^2}(\\cos(\\theta_1+2\\theta_2)+i\\sin(\\theta_1+2\\theta_2))=\\frac{2}{25}(\\cos\\frac{\\pi}{5}+i\\sin\\frac{\\pi}{5})$$ because if $z=r(\\cos\\theta+i\\sin\\theta)$, then $\\bar{z}=r(\\cos\\theta-i\\sin\\theta)$. Next apply the standard rules \\begin{gather} \\cos\\theta-i\\sin\\theta=(\\cos\\theta+i\\sin\\theta)^{-1}\\\\[4px] (\\cos\\alpha+i\\sin\\alpha)(\\cos\\beta+i\\sin\\beta)= \\cos(\\alpha+\\beta)+i\\sin(\\alpha+\\beta) \\end{gather} ### Further notes. The equality $$\\frac{z_1z_2^2}{\\overline{z_2}^{\\,2}z_2^2}=\\frac{r_1r_2^2}{|z_2|^4}$$ is generally false, because there's no reason for $z_1z_2^2$ to be real. When you expand $\\dfrac{z_1}{\\overline{z_2}^{\\,2}}$, you should write $$\\frac{r_1(\\cos\\theta_1+i\\sin\\theta_1)} {\\bigl(\\overline{r_2(\\cos\\theta_2+i\\sin\\theta_2}\\bigr)^2} = \\frac{r_1(\\cos\\theta_1+i\\sin\\theta_1)} {(r_2(\\cos\\theta_2-i\\sin\\theta_2)^2} = \\frac{r_1(\\cos\\theta_1+i\\sin\\theta_1)} {r_2^2(\\cos2\\theta_2-i\\sin2\\theta_2)}$$ • Can you please explain why $\\ast\\ast$ equality is a mistake? As far","+0 # Asymptote coding +3 37 2 +538 Hello! I was answering a question earlier, and wanted to include a graph. Since Asympotote coding makes it look clean, I tried inserting this code: [asy] size(250); import TrigMacros; real theta = 45; pair A = (3,0); pair B = 3*dir(theta); pair O = (0,0); rr_cartesian_axes(-5, 5, -5, 5, usegrid = false); draw(Circle(O, 3)); draw(O--3*dir(60)); dot(\"$\\left(\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right)$\", 3*dir(30), E); draw(O--3*dir(30)); dot(\"$\\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(60), N); draw(O--3*dir(45)); dot(\"$\\left(\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(45), dir(45)); draw(O--3*dir(150)); dot(\"$\\left(-\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right)$\", 3*dir(150), W); draw(O--3*dir(120)); dot(\"$\\left(-\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(120), N); draw(O--3*dir(135)); dot(\"$\\left(-\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(135), dir(135)); draw(O--3*dir(-60)); dot(\"$\\left(\\frac{\\sqrt{3}}{2}, -\\frac{1}{2}\\right)$\", 3*dir(-30), E); draw(O--3*dir(-30)); dot(\"$\\left(\\frac{1}{2}, -\\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(-60), -N); draw(O--3*dir(-45)); dot(\"$\\left(\\frac{\\sqrt{2}}{2}, -\\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(-45), dir(-45)); draw(O--3*dir(-150)); dot(\"$\\left(-\\frac{\\sqrt{3}}{2}, -\\frac{1}{2}\\right)$\", 3*dir(-150), W); draw(O--3*dir(-120)); dot(\"$\\left(-\\frac{1}{2}, -\\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(-120), -N); draw(O--3*dir(-135)); dot(\"$\\left(-\\frac{\\sqrt{2}}{2},- \\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(-135), dir(-135)); dot(\"$(1,0)$\", A, NE); [/asy] But, it did not show up. (it's a unit circle) Does this site support Asymptote, or do I need to use desmos? Thanks! off-topic Aug 2, 2020 #1 +128 +1 I believe that this site does not support asymptote, as I have tried to use it in the past. If you try to take a screenshot, or maybe as you said desmos will work. Hope this helps! Aug 2, 2020 #2 +538 +3 Okay! ilorty Aug 2, 2020","multiply the numerator and denominator by the conjugate of the number in the denominator. $$\\frac{\\textrm{cis} 75^o - \\textrm{cis} 155^o}{ 1 - \\cos{8^o} -i \\sin{8^o}} \\frac{1-\\cos{8^o} + i \\sin{8^o}}{1-\\cos{8^o} + i \\sin{8^o}}$$ This will give us a completely real denominator. And from there, we just have a lot of algebra to grind out to get the remaining expression down to a form of $r \\textrm{cis}{\\theta}$. I hope you find the algebra to be doable. The problem is basically a vector addition problem. $cis\\theta = cos\\theta + i sin\\theta$. So, $cis 75 + cis 83 + ... = cos 75 + cos 83 + ... + i(sin 75 + sin 83 + ...)$ Actually doing the math would be difficult, but note that you're adding a bunch of vectors of equal length, each 8 degrees apart. They form an arc and what you want is the average direction of the arc since that is also the net direction of the vectors. Hope that gets you started. I can comment further if necessary. this is a geometric series $$cis(75^\\circ) +cis(75^\\circ + 8^\\circ)+\\cdots + cis(75^\\circ+9\\times 8^\\circ)=cis(75^\\circ)\\frac{1-cis(80^\\circ)}{1-cis(8^\\circ)}$$ i hope you can take it from here.","- \\alpha_n)^{e_n}$$ for some positive integers $e_i$ and distinct complex numbers $\\alpha_i$, in a unique way (up to ordering of the factors). The fact that every nonzero complex number has $n$ nth roots is a little easier, though, as long as you know that every nonzero complex number $z$ can be represented in a unique way in the form $re^{i\\theta}$ for some $\\theta \\in \\left[0, 2\\pi), \\ \\mathbb{R} \\ni r > 0$, and that $e^{i\\theta}$ is 2$\\pi$-periodic. Then you can write $$z = re^{i\\theta} \\Longrightarrow \\left(\\sqrt[n]{r}e^{i\\left(\\frac{\\theta + 2k\\pi}{n}\\right)}\\right)^n = z, \\ \\mathbb{Z} \\ni k \\in \\left[0, n-1\\right]$$ ie. z has at least $n$ nth roots. It remains to be proved that there are exactly n of them, but you can work on that on your own~ Last edited: Apr 29, 2005 12. Apr 30, 2005 ### whozum From Orion1: $$w_1 = \\text{cis} \\; 225^{\\circ} = ( \\cos 225^{\\circ} + i \\sin 225^{\\circ})$$ Is the cis() function equivalent to cos() + i sin() ? 13. Apr 30, 2005 ### jcsd Yes it is, but no-one bothers to use it any more as cis x = e^ix which is just as easy to write. 14. Apr 30, 2005 ### robert Ihnot Orion1: Any root of i can be solved using De Moivre's Theorem. I just want to add that, if","and suppose that P=r(costheta+isintheta) where $0 \\lt r$ and $0 \\leq \\theta \\lt 360$ find $\\theta$ • is 107 • is 276 • is 840 • cannot be determined from the given information ### Key Concepts Equations Complex Numbers Integers AIME, 1996, Question 11 Complex Numbers from A to Z by Titu Andreescue ## Try with Hints First hint here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\\frac{(z^{5}-1)}{(z-1)}$=$\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0 Second Hint gives $z^{5}=1 for z\\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \\neq 1$ gives z=$\\frac{1+-(-3)^\\frac{1}{2}}{2}$=$cis60,300$ where cis$\\theta$=cos$\\theta$+isin$\\theta$ Final Step taking $0 \\lt theta \\lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276. . ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed.","\\theta & =-0.7\\end{align*} Ref angle: $\\scriptsize \\theta ={{44.4}^\\circ}$ $\\scriptsize \\sin \\theta \\lt 0$: $\\scriptsize \\theta ={{180}^\\circ}+{{44.4}^\\circ}={{224.4}^\\circ}\\text{ or }\\theta ={{360}^\\circ}-{{44.4}^\\circ}={{315.6}^\\circ}$ General solution: $\\scriptsize \\theta ={{224.4}^\\circ}+k\\text{.36}{{\\text{0}}^\\circ}\\text{ or }\\theta ={{315.6}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\text{ }$ 2. . \\scriptsize \\begin{align*}2\\tan\\theta&=0.279\\\\\\therefore \\tan\\theta&=0.1395\\end{align*} Ref angle: $\\scriptsize \\theta ={{8.0}^\\circ}$ $\\scriptsize \\tan \\theta \\gt 0:$ $\\scriptsize \\theta ={{8.0}^\\circ}\\text{ or }\\theta ={{180}^\\circ}+{{8.0}^\\circ}={{188}^\\circ}$ General solution: $\\scriptsize \\theta ={{8.0}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\text{ }$ Specific solution for $\\scriptsize \\theta \\in [{{0}^\\circ},{{360}^\\circ}]$: $\\scriptsize \\theta ={{8}^\\circ}+0\\times {{180}^\\circ}={{8}^\\circ}$ or $\\scriptsize \\theta ={{8.0}^\\circ}+1\\times {{180}^\\circ}={{188.0}^\\circ}$ 3. . \\scriptsize \\begin{align*}5\\cos2\\theta&=-2.7\\\\\\therefore \\cos2\\theta&=-0.54\\end{align*} Ref angle: $\\scriptsize 2\\theta ={{57.3}^\\circ}$ $\\scriptsize \\cos 2\\theta \\lt 0$: $\\scriptsize 2\\theta ={{180}^\\circ}-{{57.3}^\\circ}={{122.7}^\\circ}\\text{ or }2\\theta ={{180}^\\circ}+{{57.3}^\\circ}={{237.3}^\\circ}$ General solution: \\scriptsize \\begin{align*}2\\theta ={{122.7}^\\circ}+k{{.360}^\\circ}\\text{ or }2\\theta ={{237.3}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\text{ }\\\\\\therefore \\theta ={{61.35}^\\circ}+k{{.180}^\\circ}\\text{ or }\\theta ={{118.65}^\\circ}+{{180}^\\circ},k\\in \\mathbb{Z}\\text{ }\\end{align*} Specific solution for $\\scriptsize \\theta \\in [{{0}^\\circ},{{360}^\\circ}]$: $\\scriptsize \\theta ={{61.35}^\\circ}+0\\times {{180}^\\circ}={{61.35}^\\circ}$ or $\\scriptsize \\theta ={{61.35}^\\circ}+1\\times {{180}^\\circ}={{241.35}^\\circ}$ or $\\scriptsize \\theta ={{118.65}^\\circ}+0\\times {{180}^\\circ}={{118.65}^\\circ}$ or $\\scriptsize \\theta ={{118.65}^\\circ}+1\\times {{180}^\\circ}={{298.65}^\\circ}$ 4. $\\scriptsize \\tan (3\\theta -{{48}^\\circ})=3.2$ Ref angle: $\\scriptsize 3\\theta -{{48}^\\circ}={{72.6}^\\circ}$ $\\scriptsize 3\\theta -{{48}^\\circ}={{72.6}^\\circ}\\text{ or }3\\theta -{{48}^\\circ}={{180}^\\circ}+{{72.6}^\\circ}={{252.6}^\\circ}$: General solution: \\scriptsize \\begin{align*}3\\theta -{{48}^\\circ} & ={{72.6}^\\circ}+k{{.180}^\\circ}\\text{ or }3\\theta -{{48}^\\circ}={{252.6}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\text{ }\\\\\\therefore 3\\theta & ={{120.6}^\\circ}+k{{.180}^\\circ}\\text{ or }3\\theta ={{300.6}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\text{ }\\\\\\therefore \\theta & ={{40.2}^\\circ}+k{{.60}^\\circ}\\text{ or }\\theta ={{100.2}^\\circ}+k{{.60}^\\circ},k\\in \\mathbb{Z}\\text{ }\\end{align*} But $\\scriptsize {{40.2}^\\circ}+1\\times {{60}^\\circ}={{100.2}^\\circ}$ Therefore, simplest general solution is $\\scriptsize \\theta ={{40.2}^\\circ}+k{{.60}^\\circ},k\\in \\mathbb{Z}\\text{ }$ Specific solution for $\\scriptsize \\theta \\in [{{0}^\\circ},{{180}^\\circ}]$: $\\scriptsize \\theta ={{40.2}^\\circ}+0\\times {{60}^\\circ}={{40.2}^\\circ}$ or $\\scriptsize \\theta ={{40.2}^\\circ}+1\\times {{60}^\\circ}={{100.2}^\\circ}$ or $\\scriptsize \\theta ={{40.2}^\\circ}+2\\times {{60}^\\circ}={{160.2}^\\circ}$ 5. .","product of the roots of $z^{6}+z^{4}+z^{2}+1=0$ that have a positive imaginary part and suppose that P=r(costheta+isintheta) where $0 \\lt r$ and $0 \\leq \\theta \\lt 360$ find $\\theta$ • is 107 • is 276 • is 840 • cannot be determined from the given information ### Key Concepts Equations Complex Numbers Integers AIME, 1996, Question 11 Complex Numbers from A to Z by Titu Andreescue ## Try with Hints First hint here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\\frac{(z^{5}-1)}{(z-1)}$=$\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0 Second Hint gives $z^{5}=1 for z\\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \\neq 1$ gives z=$\\frac{1+-(-3)^\\frac{1}{2}}{2}$=$cis60,300$ where cis$\\theta$=cos$\\theta$+isin$\\theta$ Final Step taking $0 \\lt theta \\lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276. . Categories ## Roots of Equation and Vieta’s formula | AIME I, 1996 Problem 5 Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula. ## Roots of Equation and Vieta’s formula – AIME I, 1996 Suppose that the roots of $x^{3}+3x^{2}+4x-11=0$ are a,b and c and that the roots of $x^{3}+rx^{2}+sx+t=0$ are a+b,b+c and c+a, find t. • is 107 • is 23 • is 840 • cannot be determined from the given information ### Key Concepts Functions Roots of Equation Vieta s formula AIME I, 1996, Question 5 Polynomials by Barbeau ##","form. 1. $$5 \\; cis \\; 270^{\\circ}$$ 2. $$2 \\; cis \\; 30^{\\circ}$$ 3. $$−4 \\; cis \\; \\dfrac{\\pi}{4}$$ 4. $$6 \\; cis \\; \\dfrac{\\pi}{3}$$ 5. $$2 \\; cis \\; \\dfrac{5\\pi}{2}$$ Translate the following complex numbers from rectangular form into trigonometric polar form. 6. $$2−i$$ 7. $$5+12i$$ 8. $$6i+8$$ 9. $$i$$ Complete the following calculations and simplify. 10. $$2 \\; cis \\; 22^{\\circ} \\cdot \\dfrac{1}{5} \\; cis \\; 15^{\\circ} \\cdot 3 \\; cis \\; 95^{\\circ}$$ 11. $$9 \\; cis \\; 98^{\\circ} \\div 3 \\; cis \\; 12^{\\circ}$$ 12. $$15 \\; cis \\; \\dfrac{\\pi}{4}\\cdot 2 \\; cis \\; \\dfrac{\\pi}{6}$$ 13. $$−2 \\; cis \\; \\dfrac{2\\pi}{3}\\div 15 \\; cis \\; \\dfrac{7 \\pi}{6}$$ Let $$z_1 =r_1 \\cdot \\; cis \\; \\theta_1$$ and $$z_2 =r_2 \\cdot \\; cis \\; \\theta_2$$ with r_2 \\neq 0\\). 14. Use the trigonometric sum and difference identities to prove that $$z_1 \\cdot z_2 =r_1 \\cdot r_2 \\cdot \\; cis \\; (\\theta_1 +\\theta_2)$$. 15. Use the trigonometric sum and difference identities to prove that $$z_1 \\div z_2 =r_1 r_2 \\cdot \\; cis \\; (\\theta_1 −\\theta_2)$$. To see the Review answers, open this PDF file and look for section 11.3. ## Vocabulary Term Definition $$r\\; cis \\;\\theta$$ $$r\\; cis \\;\\theta$$ is shorthand for the expression $$r\\cos \\theta +ri\\sin \\theta$$. complex plane The complex plane is the graphical representation of","solution is correct. ### Example 2.2 Determine the general solution for $\\scriptsize 7\\cos 2\\theta +4=0$. Solution \\scriptsize \\displaystyle \\begin{align*}7\\cos 2\\theta +4&=0\\\\\\therefore \\cos 2\\theta &=-\\displaystyle \\frac{4}{7}\\end{align*} Step 1: Use a calculator to determine the reference angle \\scriptsize \\displaystyle \\begin{align*}\\cos 2\\theta &=-\\displaystyle \\frac{4}{7}\\\\\\therefore 2\\theta &={{124.8}^\\circ}\\end{align*} Note: We keep working with the reference angle of $\\scriptsize 2\\theta$ until we generate the general solution. Step 2: Use the CAST diagram to determine any other possible solutions Our equation is $\\scriptsize \\displaystyle \\cos 2\\theta =-\\displaystyle \\frac{4}{7}$. $\\scriptsize \\cos 2\\theta \\lt 0$. Cosine is negative in the second and third quadrants. Our reference angle is in the second quadrant. Second quadrant: $\\scriptsize 2\\theta ={{124.8}^\\circ}$ Third quadrant: $\\scriptsize \\cos ({{360}^\\circ}-\\theta )=\\cos \\theta$ $\\scriptsize 2\\theta ={{360}^\\circ}-{{124.8}^\\circ}={{235.2}^\\circ}$. Step 3: Generate the general solution \\scriptsize \\begin{align*}2\\theta & ={{124.8}^\\circ}+k{{.360}^\\circ}\\text{ or 2}\\theta ={{235.2}^\\circ}+k{{.360}^\\circ},k\\in \\mathbb{Z}\\text{ }\\\\\\therefore \\theta & ={{62.4}^\\circ}+k{{.180}^\\circ}\\text{ or }\\theta ={{117.6}^\\circ}+k{{.180}^\\circ},k\\in \\mathbb{Z}\\text{ }\\end{align*} Step 4: Check your general solution $\\scriptsize k=2$: $\\scriptsize \\therefore \\theta ={{62.4}^\\circ}+2\\times {{180}^\\circ}={{422.4}^\\circ}\\text{ or }\\theta ={{117.6}^\\circ}+2\\times {{180}^\\circ}={{477.6}^\\circ}$ $\\scriptsize \\displaystyle \\cos (2\\times {{422.4}^\\circ})=-\\displaystyle \\frac{4}{7}$ $\\scriptsize \\displaystyle \\cos (2\\times {{477.6}^\\circ})=-\\displaystyle \\frac{4}{7}$ ### Take note! The general solutions for equations involving the three basic trigonometric ratios can be written as follows: If $\\scriptsize \\sin \\theta =x$ then: $\\scriptsize \\theta =\\left( {{{{\\sin }}^{{-1}}}x+k{{{.360}}^\\circ}} \\right)\\text{ or }\\theta =\\left( {\\left( {{{{180}}^\\circ}-{{{\\sin }}^{{-1}}}x} \\right)+k{{{.360}}^\\circ}} \\right),k\\in \\mathbb{Z}$ If $\\scriptsize \\cos \\theta =x$ then: $\\scriptsize \\theta","# Solving a quadratic equation with complex coefficient Express $z4$=-$\\sqrt{3}$+i in polar form. Hence solve the equation $Z^2$=$z4$ for $z$ a complex number. You may leave the answer in polar form. My answer: $z4$ in polar form is 2cis-30$^{\\circ}$ and thats as far as I have gotten. I have seen this question asked on other sites, but the answers with working solutions are too complex for the level this is at. This is a complex number topic assignment on a basic mathematics course, so I can only solve this question using the basic material that has been covered already in the course. There is an example in the notes where they use De Moivre's theorem to solve it. • $$-\\sqrt3+i=2\\,\\text{cis}\\left(\\arctan\\frac1{-\\sqrt3}\\right)=2\\,\\text{cis}\\,( 150^\\circ)$$ and not $\\;-30^\\circ\\;$ , since here cosine is negative and sine positive. – Timbuc Jun 30 '15 at 13:03 • $(a{\\rm\\ cis\\ }b)^2=a^2{\\rm\\ cis\\ }2b$ should give you an idea for how to take square roots of a number given in cis form. – Gerry Myerson Jun 30 '15 at 13:12 We have $z_4=2\\mbox{cis}(150^\\circ+360^\\circ\\times k), \\ k=...-2,-1,0,1,2,...$, that is $\\mbox{cis(.)}$ is periodic with period $360^\\circ$. Then $z^2=z_4$ requires that $$z=\\sqrt{2}\\mbox{cis}\\left(\\frac{150^\\circ+360^\\circ \\times k}{2}\\right)\\ k=...-2,-1,0,1,2,...$$ but there are only two distinct values for $z$ which correspond to $k=0,1$. Hence the two solutions are:$z=\\sqrt{2}\\mbox{cis}(75^\\circ)$ and $z=\\sqrt{2}\\mbox{cis}(225^\\circ)$","form: $\\scriptsize z=13\\text{cis}{{292.62}^\\circ}$ 2. $\\scriptsize z=-7+2i$ Find $\\scriptsize r$: \\scriptsize \\begin{align*}r & =\\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\\\ & =\\sqrt{{{{{(-7)}}^{2}}+{{2}^{2}}}}\\\\& =\\sqrt{{49+4}}\\\\ & =\\sqrt{{53}}\\end{align*} Find $\\scriptsize \\theta$ ($\\scriptsize \\theta$ is in the second quadrant): $\\scriptsize \\cos \\theta =\\displaystyle \\frac{x}{r}=\\displaystyle \\frac{{-7}}{{\\sqrt{{53}}}}$ Work with $\\scriptsize \\cos \\theta =\\displaystyle \\frac{7}{{\\sqrt{{53}}}}$ to find a referance angle. \\scriptsize \\begin{align*}\\therefore \\text{reference angle} & ={{15.95}^\\circ}\\\\\\therefore \\theta & ={{180}^\\circ}-{{15.95}^\\circ}={{164.05}^\\circ}\\end{align*} Polar form: $\\scriptsize z=\\sqrt{{53}}\\text{cis164}\\text{.0}{{\\text{5}}^\\circ}$ Back to Exercise 3.1 ### Exercise 3.2 1. $\\scriptsize z=6\\text{cis}{{45}^\\circ}$ $\\scriptsize {{45}^\\circ}$ is a special angle and the complex number is in the first quadrant. $\\scriptsize \\cos {{45}^\\circ}=\\displaystyle \\frac{1}{{\\sqrt{2}}}$ $\\scriptsize \\sin {{45}^\\circ}=\\displaystyle \\frac{1}{{\\sqrt{2}}}$ \\scriptsize \\displaystyle \\begin{align*}z & =6(\\cos {{45}^\\circ}+i\\sin {{45}^\\circ})\\\\ & =6\\left( {\\displaystyle \\frac{1}{{\\sqrt{2}}}+\\displaystyle \\frac{1}{{\\sqrt{2}}}i} \\right)\\\\ & =\\displaystyle \\frac{6}{{\\sqrt{2}}}+\\displaystyle \\frac{6}{{\\sqrt{2}}}i\\end{align*} The complex number in standard form is $\\scriptsize \\displaystyle z=\\displaystyle \\frac{6}{{\\sqrt{2}}}+\\displaystyle \\frac{6}{{\\sqrt{2}}}i$. 2. $\\scriptsize z=\\sqrt{5}\\text{cis21}{{\\text{0}}^\\circ}$ $\\scriptsize {{180}^\\circ}+{{30}^\\circ}={{210}^\\circ}$. Therefore, we use the special angle of $\\scriptsize {{30}^\\circ}$ in the third quadrant. $\\scriptsize \\cos {{210}^\\circ}=\\cos ({{180}^\\circ}+{{30}^\\circ})=-\\cos {{30}^\\circ}=-\\displaystyle \\frac{{\\sqrt{3}}}{2}$ $\\scriptsize \\sin {{210}^\\circ}=\\sin ({{180}^\\circ}+{{30}^\\circ})=-\\sin {{30}^\\circ}=-\\displaystyle \\frac{1}{2}$ \\scriptsize \\displaystyle \\begin{align*}z & =\\sqrt{5}(\\cos {{210}^\\circ}+i\\sin {{210}^\\circ})\\\\ & =\\sqrt{5}\\left( {-\\displaystyle \\frac{{\\sqrt{3}}}{2}-\\displaystyle \\frac{1}{2}i} \\right)\\quad \\text{Both cosine and sine are negative in the third quadrant}\\\\ & =-\\displaystyle \\frac{{\\sqrt{{15}}}}{2}-\\displaystyle \\frac{{\\sqrt{5}}}{2}i\\end{align*} The complex number in standard form is $\\scriptsize \\displaystyle z=-\\displaystyle \\frac{{\\sqrt{{15}}}}{2}-\\displaystyle \\frac{{\\sqrt{5}}}{2}i$. 3. $\\scriptsize z=2\\text{cis}{{40}^\\circ}$ $\\scriptsize {{40}^\\circ}$ is not a special angle. The complex number is in the first quadrant. $\\scriptsize \\cos {{40}^\\circ}=0.7660$ $\\scriptsize \\sin {{40}^\\circ}=0.6428$ \\scriptsize \\displaystyle \\begin{align*}z &","? \\end{array}$$ Example 5 What are the two square roots of i? Solution Let $$\\ z=\\sqrt{0+i}$$. $$\\ r=1, \\theta=\\pi / 2 \\text { or } z=\\left[1 \\times \\operatorname{cis} \\frac{\\pi}{2}\\right]^{1 / 2}$$ Utilizing De Moivre’s theorem: $$\\ z_{1}=\\left[1 \\times \\operatorname{cis} \\frac{\\pi}{4}\\right] \\text { or } z_{2}=\\left[1 \\times \\operatorname{cis} \\frac{5 \\pi}{4}\\right]$$ $$\\ z_{1}=1\\left(\\cos \\frac{\\pi}{4}+i \\sin \\frac{\\pi}{4}\\right) \\text { or } z_{2}=1\\left(\\cos \\frac{5 \\pi}{4}+i \\sin \\frac{5 \\pi}{4}\\right)$$ $$\\ z_{1}=0.707+0.707 i \\text { or } z_{2}=-0.707-0.707 i$$ Check for z1 solution: $$\\ (0.707+0.707 i)^{2}=i ?$$ $$\\ 0.500+0.500 i+0.500 i+0.500 i^{2}=0.500+i+0.500(-1) \\text { or } i$$ Example 6 Calculate $$\\ \\sqrt[4]{(1+0 i)}$$. What are the four fourth roots of 1? Solution Let $$\\ z=1 \\text { or } z=1+0 i$$. Then the problem becomes find $$\\ z^{1 / 4}=(1+0 i)^{1 / 4}$$. Since $$\\ r=1 \\theta=0, z^{1 / 4}=[1 \\times \\operatorname{cis} 0]^{1 / 4}$$ with $$\\ z_{1}=1^{1 / 4}\\left(\\cos \\frac{0}{4}+i \\sin \\frac{0}{4}\\right)$$ or $$\\ 1(1+0)$$ or $$\\ 1$$ That root is not a surprise. Now use De Moivre’s to find the other roots: $$\\ z_{2}=1^{1 / 4}\\left[\\cos \\left(0+\\frac{\\pi}{2}\\right)+i \\sin \\left(0+\\frac{\\pi}{2}\\right)\\right]$$ Since there are 4 roots, dividing $$\\ 2 \\pi$$ by 4 yields $$\\ 0.5 \\pi$$ or $$\\ 0+i$$ or just $$\\ i$$ $$\\ z_{3}=1^{1 / 4}\\left[\\cos \\left(0+\\frac{2 \\pi}{2}\\right)+i \\sin \\left(0+\\frac{2 \\pi}{2}\\right)\\right]$$ which yields $$\\ z_{3}=-1$$ Finally, $$\\ z_{4}=1^{1 / 4}\\left[\\cos \\left(0+\\frac{3 \\pi}{2}\\right)+i \\sin","+0 # help +1 50 1 The number $$\\text{cis } 75^\\circ + \\text{cis } 83^\\circ + \\text{cis } 91^\\circ + \\dots + \\text{cis } 147^\\circ$$ is expressed in the form $$r \\, \\text{cis } \\theta$$, where $$0 \\le \\theta < 360^\\circ$$. Find $$\\theta$$ in degrees. Dec 20, 2019","told to do so, we leave this as a good, but messy, exercise. 4. To find the five fifth roots of $$1$$, we write $$1 = 1 \\text{cis}(0)$$. We have $$r = 1$$, $$\\theta = 0$$ and $$n = 5$$. Since $$\\sqrt[5]{1} = 1$$, the roots are $$w_{\\text{\\tiny$$0\\)}} = \\text{cis}(0) = 1\\), $$w_{\\text{\\tiny$$1\\)}} = \\text{cis}\\left(\\frac{2\\pi}{5}\\right)\\), $$w_{\\text{\\tiny$$2\\)}} = \\text{cis}\\left(\\frac{4\\pi}{5}\\right)\\), $$w_{\\text{\\tiny$$3\\)}} = \\text{cis}\\left(\\frac{6\\pi}{5}\\right)\\) and $$w_{\\text{\\tiny$$4\\)}} = \\text{cis}\\left(\\frac{8\\pi}{5}\\right)\\). The situation here is even graver than in the previous example, since we have not developed any identities to help us determine the cosine or sine of $$\\frac{2\\pi}{5}$$. At this stage, we could approximate our answers using a calculator, and we leave this as an exercise. Now that we have done some computations using Theorem \\ref{nthrootscomplexthm}, we take a step back to look at things geometrically. Essentially, Theorem \\ref{nthrootscomplexthm} says that to find the $$n^{\\text{th}}$$ roots of a complex number, we first take the $$n^{\\text{th}}$$ root of the modulus and divide the argument by $$n$$. This gives the first root $$w_{\\text{\\tiny$$0\\)}}\\). Each succeessive root is found by adding $$\\frac{2\\pi}{n}$$ to the argument, which amounts to rotating $$w_{\\text{\\tiny$$0\\)}}\\) by $$\\frac{2\\pi}{n}$$ radians. This results in $$n$$ roots, spaced equally around the complex plane. As an example of this, we plot our answers to number \\ref{fourthrootsneg16} in Example \\ref{nthrootscomplexex} below. We have only glimpsed at the","\\right ]^4$ c. $\\left (\\sqrt{3} - i \\right )^6$ d. The 3 complex cube roots of 1 + i e. The 4 complex fourth roots of - 16i f. The five complex fifth roots of i 1. a. $\\frac{-1}{2} + \\frac{5}{2}i$ b. 37 c. $\\frac{1}{2} + \\frac{-\\sqrt{3}}{2}i$ d. $4\\sqrt{2}(\\mbox{cos} \\ 15^\\circ + i \\ \\mbox{sin} \\ 15^\\circ)$ e. $8\\ \\mbox{cis}(60^\\circ)$ f. $4\\ \\mbox{cis}\\left ( \\frac{9\\pi}{40} \\right )$ g. $\\frac{1}{3}\\ \\mbox{cis}(240^\\circ)$ h. $\\frac{3}{4}\\ \\mbox{cis}(220^\\circ)$ 2. a. $- \\frac{27}{2} - \\frac{27\\sqrt{3}}{2} i$ b. $-2\\sqrt{2} - 2\\sqrt{2}i$ c. $-64$ d. $\\left ( \\sqrt[6]{2} \\right ) \\mbox{cis} \\ 15^\\circ, \\ \\sqrt[6]{2} \\ \\mbox{cis} \\ 135^\\circ, \\ \\sqrt[6]{2} \\ \\mbox{cis} \\ 255^\\circ,$ e. 2cis 67.5o, 2cis 157.5o, 2cis 247.5o, 2cis 337.5o f. cis 18o, cis 90o, cis 162o, cis 234o, cis 306o Feb 23, 2012 Jun 08, 2015"],"string":"[\n \"# Complex Numbers and exponential form and roots The roots of $z^7 = -\\\\frac{1}{\\\\sqrt{2}} - \\\\frac{i}{\\\\sqrt{2}}$ are $\\\\text{cis } \\\\theta_1, \\\\text{cis } \\\\theta_2, \\\\dots, \\\\text{cis } \\\\theta_7,$ where $0^\\\\circ \\\\le \\\\theta_k < 360^\\\\circ$for all $1 \\\\le k \\\\le 7$. Find $\\\\theta_1 + \\\\theta_2 + \\\\dots + \\\\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\\\left(5 \\\\pi i/4 \\\\right)$. How should I simplify? Thanks - What is $\\\\text{cis}$? Do you by chance mean $\\\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\\\theta)+isin(\\\\theta)$ – Rod Jan 24 '14 at 16:27 @AlexR: \\\"cis\\\" is a commonly used abbreviation for \\\"cosine plus $i$ times the sine.\\\" In other words, $$\\\\operatorname{cis}\\\\theta=\\\\cos\\\\theta+i\\\\sin\\\\theta=e^{i\\\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis}(7\\\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis} 225^\\\\circ,$$ as you've already determined, and so we need $$\\\\operatorname{cis}(7\\\\theta_k)=\\\\operatorname{cis} 225^\\\\circ\\\\\\\\\\\\frac{\\\\operatorname{cis}(7\\\\theta_k)}{\\\\operatorname{cis} 225^\\\\circ}=1\\\\\\\\\\\\operatorname{cis}(7\\\\theta_k-225^\\\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? - The roots are $e^{2k\\\\pi i/7}$ ($k=0,\\\\ldots,6$). Write them as $r^k$ where $r = e^{2\\\\pi i/7}$. So the sum is (using the simple formula for summing a finite geometric series) $$\\\\sum_{k=0}^6 r^k = \\\\frac{1-r^7}{1-r} = \\\\frac{1 - e^{2\\\\pi i}}{1-e^{2\\\\pi i/7}} = 0$$. - This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not\",\n \"# Complex Numbers and exponential form and roots The roots of $z^7 = -\\\\frac{1}{\\\\sqrt{2}} - \\\\frac{i}{\\\\sqrt{2}}$ are $\\\\text{cis } \\\\theta_1, \\\\text{cis } \\\\theta_2, \\\\dots, \\\\text{cis } \\\\theta_7,$ where $0^\\\\circ \\\\le \\\\theta_k < 360^\\\\circ$for all $1 \\\\le k \\\\le 7$. Find $\\\\theta_1 + \\\\theta_2 + \\\\dots + \\\\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\\\left(5 \\\\pi i/4 \\\\right)$. How should I simplify? Thanks - What is $\\\\text{cis}$? Do you by chance mean $\\\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\\\theta)+isin(\\\\theta)$ – Mr.Fry Jan 24 '14 at 16:27 @AlexR: \\\"cis\\\" is a commonly used abbreviation for \\\"cosine plus $i$ times the sine.\\\" In other words, $$\\\\operatorname{cis}\\\\theta=\\\\cos\\\\theta+i\\\\sin\\\\theta=e^{i\\\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis}(7\\\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\\\operatorname{cis}\\\\theta_k)^7=\\\\operatorname{cis} 225^\\\\circ,$$ as you've already determined, and so we need $$\\\\operatorname{cis}(7\\\\theta_k)=\\\\operatorname{cis} 225^\\\\circ\\\\\\\\\\\\frac{\\\\operatorname{cis}(7\\\\theta_k)}{\\\\operatorname{cis} 225^\\\\circ}=1\\\\\\\\\\\\operatorname{cis}(7\\\\theta_k-225^\\\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? If I am understanding correctly...for $k=1$, we take $$\\\\operatorname{cis}(7\\\\theta_1)=\\\\operatorname{cis} 225^\\\\circ$$ and then somehow determine $\\\\theta_1$, and repeat for $k=2,3,4,5,6,7$? – Mathy Person Jul 18 '15 at 21:59 No, that's getting further from the answer. Rather, think about the solutions set for the equation $$\\\\operatorname{cis}\\\\theta=1. This will let you get rid of the \\\\operatorname{cis}, and find all solutions to the original\",\n \"equation. Then you can find the seven solutions in the desired interval. – Cameron Buie Jul 18 '15 at 22:10 I'm not sure exactly what you mean. Can you elaborate? – Mathy Person Jul 18 '15 at 22:18 Well, what are the solutions to$$\\\\operatorname{cis}\\\\theta=1,$$or, more usefully put,$$\\\\cos\\\\theta+i\\\\sin\\\\theta=1+0i$$– Cameron Buie Jul 18 '15 at 22:24 \\\\cos \\\\theta = 1 and \\\\sin \\\\theta = 0, so \\\\theta = 360 degrees/2\\\\pi. – Mathy Person Jul 18 '15 at 22:30 The roots are e^{2k\\\\pi i/7} (k=0,\\\\ldots,6). Write them as r^k where r = e^{2\\\\pi i/7}. So the sum is (using the simple formula for summing a finite geometric series)$$\\\\sum_{k=0}^6 r^k = \\\\frac{1-r^7}{1-r} = \\\\frac{1 - e^{2\\\\pi i}}{1-e^{2\\\\pi i/7}} = 0. This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not summing up the roots, but rather the arguments of said roots lying in $[0^\\\\circ,360^\\\\circ).$ – Cameron Buie Jan 24 '14 at 17:15 $\\\\def\\\\cis{\\\\operatorname{cis}}$Let's consider a more general problem. We have $a=r\\\\cis\\\\alpha$ and we write its $n$th roots as \\\\begin{gather} \\\\sqrt[n]{r}\\\\cis\\\\frac{\\\\beta}{n}\\\\\\\\ \\\\sqrt[n]{r}\\\\cis\\\\left(\\\\frac{\\\\beta}{n}+\\\\frac{2\\\\pi}{n}\\\\right)\\\\\\\\ \\\\sqrt[n]{r}\\\\cis\\\\left(\\\\frac{\\\\beta}{n}+2\\\\frac{2\\\\pi}{n}\\\\right)\\\\\\\\ \\\\dots\\\\\\\\ \\\\sqrt[n]{r}\\\\cis\\\\left(\\\\frac{\\\\beta}{n}+(n-1)\\\\frac{2\\\\pi}{n}\\\\right) \\\\end{gather} Can you sum up all those angles? Use $360$ instead of $2\\\\pi$ if you so prefer or are forced to.\",\n \"# Difference between revisions of \\\"1996 AIME Problems/Problem 11\\\" ## Problem Let $\\\\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\\\\mathrm {P}=r(\\\\cos{\\\\theta^{\\\\circ}}+i\\\\sin{\\\\theta^{\\\\circ}})$, where $0 and $0\\\\leq \\\\theta <360$. Find $\\\\theta$. ## Solution 1 $\\\\begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \\\\frac{z^5-1}{z-1}\\\\\\\\ 0 &=& \\\\frac{(z^5 - 1)(z(z-1)+1)}{z-1} = \\\\frac{(z^2-z+1)(z^5-1)}{z-1} \\\\end{eqnarray*}$ Thus $z^5 = 1, z \\\\neq 1 \\\\Longrightarrow z = \\\\mathrm{cis}\\\\ 72, 144, 216, 288$, or $z^2 - z + 1 = 0 \\\\Longrightarrow z = \\\\frac{1 \\\\pm \\\\sqrt{-3}}{2} = \\\\mathrm{cis}\\\\ 60, 300$ (see cis). Discarding the roots with negative imaginary parts (leaving us with $\\\\mathrm{cis} \\\\theta,\\\\ 0 < \\\\theta < 180$), we are left with $\\\\mathrm{cis}\\\\ 60, 72, 144$; their product is $P = \\\\mathrm{cis} (60 + 72 + 144) = \\\\mathrm{cis} \\\\boxed{276}$. ## Solution 2 Let $w =$ the fifth roots of unity, except for $1$. Then $w^6 + w^4 + w^3 + w^2 + 1 = w^4 + w^3 + w^2 + w + 1 = 0$, and since both sides have the fifth roots of unity as roots, we have $z^4 + z^3 + z^2 + z + 1 | z^6 + z^4 + z^3 + z^2 + 1$. Long\",\n \"+ \\\\frac{2k\\\\pi}{3}$$ $$z = (\\\\sqrt[3]{2})cis(-\\\\frac{\\\\pi}{18} + \\\\frac{2k\\\\pi}{3})$$ $$OR = (\\\\sqrt[3]{2})cis(-\\\\frac{\\\\pi}{18})$$ $$OR = (\\\\sqrt[3]{2})cis(-\\\\frac{\\\\pi}{18} + \\\\frac{2\\\\pi}{3})$$ $$OR = (\\\\sqrt[3]{2})cis(-\\\\frac{\\\\pi}{18} + \\\\frac{4\\\\pi}{3})$$ Should they be written like this instead, seeing as I've used the exponential form earlier: $$z = (\\\\sqrt[3]{2})e^{i(-\\\\frac{\\\\pi}{18} + \\\\frac{2k\\\\pi}{3})}$$ $$OR = (\\\\sqrt[3]{2})e^{i(-\\\\frac{\\\\pi}{18})}$$ $$OR = (\\\\sqrt[3]{2})e^{i(-\\\\frac{\\\\pi}{18} + \\\\frac{2\\\\pi}{3})}$$ $$OR = (\\\\sqrt[3]{2})e^{i(-\\\\frac{\\\\pi}{18} + \\\\frac{4\\\\pi}{3})}$$ Would that be $$(\\\\sqrt3 - i)^{\\\\frac{1}{3}}$$? 5. Feb 28, 2010 ### Staff: Mentor Well, yes, but that doesn't tell your what the three cube roots are. 6. Feb 28, 2010 ### Char. Limit I W-A'd it... and only got 1 answer. 7. Mar 1, 2010 ### HallsofIvy \\\"$cis(\\\\theta)$\\\", which is short for $cos(\\\\theta)+ i sin(\\\\theta)= e^{i\\\\theta}$ is primarily an engineering notation. Which is better to use depends on what kind of course this is. If it is a mathematics course being taught be a mathematician, I would say definitely use $e^{i\\\\theta}$.\",\n \"we obtain: $\\\\theta_1=\\\\theta_2=\\\\frac{7\\\\pi}{5}$ Your $*$ is true by the De Moivre's rule. Your $**$ is true because $\\\\frac{cis\\\\theta}{cis\\\\phi}=cis(\\\\theta-\\\\phi)$ and $\\\\overline{cis\\\\theta}=cis(-\\\\theta)$. $\\\\frac{cis\\\\theta}{cis\\\\phi}=cis(\\\\theta-\\\\phi)$ because $$\\\\frac{cis\\\\theta}{cis\\\\phi}=\\\\frac{cis\\\\theta\\\\overline{cis\\\\phi}}{cis\\\\phi\\\\overline{cis\\\\phi}}=cis(\\\\theta-\\\\phi)=$$ • Please read the last line in the OP. I want to understand the points $(\\\\ast)$ and $(\\\\ast\\\\ast)$, I know the final answer. – Yos Jul 7 '17 at 7:45 • @Yos I fixed. See now. – Michael Rozenberg Jul 7 '17 at 7:56 • Why is $\\\\frac{cis\\\\theta_1}{cis\\\\theta_2}=cis(\\\\theta_1-\\\\theta_2)$? Is there an identity for that? – Yos Jul 7 '17 at 8:01 • @Yos See now please. – Michael Rozenberg Jul 7 '17 at 8:18 • @Yos Yes, but I think it's better to use the rule: $\\\\frac{cis\\\\theta}{cis\\\\phi}=cis(\\\\theta-\\\\phi)$ – Michael Rozenberg Jul 7 '17 at 8:28 The line where you have $(*)$ should better be $$\\\\left|\\\\frac{z_1}{\\\\overline{z_2}^{\\\\,2}}\\\\right|= \\\\frac{|z_1|}{\\\\bigl|\\\\overline{z_2}^{\\\\,2}\\\\bigr|}=\\\\frac{r_1}{r_2^2}$$ There is no reason for the equality before the one you mark with $(*)$. The equality you mark with $({*}{*})$ is indeed wrong and it should be $$\\\\frac{r_1(\\\\cos\\\\theta_1+i\\\\sin\\\\theta_1)}{r_2^2(\\\\cos2\\\\theta_2-i\\\\sin2\\\\theta_2)}=\\\\frac{r_1}{r_2^2}(\\\\cos(\\\\theta_1+2\\\\theta_2)+i\\\\sin(\\\\theta_1+2\\\\theta_2))=\\\\frac{2}{25}(\\\\cos\\\\frac{\\\\pi}{5}+i\\\\sin\\\\frac{\\\\pi}{5})$$ because if $z=r(\\\\cos\\\\theta+i\\\\sin\\\\theta)$, then $\\\\bar{z}=r(\\\\cos\\\\theta-i\\\\sin\\\\theta)$. Next apply the standard rules \\\\begin{gather} \\\\cos\\\\theta-i\\\\sin\\\\theta=(\\\\cos\\\\theta+i\\\\sin\\\\theta)^{-1}\\\\\\\\[4px] (\\\\cos\\\\alpha+i\\\\sin\\\\alpha)(\\\\cos\\\\beta+i\\\\sin\\\\beta)= \\\\cos(\\\\alpha+\\\\beta)+i\\\\sin(\\\\alpha+\\\\beta) \\\\end{gather} ### Further notes. The equality $$\\\\frac{z_1z_2^2}{\\\\overline{z_2}^{\\\\,2}z_2^2}=\\\\frac{r_1r_2^2}{|z_2|^4}$$ is generally false, because there's no reason for $z_1z_2^2$ to be real. When you expand $\\\\dfrac{z_1}{\\\\overline{z_2}^{\\\\,2}}$, you should write $$\\\\frac{r_1(\\\\cos\\\\theta_1+i\\\\sin\\\\theta_1)} {\\\\bigl(\\\\overline{r_2(\\\\cos\\\\theta_2+i\\\\sin\\\\theta_2}\\\\bigr)^2} = \\\\frac{r_1(\\\\cos\\\\theta_1+i\\\\sin\\\\theta_1)} {(r_2(\\\\cos\\\\theta_2-i\\\\sin\\\\theta_2)^2} = \\\\frac{r_1(\\\\cos\\\\theta_1+i\\\\sin\\\\theta_1)} {r_2^2(\\\\cos2\\\\theta_2-i\\\\sin2\\\\theta_2)}$$ • Can you please explain why $\\\\ast\\\\ast$ equality is a mistake? As far\",\n \"+0 # Asymptote coding +3 37 2 +538 Hello! I was answering a question earlier, and wanted to include a graph. Since Asympotote coding makes it look clean, I tried inserting this code: [asy] size(250); import TrigMacros; real theta = 45; pair A = (3,0); pair B = 3*dir(theta); pair O = (0,0); rr_cartesian_axes(-5, 5, -5, 5, usegrid = false); draw(Circle(O, 3)); draw(O--3*dir(60)); dot(\\\"$\\\\left(\\\\frac{\\\\sqrt{3}}{2}, \\\\frac{1}{2}\\\\right)$\\\", 3*dir(30), E); draw(O--3*dir(30)); dot(\\\"$\\\\left(\\\\frac{1}{2}, \\\\frac{\\\\sqrt{3}}{2}\\\\right)$\\\", 3*dir(60), N); draw(O--3*dir(45)); dot(\\\"$\\\\left(\\\\frac{\\\\sqrt{2}}{2}, \\\\frac{\\\\sqrt{2}}{2}\\\\right)$\\\", 3*dir(45), dir(45)); draw(O--3*dir(150)); dot(\\\"$\\\\left(-\\\\frac{\\\\sqrt{3}}{2}, \\\\frac{1}{2}\\\\right)$\\\", 3*dir(150), W); draw(O--3*dir(120)); dot(\\\"$\\\\left(-\\\\frac{1}{2}, \\\\frac{\\\\sqrt{3}}{2}\\\\right)$\\\", 3*dir(120), N); draw(O--3*dir(135)); dot(\\\"$\\\\left(-\\\\frac{\\\\sqrt{2}}{2}, \\\\frac{\\\\sqrt{2}}{2}\\\\right)$\\\", 3*dir(135), dir(135)); draw(O--3*dir(-60)); dot(\\\"$\\\\left(\\\\frac{\\\\sqrt{3}}{2}, -\\\\frac{1}{2}\\\\right)$\\\", 3*dir(-30), E); draw(O--3*dir(-30)); dot(\\\"$\\\\left(\\\\frac{1}{2}, -\\\\frac{\\\\sqrt{3}}{2}\\\\right)$\\\", 3*dir(-60), -N); draw(O--3*dir(-45)); dot(\\\"$\\\\left(\\\\frac{\\\\sqrt{2}}{2}, -\\\\frac{\\\\sqrt{2}}{2}\\\\right)$\\\", 3*dir(-45), dir(-45)); draw(O--3*dir(-150)); dot(\\\"$\\\\left(-\\\\frac{\\\\sqrt{3}}{2}, -\\\\frac{1}{2}\\\\right)$\\\", 3*dir(-150), W); draw(O--3*dir(-120)); dot(\\\"$\\\\left(-\\\\frac{1}{2}, -\\\\frac{\\\\sqrt{3}}{2}\\\\right)$\\\", 3*dir(-120), -N); draw(O--3*dir(-135)); dot(\\\"$\\\\left(-\\\\frac{\\\\sqrt{2}}{2},- \\\\frac{\\\\sqrt{2}}{2}\\\\right)$\\\", 3*dir(-135), dir(-135)); dot(\\\"$(1,0)$\\\", A, NE); [/asy] But, it did not show up. (it's a unit circle) Does this site support Asymptote, or do I need to use desmos? Thanks! off-topic Aug 2, 2020 #1 +128 +1 I believe that this site does not support asymptote, as I have tried to use it in the past. If you try to take a screenshot, or maybe as you said desmos will work. Hope this helps! Aug 2, 2020 #2 +538 +3 Okay! ilorty Aug 2, 2020\",\n \"multiply the numerator and denominator by the conjugate of the number in the denominator. $$\\\\frac{\\\\textrm{cis} 75^o - \\\\textrm{cis} 155^o}{ 1 - \\\\cos{8^o} -i \\\\sin{8^o}} \\\\frac{1-\\\\cos{8^o} + i \\\\sin{8^o}}{1-\\\\cos{8^o} + i \\\\sin{8^o}}$$ This will give us a completely real denominator. And from there, we just have a lot of algebra to grind out to get the remaining expression down to a form of $r \\\\textrm{cis}{\\\\theta}$. I hope you find the algebra to be doable. The problem is basically a vector addition problem. $cis\\\\theta = cos\\\\theta + i sin\\\\theta$. So, $cis 75 + cis 83 + ... = cos 75 + cos 83 + ... + i(sin 75 + sin 83 + ...)$ Actually doing the math would be difficult, but note that you're adding a bunch of vectors of equal length, each 8 degrees apart. They form an arc and what you want is the average direction of the arc since that is also the net direction of the vectors. Hope that gets you started. I can comment further if necessary. this is a geometric series $$cis(75^\\\\circ) +cis(75^\\\\circ + 8^\\\\circ)+\\\\cdots + cis(75^\\\\circ+9\\\\times 8^\\\\circ)=cis(75^\\\\circ)\\\\frac{1-cis(80^\\\\circ)}{1-cis(8^\\\\circ)}$$ i hope you can take it from here.\",\n \"- \\\\alpha_n)^{e_n}$$ for some positive integers $e_i$ and distinct complex numbers $\\\\alpha_i$, in a unique way (up to ordering of the factors). The fact that every nonzero complex number has $n$ nth roots is a little easier, though, as long as you know that every nonzero complex number $z$ can be represented in a unique way in the form $re^{i\\\\theta}$ for some $\\\\theta \\\\in \\\\left[0, 2\\\\pi), \\\\ \\\\mathbb{R} \\\\ni r > 0$, and that $e^{i\\\\theta}$ is 2$\\\\pi$-periodic. Then you can write $$z = re^{i\\\\theta} \\\\Longrightarrow \\\\left(\\\\sqrt[n]{r}e^{i\\\\left(\\\\frac{\\\\theta + 2k\\\\pi}{n}\\\\right)}\\\\right)^n = z, \\\\ \\\\mathbb{Z} \\\\ni k \\\\in \\\\left[0, n-1\\\\right]$$ ie. z has at least $n$ nth roots. It remains to be proved that there are exactly n of them, but you can work on that on your own~ Last edited: Apr 29, 2005 12. Apr 30, 2005 ### whozum From Orion1: $$w_1 = \\\\text{cis} \\\\; 225^{\\\\circ} = ( \\\\cos 225^{\\\\circ} + i \\\\sin 225^{\\\\circ})$$ Is the cis() function equivalent to cos() + i sin() ? 13. Apr 30, 2005 ### jcsd Yes it is, but no-one bothers to use it any more as cis x = e^ix which is just as easy to write. 14. Apr 30, 2005 ### robert Ihnot Orion1: Any root of i can be solved using De Moivre's Theorem. I just want to add that, if\",\n \"and suppose that P=r(costheta+isintheta) where $0 \\\\lt r$ and $0 \\\\leq \\\\theta \\\\lt 360$ find $\\\\theta$ • is 107 • is 276 • is 840 • cannot be determined from the given information ### Key Concepts Equations Complex Numbers Integers AIME, 1996, Question 11 Complex Numbers from A to Z by Titu Andreescue ## Try with Hints First hint here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\\\\frac{(z^{5}-1)}{(z-1)}$=$\\\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\\\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0 Second Hint gives $z^{5}=1 for z\\\\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \\\\neq 1$ gives z=$\\\\frac{1+-(-3)^\\\\frac{1}{2}}{2}$=$cis60,300$ where cis$\\\\theta$=cos$\\\\theta$+isin$\\\\theta$ Final Step taking $0 \\\\lt theta \\\\lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276. . ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed.\",\n \"\\\\theta & =-0.7\\\\end{align*} Ref angle: $\\\\scriptsize \\\\theta ={{44.4}^\\\\circ}$ $\\\\scriptsize \\\\sin \\\\theta \\\\lt 0$: $\\\\scriptsize \\\\theta ={{180}^\\\\circ}+{{44.4}^\\\\circ}={{224.4}^\\\\circ}\\\\text{ or }\\\\theta ={{360}^\\\\circ}-{{44.4}^\\\\circ}={{315.6}^\\\\circ}$ General solution: $\\\\scriptsize \\\\theta ={{224.4}^\\\\circ}+k\\\\text{.36}{{\\\\text{0}}^\\\\circ}\\\\text{ or }\\\\theta ={{315.6}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }$ 2. . \\\\scriptsize \\\\begin{align*}2\\\\tan\\\\theta&=0.279\\\\\\\\\\\\therefore \\\\tan\\\\theta&=0.1395\\\\end{align*} Ref angle: $\\\\scriptsize \\\\theta ={{8.0}^\\\\circ}$ $\\\\scriptsize \\\\tan \\\\theta \\\\gt 0:$ $\\\\scriptsize \\\\theta ={{8.0}^\\\\circ}\\\\text{ or }\\\\theta ={{180}^\\\\circ}+{{8.0}^\\\\circ}={{188}^\\\\circ}$ General solution: $\\\\scriptsize \\\\theta ={{8.0}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }$ Specific solution for $\\\\scriptsize \\\\theta \\\\in [{{0}^\\\\circ},{{360}^\\\\circ}]$: $\\\\scriptsize \\\\theta ={{8}^\\\\circ}+0\\\\times {{180}^\\\\circ}={{8}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{8.0}^\\\\circ}+1\\\\times {{180}^\\\\circ}={{188.0}^\\\\circ}$ 3. . \\\\scriptsize \\\\begin{align*}5\\\\cos2\\\\theta&=-2.7\\\\\\\\\\\\therefore \\\\cos2\\\\theta&=-0.54\\\\end{align*} Ref angle: $\\\\scriptsize 2\\\\theta ={{57.3}^\\\\circ}$ $\\\\scriptsize \\\\cos 2\\\\theta \\\\lt 0$: $\\\\scriptsize 2\\\\theta ={{180}^\\\\circ}-{{57.3}^\\\\circ}={{122.7}^\\\\circ}\\\\text{ or }2\\\\theta ={{180}^\\\\circ}+{{57.3}^\\\\circ}={{237.3}^\\\\circ}$ General solution: \\\\scriptsize \\\\begin{align*}2\\\\theta ={{122.7}^\\\\circ}+k{{.360}^\\\\circ}\\\\text{ or }2\\\\theta ={{237.3}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\\\\\\\\\therefore \\\\theta ={{61.35}^\\\\circ}+k{{.180}^\\\\circ}\\\\text{ or }\\\\theta ={{118.65}^\\\\circ}+{{180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\end{align*} Specific solution for $\\\\scriptsize \\\\theta \\\\in [{{0}^\\\\circ},{{360}^\\\\circ}]$: $\\\\scriptsize \\\\theta ={{61.35}^\\\\circ}+0\\\\times {{180}^\\\\circ}={{61.35}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{61.35}^\\\\circ}+1\\\\times {{180}^\\\\circ}={{241.35}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{118.65}^\\\\circ}+0\\\\times {{180}^\\\\circ}={{118.65}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{118.65}^\\\\circ}+1\\\\times {{180}^\\\\circ}={{298.65}^\\\\circ}$ 4. $\\\\scriptsize \\\\tan (3\\\\theta -{{48}^\\\\circ})=3.2$ Ref angle: $\\\\scriptsize 3\\\\theta -{{48}^\\\\circ}={{72.6}^\\\\circ}$ $\\\\scriptsize 3\\\\theta -{{48}^\\\\circ}={{72.6}^\\\\circ}\\\\text{ or }3\\\\theta -{{48}^\\\\circ}={{180}^\\\\circ}+{{72.6}^\\\\circ}={{252.6}^\\\\circ}$: General solution: \\\\scriptsize \\\\begin{align*}3\\\\theta -{{48}^\\\\circ} & ={{72.6}^\\\\circ}+k{{.180}^\\\\circ}\\\\text{ or }3\\\\theta -{{48}^\\\\circ}={{252.6}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\\\\\\\\\therefore 3\\\\theta & ={{120.6}^\\\\circ}+k{{.180}^\\\\circ}\\\\text{ or }3\\\\theta ={{300.6}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\\\\\\\\\therefore \\\\theta & ={{40.2}^\\\\circ}+k{{.60}^\\\\circ}\\\\text{ or }\\\\theta ={{100.2}^\\\\circ}+k{{.60}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\end{align*} But $\\\\scriptsize {{40.2}^\\\\circ}+1\\\\times {{60}^\\\\circ}={{100.2}^\\\\circ}$ Therefore, simplest general solution is $\\\\scriptsize \\\\theta ={{40.2}^\\\\circ}+k{{.60}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }$ Specific solution for $\\\\scriptsize \\\\theta \\\\in [{{0}^\\\\circ},{{180}^\\\\circ}]$: $\\\\scriptsize \\\\theta ={{40.2}^\\\\circ}+0\\\\times {{60}^\\\\circ}={{40.2}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{40.2}^\\\\circ}+1\\\\times {{60}^\\\\circ}={{100.2}^\\\\circ}$ or $\\\\scriptsize \\\\theta ={{40.2}^\\\\circ}+2\\\\times {{60}^\\\\circ}={{160.2}^\\\\circ}$ 5. .\",\n \"product of the roots of $z^{6}+z^{4}+z^{2}+1=0$ that have a positive imaginary part and suppose that P=r(costheta+isintheta) where $0 \\\\lt r$ and $0 \\\\leq \\\\theta \\\\lt 360$ find $\\\\theta$ • is 107 • is 276 • is 840 • cannot be determined from the given information ### Key Concepts Equations Complex Numbers Integers AIME, 1996, Question 11 Complex Numbers from A to Z by Titu Andreescue ## Try with Hints First hint here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\\\\frac{(z^{5}-1)}{(z-1)}$=$\\\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\\\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0 Second Hint gives $z^{5}=1 for z\\\\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \\\\neq 1$ gives z=$\\\\frac{1+-(-3)^\\\\frac{1}{2}}{2}$=$cis60,300$ where cis$\\\\theta$=cos$\\\\theta$+isin$\\\\theta$ Final Step taking $0 \\\\lt theta \\\\lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276. . Categories ## Roots of Equation and Vieta’s formula | AIME I, 1996 Problem 5 Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula. ## Roots of Equation and Vieta’s formula – AIME I, 1996 Suppose that the roots of $x^{3}+3x^{2}+4x-11=0$ are a,b and c and that the roots of $x^{3}+rx^{2}+sx+t=0$ are a+b,b+c and c+a, find t. • is 107 • is 23 • is 840 • cannot be determined from the given information ### Key Concepts Functions Roots of Equation Vieta s formula AIME I, 1996, Question 5 Polynomials by Barbeau ##\",\n \"form. 1. $$5 \\\\; cis \\\\; 270^{\\\\circ}$$ 2. $$2 \\\\; cis \\\\; 30^{\\\\circ}$$ 3. $$−4 \\\\; cis \\\\; \\\\dfrac{\\\\pi}{4}$$ 4. $$6 \\\\; cis \\\\; \\\\dfrac{\\\\pi}{3}$$ 5. $$2 \\\\; cis \\\\; \\\\dfrac{5\\\\pi}{2}$$ Translate the following complex numbers from rectangular form into trigonometric polar form. 6. $$2−i$$ 7. $$5+12i$$ 8. $$6i+8$$ 9. $$i$$ Complete the following calculations and simplify. 10. $$2 \\\\; cis \\\\; 22^{\\\\circ} \\\\cdot \\\\dfrac{1}{5} \\\\; cis \\\\; 15^{\\\\circ} \\\\cdot 3 \\\\; cis \\\\; 95^{\\\\circ}$$ 11. $$9 \\\\; cis \\\\; 98^{\\\\circ} \\\\div 3 \\\\; cis \\\\; 12^{\\\\circ}$$ 12. $$15 \\\\; cis \\\\; \\\\dfrac{\\\\pi}{4}\\\\cdot 2 \\\\; cis \\\\; \\\\dfrac{\\\\pi}{6}$$ 13. $$−2 \\\\; cis \\\\; \\\\dfrac{2\\\\pi}{3}\\\\div 15 \\\\; cis \\\\; \\\\dfrac{7 \\\\pi}{6}$$ Let $$z_1 =r_1 \\\\cdot \\\\; cis \\\\; \\\\theta_1$$ and $$z_2 =r_2 \\\\cdot \\\\; cis \\\\; \\\\theta_2$$ with r_2 \\\\neq 0\\\\). 14. Use the trigonometric sum and difference identities to prove that $$z_1 \\\\cdot z_2 =r_1 \\\\cdot r_2 \\\\cdot \\\\; cis \\\\; (\\\\theta_1 +\\\\theta_2)$$. 15. Use the trigonometric sum and difference identities to prove that $$z_1 \\\\div z_2 =r_1 r_2 \\\\cdot \\\\; cis \\\\; (\\\\theta_1 −\\\\theta_2)$$. To see the Review answers, open this PDF file and look for section 11.3. ## Vocabulary Term Definition $$r\\\\; cis \\\\;\\\\theta$$ $$r\\\\; cis \\\\;\\\\theta$$ is shorthand for the expression $$r\\\\cos \\\\theta +ri\\\\sin \\\\theta$$. complex plane The complex plane is the graphical representation of\",\n \"solution is correct. ### Example 2.2 Determine the general solution for $\\\\scriptsize 7\\\\cos 2\\\\theta +4=0$. Solution \\\\scriptsize \\\\displaystyle \\\\begin{align*}7\\\\cos 2\\\\theta +4&=0\\\\\\\\\\\\therefore \\\\cos 2\\\\theta &=-\\\\displaystyle \\\\frac{4}{7}\\\\end{align*} Step 1: Use a calculator to determine the reference angle \\\\scriptsize \\\\displaystyle \\\\begin{align*}\\\\cos 2\\\\theta &=-\\\\displaystyle \\\\frac{4}{7}\\\\\\\\\\\\therefore 2\\\\theta &={{124.8}^\\\\circ}\\\\end{align*} Note: We keep working with the reference angle of $\\\\scriptsize 2\\\\theta$ until we generate the general solution. Step 2: Use the CAST diagram to determine any other possible solutions Our equation is $\\\\scriptsize \\\\displaystyle \\\\cos 2\\\\theta =-\\\\displaystyle \\\\frac{4}{7}$. $\\\\scriptsize \\\\cos 2\\\\theta \\\\lt 0$. Cosine is negative in the second and third quadrants. Our reference angle is in the second quadrant. Second quadrant: $\\\\scriptsize 2\\\\theta ={{124.8}^\\\\circ}$ Third quadrant: $\\\\scriptsize \\\\cos ({{360}^\\\\circ}-\\\\theta )=\\\\cos \\\\theta$ $\\\\scriptsize 2\\\\theta ={{360}^\\\\circ}-{{124.8}^\\\\circ}={{235.2}^\\\\circ}$. Step 3: Generate the general solution \\\\scriptsize \\\\begin{align*}2\\\\theta & ={{124.8}^\\\\circ}+k{{.360}^\\\\circ}\\\\text{ or 2}\\\\theta ={{235.2}^\\\\circ}+k{{.360}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\\\\\\\\\therefore \\\\theta & ={{62.4}^\\\\circ}+k{{.180}^\\\\circ}\\\\text{ or }\\\\theta ={{117.6}^\\\\circ}+k{{.180}^\\\\circ},k\\\\in \\\\mathbb{Z}\\\\text{ }\\\\end{align*} Step 4: Check your general solution $\\\\scriptsize k=2$: $\\\\scriptsize \\\\therefore \\\\theta ={{62.4}^\\\\circ}+2\\\\times {{180}^\\\\circ}={{422.4}^\\\\circ}\\\\text{ or }\\\\theta ={{117.6}^\\\\circ}+2\\\\times {{180}^\\\\circ}={{477.6}^\\\\circ}$ $\\\\scriptsize \\\\displaystyle \\\\cos (2\\\\times {{422.4}^\\\\circ})=-\\\\displaystyle \\\\frac{4}{7}$ $\\\\scriptsize \\\\displaystyle \\\\cos (2\\\\times {{477.6}^\\\\circ})=-\\\\displaystyle \\\\frac{4}{7}$ ### Take note! The general solutions for equations involving the three basic trigonometric ratios can be written as follows: If $\\\\scriptsize \\\\sin \\\\theta =x$ then: $\\\\scriptsize \\\\theta =\\\\left( {{{{\\\\sin }}^{{-1}}}x+k{{{.360}}^\\\\circ}} \\\\right)\\\\text{ or }\\\\theta =\\\\left( {\\\\left( {{{{180}}^\\\\circ}-{{{\\\\sin }}^{{-1}}}x} \\\\right)+k{{{.360}}^\\\\circ}} \\\\right),k\\\\in \\\\mathbb{Z}$ If $\\\\scriptsize \\\\cos \\\\theta =x$ then: $\\\\scriptsize \\\\theta\",\n \"# Solving a quadratic equation with complex coefficient Express $z4$=-$\\\\sqrt{3}$+i in polar form. Hence solve the equation $Z^2$=$z4$ for $z$ a complex number. You may leave the answer in polar form. My answer: $z4$ in polar form is 2cis-30$^{\\\\circ}$ and thats as far as I have gotten. I have seen this question asked on other sites, but the answers with working solutions are too complex for the level this is at. This is a complex number topic assignment on a basic mathematics course, so I can only solve this question using the basic material that has been covered already in the course. There is an example in the notes where they use De Moivre's theorem to solve it. • $$-\\\\sqrt3+i=2\\\\,\\\\text{cis}\\\\left(\\\\arctan\\\\frac1{-\\\\sqrt3}\\\\right)=2\\\\,\\\\text{cis}\\\\,( 150^\\\\circ)$$ and not $\\\\;-30^\\\\circ\\\\;$ , since here cosine is negative and sine positive. – Timbuc Jun 30 '15 at 13:03 • $(a{\\\\rm\\\\ cis\\\\ }b)^2=a^2{\\\\rm\\\\ cis\\\\ }2b$ should give you an idea for how to take square roots of a number given in cis form. – Gerry Myerson Jun 30 '15 at 13:12 We have $z_4=2\\\\mbox{cis}(150^\\\\circ+360^\\\\circ\\\\times k), \\\\ k=...-2,-1,0,1,2,...$, that is $\\\\mbox{cis(.)}$ is periodic with period $360^\\\\circ$. Then $z^2=z_4$ requires that $$z=\\\\sqrt{2}\\\\mbox{cis}\\\\left(\\\\frac{150^\\\\circ+360^\\\\circ \\\\times k}{2}\\\\right)\\\\ k=...-2,-1,0,1,2,...$$ but there are only two distinct values for $z$ which correspond to $k=0,1$. Hence the two solutions are:$z=\\\\sqrt{2}\\\\mbox{cis}(75^\\\\circ)$ and $z=\\\\sqrt{2}\\\\mbox{cis}(225^\\\\circ)$\",\n \"form: $\\\\scriptsize z=13\\\\text{cis}{{292.62}^\\\\circ}$ 2. $\\\\scriptsize z=-7+2i$ Find $\\\\scriptsize r$: \\\\scriptsize \\\\begin{align*}r & =\\\\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\\\\\\\ & =\\\\sqrt{{{{{(-7)}}^{2}}+{{2}^{2}}}}\\\\\\\\& =\\\\sqrt{{49+4}}\\\\\\\\ & =\\\\sqrt{{53}}\\\\end{align*} Find $\\\\scriptsize \\\\theta$ ($\\\\scriptsize \\\\theta$ is in the second quadrant): $\\\\scriptsize \\\\cos \\\\theta =\\\\displaystyle \\\\frac{x}{r}=\\\\displaystyle \\\\frac{{-7}}{{\\\\sqrt{{53}}}}$ Work with $\\\\scriptsize \\\\cos \\\\theta =\\\\displaystyle \\\\frac{7}{{\\\\sqrt{{53}}}}$ to find a referance angle. \\\\scriptsize \\\\begin{align*}\\\\therefore \\\\text{reference angle} & ={{15.95}^\\\\circ}\\\\\\\\\\\\therefore \\\\theta & ={{180}^\\\\circ}-{{15.95}^\\\\circ}={{164.05}^\\\\circ}\\\\end{align*} Polar form: $\\\\scriptsize z=\\\\sqrt{{53}}\\\\text{cis164}\\\\text{.0}{{\\\\text{5}}^\\\\circ}$ Back to Exercise 3.1 ### Exercise 3.2 1. $\\\\scriptsize z=6\\\\text{cis}{{45}^\\\\circ}$ $\\\\scriptsize {{45}^\\\\circ}$ is a special angle and the complex number is in the first quadrant. $\\\\scriptsize \\\\cos {{45}^\\\\circ}=\\\\displaystyle \\\\frac{1}{{\\\\sqrt{2}}}$ $\\\\scriptsize \\\\sin {{45}^\\\\circ}=\\\\displaystyle \\\\frac{1}{{\\\\sqrt{2}}}$ \\\\scriptsize \\\\displaystyle \\\\begin{align*}z & =6(\\\\cos {{45}^\\\\circ}+i\\\\sin {{45}^\\\\circ})\\\\\\\\ & =6\\\\left( {\\\\displaystyle \\\\frac{1}{{\\\\sqrt{2}}}+\\\\displaystyle \\\\frac{1}{{\\\\sqrt{2}}}i} \\\\right)\\\\\\\\ & =\\\\displaystyle \\\\frac{6}{{\\\\sqrt{2}}}+\\\\displaystyle \\\\frac{6}{{\\\\sqrt{2}}}i\\\\end{align*} The complex number in standard form is $\\\\scriptsize \\\\displaystyle z=\\\\displaystyle \\\\frac{6}{{\\\\sqrt{2}}}+\\\\displaystyle \\\\frac{6}{{\\\\sqrt{2}}}i$. 2. $\\\\scriptsize z=\\\\sqrt{5}\\\\text{cis21}{{\\\\text{0}}^\\\\circ}$ $\\\\scriptsize {{180}^\\\\circ}+{{30}^\\\\circ}={{210}^\\\\circ}$. Therefore, we use the special angle of $\\\\scriptsize {{30}^\\\\circ}$ in the third quadrant. $\\\\scriptsize \\\\cos {{210}^\\\\circ}=\\\\cos ({{180}^\\\\circ}+{{30}^\\\\circ})=-\\\\cos {{30}^\\\\circ}=-\\\\displaystyle \\\\frac{{\\\\sqrt{3}}}{2}$ $\\\\scriptsize \\\\sin {{210}^\\\\circ}=\\\\sin ({{180}^\\\\circ}+{{30}^\\\\circ})=-\\\\sin {{30}^\\\\circ}=-\\\\displaystyle \\\\frac{1}{2}$ \\\\scriptsize \\\\displaystyle \\\\begin{align*}z & =\\\\sqrt{5}(\\\\cos {{210}^\\\\circ}+i\\\\sin {{210}^\\\\circ})\\\\\\\\ & =\\\\sqrt{5}\\\\left( {-\\\\displaystyle \\\\frac{{\\\\sqrt{3}}}{2}-\\\\displaystyle \\\\frac{1}{2}i} \\\\right)\\\\quad \\\\text{Both cosine and sine are negative in the third quadrant}\\\\\\\\ & =-\\\\displaystyle \\\\frac{{\\\\sqrt{{15}}}}{2}-\\\\displaystyle \\\\frac{{\\\\sqrt{5}}}{2}i\\\\end{align*} The complex number in standard form is $\\\\scriptsize \\\\displaystyle z=-\\\\displaystyle \\\\frac{{\\\\sqrt{{15}}}}{2}-\\\\displaystyle \\\\frac{{\\\\sqrt{5}}}{2}i$. 3. $\\\\scriptsize z=2\\\\text{cis}{{40}^\\\\circ}$ $\\\\scriptsize {{40}^\\\\circ}$ is not a special angle. The complex number is in the first quadrant. $\\\\scriptsize \\\\cos {{40}^\\\\circ}=0.7660$ $\\\\scriptsize \\\\sin {{40}^\\\\circ}=0.6428$ \\\\scriptsize \\\\displaystyle \\\\begin{align*}z &\",\n \"? \\\\end{array}$$ Example 5 What are the two square roots of i? Solution Let $$\\\\ z=\\\\sqrt{0+i}$$. $$\\\\ r=1, \\\\theta=\\\\pi / 2 \\\\text { or } z=\\\\left[1 \\\\times \\\\operatorname{cis} \\\\frac{\\\\pi}{2}\\\\right]^{1 / 2}$$ Utilizing De Moivre’s theorem: $$\\\\ z_{1}=\\\\left[1 \\\\times \\\\operatorname{cis} \\\\frac{\\\\pi}{4}\\\\right] \\\\text { or } z_{2}=\\\\left[1 \\\\times \\\\operatorname{cis} \\\\frac{5 \\\\pi}{4}\\\\right]$$ $$\\\\ z_{1}=1\\\\left(\\\\cos \\\\frac{\\\\pi}{4}+i \\\\sin \\\\frac{\\\\pi}{4}\\\\right) \\\\text { or } z_{2}=1\\\\left(\\\\cos \\\\frac{5 \\\\pi}{4}+i \\\\sin \\\\frac{5 \\\\pi}{4}\\\\right)$$ $$\\\\ z_{1}=0.707+0.707 i \\\\text { or } z_{2}=-0.707-0.707 i$$ Check for z1 solution: $$\\\\ (0.707+0.707 i)^{2}=i ?$$ $$\\\\ 0.500+0.500 i+0.500 i+0.500 i^{2}=0.500+i+0.500(-1) \\\\text { or } i$$ Example 6 Calculate $$\\\\ \\\\sqrt[4]{(1+0 i)}$$. What are the four fourth roots of 1? Solution Let $$\\\\ z=1 \\\\text { or } z=1+0 i$$. Then the problem becomes find $$\\\\ z^{1 / 4}=(1+0 i)^{1 / 4}$$. Since $$\\\\ r=1 \\\\theta=0, z^{1 / 4}=[1 \\\\times \\\\operatorname{cis} 0]^{1 / 4}$$ with $$\\\\ z_{1}=1^{1 / 4}\\\\left(\\\\cos \\\\frac{0}{4}+i \\\\sin \\\\frac{0}{4}\\\\right)$$ or $$\\\\ 1(1+0)$$ or $$\\\\ 1$$ That root is not a surprise. Now use De Moivre’s to find the other roots: $$\\\\ z_{2}=1^{1 / 4}\\\\left[\\\\cos \\\\left(0+\\\\frac{\\\\pi}{2}\\\\right)+i \\\\sin \\\\left(0+\\\\frac{\\\\pi}{2}\\\\right)\\\\right]$$ Since there are 4 roots, dividing $$\\\\ 2 \\\\pi$$ by 4 yields $$\\\\ 0.5 \\\\pi$$ or $$\\\\ 0+i$$ or just $$\\\\ i$$ $$\\\\ z_{3}=1^{1 / 4}\\\\left[\\\\cos \\\\left(0+\\\\frac{2 \\\\pi}{2}\\\\right)+i \\\\sin \\\\left(0+\\\\frac{2 \\\\pi}{2}\\\\right)\\\\right]$$ which yields $$\\\\ z_{3}=-1$$ Finally, $$\\\\ z_{4}=1^{1 / 4}\\\\left[\\\\cos \\\\left(0+\\\\frac{3 \\\\pi}{2}\\\\right)+i \\\\sin\",\n \"+0 # help +1 50 1 The number $$\\\\text{cis } 75^\\\\circ + \\\\text{cis } 83^\\\\circ + \\\\text{cis } 91^\\\\circ + \\\\dots + \\\\text{cis } 147^\\\\circ$$ is expressed in the form $$r \\\\, \\\\text{cis } \\\\theta$$, where $$0 \\\\le \\\\theta < 360^\\\\circ$$. Find $$\\\\theta$$ in degrees. Dec 20, 2019\",\n \"told to do so, we leave this as a good, but messy, exercise. 4. To find the five fifth roots of $$1$$, we write $$1 = 1 \\\\text{cis}(0)$$. We have $$r = 1$$, $$\\\\theta = 0$$ and $$n = 5$$. Since $$\\\\sqrt[5]{1} = 1$$, the roots are $$w_{\\\\text{\\\\tiny$$0\\\\)}} = \\\\text{cis}(0) = 1\\\\), $$w_{\\\\text{\\\\tiny$$1\\\\)}} = \\\\text{cis}\\\\left(\\\\frac{2\\\\pi}{5}\\\\right)\\\\), $$w_{\\\\text{\\\\tiny$$2\\\\)}} = \\\\text{cis}\\\\left(\\\\frac{4\\\\pi}{5}\\\\right)\\\\), $$w_{\\\\text{\\\\tiny$$3\\\\)}} = \\\\text{cis}\\\\left(\\\\frac{6\\\\pi}{5}\\\\right)\\\\) and $$w_{\\\\text{\\\\tiny$$4\\\\)}} = \\\\text{cis}\\\\left(\\\\frac{8\\\\pi}{5}\\\\right)\\\\). The situation here is even graver than in the previous example, since we have not developed any identities to help us determine the cosine or sine of $$\\\\frac{2\\\\pi}{5}$$. At this stage, we could approximate our answers using a calculator, and we leave this as an exercise. Now that we have done some computations using Theorem \\\\ref{nthrootscomplexthm}, we take a step back to look at things geometrically. Essentially, Theorem \\\\ref{nthrootscomplexthm} says that to find the $$n^{\\\\text{th}}$$ roots of a complex number, we first take the $$n^{\\\\text{th}}$$ root of the modulus and divide the argument by $$n$$. This gives the first root $$w_{\\\\text{\\\\tiny$$0\\\\)}}\\\\). Each succeessive root is found by adding $$\\\\frac{2\\\\pi}{n}$$ to the argument, which amounts to rotating $$w_{\\\\text{\\\\tiny$$0\\\\)}}\\\\) by $$\\\\frac{2\\\\pi}{n}$$ radians. This results in $$n$$ roots, spaced equally around the complex plane. As an example of this, we plot our answers to number \\\\ref{fourthrootsneg16} in Example \\\\ref{nthrootscomplexex} below. We have only glimpsed at the\",\n \"\\\\right ]^4$ c. $\\\\left (\\\\sqrt{3} - i \\\\right )^6$ d. The 3 complex cube roots of 1 + i e. The 4 complex fourth roots of - 16i f. The five complex fifth roots of i 1. a. $\\\\frac{-1}{2} + \\\\frac{5}{2}i$ b. 37 c. $\\\\frac{1}{2} + \\\\frac{-\\\\sqrt{3}}{2}i$ d. $4\\\\sqrt{2}(\\\\mbox{cos} \\\\ 15^\\\\circ + i \\\\ \\\\mbox{sin} \\\\ 15^\\\\circ)$ e. $8\\\\ \\\\mbox{cis}(60^\\\\circ)$ f. $4\\\\ \\\\mbox{cis}\\\\left ( \\\\frac{9\\\\pi}{40} \\\\right )$ g. $\\\\frac{1}{3}\\\\ \\\\mbox{cis}(240^\\\\circ)$ h. $\\\\frac{3}{4}\\\\ \\\\mbox{cis}(220^\\\\circ)$ 2. a. $- \\\\frac{27}{2} - \\\\frac{27\\\\sqrt{3}}{2} i$ b. $-2\\\\sqrt{2} - 2\\\\sqrt{2}i$ c. $-64$ d. $\\\\left ( \\\\sqrt[6]{2} \\\\right ) \\\\mbox{cis} \\\\ 15^\\\\circ, \\\\ \\\\sqrt[6]{2} \\\\ \\\\mbox{cis} \\\\ 135^\\\\circ, \\\\ \\\\sqrt[6]{2} \\\\ \\\\mbox{cis} \\\\ 255^\\\\circ,$ e. 2cis 67.5o, 2cis 157.5o, 2cis 247.5o, 2cis 337.5o f. cis 18o, cis 90o, cis 162o, cis 234o, cis 306o Feb 23, 2012 Jun 08, 2015\"\n]"}}},{"rowIdx":6938,"cells":{"problem":{"kind":"string","value":"Simplify $\\cot 10 + \\tan 5.$\n\nEnter your answer as a trigonometric function evaluated at an integer, such as \"sin 7\"."},"level":{"kind":"string","value":"Level 4"},"type":{"kind":"string","value":"Precalculus"},"solution":{"kind":"string","value":"We can write\n\\[\\cot 10 + \\tan 5 = \\frac{\\cos 10}{\\sin 10} + \\frac{\\sin 5}{\\cos 5} = \\frac{\\cos 10 \\cos 5 + \\sin 5 \\sin 10}{\\sin 10 \\cos 5}.\\]From the angle subtraction formula, the numerator is equal to $\\cos (10 - 5) = \\cos 5,$ so\n\\[\\frac{\\cos 10 \\cos 5 + \\sin 5 \\sin 10}{\\sin 10 \\cos 5} = \\frac{\\cos 5}{\\sin 10 \\cos 5} = \\boxed{\\csc 10}.\\]"},"answer":{"kind":"string","value":"\\csc 10"},"p_retrievals":{"kind":"list like","value":["= \\ = \\tan = \\ = \\cot = = \\ = \\sec = \\ = \\csc = \\ = \\ Evaluate the function without using a calculator.Question 12. sin \\ Explanation: = \\ = \\ 2 = \\sin \\ = sin\\ = \\ Question 15.sec \\ Explanation: = \\ = 4 \\ = \\sec \\ = sec\\ = 2 9.4 Graphing Sine and Cosine Functions Identify the amplitude and period of the function. Then graph the function and describe the graph of g as a transformation of the graph of the parent function.Question 16. 9.5 Graphing Other Trigonometric Functions Question 23. Question 26.g = sec \\x Don’t Miss: Algebraic Proofs Worksheets","# Math Help - Simplify the Trig Equation 1. ## Simplify the Trig Equation cot(x)sin(x)+cot(x) (cos(x)/sin(x))*sin(x) + cot(x) cos(x) + cot(x) Can this go further? 2. ## Re: Simplify the Trig Equation There are other forms, but none simpler in my opinion.","Math Help - Simplify Trig 1. Simplify Trig Please take a look at the attached question, thank you! 2. Recall that: $\\tan^2{x} = \\sec^2{x} - 1$ Notice that $1-\\sec^2{x} = -(\\sec^2{x}-1)$ You end up with $-\\cot{x}\\sin{x}$...Remember that $\\cot{x} = \\frac{\\cos{x}}{\\sin{x}}$","# Simplify Trig $\\tan^2{x} = \\sec^2{x} - 1$ Notice that $1-\\sec^2{x} = -(\\sec^2{x}-1)$ You end up with $-\\cot{x}\\sin{x}$...Remember that $\\cot{x} = \\frac{\\cos{x}}{\\sin{x}}$","# Math Help - Simplifying Trigonometric Equations 1. ## Simplifying Trigonometric Equations Hi all! I know this is probably a fairly simple concept for Trigonometry, but I can't quite remember how to do them. I'm just trying to make sure I know them well enough so that I can do them on a test I have in a few days. It would be awesome if you could just get me started on them. 1. $cot(x)sin(x) = \\frac{\\sqrt{3}}{2}$ 2. $tan(x)$ $cos(x) = 1$ 3. $sin^2(x)$ + $cos^2(x) + cos(x)$ = 1 4. [ $cot(x)$ $(\\frac{sin(x)}{cos(x)})$] • $sec^2(x)$ = $\\frac{1}{2}$ 2. Originally Posted by sar Hi all! I know this is probably a fairly simple concept for Trigonometry, but I can't quite remember how to do them. I'm just trying to make sure I know them well enough so that I can do them on a test I have in a few days. It would be awesome if you could just get me started on them. 1. $cot(x)sin(x) = \\frac{\\sqrt{3}}{2}$ It is normally best to write all of your sec(x), csc(x), tan(x), and cot(x) in terms of sin(x) and cos(x). That way, if anything cancels, you can simplify the expression. $\\frac{cos(x)}{sin(x)} • sin(x) = \\frac{\\sqrt{3}}{2}$ Hey, there's a sin(x) on the top as well as the bottom! So, as long","activity… check out slide 1 Hi Daniel, thank you for your post! The function I’m trying to check is a trig function (specifically the answer should be the sine function in this case). I tried modifying the code like this, but the CL gives an error message saying \"Unknown function ‘sin’ \". answer = simpleFunction(this.latex) check = answer.evaluateAt(0)=numericValue(\\sin(0)) and answer.evaluateAt(1)=numericValue(\\sin(1)) Note that CL performs trigonometric functions in radians. If this doesn’t work try { } instead of ( ) for the sin functions.","How do you simplify trigonometric functions? How do you simplify trigonometric functions like $\\tan(x)\\sec(x)$ or $\\csc(x)\\cot(x)$ as well as other equations like $\\frac{\\tan(x)}{\\sec(x)}$ and so on? And could you explain why you are doing the steps so I can understand it a little better and be able to do these on my own. What about $\\sec(x)\\cot(x)$? Thank you. • What do you mean exactly with simplify? $\\tan(x)$ is a definition that you can read on Wikipedia or somewhere else. – BIS HD Mar 14 '14 at 13:45 • $$\\sec x \\cot x = \\require{cancel} \\dfrac 1{\\cancel{\\cos x}}\\cdot \\dfrac{\\cancel{\\cos x}}{\\sin x} = \\dfrac 1{\\sin x} = \\csc x$$ – Namaste Mar 14 '14 at 13:56 • @BISHD Its exactly as I have seen the questions on my assignments it is worded \"What is a simpler way to write...\" or \"Which of the following is ... in a simplified form\"?\" – user135435 Mar 14 '14 at 13:58 • Not every trigonometric expression simplifies. – steven gregory Mar 4 '16 at 20:17 It pretty much boils down to using the definitions of $\\tan x, \\sec x, \\csc x, \\cot x$. For example $$\\tan x \\sec x = \\dfrac{\\sin x}{\\cos x}\\cdot \\dfrac 1{\\cos x} = \\dfrac {\\sin x}{\\cos^2 x}$$ $$\\csc x \\cot x = \\dfrac{1}{\\sin x} \\cdot \\dfrac{\\cos x}{\\sin x} = \\dfrac{\\cos","into & = 2 D: show that +6 & (i. . a product). Hence & +6 Problem: Evaluate = 0. Thus when = 2 D : , N& + 9 = 0 gives: 18 D : = 4=D 2? . & = ? 6 D : +9 = 0 + 2D : when ? = 0 given +6 + 9 = 0. ? 12D : and ? 36 D : & & = 18 D : + 12D : ? 12D : + 1. 7. 6 sec : Differentiation of Inverse Trigonometric Functions = cosec : = sin: , y = cos : , The inverse trigonometric functions: = cos , and = tan , = cot , correspond respectively to the trigonometric functions: = sec and = cosec . = tan: , = cot : = sin , , = The inverse trigonometric functions do not fall into any of the categories (product or quotient) so far considered. Therefore new techniques are involved to differentiate them. It can be shown that if = ( ), then = tan: = f( ) and , = tan and = ? Example: If Problem: Differentiate the remaining inverse trigonometric functions. = =D = = 1 + C’9 ? . =1+ .? = ? = . [ ; 8 1. 7.","A and cot A. Problem 2: In the figure below, find c. Problem 3: If x is an acute angle of a right triangle and sin x = 3 / 7, find the exact value of the trigonometric functions cos x and cot x. Problem 4: Find the exact values of x and y. Problem 5: If x is an acute angle and tan x = 5, find the exact value of the trigonometric functions sin x and cos x. More references on solving problems related to trigonometry and geometry.","Login # The value of expression tan^(- 1)((sqrt(2))/2)+sin^(- 1)((sqrt(5))/5)-cos^(- 1)((sqrt(10))/10) Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free, UPLOAD PHOTO AND GET THE ANSWER NOW! Click here to get PDF DOWNLOAD for all questions and answers of this chapter - ARIHANT MATHS Class 12 INVERSE TRIGONOMETRIC FUNCTIONS Click here to get PDF DOWNLOAD for all questions and answers of this Book - ARIHANT MATHS Class 12 MATHS Text Solution cot^(-1) ((1+sqrt2)/(1-sqrt2))cot^(-1) ((sqrt2 + 1)/(sqrt2-1))- pi + cot^(-1) ((1+sqrt2)/(1 - sqrt2))pi - cot^(-1) ((1-sqrt2)/(1 + sqrt2)) Answer : C Answer Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Show More","anonymous 5 years ago The value of one trigonometric function is give 0 0","# Math Help - Co-functions help needed.. 1. ## Co-functions help needed.. Hey guys, Due to the lecturer not knowing himself, I don't know how exactly to approach some of the following problems. Simplify: 1. sec(90°+x)sin(180°+x) + sin(-x) 2. __sin(-x)_ ______1______ __tanx___ sin(180°+x) + secx.sin(90°+x) - cot(90°+x) Solve for x: 1. 3cot(90°+x) = tanx.sinx 2. tan(90°-x) = sec(3x+10°)cotx I feel a bit stupid asking, BUT then, the only stupid question is the one not asked! Riaan. 2. ## Re: Co-functions help needed.. Originally Posted by riaanj Simplify: 1. sec(90°+x)sin(180°+x) + sin(-x) The first thing I usually do is to get everything in terms of sine and cosine. sec(x) = 1/cos(x), and we have the sum of angles formula: $cos(a + b) = cos(a) ~ cos(b) - sin(a) ~ sin(b)$ So what is $cos( 90 + x)$? The same goes for sine: $sin(a + b) = sin(a) ~ cos(b) + sin(b) ~ cos(a)$ See what you can do with this. If you need more help with it, simply ask. -Dan 3. ## Re: Co-functions help needed.. Ok, so I had class tonight and I'm gonna try this one quick.. sec(90°+x)sin(180°+x) + sin(-x) = -cscx(-sinx)+(-sinx) = -1/sinx(-sinx)-sinx = 1-sinx I'll try the rest after I had some sleep..","For example, sin^4(x)+cos^4(x), (tan(x/2))/(1+cot(x)) are trigonometric expressions. To transform trigonometric expressions properties of trigonometric functions and trigonometric formulas are used.","Question - Inverse Trigonometric Functions (Simplification and Examples) Account Register Share Books Shortlist Question If tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4 ,find the value of x Solution You need to to view the solution Is there an error in this question or solution? S","+ bi form: -2+ 3i 5 7 5 7i Your answer must be input in a + bi form, … I need help with this question. Please explain step by step Consider the point: (r13) = l: _ 51 _ 4’5)- 011 the graph below, drag the purple point to the location I: — 5, — 4.5), and then select t… Prove the following identity. Explain your thought process. Please show full steps and solutions. Thanks! 1) cot 2 x = (cos 2 x) / (1 – cos 2 x) 2) Solve the following trigonometric equation for 0 ?… cos Ѳ / (1- tan Ѳ) + sin Ѳ / (1-cot Ѳ) = sin Ѳ + cos Ѳ I need to know how to verify this practice problem from my textbook I need help figuring out some questions about graphing. Consider the graph y=3 sin 6 (x – pi/2) + 5. You should start by graphing this on your graphing calculator. Why is the graph taller than a base … Find the value of a and c if the angle at A is π/6. The value of a and c are _____ and ________ Hi, I needed help with these FOUR problems. Thank you. Question 5 (4 points} (Sec 4.3, like HW #43) A plane is flying at a","# Simplify the expression to the sum of two trigonometric functions (tan^2(x-1))cot(x)/sin(x)-cos(x)? • Simplify the expression to the sum of two trigonometric functions (tan^2(x-1))cot(x)/sin(x)-cos(x)? ... x) = sin { tan ( sqrt x^3 + 6 ) } 4) f(x) = {sec^2(100x) - tan^2 ... following trigonometric functions: sin(13pi ... 90+x) 1.A.SIMPLIFY sin(x-360)sin ... Positive: 36 % Prove\\:\\cot(2x)=\\frac{1-\\tan^2(x)} ... ... More resources ... for the sum and difference of 2 angles, ... Simplify a Trigonometric Expression ... Positive: 33 % ### More resources These calculations soon came to be defined as the trigonometric functions and today are pervasive in both pure and applied mathematics: ... Positive: 36 % Search the history of over 472 billion pages on the Internet. search Search the Wayback Machine Positive: 31 % Full text of \"Five-place Logarithmic and Trigonometric Tables\" See other formats ...","for all N£N 4. ### Maths 1) Simplification (trigonometric ratios of allied angles)? Simplify : i) [cos^2 (2x180-A). tan^2 (180+A). sec^2 (180-A)] / [ sin^2 (3x180+ A). cosec^2 (-A). cot^2 (90+A)] ii)[ cos(90+A). sec(270+A). sin(180-A) ] / [ cosec (-A)."],"string":"[\n \"= \\\\ = \\\\tan = \\\\ = \\\\cot = = \\\\ = \\\\sec = \\\\ = \\\\csc = \\\\ = \\\\ Evaluate the function without using a calculator.Question 12. sin \\\\ Explanation: = \\\\ = \\\\ 2 = \\\\sin \\\\ = sin\\\\ = \\\\ Question 15.sec \\\\ Explanation: = \\\\ = 4 \\\\ = \\\\sec \\\\ = sec\\\\ = 2 9.4 Graphing Sine and Cosine Functions Identify the amplitude and period of the function. Then graph the function and describe the graph of g as a transformation of the graph of the parent function.Question 16. 9.5 Graphing Other Trigonometric Functions Question 23. Question 26.g = sec \\\\x Don’t Miss: Algebraic Proofs Worksheets\",\n \"# Math Help - Simplify the Trig Equation 1. ## Simplify the Trig Equation cot(x)sin(x)+cot(x) (cos(x)/sin(x))*sin(x) + cot(x) cos(x) + cot(x) Can this go further? 2. ## Re: Simplify the Trig Equation There are other forms, but none simpler in my opinion.\",\n \"Math Help - Simplify Trig 1. Simplify Trig Please take a look at the attached question, thank you! 2. Recall that: $\\\\tan^2{x} = \\\\sec^2{x} - 1$ Notice that $1-\\\\sec^2{x} = -(\\\\sec^2{x}-1)$ You end up with $-\\\\cot{x}\\\\sin{x}$...Remember that $\\\\cot{x} = \\\\frac{\\\\cos{x}}{\\\\sin{x}}$\",\n \"# Simplify Trig $\\\\tan^2{x} = \\\\sec^2{x} - 1$ Notice that $1-\\\\sec^2{x} = -(\\\\sec^2{x}-1)$ You end up with $-\\\\cot{x}\\\\sin{x}$...Remember that $\\\\cot{x} = \\\\frac{\\\\cos{x}}{\\\\sin{x}}$\",\n \"# Math Help - Simplifying Trigonometric Equations 1. ## Simplifying Trigonometric Equations Hi all! I know this is probably a fairly simple concept for Trigonometry, but I can't quite remember how to do them. I'm just trying to make sure I know them well enough so that I can do them on a test I have in a few days. It would be awesome if you could just get me started on them. 1. $cot(x)sin(x) = \\\\frac{\\\\sqrt{3}}{2}$ 2. $tan(x)$ $cos(x) = 1$ 3. $sin^2(x)$ + $cos^2(x) + cos(x)$ = 1 4. [ $cot(x)$ $(\\\\frac{sin(x)}{cos(x)})$] • $sec^2(x)$ = $\\\\frac{1}{2}$ 2. Originally Posted by sar Hi all! I know this is probably a fairly simple concept for Trigonometry, but I can't quite remember how to do them. I'm just trying to make sure I know them well enough so that I can do them on a test I have in a few days. It would be awesome if you could just get me started on them. 1. $cot(x)sin(x) = \\\\frac{\\\\sqrt{3}}{2}$ It is normally best to write all of your sec(x), csc(x), tan(x), and cot(x) in terms of sin(x) and cos(x). That way, if anything cancels, you can simplify the expression. $\\\\frac{cos(x)}{sin(x)} • sin(x) = \\\\frac{\\\\sqrt{3}}{2}$ Hey, there's a sin(x) on the top as well as the bottom! So, as long\",\n \"activity… check out slide 1 Hi Daniel, thank you for your post! The function I’m trying to check is a trig function (specifically the answer should be the sine function in this case). I tried modifying the code like this, but the CL gives an error message saying \\\"Unknown function ‘sin’ \\\". answer = simpleFunction(this.latex) check = answer.evaluateAt(0)=numericValue(\\\\sin(0)) and answer.evaluateAt(1)=numericValue(\\\\sin(1)) Note that CL performs trigonometric functions in radians. If this doesn’t work try { } instead of ( ) for the sin functions.\",\n \"How do you simplify trigonometric functions? How do you simplify trigonometric functions like $\\\\tan(x)\\\\sec(x)$ or $\\\\csc(x)\\\\cot(x)$ as well as other equations like $\\\\frac{\\\\tan(x)}{\\\\sec(x)}$ and so on? And could you explain why you are doing the steps so I can understand it a little better and be able to do these on my own. What about $\\\\sec(x)\\\\cot(x)$? Thank you. • What do you mean exactly with simplify? $\\\\tan(x)$ is a definition that you can read on Wikipedia or somewhere else. – BIS HD Mar 14 '14 at 13:45 • $$\\\\sec x \\\\cot x = \\\\require{cancel} \\\\dfrac 1{\\\\cancel{\\\\cos x}}\\\\cdot \\\\dfrac{\\\\cancel{\\\\cos x}}{\\\\sin x} = \\\\dfrac 1{\\\\sin x} = \\\\csc x$$ – Namaste Mar 14 '14 at 13:56 • @BISHD Its exactly as I have seen the questions on my assignments it is worded \\\"What is a simpler way to write...\\\" or \\\"Which of the following is ... in a simplified form\\\"?\\\" – user135435 Mar 14 '14 at 13:58 • Not every trigonometric expression simplifies. – steven gregory Mar 4 '16 at 20:17 It pretty much boils down to using the definitions of $\\\\tan x, \\\\sec x, \\\\csc x, \\\\cot x$. For example $$\\\\tan x \\\\sec x = \\\\dfrac{\\\\sin x}{\\\\cos x}\\\\cdot \\\\dfrac 1{\\\\cos x} = \\\\dfrac {\\\\sin x}{\\\\cos^2 x}$$ $$\\\\csc x \\\\cot x = \\\\dfrac{1}{\\\\sin x} \\\\cdot \\\\dfrac{\\\\cos x}{\\\\sin x} = \\\\dfrac{\\\\cos\",\n \"into & = 2 D: show that +6 & (i. . a product). Hence & +6 Problem: Evaluate = 0. Thus when = 2 D : , N& + 9 = 0 gives: 18 D : = 4=D 2? . & = ? 6 D : +9 = 0 + 2D : when ? = 0 given +6 + 9 = 0. ? 12D : and ? 36 D : & & = 18 D : + 12D : ? 12D : + 1. 7. 6 sec : Differentiation of Inverse Trigonometric Functions = cosec : = sin: , y = cos : , The inverse trigonometric functions: = cos , and = tan , = cot , correspond respectively to the trigonometric functions: = sec and = cosec . = tan: , = cot : = sin , , = The inverse trigonometric functions do not fall into any of the categories (product or quotient) so far considered. Therefore new techniques are involved to differentiate them. It can be shown that if = ( ), then = tan: = f( ) and , = tan and = ? Example: If Problem: Differentiate the remaining inverse trigonometric functions. = =D = = 1 + C’9 ? . =1+ .? = ? = . [ ; 8 1. 7.\",\n \"A and cot A. Problem 2: In the figure below, find c. Problem 3: If x is an acute angle of a right triangle and sin x = 3 / 7, find the exact value of the trigonometric functions cos x and cot x. Problem 4: Find the exact values of x and y. Problem 5: If x is an acute angle and tan x = 5, find the exact value of the trigonometric functions sin x and cos x. More references on solving problems related to trigonometry and geometry.\",\n \"Login # The value of expression tan^(- 1)((sqrt(2))/2)+sin^(- 1)((sqrt(5))/5)-cos^(- 1)((sqrt(10))/10) Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free, UPLOAD PHOTO AND GET THE ANSWER NOW! Click here to get PDF DOWNLOAD for all questions and answers of this chapter - ARIHANT MATHS Class 12 INVERSE TRIGONOMETRIC FUNCTIONS Click here to get PDF DOWNLOAD for all questions and answers of this Book - ARIHANT MATHS Class 12 MATHS Text Solution cot^(-1) ((1+sqrt2)/(1-sqrt2))cot^(-1) ((sqrt2 + 1)/(sqrt2-1))- pi + cot^(-1) ((1+sqrt2)/(1 - sqrt2))pi - cot^(-1) ((1-sqrt2)/(1 + sqrt2)) Answer : C Answer Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Show More\",\n \"anonymous 5 years ago The value of one trigonometric function is give 0 0\",\n \"# Math Help - Co-functions help needed.. 1. ## Co-functions help needed.. Hey guys, Due to the lecturer not knowing himself, I don't know how exactly to approach some of the following problems. Simplify: 1. sec(90°+x)sin(180°+x) + sin(-x) 2. __sin(-x)_ ______1______ __tanx___ sin(180°+x) + secx.sin(90°+x) - cot(90°+x) Solve for x: 1. 3cot(90°+x) = tanx.sinx 2. tan(90°-x) = sec(3x+10°)cotx I feel a bit stupid asking, BUT then, the only stupid question is the one not asked! Riaan. 2. ## Re: Co-functions help needed.. Originally Posted by riaanj Simplify: 1. sec(90°+x)sin(180°+x) + sin(-x) The first thing I usually do is to get everything in terms of sine and cosine. sec(x) = 1/cos(x), and we have the sum of angles formula: $cos(a + b) = cos(a) ~ cos(b) - sin(a) ~ sin(b)$ So what is $cos( 90 + x)$? The same goes for sine: $sin(a + b) = sin(a) ~ cos(b) + sin(b) ~ cos(a)$ See what you can do with this. If you need more help with it, simply ask. -Dan 3. ## Re: Co-functions help needed.. Ok, so I had class tonight and I'm gonna try this one quick.. sec(90°+x)sin(180°+x) + sin(-x) = -cscx(-sinx)+(-sinx) = -1/sinx(-sinx)-sinx = 1-sinx I'll try the rest after I had some sleep..\",\n \"For example, sin^4(x)+cos^4(x), (tan(x/2))/(1+cot(x)) are trigonometric expressions. To transform trigonometric expressions properties of trigonometric functions and trigonometric formulas are used.\",\n \"Question - Inverse Trigonometric Functions (Simplification and Examples) Account Register Share Books Shortlist Question If tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4 ,find the value of x Solution You need to to view the solution Is there an error in this question or solution? S\",\n \"+ bi form: -2+ 3i 5 7 5 7i Your answer must be input in a + bi form, … I need help with this question. Please explain step by step Consider the point: (r13) = l: _ 51 _ 4’5)- 011 the graph below, drag the purple point to the location I: — 5, — 4.5), and then select t… Prove the following identity. Explain your thought process. Please show full steps and solutions. Thanks! 1) cot 2 x = (cos 2 x) / (1 – cos 2 x) 2) Solve the following trigonometric equation for 0 ?… cos Ѳ / (1- tan Ѳ) + sin Ѳ / (1-cot Ѳ) = sin Ѳ + cos Ѳ I need to know how to verify this practice problem from my textbook I need help figuring out some questions about graphing. Consider the graph y=3 sin 6 (x – pi/2) + 5. You should start by graphing this on your graphing calculator. Why is the graph taller than a base … Find the value of a and c if the angle at A is π/6. The value of a and c are _____ and ________ Hi, I needed help with these FOUR problems. Thank you. Question 5 (4 points} (Sec 4.3, like HW #43) A plane is flying at a\",\n \"# Simplify the expression to the sum of two trigonometric functions (tan^2(x-1))cot(x)/sin(x)-cos(x)? • Simplify the expression to the sum of two trigonometric functions (tan^2(x-1))cot(x)/sin(x)-cos(x)? ... x) = sin { tan ( sqrt x^3 + 6 ) } 4) f(x) = {sec^2(100x) - tan^2 ... following trigonometric functions: sin(13pi ... 90+x) 1.A.SIMPLIFY sin(x-360)sin ... Positive: 36 % Prove\\\\:\\\\cot(2x)=\\\\frac{1-\\\\tan^2(x)} ... ... More resources ... for the sum and difference of 2 angles, ... Simplify a Trigonometric Expression ... Positive: 33 % ### More resources These calculations soon came to be defined as the trigonometric functions and today are pervasive in both pure and applied mathematics: ... Positive: 36 % Search the history of over 472 billion pages on the Internet. search Search the Wayback Machine Positive: 31 % Full text of \\\"Five-place Logarithmic and Trigonometric Tables\\\" See other formats ...\",\n \"for all N£N 4. ### Maths 1) Simplification (trigonometric ratios of allied angles)? Simplify : i) [cos^2 (2x180-A). tan^2 (180+A). sec^2 (180-A)] / [ sin^2 (3x180+ A). cosec^2 (-A). cot^2 (90+A)] ii)[ cos(90+A). sec(270+A). sin(180-A) ] / [ cosec (-A).\"\n]"},"s_retrievals":{"kind":"list like","value":["90o – θ ) = sin θ 17. tan ( 90o – θ ) = cot θ 18. cot ( 90o – θ ) = tan θ 19. sec ( 90o – θ ) = cosec θ 20. cosec ( 90o – θ ) = sec θ 21. sin 00 = 0, cos 00 = 1 and tan 00 = 0 22. cosec 00 = Not defined sec 00 = 1 and cot 00 = Not Defined 23. sin 300 = $\\frac{1}{2}$, cos 300 = $\\frac{\\sqrt{3}}{2}$ and tan 300 = $\\frac{1}{\\sqrt{3}}$ 24. cosec 300 = 2, sec 300 = $\\frac{2}{\\sqrt{3}}$ and cot 300 = $\\sqrt{3}$ 25. sin 450 = $\\frac{1}{\\sqrt{2}}$, cos450 = $\\frac{1}{\\sqrt{2}}$ and tan 450 = 1 26. cosec 450 = $\\sqrt{2}$, sec 450 = $\\sqrt{2}$, and cot 450 = 1 27. sin 600 =$\\frac{\\sqrt{3}}{2}$ , cos 600 = $\\frac{1}{2}$, and tan 600 =$\\sqrt{3}$ 28. cosec 600 = $\\frac{2}{\\sqrt{3}}$, sec 600 = 2 and cot 600 = $\\frac{1}{\\sqrt{3}}$ 29. sin 900 = 1, cos 900 = 0 and tan 00 = Not Defined 30. cosec 900 = 1sec 900 = Not Defined and cot 900 = 0","# How do you evaluate cot (arcsin(5/6))? Sep 4, 2015 $\\frac{\\sqrt{11}}{5}$ #### Explanation: Let arc $\\sin \\left(\\frac{5}{6}\\right) = \\theta$ $\\sin \\theta = \\frac{5}{6}$, that means $\\csc \\theta = \\frac{6}{5}$ ${\\cot}^{2} \\theta = {\\csc}^{2} \\theta - 1 = \\frac{11}{25}$ $\\cot \\theta = \\frac{\\sqrt{11}}{5}$, that means $\\theta = a r c \\cot \\left(\\frac{\\sqrt{11}}{5}\\right)$ Hence $\\cot a r c \\sin \\left(\\frac{5}{6}\\right) = \\cot a r c \\cot \\left(\\frac{\\sqrt{11}}{5}\\right) = \\frac{\\sqrt{11}}{5}$","# How do you find the value of cot 90? Jul 11, 2015 $\\cot \\left(90\\right) = 0$ #### Explanation: Recall that $\\cot \\left(\\theta\\right) = \\frac{1}{\\tan} \\left(\\theta\\right)$ and that $\\tan \\left(\\theta\\right) = \\sin \\frac{\\theta}{\\cos} \\left(\\theta\\right)$ So, $\\cot \\left(\\theta\\right) = \\frac{1}{\\tan} \\left(\\theta\\right) = \\frac{1}{\\sin \\frac{\\theta}{\\cos} \\left(\\theta\\right)} = \\cos \\frac{\\theta}{\\sin} \\left(\\theta\\right)$ Now, let's just put in 90 degrees for $\\theta$ $\\cot \\left(\\theta\\right) = \\cos \\frac{\\theta}{\\sin} \\left(\\theta\\right)$ $\\cot \\left(90\\right) = \\cos \\frac{90}{\\sin} \\left(90\\right)$ Recall, from the unit circle (below) that $\\sin \\left(90\\right) = 1$ and $\\cos \\left(90\\right) = 0$: So, $\\cot \\left(90\\right) = \\cos \\frac{90}{\\sin} \\left(90\\right)$ $\\cot \\left(90\\right) = \\frac{0}{1} = 0$","cos 125° + cos 204° + cos 300° = $\\frac{1}{2}$ Ex. 5.30 | Q 2.4 | Page 39 Prove that: tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0 Ex. 5.30 | Q 2.5 | Page 39 Prove that:cos 570° sin 510° + sin (−330°) cos (−390°) = 0 Ex. 5.30 | Q 2.6 | Page 39 Prove that: $\\tan\\frac{11\\pi}{3} - 2\\sin\\frac{4\\pi}{6} - \\frac{3}{4} {cosec}^2 \\frac{\\pi}{4} + 4 \\cos^2 \\frac{17\\pi}{6} = \\frac{3 - 4\\sqrt{3}}{2}$ Ex. 5.30 | Q 2.7 | Page 39 Prove that: $3\\sin\\frac{\\pi}{6}\\sec\\frac{\\pi}{3} - 4\\sin\\frac{5\\pi}{6}\\cot\\frac{\\pi}{4} = 1$ Ex. 5.30 | Q 3.1 | Page 39 Prove that: $\\frac{\\cos (2\\pi + x) cosec (2\\pi + x) \\tan (\\pi/2 + x)}{\\sec(\\pi/2 + x)\\cos x \\cot(\\pi + x)} = 1$ Ex. 5.30 | Q 3.2 | Page 39 Prove that $\\frac{cosec(90^\\circ + x) + \\cot(450^\\circ + x)}{cosec(90^\\circ - x) + \\tan(180^\\circ - x)} + \\frac{\\tan(180^\\circ + x) + \\sec(180^\\circ - x)}{\\tan(360^\\circ + x) - \\sec( - x)} = 2$ Ex. 5.30 | Q 3.3 | Page 39 Prove that $\\frac{\\sin(180^\\circ + x) \\cos(90^\\circ + x) \\tan(270^\\circ - x) \\cot(360^\\circ - x)}{\\sin(360^\\circ - x) \\cos(360^\\circ + x) cosec( - x) \\sin(270^\\circ + x)} = 1$ Ex. 5.30 | Q 3.4 | Page 39 Prove that $\\left\\{ 1 + \\cot x - \\sec\\left( \\frac{\\pi}{2} + x \\right) \\right\\}\\left\\{ 1 +","33 | Page 59 The value of $\\frac{\\tan 55°}{\\cot 35°}$ + cot 1° cot 2° cot 3° .... cot 90°, is • −2 • 2 • 1 • 0 Q 33 | Page 59 The value of $\\frac{\\tan 55°}{\\cot 35°}$ + cot 1° cot 2° cot 3° .... cot 90°, is • −2 • 2 • 1 • 0 Q 34 | Page 59 In the following figure the value of cos ϕ is • $\\frac{5}{4}$ • $\\frac{5}{3}$ • $\\frac{3}{5}$ • $\\frac{4}{5}$ Q 34 | Page 59 In the following figure the value of cos ϕ is • $\\frac{5}{4}$ • $\\frac{5}{3}$ • $\\frac{3}{5}$ • $\\frac{4}{5}$ Q 35 | Page 59 In the following Figure. AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ. • $\\frac{12}{5}$ • $\\frac{5}{12}$ • $\\frac{13}{12}$ • $\\frac{12}{13}$ Q 35 | Page 59 In the following Figure. AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ. • $\\frac{12}{5}$ • $\\frac{5}{12}$ • $\\frac{13}{12}$ • $\\frac{12}{13}$ ## Chapter 10: Trigonometric Ratios Ex. 10.10Ex. 10.20Ex. 10.30Others ## RD Sharma solutions for Class 10 Mathematics chapter 10 - Trigonometric Ratios RD Sharma solutions for Class 10 Maths chapter 10 (Trigonometric Ratios) include all questions with solution and detail explanation. This will clear students","B+cot A}\\\\cot (A-B)=\\frac{cot A cot B+1}{cot B-cot A}$ Some additional formulas for sum and product of angles: $\\cos (A + B) \\cos (A – B) = \\cos^{2}A – \\sin^{2}B = \\cos^{2}B – \\sin^{2}A$ $\\sin (A + B) \\sin (A – B) = \\sin^{2}A – \\sin^{2}B = \\cos^{2}B – \\cos^{2}A$ $\\sin A+\\sin B = 2\\sin\\frac{A+B}{2}\\cos\\frac{A-B}{2}$ Formulas for twice of the angles: $\\sin 2A = 2 \\sin A \\cos A = \\frac{2\\tan A}{1+\\tan^{2}A}$ $\\cos 2A = \\cos^{A} – \\sin^{2}A = 1 – 2sin^{2}A = 2cos^{2}A – 1 = \\frac{1-\\tan^{2}A}{1 + \\tan^{2}A}$ $\\tan 2A =\\frac{2 \\tan A}{1 – \\tan^{2}A}$ Formulas for thrice of the angles: $\\sin 3A = 3\\sin A – 4\\sin^{3}A = 4\\sin(60^{\\circ}-A).\\sin A .\\sin( 60^{\\circ}+A)$ $\\cos 3A = 4\\cos^{3}A – 3\\cos A = 4\\cos\\left ( 60^{\\circ}-A \\right ).\\cos A . \\cos\\left ( 60^{\\circ} +A\\right )$ $\\tan 3A = \\frac{3\\tan A – \\tan^{3}A}{1-3\\tan^{2}A} = \\tan\\left ( 60^{\\circ}-A \\right ).\\tan A . \\tan\\left ( 60^{\\circ}+A\\right )$ Also check: Example Example: Find the maximum value of $\\cos^{2}\\cos\\theta + \\sin^{2} \\sin\\theta$ for any real value of $\\theta$ The maximum value of $\\cos^{2}\\cos\\theta = 1$ and $\\sin^{2}\\sin\\theta$ is $\\sin^{2} 1$ where both exist for $\\theta = \\frac{\\pi}{2}$. Therefore maximum value will be $1+ \\sin^{2}1$.. 1. chirasmita nanda helped me a lot. thank you byju’s 🙂 2. Devanshu Thanks guyss for such useful quick notes","$\\begin{array}{ccc}\\cos5x &=& \\cos^5\\!x - 10\\cos^3\\!x\\sin^2\\!x + 5\\cos x\\sin^4\\!x \\\\ \\sin5x &=& 5\\cos^4\\!x\\sin x - 10\\cos^2\\!x\\sin^3\\!x + \\sin^5\\!x \\end{array}$ We have: . $\\tan5x \\;=\\;\\dfrac{\\sin5x}{\\cos5x}$ . . . . . . . . . . . . . $=\\;\\dfrac{5\\cos^4\\!x\\sin x - 10\\cos^2\\!x\\sin^3\\!x + \\sin^5\\!x}{\\cos^5\\!x - 10\\cos^3\\!x\\sin^2\\!x + 5\\cos x\\sin^4\\!x}$ Divide numerator and denominator by $\\cos^5\\!x$ . . $\\displaystyle \\tan5x \\;=\\;\\frac{\\;\\dfrac{5\\cos^4\\!x\\sin x}{\\cos^5\\!x} - \\dfrac{10\\cos^2\\!x\\sin^3\\!x}{\\cos^5\\!x} + \\dfrac{\\sin^5\\!x}{\\cos^5\\!x}\\;} {\\dfrac{\\cos^5\\!x}{\\cos^5\\!x} - \\dfrac{10\\cos^3\\!x\\sin^2\\!x}{\\cos^5\\!x} + \\dfrac{5\\cos x\\sin^4\\!x}{\\cos^5\\!x}}$ . . $\\boxed{\\tan5x \\;=\\;\\dfrac{5\\tan x - 10\\tan^3\\!x + \\tan^5\\!x}{1 - 10\\tan^2\\!x + 5\\tan^4\\!x}}$","the swap unit button first. Besides cos0.5, similar trigonometric calculations on our site include, but are not limited, to: The identities of cosine 0.5 are as follows: cos0.5 = sin (π/2 + 0.5) = sin 2.0707963267949 = sin (π/2 – 0.5) = sin 1.0707963267949 -cos0.5 = cos (π + 0.5) = cos 3.64159265358979 = cos (π – 0.5) = cos 2.64159265358979 Note that cos0.5 is periodic: cos (0.5 + n × 2π) = cos 0.5, n$\\hspace{5px} \\in \\hspace{5px} \\mathbb{Z}$. There are more formulas for the double angle (2 × 0.5), half angle ((0.5/2)) as well as the sum, difference and products of two angles such as 0.5 and β. You can locate all of them in the respective article found in the header menu. To find everything about cos -0.5 click the link. And here is all about sin 0.5, including, for instance, a converter. In terms of the other five trigonometric functions, cos of 0.5 = • $\\pm \\sqrt{1-\\sin^{2} 0.5 }$ • $\\pm\\frac{1}{\\sqrt{1 + \\tan^{2} 0.5}}$ • $\\pm\\frac{\\cot 0.5}{\\sqrt{1 + \\cot^{2} 0.5}}$ • $\\frac{1}{\\sec 0.5}$ • $\\pm\\frac{\\sqrt{\\csc^{2} (0.5) – 1} }{\\csc 0.5}$ As the cosine function is the reciprocal of the secant function, 1 / sec 0.5 = cos0.5. In the next part we discuss the trigonometric significance of cos0.5, and there you can also learn what","\\frac { \\cos 7x + \\cos 5x } { \\sin 7x - \\sin 5x } = \\cot x? How do you prove that \\frac { \\cos 7x + \\cos 5x } { \\sin 7x - \\sin 5x } = \\cot x?...","your drawing made a false assumption that you then used to get a false answer. – Anon Oct 5 '16 at 16:48 by dividing both to $\\sin { x } \\cos { y }$ and consider the fact that $\\cot { x } =2$ we get $$5\\sin { x\\cos { y } +4\\cos { x } \\sin { y } =0, } \\\\ 5+4\\frac { \\cos { x } \\sin { y } }{ \\sin { x } \\cos { y } } =0\\\\ 5+4\\cot { x } \\tan { y } =0\\\\ 5+\\frac { 8 }{ \\cot { y } } =0\\\\ \\cot { y } =-\\frac { 8 }{ 5 }$$ • @McFry,i fixed thank you – haqnatural Oct 5 '16 at 16:21 • I know how to do it, but I'd like to know why the method I originally took is incorrect? If possible... – user375494 Oct 5 '16 at 16:21","seeing it yet? I see that because that is what I did to work out a hexagon and the larger even numbered polygons. My problem is that I have worked out 3 numbers for the Pentagon by working it out 2 different ways and so I am interested in how NateTG and Holly work it out. Bascially I would like someone to apply their equations to a simple shape (e.g. a sqaure with sides of 10cm) and then see if they get it right and how they did it so I can apply it, learn how to do it and then apply it myself. Cheers. P: 1,116 Please people. I need a simple example so I can see how the equations work. The Bob (2004 ©) HW Helper P: 2,538 Quote by The Bob Please people. I need a simple example so I can see how the equations work. The Bob (2004 ©) Ok - working in degrees for you: A square with 10 cm sides: n=4 l=10 $$A=\\frac{1}{4} nl^2 \\cot(\\frac{180}{n})=\\frac{1}{4} 4 (10^2) \\cot(\\frac{180}{4})=100$$ A 73-gon with 10 cm sides n=73 l=10 $$A=\\frac{1}{4}nl^2 \\cot(\\frac{180}{n})=\\frac{1}{4} 73 (10^2) \\cot(\\frac{180}{73})\\approx42380$$ P: 1,116 Quote by NateTG Ok - working in degrees for you: A square with 10 cm sides: n=4 l=10 $$A=\\frac{1}{4} nl^2 \\cot(\\frac{180}{n})=\\frac{1}{4} 4 (10^2) \\cot(\\frac{180}{4})=100$$ Cheers NateTG. That is a","How do you find tan and cot if cot = 12/5 and sin <0? May 2, 2015 Easy steps I assume you have to calculate tan and cos $\\tan \\theta = \\frac{1}{\\cot} \\theta$ $\\tan \\theta = \\frac{5}{12}$ $\\sin \\frac{\\theta}{\\cos} \\theta = \\frac{5}{12}$ $12 \\sin \\theta - 5 \\cos \\theta = 0$ ${\\sin}^{2} \\theta + {\\cos}^{2} \\theta = 1$ Assume sin theta =x ; cos theta = y ${x}^{2} + {y}^{2} = 1$ $12 x - 5 y = 0$ $y = \\frac{12 x}{5}$ ${x}^{2} + \\frac{144 {x}^{2}}{25} = 1$ ${x}^{2} \\left(1 + \\frac{144}{25}\\right) = 1$ ${x}^{2} = \\frac{1}{\\frac{169}{25}}$ ${x}^{2} = {\\left(\\frac{5}{13}\\right)}^{2}$ $x = \\pm \\frac{5}{13}$ $\\sin \\theta = \\pm \\frac{5}{13}$ As $\\sin \\theta < 0 ,$ $\\sin \\theta = - \\frac{5}{13}$ $\\cos \\theta = - \\frac{12}{13}$","# Trig equations • April 5th 2010, 08:39 AM darksupernova Trig equations Hello, I am having a bit of trouble solving this eqn: tan (x) + cot (x) = 2sec (x) For all angles between 0 and 360 degrees. Thank you! Max • April 5th 2010, 09:33 AM Sudharaka Quote: Originally Posted by darksupernova Hello, I am having a bit of trouble solving this eqn: tan (x) + cot (x) = 2sec (x) For all angles between 0 and 360 degrees. Thank you! Max Dear darksupernova, $tanx+cotx=2secx$ $\\frac{1}{sin(x)cos(x)}=2secx$ $2sinx=1$ $sinx=\\frac{1}{2}$ Since x is inbetween 0 and 360 degrees, $x=30^0$ • April 5th 2010, 10:27 AM darksupernova Ah very clever! Thank you very much! Im now a but stuck on this: tan (x) + 3cot (x) = 5sec (x) Very similar but the cos^2 {x) + sin^2 (X) = 1 trick doesnt work? • April 5th 2010, 10:38 AM Quote: Originally Posted by darksupernova Hello, I am having a bit of trouble solving this eqn: tan (x) + cot (x) = 2sec (x) For all angles between 0 and 360 degrees. Thank you! Max $tanx+cotx=\\frac{sinx}{cosx}+\\frac{cosx}{sinx}=\\fra c{sinx}{sinx}\\ \\frac{sinx}{cosx}+\\frac{cosx}{cosx}\\ \\frac{cosx}{sinx}$ $=\\frac{sin^2x+cos^2x}{sinxcosx}$ $\\Rightarrow\\ \\frac{1}{sinxcosx}=\\frac{2}{cosx}\\ \\Rightarrow\\ \\frac{1}{sinx}=2\\ \\Rightarrow\\ sinx=\\frac{1}{2}$ The vertical co-ordinate of the angle is 0.5 in the unit circle, centre (0,0). This corresponds to two angles, $x=sin^{-1}\\left(\\frac{1}{2}\\right),\\ x={\\pi}-sin^{-1}\\left(\\frac{1}{2}\\right)$ $x=30^o,\\ x=150^o$ •","30 - (\\sin 30 - \\sin 10) \\\\ = \\sin 30 +\\sin 10 = 2 \\sin 20 \\cos 10$$ Therefore, $$\\tan \\alpha = \\frac{2 \\sin 10 \\sin 20}{1 - 2 \\sin 10 \\cos 20} = \\frac{2\\sin 10 \\sin 20}{2 \\cos 10 \\sin 20} = \\tan 10$$ Now, since $$0 < \\alpha < 180$$, we get that $$\\alpha = 10$$. From here, $$80-\\alpha = 70$$ is the desired angle.","# Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 682: 101 See graph. a. $y=tan(\\frac{x}{2})$ b. See explanations. #### Work Step by Step Graph the equation as shown in the figure. a. The graph appears to be the same as $y=tan(\\frac{x}{2})$ b. Using the double angle formulas, we have $y=csc(x)-cot(x)=\\frac{1}{sin(x)}-\\frac{cos(x)}{sin(x)}=\\frac{1-cos(x)}{sin(x)}=\\frac{1-(1-2sin^2(x/2))}{2sin(x/2)cos(x/2)}=tan(\\frac{x}{2})$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.","n=0 and works very well for any negative value of n. I made a mistake, I meant that it equals 5.7*10^(-n+1). Sorry about that. I don't know if this is a co-incidence but I had noticed the same pattern a few days ago when I was plugging in 89, 89.9, 89.99, 89.999 etc in my calculater to notice the changes in values of tanx. And, every time the same digits appeared that you've written with the decimal point displaced. I ignored it and didn't take the trouble of making a formula. It might be a co-incidence that you're uploading this today. I don't know how this is working. But, it's an approximation mostly for negative values of n and so, it works for a very small range. There are only two non-negative values values of n we can try. Sure you didn't mean ##\\tan(90° - 10^{-n}) \\approx 5.7_3 \\cdot 10^{n+1}\\; \\;(n \\in \\mathbb{N})\\;##? It's pretty much the same. But yours is a bit more rigorous, so I guess it's better. Although it also works for 80 degrees (n=-1 in your definition). I'm guessing it has something to do with the Taylor series. $$\\tan(\\frac{\\pi}{2}-\\theta) = \\cot \\theta$$ The Taylor series for cotangent is: $$\\cot \\theta = \\frac{1}{\\theta} - \\frac{1}{3}\\theta - \\frac{1}{45}\\theta^3 + \\cdots$$ I'm assuming you were working in","anonymous 5 years ago write an expression equal to 1/tan^2x $\\frac{1}{\\tan^2x} = \\cot^2x = \\frac{\\cos^2x}{\\sin^2x}$ ^_^","How do you find the value of cot ((5pi/3) using the double angle or half angle identity? Nov 11, 2016 $- \\frac{\\sqrt{3}}{3}$ Explanation: Trig table and unit circle --> $\\cot \\left(\\frac{5 \\pi}{3}\\right) = \\frac{1}{\\tan} \\left(\\frac{5 \\pi}{3}\\right)$ Find $\\tan \\left(\\frac{5 \\pi}{3}\\right)$ $\\tan \\left(\\frac{5 \\pi}{3}\\right) = \\tan \\left(\\frac{2 \\pi}{3} + \\pi\\right) = \\tan \\left(\\frac{2 \\pi}{3}\\right) = - \\sqrt{3}$ There for: $\\cot \\left(\\frac{5 \\pi}{3}\\right) = - \\frac{1}{\\sqrt{3}} = - \\frac{\\sqrt{3}}{3}$","# Thread: trig compound angle formulae and proofs 1. ## trig compound angle formulae and proofs Hi I've got some maths homework and stuck on the last 3 questions. 1. Given that sinA=3/5 and cosB=12/13, where A is obtuse and B is acute, find exact values of a) cos(A+B) b)cot (A-B) 2. Prove cot(A+B)is identical to cotAcotB-1 cotA+cotB 3. prove that tanA+tanB is identical to sin(A+B) cosAcosB I have tried all these and just get stuck . any help would be really appreciative thankyou 2. 1) $\\cos A=-\\sqrt{1-\\sin^2A}, \\ \\sin B=\\sqrt{1-\\cos^2B}$ $\\cos(A+B)=\\cos A\\cos B-\\sin A\\sin B$ $\\cot(A-B)=\\frac{\\cos(A-B)}{\\sin(A-B)}=\\frac{\\cos A\\cos B+\\sin A\\sin B}{\\sin A\\cos B-\\sin B\\cos A}$ 2) $\\cos(A+B)=\\frac{\\cos(A+B)}{\\sin(A+B)}=\\frac{\\cos A\\cos B-\\sin A\\sin B}{\\sin A\\cos B+\\sin B\\cos A}=$ $=\\frac{\\frac{\\cos A\\cos B}{\\sin A\\sin B}-1}{\\frac{\\sin A\\cos B}{\\sin B\\sin A}+\\frac{\\sin B\\cos A}{\\sin A\\sin B}}=\\frac{\\cot A\\cot B-1}{\\cot A+cot B}$ 3) $\\tan A+\\tan B=\\frac{\\sin A}{\\cos A}+\\frac{\\sin B}{\\cos B}=\\frac{\\sin A\\cos B+\\sin B\\cos A}{\\cos A\\cos B}=\\frac{\\sin(A+B)}{\\cos A\\cos B}$ 3. ## thankyou thanks so much. not sure if i get questions 1 yet though, i'll try it.","Sec, Cosec, Cot Printable View • May 24th 2008, 03:14 AM jarald83 Sec, Cosec, Cot how to calculate the following value? Cosec 30° cot 60° sec 310° cosec 120° ° cot 600° and -cot 130° (Doh)(Crying) • May 24th 2008, 03:44 AM wingless $\\tan x = \\frac{\\sin x}{\\cos x}$ $\\cot x = \\frac{\\cos x}{\\sin x}$ $\\csc x = \\frac{1}{\\sin x}$ $\\sec x = \\frac{1}{\\cos x}$ • May 24th 2008, 03:58 AM nandu11 u need to know the simple sin, cos and tan trigo ratios for 0, 30, 45, 60 and 90 to do these types of questions and then use the formulas like cosecx=1/sinx, cotx=cosx/sinx etcc"],"string":"[\n \"90o – θ ) = sin θ 17. tan ( 90o – θ ) = cot θ 18. cot ( 90o – θ ) = tan θ 19. sec ( 90o – θ ) = cosec θ 20. cosec ( 90o – θ ) = sec θ 21. sin 00 = 0, cos 00 = 1 and tan 00 = 0 22. cosec 00 = Not defined sec 00 = 1 and cot 00 = Not Defined 23. sin 300 = $\\\\frac{1}{2}$, cos 300 = $\\\\frac{\\\\sqrt{3}}{2}$ and tan 300 = $\\\\frac{1}{\\\\sqrt{3}}$ 24. cosec 300 = 2, sec 300 = $\\\\frac{2}{\\\\sqrt{3}}$ and cot 300 = $\\\\sqrt{3}$ 25. sin 450 = $\\\\frac{1}{\\\\sqrt{2}}$, cos450 = $\\\\frac{1}{\\\\sqrt{2}}$ and tan 450 = 1 26. cosec 450 = $\\\\sqrt{2}$, sec 450 = $\\\\sqrt{2}$, and cot 450 = 1 27. sin 600 =$\\\\frac{\\\\sqrt{3}}{2}$ , cos 600 = $\\\\frac{1}{2}$, and tan 600 =$\\\\sqrt{3}$ 28. cosec 600 = $\\\\frac{2}{\\\\sqrt{3}}$, sec 600 = 2 and cot 600 = $\\\\frac{1}{\\\\sqrt{3}}$ 29. sin 900 = 1, cos 900 = 0 and tan 00 = Not Defined 30. cosec 900 = 1sec 900 = Not Defined and cot 900 = 0\",\n \"# How do you evaluate cot (arcsin(5/6))? Sep 4, 2015 $\\\\frac{\\\\sqrt{11}}{5}$ #### Explanation: Let arc $\\\\sin \\\\left(\\\\frac{5}{6}\\\\right) = \\\\theta$ $\\\\sin \\\\theta = \\\\frac{5}{6}$, that means $\\\\csc \\\\theta = \\\\frac{6}{5}$ ${\\\\cot}^{2} \\\\theta = {\\\\csc}^{2} \\\\theta - 1 = \\\\frac{11}{25}$ $\\\\cot \\\\theta = \\\\frac{\\\\sqrt{11}}{5}$, that means $\\\\theta = a r c \\\\cot \\\\left(\\\\frac{\\\\sqrt{11}}{5}\\\\right)$ Hence $\\\\cot a r c \\\\sin \\\\left(\\\\frac{5}{6}\\\\right) = \\\\cot a r c \\\\cot \\\\left(\\\\frac{\\\\sqrt{11}}{5}\\\\right) = \\\\frac{\\\\sqrt{11}}{5}$\",\n \"# How do you find the value of cot 90? Jul 11, 2015 $\\\\cot \\\\left(90\\\\right) = 0$ #### Explanation: Recall that $\\\\cot \\\\left(\\\\theta\\\\right) = \\\\frac{1}{\\\\tan} \\\\left(\\\\theta\\\\right)$ and that $\\\\tan \\\\left(\\\\theta\\\\right) = \\\\sin \\\\frac{\\\\theta}{\\\\cos} \\\\left(\\\\theta\\\\right)$ So, $\\\\cot \\\\left(\\\\theta\\\\right) = \\\\frac{1}{\\\\tan} \\\\left(\\\\theta\\\\right) = \\\\frac{1}{\\\\sin \\\\frac{\\\\theta}{\\\\cos} \\\\left(\\\\theta\\\\right)} = \\\\cos \\\\frac{\\\\theta}{\\\\sin} \\\\left(\\\\theta\\\\right)$ Now, let's just put in 90 degrees for $\\\\theta$ $\\\\cot \\\\left(\\\\theta\\\\right) = \\\\cos \\\\frac{\\\\theta}{\\\\sin} \\\\left(\\\\theta\\\\right)$ $\\\\cot \\\\left(90\\\\right) = \\\\cos \\\\frac{90}{\\\\sin} \\\\left(90\\\\right)$ Recall, from the unit circle (below) that $\\\\sin \\\\left(90\\\\right) = 1$ and $\\\\cos \\\\left(90\\\\right) = 0$: So, $\\\\cot \\\\left(90\\\\right) = \\\\cos \\\\frac{90}{\\\\sin} \\\\left(90\\\\right)$ $\\\\cot \\\\left(90\\\\right) = \\\\frac{0}{1} = 0$\",\n \"cos 125° + cos 204° + cos 300° = $\\\\frac{1}{2}$ Ex. 5.30 | Q 2.4 | Page 39 Prove that: tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0 Ex. 5.30 | Q 2.5 | Page 39 Prove that:cos 570° sin 510° + sin (−330°) cos (−390°) = 0 Ex. 5.30 | Q 2.6 | Page 39 Prove that: $\\\\tan\\\\frac{11\\\\pi}{3} - 2\\\\sin\\\\frac{4\\\\pi}{6} - \\\\frac{3}{4} {cosec}^2 \\\\frac{\\\\pi}{4} + 4 \\\\cos^2 \\\\frac{17\\\\pi}{6} = \\\\frac{3 - 4\\\\sqrt{3}}{2}$ Ex. 5.30 | Q 2.7 | Page 39 Prove that: $3\\\\sin\\\\frac{\\\\pi}{6}\\\\sec\\\\frac{\\\\pi}{3} - 4\\\\sin\\\\frac{5\\\\pi}{6}\\\\cot\\\\frac{\\\\pi}{4} = 1$ Ex. 5.30 | Q 3.1 | Page 39 Prove that: $\\\\frac{\\\\cos (2\\\\pi + x) cosec (2\\\\pi + x) \\\\tan (\\\\pi/2 + x)}{\\\\sec(\\\\pi/2 + x)\\\\cos x \\\\cot(\\\\pi + x)} = 1$ Ex. 5.30 | Q 3.2 | Page 39 Prove that $\\\\frac{cosec(90^\\\\circ + x) + \\\\cot(450^\\\\circ + x)}{cosec(90^\\\\circ - x) + \\\\tan(180^\\\\circ - x)} + \\\\frac{\\\\tan(180^\\\\circ + x) + \\\\sec(180^\\\\circ - x)}{\\\\tan(360^\\\\circ + x) - \\\\sec( - x)} = 2$ Ex. 5.30 | Q 3.3 | Page 39 Prove that $\\\\frac{\\\\sin(180^\\\\circ + x) \\\\cos(90^\\\\circ + x) \\\\tan(270^\\\\circ - x) \\\\cot(360^\\\\circ - x)}{\\\\sin(360^\\\\circ - x) \\\\cos(360^\\\\circ + x) cosec( - x) \\\\sin(270^\\\\circ + x)} = 1$ Ex. 5.30 | Q 3.4 | Page 39 Prove that $\\\\left\\\\{ 1 + \\\\cot x - \\\\sec\\\\left( \\\\frac{\\\\pi}{2} + x \\\\right) \\\\right\\\\}\\\\left\\\\{ 1 +\",\n \"33 | Page 59 The value of $\\\\frac{\\\\tan 55°}{\\\\cot 35°}$ + cot 1° cot 2° cot 3° .... cot 90°, is • −2 • 2 • 1 • 0 Q 33 | Page 59 The value of $\\\\frac{\\\\tan 55°}{\\\\cot 35°}$ + cot 1° cot 2° cot 3° .... cot 90°, is • −2 • 2 • 1 • 0 Q 34 | Page 59 In the following figure the value of cos ϕ is • $\\\\frac{5}{4}$ • $\\\\frac{5}{3}$ • $\\\\frac{3}{5}$ • $\\\\frac{4}{5}$ Q 34 | Page 59 In the following figure the value of cos ϕ is • $\\\\frac{5}{4}$ • $\\\\frac{5}{3}$ • $\\\\frac{3}{5}$ • $\\\\frac{4}{5}$ Q 35 | Page 59 In the following Figure. AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ. • $\\\\frac{12}{5}$ • $\\\\frac{5}{12}$ • $\\\\frac{13}{12}$ • $\\\\frac{12}{13}$ Q 35 | Page 59 In the following Figure. AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ. • $\\\\frac{12}{5}$ • $\\\\frac{5}{12}$ • $\\\\frac{13}{12}$ • $\\\\frac{12}{13}$ ## Chapter 10: Trigonometric Ratios Ex. 10.10Ex. 10.20Ex. 10.30Others ## RD Sharma solutions for Class 10 Mathematics chapter 10 - Trigonometric Ratios RD Sharma solutions for Class 10 Maths chapter 10 (Trigonometric Ratios) include all questions with solution and detail explanation. This will clear students\",\n \"B+cot A}\\\\\\\\cot (A-B)=\\\\frac{cot A cot B+1}{cot B-cot A}$ Some additional formulas for sum and product of angles: $\\\\cos (A + B) \\\\cos (A – B) = \\\\cos^{2}A – \\\\sin^{2}B = \\\\cos^{2}B – \\\\sin^{2}A$ $\\\\sin (A + B) \\\\sin (A – B) = \\\\sin^{2}A – \\\\sin^{2}B = \\\\cos^{2}B – \\\\cos^{2}A$ $\\\\sin A+\\\\sin B = 2\\\\sin\\\\frac{A+B}{2}\\\\cos\\\\frac{A-B}{2}$ Formulas for twice of the angles: $\\\\sin 2A = 2 \\\\sin A \\\\cos A = \\\\frac{2\\\\tan A}{1+\\\\tan^{2}A}$ $\\\\cos 2A = \\\\cos^{A} – \\\\sin^{2}A = 1 – 2sin^{2}A = 2cos^{2}A – 1 = \\\\frac{1-\\\\tan^{2}A}{1 + \\\\tan^{2}A}$ $\\\\tan 2A =\\\\frac{2 \\\\tan A}{1 – \\\\tan^{2}A}$ Formulas for thrice of the angles: $\\\\sin 3A = 3\\\\sin A – 4\\\\sin^{3}A = 4\\\\sin(60^{\\\\circ}-A).\\\\sin A .\\\\sin( 60^{\\\\circ}+A)$ $\\\\cos 3A = 4\\\\cos^{3}A – 3\\\\cos A = 4\\\\cos\\\\left ( 60^{\\\\circ}-A \\\\right ).\\\\cos A . \\\\cos\\\\left ( 60^{\\\\circ} +A\\\\right )$ $\\\\tan 3A = \\\\frac{3\\\\tan A – \\\\tan^{3}A}{1-3\\\\tan^{2}A} = \\\\tan\\\\left ( 60^{\\\\circ}-A \\\\right ).\\\\tan A . \\\\tan\\\\left ( 60^{\\\\circ}+A\\\\right )$ Also check: Example Example: Find the maximum value of $\\\\cos^{2}\\\\cos\\\\theta + \\\\sin^{2} \\\\sin\\\\theta$ for any real value of $\\\\theta$ The maximum value of $\\\\cos^{2}\\\\cos\\\\theta = 1$ and $\\\\sin^{2}\\\\sin\\\\theta$ is $\\\\sin^{2} 1$ where both exist for $\\\\theta = \\\\frac{\\\\pi}{2}$. Therefore maximum value will be $1+ \\\\sin^{2}1$.. 1. chirasmita nanda helped me a lot. thank you byju’s 🙂 2. Devanshu Thanks guyss for such useful quick notes\",\n \"$\\\\begin{array}{ccc}\\\\cos5x &=& \\\\cos^5\\\\!x - 10\\\\cos^3\\\\!x\\\\sin^2\\\\!x + 5\\\\cos x\\\\sin^4\\\\!x \\\\\\\\ \\\\sin5x &=& 5\\\\cos^4\\\\!x\\\\sin x - 10\\\\cos^2\\\\!x\\\\sin^3\\\\!x + \\\\sin^5\\\\!x \\\\end{array}$ We have: . $\\\\tan5x \\\\;=\\\\;\\\\dfrac{\\\\sin5x}{\\\\cos5x}$ . . . . . . . . . . . . . $=\\\\;\\\\dfrac{5\\\\cos^4\\\\!x\\\\sin x - 10\\\\cos^2\\\\!x\\\\sin^3\\\\!x + \\\\sin^5\\\\!x}{\\\\cos^5\\\\!x - 10\\\\cos^3\\\\!x\\\\sin^2\\\\!x + 5\\\\cos x\\\\sin^4\\\\!x}$ Divide numerator and denominator by $\\\\cos^5\\\\!x$ . . $\\\\displaystyle \\\\tan5x \\\\;=\\\\;\\\\frac{\\\\;\\\\dfrac{5\\\\cos^4\\\\!x\\\\sin x}{\\\\cos^5\\\\!x} - \\\\dfrac{10\\\\cos^2\\\\!x\\\\sin^3\\\\!x}{\\\\cos^5\\\\!x} + \\\\dfrac{\\\\sin^5\\\\!x}{\\\\cos^5\\\\!x}\\\\;} {\\\\dfrac{\\\\cos^5\\\\!x}{\\\\cos^5\\\\!x} - \\\\dfrac{10\\\\cos^3\\\\!x\\\\sin^2\\\\!x}{\\\\cos^5\\\\!x} + \\\\dfrac{5\\\\cos x\\\\sin^4\\\\!x}{\\\\cos^5\\\\!x}}$ . . $\\\\boxed{\\\\tan5x \\\\;=\\\\;\\\\dfrac{5\\\\tan x - 10\\\\tan^3\\\\!x + \\\\tan^5\\\\!x}{1 - 10\\\\tan^2\\\\!x + 5\\\\tan^4\\\\!x}}$\",\n \"the swap unit button first. Besides cos0.5, similar trigonometric calculations on our site include, but are not limited, to: The identities of cosine 0.5 are as follows: cos0.5 = sin (π/2 + 0.5) = sin 2.0707963267949 = sin (π/2 – 0.5) = sin 1.0707963267949 -cos0.5 = cos (π + 0.5) = cos 3.64159265358979 = cos (π – 0.5) = cos 2.64159265358979 Note that cos0.5 is periodic: cos (0.5 + n × 2π) = cos 0.5, n$\\\\hspace{5px} \\\\in \\\\hspace{5px} \\\\mathbb{Z}$. There are more formulas for the double angle (2 × 0.5), half angle ((0.5/2)) as well as the sum, difference and products of two angles such as 0.5 and β. You can locate all of them in the respective article found in the header menu. To find everything about cos -0.5 click the link. And here is all about sin 0.5, including, for instance, a converter. In terms of the other five trigonometric functions, cos of 0.5 = • $\\\\pm \\\\sqrt{1-\\\\sin^{2} 0.5 }$ • $\\\\pm\\\\frac{1}{\\\\sqrt{1 + \\\\tan^{2} 0.5}}$ • $\\\\pm\\\\frac{\\\\cot 0.5}{\\\\sqrt{1 + \\\\cot^{2} 0.5}}$ • $\\\\frac{1}{\\\\sec 0.5}$ • $\\\\pm\\\\frac{\\\\sqrt{\\\\csc^{2} (0.5) – 1} }{\\\\csc 0.5}$ As the cosine function is the reciprocal of the secant function, 1 / sec 0.5 = cos0.5. In the next part we discuss the trigonometric significance of cos0.5, and there you can also learn what\",\n \"\\\\frac { \\\\cos 7x + \\\\cos 5x } { \\\\sin 7x - \\\\sin 5x } = \\\\cot x? How do you prove that \\\\frac { \\\\cos 7x + \\\\cos 5x } { \\\\sin 7x - \\\\sin 5x } = \\\\cot x?...\",\n \"your drawing made a false assumption that you then used to get a false answer. – Anon Oct 5 '16 at 16:48 by dividing both to $\\\\sin { x } \\\\cos { y }$ and consider the fact that $\\\\cot { x } =2$ we get $$5\\\\sin { x\\\\cos { y } +4\\\\cos { x } \\\\sin { y } =0, } \\\\\\\\ 5+4\\\\frac { \\\\cos { x } \\\\sin { y } }{ \\\\sin { x } \\\\cos { y } } =0\\\\\\\\ 5+4\\\\cot { x } \\\\tan { y } =0\\\\\\\\ 5+\\\\frac { 8 }{ \\\\cot { y } } =0\\\\\\\\ \\\\cot { y } =-\\\\frac { 8 }{ 5 }$$ • @McFry,i fixed thank you – haqnatural Oct 5 '16 at 16:21 • I know how to do it, but I'd like to know why the method I originally took is incorrect? If possible... – user375494 Oct 5 '16 at 16:21\",\n \"seeing it yet? I see that because that is what I did to work out a hexagon and the larger even numbered polygons. My problem is that I have worked out 3 numbers for the Pentagon by working it out 2 different ways and so I am interested in how NateTG and Holly work it out. Bascially I would like someone to apply their equations to a simple shape (e.g. a sqaure with sides of 10cm) and then see if they get it right and how they did it so I can apply it, learn how to do it and then apply it myself. Cheers. P: 1,116 Please people. I need a simple example so I can see how the equations work. The Bob (2004 ©) HW Helper P: 2,538 Quote by The Bob Please people. I need a simple example so I can see how the equations work. The Bob (2004 ©) Ok - working in degrees for you: A square with 10 cm sides: n=4 l=10 $$A=\\\\frac{1}{4} nl^2 \\\\cot(\\\\frac{180}{n})=\\\\frac{1}{4} 4 (10^2) \\\\cot(\\\\frac{180}{4})=100$$ A 73-gon with 10 cm sides n=73 l=10 $$A=\\\\frac{1}{4}nl^2 \\\\cot(\\\\frac{180}{n})=\\\\frac{1}{4} 73 (10^2) \\\\cot(\\\\frac{180}{73})\\\\approx42380$$ P: 1,116 Quote by NateTG Ok - working in degrees for you: A square with 10 cm sides: n=4 l=10 $$A=\\\\frac{1}{4} nl^2 \\\\cot(\\\\frac{180}{n})=\\\\frac{1}{4} 4 (10^2) \\\\cot(\\\\frac{180}{4})=100$$ Cheers NateTG. That is a\",\n \"How do you find tan and cot if cot = 12/5 and sin <0? May 2, 2015 Easy steps I assume you have to calculate tan and cos $\\\\tan \\\\theta = \\\\frac{1}{\\\\cot} \\\\theta$ $\\\\tan \\\\theta = \\\\frac{5}{12}$ $\\\\sin \\\\frac{\\\\theta}{\\\\cos} \\\\theta = \\\\frac{5}{12}$ $12 \\\\sin \\\\theta - 5 \\\\cos \\\\theta = 0$ ${\\\\sin}^{2} \\\\theta + {\\\\cos}^{2} \\\\theta = 1$ Assume sin theta =x ; cos theta = y ${x}^{2} + {y}^{2} = 1$ $12 x - 5 y = 0$ $y = \\\\frac{12 x}{5}$ ${x}^{2} + \\\\frac{144 {x}^{2}}{25} = 1$ ${x}^{2} \\\\left(1 + \\\\frac{144}{25}\\\\right) = 1$ ${x}^{2} = \\\\frac{1}{\\\\frac{169}{25}}$ ${x}^{2} = {\\\\left(\\\\frac{5}{13}\\\\right)}^{2}$ $x = \\\\pm \\\\frac{5}{13}$ $\\\\sin \\\\theta = \\\\pm \\\\frac{5}{13}$ As $\\\\sin \\\\theta < 0 ,$ $\\\\sin \\\\theta = - \\\\frac{5}{13}$ $\\\\cos \\\\theta = - \\\\frac{12}{13}$\",\n \"# Trig equations • April 5th 2010, 08:39 AM darksupernova Trig equations Hello, I am having a bit of trouble solving this eqn: tan (x) + cot (x) = 2sec (x) For all angles between 0 and 360 degrees. Thank you! Max • April 5th 2010, 09:33 AM Sudharaka Quote: Originally Posted by darksupernova Hello, I am having a bit of trouble solving this eqn: tan (x) + cot (x) = 2sec (x) For all angles between 0 and 360 degrees. Thank you! Max Dear darksupernova, $tanx+cotx=2secx$ $\\\\frac{1}{sin(x)cos(x)}=2secx$ $2sinx=1$ $sinx=\\\\frac{1}{2}$ Since x is inbetween 0 and 360 degrees, $x=30^0$ • April 5th 2010, 10:27 AM darksupernova Ah very clever! Thank you very much! Im now a but stuck on this: tan (x) + 3cot (x) = 5sec (x) Very similar but the cos^2 {x) + sin^2 (X) = 1 trick doesnt work? • April 5th 2010, 10:38 AM Quote: Originally Posted by darksupernova Hello, I am having a bit of trouble solving this eqn: tan (x) + cot (x) = 2sec (x) For all angles between 0 and 360 degrees. Thank you! Max $tanx+cotx=\\\\frac{sinx}{cosx}+\\\\frac{cosx}{sinx}=\\\\fra c{sinx}{sinx}\\\\ \\\\frac{sinx}{cosx}+\\\\frac{cosx}{cosx}\\\\ \\\\frac{cosx}{sinx}$ $=\\\\frac{sin^2x+cos^2x}{sinxcosx}$ $\\\\Rightarrow\\\\ \\\\frac{1}{sinxcosx}=\\\\frac{2}{cosx}\\\\ \\\\Rightarrow\\\\ \\\\frac{1}{sinx}=2\\\\ \\\\Rightarrow\\\\ sinx=\\\\frac{1}{2}$ The vertical co-ordinate of the angle is 0.5 in the unit circle, centre (0,0). This corresponds to two angles, $x=sin^{-1}\\\\left(\\\\frac{1}{2}\\\\right),\\\\ x={\\\\pi}-sin^{-1}\\\\left(\\\\frac{1}{2}\\\\right)$ $x=30^o,\\\\ x=150^o$ •\",\n \"30 - (\\\\sin 30 - \\\\sin 10) \\\\\\\\ = \\\\sin 30 +\\\\sin 10 = 2 \\\\sin 20 \\\\cos 10$$ Therefore, $$\\\\tan \\\\alpha = \\\\frac{2 \\\\sin 10 \\\\sin 20}{1 - 2 \\\\sin 10 \\\\cos 20} = \\\\frac{2\\\\sin 10 \\\\sin 20}{2 \\\\cos 10 \\\\sin 20} = \\\\tan 10$$ Now, since $$0 < \\\\alpha < 180$$, we get that $$\\\\alpha = 10$$. From here, $$80-\\\\alpha = 70$$ is the desired angle.\",\n \"# Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 682: 101 See graph. a. $y=tan(\\\\frac{x}{2})$ b. See explanations. #### Work Step by Step Graph the equation as shown in the figure. a. The graph appears to be the same as $y=tan(\\\\frac{x}{2})$ b. Using the double angle formulas, we have $y=csc(x)-cot(x)=\\\\frac{1}{sin(x)}-\\\\frac{cos(x)}{sin(x)}=\\\\frac{1-cos(x)}{sin(x)}=\\\\frac{1-(1-2sin^2(x/2))}{2sin(x/2)cos(x/2)}=tan(\\\\frac{x}{2})$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.\",\n \"n=0 and works very well for any negative value of n. I made a mistake, I meant that it equals 5.7*10^(-n+1). Sorry about that. I don't know if this is a co-incidence but I had noticed the same pattern a few days ago when I was plugging in 89, 89.9, 89.99, 89.999 etc in my calculater to notice the changes in values of tanx. And, every time the same digits appeared that you've written with the decimal point displaced. I ignored it and didn't take the trouble of making a formula. It might be a co-incidence that you're uploading this today. I don't know how this is working. But, it's an approximation mostly for negative values of n and so, it works for a very small range. There are only two non-negative values values of n we can try. Sure you didn't mean ##\\\\tan(90° - 10^{-n}) \\\\approx 5.7_3 \\\\cdot 10^{n+1}\\\\; \\\\;(n \\\\in \\\\mathbb{N})\\\\;##? It's pretty much the same. But yours is a bit more rigorous, so I guess it's better. Although it also works for 80 degrees (n=-1 in your definition). I'm guessing it has something to do with the Taylor series. $$\\\\tan(\\\\frac{\\\\pi}{2}-\\\\theta) = \\\\cot \\\\theta$$ The Taylor series for cotangent is: $$\\\\cot \\\\theta = \\\\frac{1}{\\\\theta} - \\\\frac{1}{3}\\\\theta - \\\\frac{1}{45}\\\\theta^3 + \\\\cdots$$ I'm assuming you were working in\",\n \"anonymous 5 years ago write an expression equal to 1/tan^2x $\\\\frac{1}{\\\\tan^2x} = \\\\cot^2x = \\\\frac{\\\\cos^2x}{\\\\sin^2x}$ ^_^\",\n \"How do you find the value of cot ((5pi/3) using the double angle or half angle identity? Nov 11, 2016 $- \\\\frac{\\\\sqrt{3}}{3}$ Explanation: Trig table and unit circle --> $\\\\cot \\\\left(\\\\frac{5 \\\\pi}{3}\\\\right) = \\\\frac{1}{\\\\tan} \\\\left(\\\\frac{5 \\\\pi}{3}\\\\right)$ Find $\\\\tan \\\\left(\\\\frac{5 \\\\pi}{3}\\\\right)$ $\\\\tan \\\\left(\\\\frac{5 \\\\pi}{3}\\\\right) = \\\\tan \\\\left(\\\\frac{2 \\\\pi}{3} + \\\\pi\\\\right) = \\\\tan \\\\left(\\\\frac{2 \\\\pi}{3}\\\\right) = - \\\\sqrt{3}$ There for: $\\\\cot \\\\left(\\\\frac{5 \\\\pi}{3}\\\\right) = - \\\\frac{1}{\\\\sqrt{3}} = - \\\\frac{\\\\sqrt{3}}{3}$\",\n \"# Thread: trig compound angle formulae and proofs 1. ## trig compound angle formulae and proofs Hi I've got some maths homework and stuck on the last 3 questions. 1. Given that sinA=3/5 and cosB=12/13, where A is obtuse and B is acute, find exact values of a) cos(A+B) b)cot (A-B) 2. Prove cot(A+B)is identical to cotAcotB-1 cotA+cotB 3. prove that tanA+tanB is identical to sin(A+B) cosAcosB I have tried all these and just get stuck . any help would be really appreciative thankyou 2. 1) $\\\\cos A=-\\\\sqrt{1-\\\\sin^2A}, \\\\ \\\\sin B=\\\\sqrt{1-\\\\cos^2B}$ $\\\\cos(A+B)=\\\\cos A\\\\cos B-\\\\sin A\\\\sin B$ $\\\\cot(A-B)=\\\\frac{\\\\cos(A-B)}{\\\\sin(A-B)}=\\\\frac{\\\\cos A\\\\cos B+\\\\sin A\\\\sin B}{\\\\sin A\\\\cos B-\\\\sin B\\\\cos A}$ 2) $\\\\cos(A+B)=\\\\frac{\\\\cos(A+B)}{\\\\sin(A+B)}=\\\\frac{\\\\cos A\\\\cos B-\\\\sin A\\\\sin B}{\\\\sin A\\\\cos B+\\\\sin B\\\\cos A}=$ $=\\\\frac{\\\\frac{\\\\cos A\\\\cos B}{\\\\sin A\\\\sin B}-1}{\\\\frac{\\\\sin A\\\\cos B}{\\\\sin B\\\\sin A}+\\\\frac{\\\\sin B\\\\cos A}{\\\\sin A\\\\sin B}}=\\\\frac{\\\\cot A\\\\cot B-1}{\\\\cot A+cot B}$ 3) $\\\\tan A+\\\\tan B=\\\\frac{\\\\sin A}{\\\\cos A}+\\\\frac{\\\\sin B}{\\\\cos B}=\\\\frac{\\\\sin A\\\\cos B+\\\\sin B\\\\cos A}{\\\\cos A\\\\cos B}=\\\\frac{\\\\sin(A+B)}{\\\\cos A\\\\cos B}$ 3. ## thankyou thanks so much. not sure if i get questions 1 yet though, i'll try it.\",\n \"Sec, Cosec, Cot Printable View • May 24th 2008, 03:14 AM jarald83 Sec, Cosec, Cot how to calculate the following value? Cosec 30° cot 60° sec 310° cosec 120° ° cot 600° and -cot 130° (Doh)(Crying) • May 24th 2008, 03:44 AM wingless $\\\\tan x = \\\\frac{\\\\sin x}{\\\\cos x}$ $\\\\cot x = \\\\frac{\\\\cos x}{\\\\sin x}$ $\\\\csc x = \\\\frac{1}{\\\\sin x}$ $\\\\sec x = \\\\frac{1}{\\\\cos x}$ • May 24th 2008, 03:58 AM nandu11 u need to know the simple sin, cos and tan trigo ratios for 0, 30, 45, 60 and 90 to do these types of questions and then use the formulas like cosecx=1/sinx, cotx=cosx/sinx etcc\"\n]"},"ps_retrievals":{"kind":"list like","value":["## Trigonometry Formulas for class 11 Trigonometry is quite a interesting subject. Here are the useful Trigonometry Formulas for class 11 Maths Basic Formula $tan(x) = \\frac {sin(x)}{cos(x)}$ $cot(x) = \\frac {cos(x)}{sin(x)}$ Reciprocal Identities: $cosec(x) = \\frac {1}{sin(x)}$ $sec(x) =\\frac { 1}{cos(x)}$ $cot(x) = \\frac {1}{tan(x)}$ $sin(x) = \\frac {1}{cosec(x)}$ $cos(x) = \\frac {1}{sec(x)}$ $tan(x) = \\frac {1}{cot(x)}$ Pythagorean Identities: $sin^2(x) + cos^2(x) = 1$ $cot^2x +1 = cosec^2x$ $1+tan^2x = sec^2x$ Trigonometric Ratio’s of Common angles We can find the values of trigonometric ratio’s various angle Trigonometry Formula for Complementary and supplementary angles Sin and cos function 1. $cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$ 2. $cos(A-B)=cos(A)cos(B)+sin(A)sin(B)$ 3. $cos(\\pi /2 -A)=sin(A)$ 4. $sin(\\pi /2 -A)=cos(A)$ 5. $sin(A+B)=sin(A)cos(B)+sin(B)cos(A)$ 6. $sin(A-B)=sin(A)cos(B)-sin(B)cos(A)$ Tan and cot functions If none of the angles x, y and (x + y) is an odd multiple of $\\pi /2$ $tan(A+B)=\\frac{tan(A)+tan(B)}{1-tan(A)tan(B)}$ $tan(A-B)=\\frac{tan(A)-tan(B)}{1+tan(A)tan(B)}$ If none of the angles x, y and (x + y) is an multiple of $\\pi /2$ $cot(A+B)=\\frac{cot(A)cot(B)-1}{cot(A)+cot(B)}$ $cot(A-B)=\\frac{cot(A)cot(B)+1}{cot(B)-cot(A)}$ Some more Trigonometric Functions Double of x (Double Of Angles) $cos2x=cos^{^{2}}x-sin^{^{2}}x=2cos^{^{2}}x-1=1-2sin^{^{2}}x=\\frac{1-tan^{^{2}}x}{1+tan^{^{2}}x}$ $sin2x=2cos(x)sin(x)=\\frac{2tan(x)}{1+tan^{^{2}}x}$ $tan2x=\\frac{2tan(x)}{1-tan^{^{2}}x}$ Triple of x ( Triple of Angles) $sin3x=3sin(x)-4sin^{3}x$ $cos3x=4cos^{3}x-3cos(x)$ $tan(3x)=\\frac{3tanx-tan^{^{3}}x}{1-3tan^{^{2}}x}$ Sum and Difference of Angles $cos(A)+cos(B)=2cos\\frac{A+B}{2}cos\\frac{A-B}{2}$ $cos(A)-cos(B)=-2sin\\frac{A+B}{2}sin\\frac{A-B}{2}$ $sin(A)+sin(B)=2sin\\frac{A+B}{2}cos\\frac{A-B}{2}$ $sin(A)-sin(B)=2cos\\frac{A+B}{2}sin\\frac{A-B}{2}$ Half Angle Formula Power Reducing Functions Trigonometric equations Formula’s 1.$sin x = 0$ implies $x = n \\pi$, where n is any integer
2.$cos x = 0$ implies $x","# How do you find sin, cos, tan, sec, csc, and cot given (0,6)? Dec 2, 2016 Find values of trig functions #### Explanation: The point (0, 6) is located on the Oy axis. The related arc (angle) is $t = \\frac{\\pi}{2}$. sin t = 1 cos t = 0 tan t = inf. cot t = 0 $\\sec t = \\frac{1}{\\cos} =$ inf. $\\csc t = \\frac{1}{\\sin} = 1$","the swap unit button first. Besides cos0.5, similar trigonometric calculations on our site include, but are not limited, to: The identities of cosine 0.5 are as follows: cos0.5 = sin (π/2 + 0.5) = sin 2.0707963267949 = sin (π/2 – 0.5) = sin 1.0707963267949 -cos0.5 = cos (π + 0.5) = cos 3.64159265358979 = cos (π – 0.5) = cos 2.64159265358979 Note that cos0.5 is periodic: cos (0.5 + n × 2π) = cos 0.5, n$\\hspace{5px} \\in \\hspace{5px} \\mathbb{Z}$. There are more formulas for the double angle (2 × 0.5), half angle ((0.5/2)) as well as the sum, difference and products of two angles such as 0.5 and β. You can locate all of them in the respective article found in the header menu. To find everything about cos -0.5 click the link. And here is all about sin 0.5, including, for instance, a converter. In terms of the other five trigonometric functions, cos of 0.5 = • $\\pm \\sqrt{1-\\sin^{2} 0.5 }$ • $\\pm\\frac{1}{\\sqrt{1 + \\tan^{2} 0.5}}$ • $\\pm\\frac{\\cot 0.5}{\\sqrt{1 + \\cot^{2} 0.5}}$ • $\\frac{1}{\\sec 0.5}$ • $\\pm\\frac{\\sqrt{\\csc^{2} (0.5) – 1} }{\\csc 0.5}$ As the cosine function is the reciprocal of the secant function, 1 / sec 0.5 = cos0.5. In the next part we discuss the trigonometric significance of cos0.5, and there you can also learn what","Trigonometry is quite a interesting subject. Here are the useful Trigonometry Formulas for class 11 Maths Basic Formula $tan(x) = \\frac {sin(x)}{cos(x)}$ $cot(x) = \\frac {cos(x)}{sin(x)}$ Reciprocal Identities: $cosec(x) = \\frac {1}{sin(x)}$ $sec(x) =\\frac { 1}{cos(x)}$ $cot(x) = \\frac {1}{tan(x)}$ $sin(x) = \\frac {1}{cosec(x)}$ $cos(x) = \\frac {1}{sec(x)}$ $tan(x) = \\frac {1}{cot(x)}$ Pythagorean Identities: $sin^2(x) + cos^2(x) = 1$ $cot^2x +1 = cosec^2x$ $1+tan^2x = sec^2x$ Trigonometric Ratio’s of Common angles We can find the values of trigonometric ratio’s various angle Trigonometry Formula for Complementary and supplementary angles Sin and cos function 1. $cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$ 2. $cos(A-B)=cos(A)cos(B)+sin(A)sin(B)$ 3. $cos(\\pi /2 -A)=sin(A)$ 4. $sin(\\pi /2 -A)=cos(A)$ 5. $sin(A+B)=sin(A)cos(B)+sin(B)cos(A)$ 6. $sin(A-B)=sin(A)cos(B)-sin(B)cos(A)$ Tan and cot functions If none of the angles x, y and (x + y) is an odd multiple of $\\pi /2$ $tan(A+B)=\\frac{tan(A)+tan(B)}{1-tan(A)tan(B)}$ $tan(A-B)=\\frac{tan(A)-tan(B)}{1+tan(A)tan(B)}$ If none of the angles x, y and (x + y) is an multiple of $\\pi /2$ $cot(A+B)=\\frac{cot(A)cot(B)-1}{cot(A)+cot(B)}$ $cot(A-B)=\\frac{cot(A)cot(B)+1}{cot(B)-cot(A)}$ Some more Trigonometric Functions Double of x $cos2x=cos^{^{2}}x-sin^{^{2}}x=2cos^{^{2}}x-1=1-2sin^{^{2}}x=\\frac{1-tan^{^{2}}x}{1+tan^{^{2}}x}$ $sin2x=2cos(x)sin(x)=\\frac{2tan(x)}{1+tan^{^{2}}x}$ $tan2x=\\frac{2tan(x)}{1-tan^{^{2}}x}$ Triple of x $sin3x=3sin(x)-4sin^{3}x$ $cos3x=4cos^{3}x-3cos(x)$ $tan(3x)=\\frac{3tanx-tan^{^{3}}x}{1-3tan^{^{2}}x}$ Some other Important functions $cos(A)+cos(B)=2cos\\frac{A+B}{2}cos\\frac{A-B}{2}$ $cos(A)-cos(B)=-2sin\\frac{A+B}{2}sin\\frac{A-B}{2}$ $sin(A)+sin(B)=2sin\\frac{A+B}{2}cos\\frac{A-B}{2}$ $sin(A)-sin(B)=2cos\\frac{A+B}{2}sin\\frac{A-B}{2}$ Half Angle Formula Power Reducing Functions Trigonometric equations Formula’s 1.$sin x = 0$ implies $x = n \\pi$, where n is any integer
2.$cos x = 0$ implies $x = (2n + 1)(\\pi /2)$
1. $sinx =siny$ then $x=n \\pi + (-1)^{n}y$ where","1 and $\\sqrt3$ and we look from the position of the angle and say that : sin= $\\frac{opposite}{hypotenuse}$ and csc is reciprocal of that cos= $\\frac{adjacent}{hypotenuse}$ and sec is reciprocal of that tan= $\\frac{sin}{cos}$ and cot is reciprocal of that Good question!","find other five trigonometric functions calculator Step 3: … 1 - Enter the angle: in Degrees. Solution : sec θ = Hypotenuse side/Hypotenuse side sec θ = 13/5. more. For P(3,4), find the value of all six trigonometric functions, This site is protected by reCAPTCHA and the Google, (3sec(60)+2tan(45)+csc(30))/(sin^2(60)+cot^2(45)), (3sec(60)+2tan(45)+csc(30))(sin^2(60)+cot^2(45)), sin(30)*cos(45)*tan(60)=sin(45)*cos(60)*cot(30), tan(45)*sec(72)*sin(58)=cos(32)*csc(18), sin^2(34)-cot^2(46)-cos^2(56)+tan^2(44)=sec^2(62)-csc^2(28), sin(35)*sec(55)-cos(55)*csc(35)=tan(25)-cot(65), 2sin(30)+2tan(45)-3cos(60)-2cos^2(30)=0, (tan(x)+cot(x))(sec(x)-cos(x))(csc(x)-sin(x)), (tan^2(x)-cot^2(x))/(sin^2(x)-cos^2(x)), (tan^2(x)+cot^2(x))/(tan^2(x)-cot^2(x)), (tan^2(x)-cot^2(x))/(tan^2(x)+cot^2(x)), (3sin(x)+5cos(x))^2+(3cos(x)-5sin(x))^2, sin^2(x)cos(x)+sin^3(x)+cos^2(x)sin(x)+cos^3(x), (sin^4(x)-cos^4(x))/(sin^2(x)-cos^2(x)), If sin(x)=5/13 then find other trigonometry, If cos(x)=12/13 then find other trigonometry, If tan(x)=5/12 then find other trigonometry, If csc(x)=13/5 then find other trigonometry, If sec(x)=13/12 then find other trigonometry, If cot(x)=12/5 then find other trigonometry, If sin(x)=3/5 then solve cos(x)csc(x)+tan(x)sec(x), If tan(x)=1/2 then solve (1-tan^2(x))/(1+tan^2(x))+(2tan(x))/(1+tan^2(x)), If sin(x)=4/5 then solve (1-sin(x))/cos(x)+cos(x)/(1-sin(x)), If cos(x)=3/5 then solve (1-sin(x))/cos(x)+cos(x)/(1-sin(x)). In six trigonometric ratios sin, cos, tan, csc, sec and cot, if the value of one of the ratios is given, we can find the values of the other five functions. Online calculator. Please add atozmath.com to your ad blocking whitelist or disable your adblocking software. Step 2: Now click the button “Calculate Trigonometric Ratios” to get the result. Please try again using a different payment method. To evaluate trigonometric functions of other angles, we use a scientific or graphing calculator or computer software. By using this website, you agree to our Cookie Policy. By browsing this website, you agree","to the left and T to the right, so we get: omega/(2*pi) = 1/T and since 1/T = f we finally get: f = omega / (2*pi) ### How do you simplify csc theta cot theta? There are 6 basic trig functions. sin(x) = 1/csc(x) cos(x) = 1/sec(x) tan(x) = sin(x)/cos(x) or 1/cot(x) csc(x) = 1/sin(x) sec(x) = 1/cos(x) cot(x) = cos(x)/sin(x) or 1/tan(x) ---- In your problem csc(x)*cot(x) we can simplify csc(x). csc(x) = 1/sin(x) Similarly, cot(x) = cos(x)/sin(x). csc(x)*cot(x) = (1/sin[x])*(cos[x]/sin[x]) = cos(x)/sin2(x) = cos(x) * 1/sin2(x) Either of the above answers should work. In general, try converting your trig functions into sine and cosine to make things simpler. ### How do you find sin cos and tan values manually? This is so much work that it is not worthwhile to do in practice, although the formulae themselves are actually quite simple. The basic method is to use a so-called \"infinite series\". The angle must be expressed in radians. If the angle is in degrees, multiply it by (pi/180), to get the equivalent angle in radians. Then, use the formula: sin(x) = x - x3/3! + x5/5! - x7/7! + x9/9!... The individual terms become smaller and smaller, quite quickly, so the idea is to continue adding more terms until you see that the terms become","### Home > PC > Chapter 6 > Lesson 6.1.1 > Problem6-12 6-12. 1. Find the exact value of the following trig expressions. Homework Help ✎ 1. csc 2. tan 3. cot 4. sec 1. Plot the point on a unit circle. 2. Find the sin first. 3. Flip the answer for the csc. (Find the reciprocal) $\\text{Since the tan }x=\\frac{\\text{sin }x}{\\text{cos }x},$ what issues can occur for denominator values of a fraction? Plot the point finding the side of the reference angle. Find the tan first and take its reciprocal for the cot. Find the cos first and then its reciprocal for the sec.","# Indefinite Integration of a trig function How do you integrate $$1/(\\sin2x(\\tan^5x+\\cot^5x))$$ with respect to $$x$$? I tried writing tan and cot in terms of sin and cos but when I take $$\\mathrm{LCM}$$ I get powers of $$10$$ for $$\\sin$$ and $$\\cos$$ in the denominator. I can't think of how I would simplify this without the use of binomial expansion but that might make things more complicated. Is there another method I could use to approach this problem? $$\\displaystyle \\int \\frac{dx}{\\sin2x\\cdot(tan^5x+cot^5x)} = \\frac12 \\int \\frac{\\cot x \\,dx}{\\cos^2x\\cdot(\\tan^5x+\\cot^5x)}$$ $$\\displaystyle \\tan x = t, \\frac{dx}{\\cos^2x}=dt$$","Sec, Cosec, Cot Printable View • May 24th 2008, 03:14 AM jarald83 Sec, Cosec, Cot how to calculate the following value? Cosec 30° cot 60° sec 310° cosec 120° ° cot 600° and -cot 130° (Doh)(Crying) • May 24th 2008, 03:44 AM wingless $\\tan x = \\frac{\\sin x}{\\cos x}$ $\\cot x = \\frac{\\cos x}{\\sin x}$ $\\csc x = \\frac{1}{\\sin x}$ $\\sec x = \\frac{1}{\\cos x}$ • May 24th 2008, 03:58 AM nandu11 u need to know the simple sin, cos and tan trigo ratios for 0, 30, 45, 60 and 90 to do these types of questions and then use the formulas like cosecx=1/sinx, cotx=cosx/sinx etcc","90o – θ ) = sin θ 17. tan ( 90o – θ ) = cot θ 18. cot ( 90o – θ ) = tan θ 19. sec ( 90o – θ ) = cosec θ 20. cosec ( 90o – θ ) = sec θ 21. sin 00 = 0, cos 00 = 1 and tan 00 = 0 22. cosec 00 = Not defined sec 00 = 1 and cot 00 = Not Defined 23. sin 300 = $\\frac{1}{2}$, cos 300 = $\\frac{\\sqrt{3}}{2}$ and tan 300 = $\\frac{1}{\\sqrt{3}}$ 24. cosec 300 = 2, sec 300 = $\\frac{2}{\\sqrt{3}}$ and cot 300 = $\\sqrt{3}$ 25. sin 450 = $\\frac{1}{\\sqrt{2}}$, cos450 = $\\frac{1}{\\sqrt{2}}$ and tan 450 = 1 26. cosec 450 = $\\sqrt{2}$, sec 450 = $\\sqrt{2}$, and cot 450 = 1 27. sin 600 =$\\frac{\\sqrt{3}}{2}$ , cos 600 = $\\frac{1}{2}$, and tan 600 =$\\sqrt{3}$ 28. cosec 600 = $\\frac{2}{\\sqrt{3}}$, sec 600 = 2 and cot 600 = $\\frac{1}{\\sqrt{3}}$ 29. sin 900 = 1, cos 900 = 0 and tan 00 = Not Defined 30. cosec 900 = 1sec 900 = Not Defined and cot 900 = 0","sine and cos as their ratio. From the diagram taken above, the tan function will be the following. Tan a = $\\frac{Opposite}{Adjacent}$ = $\\frac{CB}{BA}$ Also, in terms of sine and cos, tan can be represented as: Tan a = $\\frac{sin\\ a}{cos\\ a}$ ### Secant, cosecant, and cotangent Secant, cosecant (csc) and cotangent are the three additional trigonometric functions which are derived from the primary functions of sine, cos, and tan. The reciprocal of sine, cos, and tan are cosecant (csc), secant (sec), and cotangent (cot) respectively. The formula of each of these functions are given as: Sec a = 1/(cos a) = $\\frac{Hypotenuse}{Adjacent}$ = $\\frac{CA}{AB}$ Cosec a = 1/(sin a) = $\\frac{Hypotenuse}{Opposite}$ = $\\frac{CA}{CB}$ cot a = 1/(tan a) = $\\frac{Adjacent}{Opposite}$ = $\\frac{BA}{CB}$ ### Trigonometric Functions Table The trigonometric table for six functions, Sin, Cos, Tan, Cosec, Sec, Cot, are as : Trigonometric Ratios/ angle= a degrees 0 ° 30 ° 45 ° 60 ° 90 ° Sin a 0 1/2 1/√2 √3/2 1 Cos a 1 √3/2 1/√2 1/2 0 Tan a 0 1/√3 1 √3 Not Defined Cosec a Not Defined 2 √2 2/√3 1 Sec a 1 2/√3 √2 2 Not Defined Cot a Not defined √3 1 1/√3 0 ### Inverse Trigonometric Functions Inverse functions are used to obtain an angle from any of","# Math Help - Simplifying Trigonometric Equations 1. ## Simplifying Trigonometric Equations Hi all! I know this is probably a fairly simple concept for Trigonometry, but I can't quite remember how to do them. I'm just trying to make sure I know them well enough so that I can do them on a test I have in a few days. It would be awesome if you could just get me started on them. 1. $cot(x)sin(x) = \\frac{\\sqrt{3}}{2}$ 2. $tan(x)$ $cos(x) = 1$ 3. $sin^2(x)$ + $cos^2(x) + cos(x)$ = 1 4. [ $cot(x)$ $(\\frac{sin(x)}{cos(x)})$] • $sec^2(x)$ = $\\frac{1}{2}$ 2. Originally Posted by sar Hi all! I know this is probably a fairly simple concept for Trigonometry, but I can't quite remember how to do them. I'm just trying to make sure I know them well enough so that I can do them on a test I have in a few days. It would be awesome if you could just get me started on them. 1. $cot(x)sin(x) = \\frac{\\sqrt{3}}{2}$ It is normally best to write all of your sec(x), csc(x), tan(x), and cot(x) in terms of sin(x) and cos(x). That way, if anything cancels, you can simplify the expression. $\\frac{cos(x)}{sin(x)} • sin(x) = \\frac{\\sqrt{3}}{2}$ Hey, there's a sin(x) on the top as well as the bottom! So, as long","anonymous 5 years ago The value of one trigonometric function is give 0 CALC > Chapter 4 > Lesson 4.5.1 > Problem4-164 4-164. 1. Rewrite the following using a single trigonometric function. You may wish to review your trigonometric identities in Chapter 1. Homework Help ✎ 1. 10 sin(3x) cos(3x) 2. sin x cos 3x − sin 3x cos x 3. cos4 x − sin4 x 4. tan x + cot x Double Angle identity (factor first). Factor first. Sum and Difference (Angle Sum) identity. Start by rewriting tan(x) and cot(x) as fractions. You will use more than one identity. $\\frac{\\text{sin}x}{\\text{cos}x}+\\frac{\\text{cos}x}{\\text{sin}x}=\\frac{\\text{sin}^{2}x+\\text{cos}^{2}x}{(\\text{cos}x)(\\text{sin}x)}=$ $\\frac{1}{\\frac{1}{2}\\text{sin}(2x)}=2\\text{csc}(2x)$","Trigonometric ratios of 360 plus theta (360 + θ) In this section we will learn trigonometric ratios of 360 plus theta (360 + θ). We know that the terminal sides of co-terminal angles always coincide and θ and (360 + θ) are co-terminal angles. Clearly, 360 + θ and θ are co-terminal angles. sin(360 + θ) = Sin[360 -(- θ)] But we know that -sin(- θ) = sinθ $\\underline{sin(360 + θ)= sin θ}$ Again, cos(360 + θ) = cos[360-(- θ)] But we know that cos(- θ) =cosθ $\\underline{cos(360 + θ)= cos θ}$ Now, tan(360 + θ) = tan[360-(- θ)] But we know that -tan(- θ) = tanθ $\\underline{tan(360 + θ)= tan θ}$ As we know that, csc θ = $\\frac{1}{sinθ}$ ∴ csc (360+θ) = $\\frac{1}{sin(360+θ)}$ But sin(360+θ) = sin(θ) ∴ csc (360+θ)= $\\frac{1}{sin θ}$ $\\underline{csc(360 + θ)= csc θ}$ And, sec θ = $\\frac{1}{cosθ}$ ∴ sec (360+θ) = $\\frac{1}{cos(360+θ)}$ But cos(360+θ) = cos(θ) ∴ sec (360+θ)= $\\frac{1}{cos θ}$ $\\underline{sec(360 + θ)= cos θ}$ Again, cot θ = $\\frac{1}{tanθ}$ ∴ cot (360+θ) = $\\frac{1}{tan(360+θ)}$ But tan(360+θ) = tan(θ) ∴ cot (360+θ)= $\\frac{1}{ tan θ}$ $\\underline{cot(360 + θ)= cot θ}$ For any positive integer 'n', angle (360 x n + θ) is co-terminal to angle θ. Therefore, for any positive integer n, we have, sin(360 x n + θ )"],"string":"[\n \"## Trigonometry Formulas for class 11 Trigonometry is quite a interesting subject. Here are the useful Trigonometry Formulas for class 11 Maths Basic Formula $tan(x) = \\\\frac {sin(x)}{cos(x)}$ $cot(x) = \\\\frac {cos(x)}{sin(x)}$ Reciprocal Identities: $cosec(x) = \\\\frac {1}{sin(x)}$ $sec(x) =\\\\frac { 1}{cos(x)}$ $cot(x) = \\\\frac {1}{tan(x)}$ $sin(x) = \\\\frac {1}{cosec(x)}$ $cos(x) = \\\\frac {1}{sec(x)}$ $tan(x) = \\\\frac {1}{cot(x)}$ Pythagorean Identities: $sin^2(x) + cos^2(x) = 1$ $cot^2x +1 = cosec^2x$ $1+tan^2x = sec^2x$ Trigonometric Ratio’s of Common angles We can find the values of trigonometric ratio’s various angle Trigonometry Formula for Complementary and supplementary angles Sin and cos function 1. $cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$ 2. $cos(A-B)=cos(A)cos(B)+sin(A)sin(B)$ 3. $cos(\\\\pi /2 -A)=sin(A)$ 4. $sin(\\\\pi /2 -A)=cos(A)$ 5. $sin(A+B)=sin(A)cos(B)+sin(B)cos(A)$ 6. $sin(A-B)=sin(A)cos(B)-sin(B)cos(A)$ Tan and cot functions If none of the angles x, y and (x + y) is an odd multiple of $\\\\pi /2$ $tan(A+B)=\\\\frac{tan(A)+tan(B)}{1-tan(A)tan(B)}$ $tan(A-B)=\\\\frac{tan(A)-tan(B)}{1+tan(A)tan(B)}$ If none of the angles x, y and (x + y) is an multiple of $\\\\pi /2$ $cot(A+B)=\\\\frac{cot(A)cot(B)-1}{cot(A)+cot(B)}$ $cot(A-B)=\\\\frac{cot(A)cot(B)+1}{cot(B)-cot(A)}$ Some more Trigonometric Functions Double of x (Double Of Angles) $cos2x=cos^{^{2}}x-sin^{^{2}}x=2cos^{^{2}}x-1=1-2sin^{^{2}}x=\\\\frac{1-tan^{^{2}}x}{1+tan^{^{2}}x}$ $sin2x=2cos(x)sin(x)=\\\\frac{2tan(x)}{1+tan^{^{2}}x}$ $tan2x=\\\\frac{2tan(x)}{1-tan^{^{2}}x}$ Triple of x ( Triple of Angles) $sin3x=3sin(x)-4sin^{3}x$ $cos3x=4cos^{3}x-3cos(x)$ $tan(3x)=\\\\frac{3tanx-tan^{^{3}}x}{1-3tan^{^{2}}x}$ Sum and Difference of Angles $cos(A)+cos(B)=2cos\\\\frac{A+B}{2}cos\\\\frac{A-B}{2}$ $cos(A)-cos(B)=-2sin\\\\frac{A+B}{2}sin\\\\frac{A-B}{2}$ $sin(A)+sin(B)=2sin\\\\frac{A+B}{2}cos\\\\frac{A-B}{2}$ $sin(A)-sin(B)=2cos\\\\frac{A+B}{2}sin\\\\frac{A-B}{2}$ Half Angle Formula Power Reducing Functions Trigonometric equations Formula’s 1.$sin x = 0$ implies $x = n \\\\pi$, where n is any integer
2.$cos x = 0$ implies $x\",\n \"# How do you find sin, cos, tan, sec, csc, and cot given (0,6)? Dec 2, 2016 Find values of trig functions #### Explanation: The point (0, 6) is located on the Oy axis. The related arc (angle) is $t = \\\\frac{\\\\pi}{2}$. sin t = 1 cos t = 0 tan t = inf. cot t = 0 $\\\\sec t = \\\\frac{1}{\\\\cos} =$ inf. $\\\\csc t = \\\\frac{1}{\\\\sin} = 1$\",\n \"the swap unit button first. Besides cos0.5, similar trigonometric calculations on our site include, but are not limited, to: The identities of cosine 0.5 are as follows: cos0.5 = sin (π/2 + 0.5) = sin 2.0707963267949 = sin (π/2 – 0.5) = sin 1.0707963267949 -cos0.5 = cos (π + 0.5) = cos 3.64159265358979 = cos (π – 0.5) = cos 2.64159265358979 Note that cos0.5 is periodic: cos (0.5 + n × 2π) = cos 0.5, n$\\\\hspace{5px} \\\\in \\\\hspace{5px} \\\\mathbb{Z}$. There are more formulas for the double angle (2 × 0.5), half angle ((0.5/2)) as well as the sum, difference and products of two angles such as 0.5 and β. You can locate all of them in the respective article found in the header menu. To find everything about cos -0.5 click the link. And here is all about sin 0.5, including, for instance, a converter. In terms of the other five trigonometric functions, cos of 0.5 = • $\\\\pm \\\\sqrt{1-\\\\sin^{2} 0.5 }$ • $\\\\pm\\\\frac{1}{\\\\sqrt{1 + \\\\tan^{2} 0.5}}$ • $\\\\pm\\\\frac{\\\\cot 0.5}{\\\\sqrt{1 + \\\\cot^{2} 0.5}}$ • $\\\\frac{1}{\\\\sec 0.5}$ • $\\\\pm\\\\frac{\\\\sqrt{\\\\csc^{2} (0.5) – 1} }{\\\\csc 0.5}$ As the cosine function is the reciprocal of the secant function, 1 / sec 0.5 = cos0.5. In the next part we discuss the trigonometric significance of cos0.5, and there you can also learn what\",\n \"Trigonometry is quite a interesting subject. Here are the useful Trigonometry Formulas for class 11 Maths Basic Formula $tan(x) = \\\\frac {sin(x)}{cos(x)}$ $cot(x) = \\\\frac {cos(x)}{sin(x)}$ Reciprocal Identities: $cosec(x) = \\\\frac {1}{sin(x)}$ $sec(x) =\\\\frac { 1}{cos(x)}$ $cot(x) = \\\\frac {1}{tan(x)}$ $sin(x) = \\\\frac {1}{cosec(x)}$ $cos(x) = \\\\frac {1}{sec(x)}$ $tan(x) = \\\\frac {1}{cot(x)}$ Pythagorean Identities: $sin^2(x) + cos^2(x) = 1$ $cot^2x +1 = cosec^2x$ $1+tan^2x = sec^2x$ Trigonometric Ratio’s of Common angles We can find the values of trigonometric ratio’s various angle Trigonometry Formula for Complementary and supplementary angles Sin and cos function 1. $cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$ 2. $cos(A-B)=cos(A)cos(B)+sin(A)sin(B)$ 3. $cos(\\\\pi /2 -A)=sin(A)$ 4. $sin(\\\\pi /2 -A)=cos(A)$ 5. $sin(A+B)=sin(A)cos(B)+sin(B)cos(A)$ 6. $sin(A-B)=sin(A)cos(B)-sin(B)cos(A)$ Tan and cot functions If none of the angles x, y and (x + y) is an odd multiple of $\\\\pi /2$ $tan(A+B)=\\\\frac{tan(A)+tan(B)}{1-tan(A)tan(B)}$ $tan(A-B)=\\\\frac{tan(A)-tan(B)}{1+tan(A)tan(B)}$ If none of the angles x, y and (x + y) is an multiple of $\\\\pi /2$ $cot(A+B)=\\\\frac{cot(A)cot(B)-1}{cot(A)+cot(B)}$ $cot(A-B)=\\\\frac{cot(A)cot(B)+1}{cot(B)-cot(A)}$ Some more Trigonometric Functions Double of x $cos2x=cos^{^{2}}x-sin^{^{2}}x=2cos^{^{2}}x-1=1-2sin^{^{2}}x=\\\\frac{1-tan^{^{2}}x}{1+tan^{^{2}}x}$ $sin2x=2cos(x)sin(x)=\\\\frac{2tan(x)}{1+tan^{^{2}}x}$ $tan2x=\\\\frac{2tan(x)}{1-tan^{^{2}}x}$ Triple of x $sin3x=3sin(x)-4sin^{3}x$ $cos3x=4cos^{3}x-3cos(x)$ $tan(3x)=\\\\frac{3tanx-tan^{^{3}}x}{1-3tan^{^{2}}x}$ Some other Important functions $cos(A)+cos(B)=2cos\\\\frac{A+B}{2}cos\\\\frac{A-B}{2}$ $cos(A)-cos(B)=-2sin\\\\frac{A+B}{2}sin\\\\frac{A-B}{2}$ $sin(A)+sin(B)=2sin\\\\frac{A+B}{2}cos\\\\frac{A-B}{2}$ $sin(A)-sin(B)=2cos\\\\frac{A+B}{2}sin\\\\frac{A-B}{2}$ Half Angle Formula Power Reducing Functions Trigonometric equations Formula’s 1.$sin x = 0$ implies $x = n \\\\pi$, where n is any integer
2.$cos x = 0$ implies $x = (2n + 1)(\\\\pi /2)$
1. $sinx =siny$ then $x=n \\\\pi + (-1)^{n}y$ where\",\n \"1 and $\\\\sqrt3$ and we look from the position of the angle and say that : sin= $\\\\frac{opposite}{hypotenuse}$ and csc is reciprocal of that cos= $\\\\frac{adjacent}{hypotenuse}$ and sec is reciprocal of that tan= $\\\\frac{sin}{cos}$ and cot is reciprocal of that Good question!\",\n \"find other five trigonometric functions calculator Step 3: … 1 - Enter the angle: in Degrees. Solution : sec θ = Hypotenuse side/Hypotenuse side sec θ = 13/5. more. For P(3,4), find the value of all six trigonometric functions, This site is protected by reCAPTCHA and the Google, (3sec(60)+2tan(45)+csc(30))/(sin^2(60)+cot^2(45)), (3sec(60)+2tan(45)+csc(30))(sin^2(60)+cot^2(45)), sin(30)*cos(45)*tan(60)=sin(45)*cos(60)*cot(30), tan(45)*sec(72)*sin(58)=cos(32)*csc(18), sin^2(34)-cot^2(46)-cos^2(56)+tan^2(44)=sec^2(62)-csc^2(28), sin(35)*sec(55)-cos(55)*csc(35)=tan(25)-cot(65), 2sin(30)+2tan(45)-3cos(60)-2cos^2(30)=0, (tan(x)+cot(x))(sec(x)-cos(x))(csc(x)-sin(x)), (tan^2(x)-cot^2(x))/(sin^2(x)-cos^2(x)), (tan^2(x)+cot^2(x))/(tan^2(x)-cot^2(x)), (tan^2(x)-cot^2(x))/(tan^2(x)+cot^2(x)), (3sin(x)+5cos(x))^2+(3cos(x)-5sin(x))^2, sin^2(x)cos(x)+sin^3(x)+cos^2(x)sin(x)+cos^3(x), (sin^4(x)-cos^4(x))/(sin^2(x)-cos^2(x)), If sin(x)=5/13 then find other trigonometry, If cos(x)=12/13 then find other trigonometry, If tan(x)=5/12 then find other trigonometry, If csc(x)=13/5 then find other trigonometry, If sec(x)=13/12 then find other trigonometry, If cot(x)=12/5 then find other trigonometry, If sin(x)=3/5 then solve cos(x)csc(x)+tan(x)sec(x), If tan(x)=1/2 then solve (1-tan^2(x))/(1+tan^2(x))+(2tan(x))/(1+tan^2(x)), If sin(x)=4/5 then solve (1-sin(x))/cos(x)+cos(x)/(1-sin(x)), If cos(x)=3/5 then solve (1-sin(x))/cos(x)+cos(x)/(1-sin(x)). In six trigonometric ratios sin, cos, tan, csc, sec and cot, if the value of one of the ratios is given, we can find the values of the other five functions. Online calculator. Please add atozmath.com to your ad blocking whitelist or disable your adblocking software. Step 2: Now click the button “Calculate Trigonometric Ratios” to get the result. Please try again using a different payment method. To evaluate trigonometric functions of other angles, we use a scientific or graphing calculator or computer software. By using this website, you agree to our Cookie Policy. By browsing this website, you agree\",\n \"to the left and T to the right, so we get: omega/(2*pi) = 1/T and since 1/T = f we finally get: f = omega / (2*pi) ### How do you simplify csc theta cot theta? There are 6 basic trig functions. sin(x) = 1/csc(x) cos(x) = 1/sec(x) tan(x) = sin(x)/cos(x) or 1/cot(x) csc(x) = 1/sin(x) sec(x) = 1/cos(x) cot(x) = cos(x)/sin(x) or 1/tan(x) ---- In your problem csc(x)*cot(x) we can simplify csc(x). csc(x) = 1/sin(x) Similarly, cot(x) = cos(x)/sin(x). csc(x)*cot(x) = (1/sin[x])*(cos[x]/sin[x]) = cos(x)/sin2(x) = cos(x) * 1/sin2(x) Either of the above answers should work. In general, try converting your trig functions into sine and cosine to make things simpler. ### How do you find sin cos and tan values manually? This is so much work that it is not worthwhile to do in practice, although the formulae themselves are actually quite simple. The basic method is to use a so-called \\\"infinite series\\\". The angle must be expressed in radians. If the angle is in degrees, multiply it by (pi/180), to get the equivalent angle in radians. Then, use the formula: sin(x) = x - x3/3! + x5/5! - x7/7! + x9/9!... The individual terms become smaller and smaller, quite quickly, so the idea is to continue adding more terms until you see that the terms become\",\n \"### Home > PC > Chapter 6 > Lesson 6.1.1 > Problem6-12 6-12. 1. Find the exact value of the following trig expressions. Homework Help ✎ 1. csc 2. tan 3. cot 4. sec 1. Plot the point on a unit circle. 2. Find the sin first. 3. Flip the answer for the csc. (Find the reciprocal) $\\\\text{Since the tan }x=\\\\frac{\\\\text{sin }x}{\\\\text{cos }x},$ what issues can occur for denominator values of a fraction? Plot the point finding the side of the reference angle. Find the tan first and take its reciprocal for the cot. Find the cos first and then its reciprocal for the sec.\",\n \"# Indefinite Integration of a trig function How do you integrate $$1/(\\\\sin2x(\\\\tan^5x+\\\\cot^5x))$$ with respect to $$x$$? I tried writing tan and cot in terms of sin and cos but when I take $$\\\\mathrm{LCM}$$ I get powers of $$10$$ for $$\\\\sin$$ and $$\\\\cos$$ in the denominator. I can't think of how I would simplify this without the use of binomial expansion but that might make things more complicated. Is there another method I could use to approach this problem? $$\\\\displaystyle \\\\int \\\\frac{dx}{\\\\sin2x\\\\cdot(tan^5x+cot^5x)} = \\\\frac12 \\\\int \\\\frac{\\\\cot x \\\\,dx}{\\\\cos^2x\\\\cdot(\\\\tan^5x+\\\\cot^5x)}$$ $$\\\\displaystyle \\\\tan x = t, \\\\frac{dx}{\\\\cos^2x}=dt$$\",\n \"Sec, Cosec, Cot Printable View • May 24th 2008, 03:14 AM jarald83 Sec, Cosec, Cot how to calculate the following value? Cosec 30° cot 60° sec 310° cosec 120° ° cot 600° and -cot 130° (Doh)(Crying) • May 24th 2008, 03:44 AM wingless $\\\\tan x = \\\\frac{\\\\sin x}{\\\\cos x}$ $\\\\cot x = \\\\frac{\\\\cos x}{\\\\sin x}$ $\\\\csc x = \\\\frac{1}{\\\\sin x}$ $\\\\sec x = \\\\frac{1}{\\\\cos x}$ • May 24th 2008, 03:58 AM nandu11 u need to know the simple sin, cos and tan trigo ratios for 0, 30, 45, 60 and 90 to do these types of questions and then use the formulas like cosecx=1/sinx, cotx=cosx/sinx etcc\",\n \"90o – θ ) = sin θ 17. tan ( 90o – θ ) = cot θ 18. cot ( 90o – θ ) = tan θ 19. sec ( 90o – θ ) = cosec θ 20. cosec ( 90o – θ ) = sec θ 21. sin 00 = 0, cos 00 = 1 and tan 00 = 0 22. cosec 00 = Not defined sec 00 = 1 and cot 00 = Not Defined 23. sin 300 = $\\\\frac{1}{2}$, cos 300 = $\\\\frac{\\\\sqrt{3}}{2}$ and tan 300 = $\\\\frac{1}{\\\\sqrt{3}}$ 24. cosec 300 = 2, sec 300 = $\\\\frac{2}{\\\\sqrt{3}}$ and cot 300 = $\\\\sqrt{3}$ 25. sin 450 = $\\\\frac{1}{\\\\sqrt{2}}$, cos450 = $\\\\frac{1}{\\\\sqrt{2}}$ and tan 450 = 1 26. cosec 450 = $\\\\sqrt{2}$, sec 450 = $\\\\sqrt{2}$, and cot 450 = 1 27. sin 600 =$\\\\frac{\\\\sqrt{3}}{2}$ , cos 600 = $\\\\frac{1}{2}$, and tan 600 =$\\\\sqrt{3}$ 28. cosec 600 = $\\\\frac{2}{\\\\sqrt{3}}$, sec 600 = 2 and cot 600 = $\\\\frac{1}{\\\\sqrt{3}}$ 29. sin 900 = 1, cos 900 = 0 and tan 00 = Not Defined 30. cosec 900 = 1sec 900 = Not Defined and cot 900 = 0\",\n \"sine and cos as their ratio. From the diagram taken above, the tan function will be the following. Tan a = $\\\\frac{Opposite}{Adjacent}$ = $\\\\frac{CB}{BA}$ Also, in terms of sine and cos, tan can be represented as: Tan a = $\\\\frac{sin\\\\ a}{cos\\\\ a}$ ### Secant, cosecant, and cotangent Secant, cosecant (csc) and cotangent are the three additional trigonometric functions which are derived from the primary functions of sine, cos, and tan. The reciprocal of sine, cos, and tan are cosecant (csc), secant (sec), and cotangent (cot) respectively. The formula of each of these functions are given as: Sec a = 1/(cos a) = $\\\\frac{Hypotenuse}{Adjacent}$ = $\\\\frac{CA}{AB}$ Cosec a = 1/(sin a) = $\\\\frac{Hypotenuse}{Opposite}$ = $\\\\frac{CA}{CB}$ cot a = 1/(tan a) = $\\\\frac{Adjacent}{Opposite}$ = $\\\\frac{BA}{CB}$ ### Trigonometric Functions Table The trigonometric table for six functions, Sin, Cos, Tan, Cosec, Sec, Cot, are as : Trigonometric Ratios/ angle= a degrees 0 ° 30 ° 45 ° 60 ° 90 ° Sin a 0 1/2 1/√2 √3/2 1 Cos a 1 √3/2 1/√2 1/2 0 Tan a 0 1/√3 1 √3 Not Defined Cosec a Not Defined 2 √2 2/√3 1 Sec a 1 2/√3 √2 2 Not Defined Cot a Not defined √3 1 1/√3 0 ### Inverse Trigonometric Functions Inverse functions are used to obtain an angle from any of\",\n \"# Math Help - Simplifying Trigonometric Equations 1. ## Simplifying Trigonometric Equations Hi all! I know this is probably a fairly simple concept for Trigonometry, but I can't quite remember how to do them. I'm just trying to make sure I know them well enough so that I can do them on a test I have in a few days. It would be awesome if you could just get me started on them. 1. $cot(x)sin(x) = \\\\frac{\\\\sqrt{3}}{2}$ 2. $tan(x)$ $cos(x) = 1$ 3. $sin^2(x)$ + $cos^2(x) + cos(x)$ = 1 4. [ $cot(x)$ $(\\\\frac{sin(x)}{cos(x)})$] • $sec^2(x)$ = $\\\\frac{1}{2}$ 2. Originally Posted by sar Hi all! I know this is probably a fairly simple concept for Trigonometry, but I can't quite remember how to do them. I'm just trying to make sure I know them well enough so that I can do them on a test I have in a few days. It would be awesome if you could just get me started on them. 1. $cot(x)sin(x) = \\\\frac{\\\\sqrt{3}}{2}$ It is normally best to write all of your sec(x), csc(x), tan(x), and cot(x) in terms of sin(x) and cos(x). That way, if anything cancels, you can simplify the expression. $\\\\frac{cos(x)}{sin(x)} • sin(x) = \\\\frac{\\\\sqrt{3}}{2}$ Hey, there's a sin(x) on the top as well as the bottom! So, as long\",\n \"anonymous 5 years ago The value of one trigonometric function is give 0 CALC > Chapter 4 > Lesson 4.5.1 > Problem4-164 4-164. 1. Rewrite the following using a single trigonometric function. You may wish to review your trigonometric identities in Chapter 1. Homework Help ✎ 1. 10 sin(3x) cos(3x) 2. sin x cos 3x − sin 3x cos x 3. cos4 x − sin4 x 4. tan x + cot x Double Angle identity (factor first). Factor first. Sum and Difference (Angle Sum) identity. Start by rewriting tan(x) and cot(x) as fractions. You will use more than one identity. $\\\\frac{\\\\text{sin}x}{\\\\text{cos}x}+\\\\frac{\\\\text{cos}x}{\\\\text{sin}x}=\\\\frac{\\\\text{sin}^{2}x+\\\\text{cos}^{2}x}{(\\\\text{cos}x)(\\\\text{sin}x)}=$ $\\\\frac{1}{\\\\frac{1}{2}\\\\text{sin}(2x)}=2\\\\text{csc}(2x)$\",\n \"Trigonometric ratios of 360 plus theta (360 + θ) In this section we will learn trigonometric ratios of 360 plus theta (360 + θ). We know that the terminal sides of co-terminal angles always coincide and θ and (360 + θ) are co-terminal angles. Clearly, 360 + θ and θ are co-terminal angles. sin(360 + θ) = Sin[360 -(- θ)] But we know that -sin(- θ) = sinθ $\\\\underline{sin(360 + θ)= sin θ}$ Again, cos(360 + θ) = cos[360-(- θ)] But we know that cos(- θ) =cosθ $\\\\underline{cos(360 + θ)= cos θ}$ Now, tan(360 + θ) = tan[360-(- θ)] But we know that -tan(- θ) = tanθ $\\\\underline{tan(360 + θ)= tan θ}$ As we know that, csc θ = $\\\\frac{1}{sinθ}$ ∴ csc (360+θ) = $\\\\frac{1}{sin(360+θ)}$ But sin(360+θ) = sin(θ) ∴ csc (360+θ)= $\\\\frac{1}{sin θ}$ $\\\\underline{csc(360 + θ)= csc θ}$ And, sec θ = $\\\\frac{1}{cosθ}$ ∴ sec (360+θ) = $\\\\frac{1}{cos(360+θ)}$ But cos(360+θ) = cos(θ) ∴ sec (360+θ)= $\\\\frac{1}{cos θ}$ $\\\\underline{sec(360 + θ)= cos θ}$ Again, cot θ = $\\\\frac{1}{tanθ}$ ∴ cot (360+θ) = $\\\\frac{1}{tan(360+θ)}$ But tan(360+θ) = tan(θ) ∴ cot (360+θ)= $\\\\frac{1}{ tan θ}$ $\\\\underline{cot(360 + θ)= cot θ}$ For any positive integer 'n', angle (360 x n + θ) is co-terminal to angle θ. Therefore, for any positive integer n, we have, sin(360 x n + θ )\"\n]"}}},{"rowIdx":6939,"cells":{"problem":{"kind":"string","value":"Compute $\\begin{pmatrix} 1 & -1 \\\\ 1 & 0 \\end{pmatrix}^3.$"},"level":{"kind":"string","value":"Level 2"},"type":{"kind":"string","value":"Precalculus"},"solution":{"kind":"string","value":"We have that\n\\begin{align*}\n\\begin{pmatrix} 1 & -1 \\\\ 1 & 0 \\end{pmatrix}^3 &= \\begin{pmatrix} 1 & -1 \\\\ 1 & 0 \\end{pmatrix} \\begin{pmatrix} 1 & -1 \\\\ 1 & 0 \\end{pmatrix} \\begin{pmatrix} 1 & -1 \\\\ 1 & 0 \\end{pmatrix} \\\\\n&= \\begin{pmatrix} 0 & -1 \\\\ 1 & -1 \\end{pmatrix} \\begin{pmatrix} 1 & -1 \\\\ 1 & 0 \\end{pmatrix} \\\\\n&= \\boxed{\\begin{pmatrix} -1 & 0 \\\\ 0 & -1 \\end{pmatrix}.}\n\\end{align*}"},"answer":{"kind":"string","value":"\\begin{pmatrix} -1 & 0 \\\\ 0 & -1 \\end{pmatrix}."},"p_retrievals":{"kind":"list like","value":["1 & 1 \\end{pmatrix}$","= \\begin{pmatrix} -1 \\\\\\\\ 1 \\end{pmatrix}$","4 \\\\ 3 \\end{pmatrix}.$ ×","\\\\ 0 & 1\\end{pmatrix}.$ -","1/2 \\\\ \\end{pmatrix}$$","\\\\0 \\\\ 1\\end{pmatrix}$$","$$\\begin{pmatrix} 1&1&1 \\\\1&0&0.5 \\\\ 0&1&0.5 \\end{pmatrix}$$","true. So $A^n =\\begin{pmatrix}a^n & 0\\\\0 & b^n\\end{pmatrix}, \\forall n \\in \\mathbb{N}$","& 1\\end{pmatrix} (b)$\\frac{1}{1-t^2}\\begin{pmatrix} -t^2 & t\\\\ -t &1\\end{pmatrix}(t\\in(1,\\infty))$","$$(1, 1) \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{12} & a_{22} \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix} = a_{11} + 2 a_{12} + a_{22}$$","6 & 0 & 5 \\end{pmatrix}$","& 1 \\\\ 1 & 0 \\end{pmatrix}$ $D = \\begin {pmatrix} -1 & 0 \\\\ 0 & 1 \\end {pmatrix}$ $P = \\begin {pmatrix} -1 & 1 \\\\ 1 & 1 \\end{pmatrix}$","\\\\ 0 \\\\ 0 \\\\ \\end{pmatrix} \\ \\ \\ \\$$","& 0 & 0 \\\\ 0 & 0 & 0 & 2 \\\\ \\end{pmatrix}\\}$$","\\leq \\begin{pmatrix}\\ldots\\\\k_i\\\\k_i\\\\\\ldots \\end{pmatrix}$$","\\end{pmatrix}$ $\\begin{pmatrix} 0 & 1\\\\ -1 & 0 \\end{pmatrix}$= $\\begin{pmatrix} 2 & 0\\\\ 0 & 1 \\end{pmatrix}$","In your particular case, $$\\quad F=\\begin{pmatrix} 0&\\ldots&0&l_{k+1,k} \\\\ 0&\\ldots&0&0 \\\\ \\vdots&&\\vdots&\\vdots \\\\ 0&\\ldots&0&0 \\\\ 0&\\ldots&0&l_{n,k} \\end{pmatrix}.$$ -","2 } \\\\ { 20 } & { 6 } \\\\ { 0 } & { 14 } \\end{pmatrix} \\$","0 & 1 & 0 & 1 & 1 \\end{pmatrix}.$$ -","3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$"],"string":"[\n \"1 & 1 \\\\end{pmatrix}$\",\n \"= \\\\begin{pmatrix} -1 \\\\\\\\\\\\\\\\ 1 \\\\end{pmatrix}$\",\n \"4 \\\\\\\\ 3 \\\\end{pmatrix}.$ ×\",\n \"\\\\\\\\ 0 & 1\\\\end{pmatrix}.$ -\",\n \"1/2 \\\\\\\\ \\\\end{pmatrix}$$\",\n \"\\\\\\\\0 \\\\\\\\ 1\\\\end{pmatrix}$$\",\n \"$$\\\\begin{pmatrix} 1&1&1 \\\\\\\\1&0&0.5 \\\\\\\\ 0&1&0.5 \\\\end{pmatrix}$$\",\n \"true. So $A^n =\\\\begin{pmatrix}a^n & 0\\\\\\\\0 & b^n\\\\end{pmatrix}, \\\\forall n \\\\in \\\\mathbb{N}$\",\n \"& 1\\\\end{pmatrix} (b)$\\\\frac{1}{1-t^2}\\\\begin{pmatrix} -t^2 & t\\\\\\\\ -t &1\\\\end{pmatrix}(t\\\\in(1,\\\\infty))$\",\n \"$$(1, 1) \\\\begin{pmatrix} a_{11} & a_{12} \\\\\\\\ a_{12} & a_{22} \\\\end{pmatrix} \\\\begin{pmatrix} 1 \\\\\\\\ 1 \\\\end{pmatrix} = a_{11} + 2 a_{12} + a_{22}$$\",\n \"6 & 0 & 5 \\\\end{pmatrix}$\",\n \"& 1 \\\\\\\\ 1 & 0 \\\\end{pmatrix}$ $D = \\\\begin {pmatrix} -1 & 0 \\\\\\\\ 0 & 1 \\\\end {pmatrix}$ $P = \\\\begin {pmatrix} -1 & 1 \\\\\\\\ 1 & 1 \\\\end{pmatrix}$\",\n \"\\\\\\\\ 0 \\\\\\\\ 0 \\\\\\\\ \\\\end{pmatrix} \\\\ \\\\ \\\\ \\\\$$\",\n \"& 0 & 0 \\\\\\\\ 0 & 0 & 0 & 2 \\\\\\\\ \\\\end{pmatrix}\\\\}$$\",\n \"\\\\leq \\\\begin{pmatrix}\\\\ldots\\\\\\\\k_i\\\\\\\\k_i\\\\\\\\\\\\ldots \\\\end{pmatrix}$$\",\n \"\\\\end{pmatrix}$ $\\\\begin{pmatrix} 0 & 1\\\\\\\\ -1 & 0 \\\\end{pmatrix}$= $\\\\begin{pmatrix} 2 & 0\\\\\\\\ 0 & 1 \\\\end{pmatrix}$\",\n \"In your particular case, $$\\\\quad F=\\\\begin{pmatrix} 0&\\\\ldots&0&l_{k+1,k} \\\\\\\\ 0&\\\\ldots&0&0 \\\\\\\\ \\\\vdots&&\\\\vdots&\\\\vdots \\\\\\\\ 0&\\\\ldots&0&0 \\\\\\\\ 0&\\\\ldots&0&l_{n,k} \\\\end{pmatrix}.$$ -\",\n \"2 } \\\\\\\\ { 20 } & { 6 } \\\\\\\\ { 0 } & { 14 } \\\\end{pmatrix} \\\\$\",\n \"0 & 1 & 0 & 1 & 1 \\\\end{pmatrix}.$$ -\",\n \"3 \\\\end{pmatrix} = \\\\begin{pmatrix}1 & 2 & 3\\\\\\\\ 2 & 4 & 6 \\\\\\\\ 3 & 6 & 9 \\\\end{pmatrix}$$\"\n]"},"s_retrievals":{"kind":"list like","value":["by clicking on the problem.) \\begin{align*} 1) \\hspace{.5cm} \\begin{pmatrix} 3 & 1 \\\\ 2 & 1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\begin{pmatrix} 1 & −1 \\\\ −2 & 3 \\end{pmatrix}\\end{align*} \\begin{align*} 2) \\hspace{.5cm} \\begin{pmatrix} 2 & 1 \\\\ 4 & 3 \\end{pmatrix} \\end{align*} Solution: \\begin{align*}\\begin{pmatrix} \\frac{3}{2} & −\\frac{1}{2} \\\\ −2 & 1 \\end{pmatrix} \\end{align*} \\begin{align*} 3) \\hspace{.5cm} \\begin{pmatrix} 0 & 1 \\\\ 5 & 7 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\begin{pmatrix} −\\frac{7}{5} & \\frac{1}{5} \\\\ 1 & 0 \\end{pmatrix}\\end{align*} \\begin{align*} 4) \\hspace{.5cm} \\begin{pmatrix} 2 & 1 \\\\ 6 & 3 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\text{not invertible} \\end{align*} \\begin{align*} 5) \\hspace{.5cm} \\begin{pmatrix} 1 & −2 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & −1 & 1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 1 & 1 \\end{pmatrix}\\end{align*} \\begin{align*} 6) \\hspace{.5cm} \\begin{pmatrix} 1 & 0 & 1 \\\\ 2 & 3 & 4 \\\\ −1 & 1 & 0 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\begin{pmatrix} −4 & 1 & −3 \\\\ −4 & 1 & −2 \\\\ 5 & −1 & 3 \\end{pmatrix}\\end{align*} \\begin{align*} 7) \\hspace{.5cm} \\begin{pmatrix} 2 & 1 & 1 \\\\ 1 & 1 & 7 \\\\ 5 & 2 & −4 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\text{not invertible} \\end{align*} \\begin{align*} 8) \\hspace{.5cm} \\begin{pmatrix} 3 & 2","be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}","k & \\ell\\end{pmatrix}\\\\ & = \\begin{pmatrix}(ae+bg)i+(af+bh)k & (ae+bg)j+(af+bh)\\ell \\\\ (ce+dg)i+(cf+dh)k & (ce+dg)j+(cf+dh)\\ell\\end{pmatrix}\\\\ & = \\begin{pmatrix}aei+bgi+afk+bhk & aej+bgj+af\\ell+bh\\ell \\\\ cei+dgi+cfk+dhk & cej+dgj+cf\\ell+dh\\ell\\end{pmatrix}\\end{align*} \\begin{align*}A\\cdot (B\\cdot C)&=\\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}\\cdot\\left ( \\begin{pmatrix}e & f \\\\ g & h\\end{pmatrix}\\cdot \\begin{pmatrix}i & j \\\\ k & \\ell\\end{pmatrix}\\right )= \\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}\\cdot\\begin{pmatrix}ei+fk & ej+f\\ell \\\\ gi+hk & gj+h\\ell\\end{pmatrix} \\\\ & = \\begin{pmatrix}a(ei+fk)+b(gi+hk) & a(ej+f\\ell)+b(gj+h\\ell) \\\\ c(ei+fk)+d(gi+hk) & c(ej+f\\ell)+d(gj+h\\ell)\\end{pmatrix} \\\\ & = \\begin{pmatrix}aei+afk+bgi+bhk & aej+af\\ell+bgj+bh\\ell \\\\ cei+cfk+dgi+dhk & cej+cf\\ell+dgj+dh\\ell\\end{pmatrix}\\\\ & = \\begin{pmatrix}aei+bgi+afk+bhk & aej+bgj+af\\ell+bh\\ell \\\\ cei+dgi+cfk+dhk & cej+dgj+cf\\ell+dh\\ell\\end{pmatrix}\\end{align*} The results are the same. Therefore it holds that $(A\\cdot B)\\cdot C=A\\cdot (B\\cdot C)$ which means the multiplication over $M_2(\\mathbb{R})$ is associative. 2. Let $A=\\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}, \\ B=\\begin{pmatrix}e & f \\\\ g & h\\end{pmatrix}$. Then we have the following: \\begin{equation*}A\\cdot B=\\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}\\cdot \\begin{pmatrix}e & f \\\\ g & h\\end{pmatrix}=\\begin{pmatrix}ae+bg & af+bh \\\\ ce+dg & cf+dh\\end{pmatrix} \\end{equation*} Then we have the following: \\begin{equation*}B\\cdot A=\\begin{pmatrix}e & f \\\\ g & h\\end{pmatrix}\\cdot \\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}=\\begin{pmatrix}ea+fc & eb+fd \\\\ ga+hc & gb+hd\\end{pmatrix} \\end{equation*} We see that $A\\cdot B\\neq B\\cdot A$, which means that the multiplication over $M_2(\\mathbb{R})$ is not commutative. 3. The neutral element in respect of the multiplication over $M_2(\\mathbb{R})$ is the identity matrix \\begin{equation*}I_2=\\begin{pmatrix}1 & 0 \\\\ 0","&=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&-7\\\\ 0&0&0&7\\end{pmatrix}\\tag{R_3=R_3-R_1}\\\\\\\\ &=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&1\\\\ 0&0&0&7\\tag{R_2=-\\frac17 R_2}\\end{pmatrix}\\\\\\\\ &=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&1\\\\ 0&0&0&0\\tag{R_3=R_3 -7R_2}\\end{pmatrix} \\end{align}","the observer). \\begin{aligned} R_x(\\theta)&=\\begin{pmatrix}1 &0 &0\\\\ 0 &\\cos\\theta &-\\sin\\theta\\\\ 0 &\\sin\\theta &\\cos\\theta \\end{pmatrix}\\\\ R_y(\\theta)&=\\begin{pmatrix}\\cos\\theta &0 &\\sin\\theta\\\\ 0 &1 &0\\\\ -\\sin\\theta &0 &\\cos\\theta \\end{pmatrix}\\\\ R_z(\\theta)&=\\begin{pmatrix} \\cos\\theta &-\\sin\\theta &0\\\\ \\sin\\theta &\\cos\\theta &0\\\\ 0 &0 &1 \\end{pmatrix} \\end{aligned} ## Example 1 Rotating $\\begin{pmatrix}1\\\\ 0\\\\ 0\\end{pmatrix}$ $45^\\circ$ anticlockwise about $z$-axis: \\begin{aligned} R_z(45^\\circ)\\begin{pmatrix}1\\\\ 0\\\\ 0\\end{pmatrix}&=\\begin{pmatrix} \\cos 45^\\circ &-\\sin 45^\\circ &0\\\\ \\sin 45^\\circ &\\cos 45^\\circ &0\\\\ 0 &0 &1 \\end{pmatrix} \\begin{pmatrix}1\\\\ 0\\\\ 0\\end{pmatrix}\\\\ &=\\begin{pmatrix}\\sqrt{2}/2\\\\\\sqrt{2}/2\\\\0\\end{pmatrix}. \\end{aligned} ## Example 2 Rotating $\\begin{pmatrix}0\\\\ 1\\\\ 0\\end{pmatrix}$ $45^\\circ$ anticlockwise about $z$-axis: \\begin{aligned} R_z(45^\\circ)\\begin{pmatrix}0\\\\ 1\\\\ 0\\end{pmatrix}&=\\begin{pmatrix} \\cos 45^\\circ &-\\sin 45^\\circ &0\\\\ \\sin 45^\\circ &\\cos 45^\\circ &0\\\\ 0 &0 &1 \\end{pmatrix} \\begin{pmatrix}0\\\\ 1\\\\ 0\\end{pmatrix}\\\\ &=\\begin{pmatrix}-\\sqrt{2}/2\\\\\\sqrt{2}/2\\\\0\\end{pmatrix}. \\end{aligned} Posted in Uncategorized | 1 Comment ## 十二平均律 12-tone Equal Temperament 《十二平均律》 是明朝 "布衣王子" 朱载堉 (1536年-1610年) 发明的,由当时在中国传教的意大利人 利马窦 (Matteo Ricci, 传给欧洲的法国数学家 Marin Mersenne (Mersenne Prime ) 。现代音乐之父 巴哈 Bach 第一个采用,制作世界第一架钢琴有12黑白键,并作曲 《Bach 12-tone Equal Temperament》。 Side Note: 1977年法国大学数学教授在课堂好奇地问我,你们祖先如何解代数?是用算盘吗?当时计算机还不流行,复杂的算法只能用Log Table 或 计算尺 (Slide Rule) 。 Bach 12-tone Equal Temperament https://zh.m.wikipedia.org/wiki/%E5%8D%81%E4%BA%8C%E5%B9%B3%E5%9D%87%E5%BE%8B View original post ## Richard Dedekind Julius Wilhelmina Richard Dedekind (6 Oct 1831 – 12 Feb 1916) – Last student of Gauss at Göttingen – Student and closed friend of Dirichlet who influenced his Mathematical education – Introduced the word Field (Körper) – Gave the first university course on Galois Theory – Developed Real Number ‘Dedekind Cut‘ in 1872 – Accomplished musician – Never married, lived with","to align the \\Rightarrows – daleif Oct 6 '16 at 7:59 This is an automatic solution, based on the eqparbox package. I defined an \\eqmathbox command, which accepts $2$ optional arguments (the end of name of the box, and the alignment: l, r or c) and one mandatory argument: the contents of the math box. \\documentclass{article} %\\usepackage{enumitem} \\usepackage{mathtools} \\usepackage{xparse} \\usepackage{eqparbox} \\NewDocumentCommand\\Eqmathbox{O{0}O{l}m}{\\eqmakebox[Form#1][#2]{\\ensuremath{ \\displaystyle#3}}} \\usepackage{siunitx} \\sisetup{group-separator={,},group-minimum-digits = 4} \\begin{document} \\begin{enumerate} \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 20 & 25 & 20 \\\\ 15 & 20 & 18 \\\\ 18 & 23 & 22 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 975 \\\\ 810 \\\\ 965 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 10 \\\\ 15 \\\\ 20 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 90 & 95 & 80 \\\\ 80 & 85 & 90 \\\\ 75 & 90 & 95 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 86 \\\\ 86 \\\\ 88 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 0.3 \\\\ 0.2 \\\\ 0.5 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] \\num{5000} & \\num{400}0 \\\\ \\num{6000} & \\num{5000} \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} = \\begin{pmatrix*}[r] \\num{37000000} \\\\ \\num{45000000 }\\\\ \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} =} \\begin{pmatrix*}[r] \\num{5000} \\\\ \\num{3000} \\end{pmatrix*}$$ \\end{enumerate} \\end{document} Here is one way. Note that you are adding a lot of unneeded","\\end{pmatrix} \\[6px] &\\hspace{2px}\\begin{pmatrix} 2 & 3 & 4 \\end{pmatrix} \\end{align*} would do the job. For some strange reason the align gave errors when leaving out the first \"&\" symbol and it gave a 2px offset. I figured you wanted some space between the two if not leave the [6px]. You can always use \\hspace{amount of whitespace} to place your second matrix in the place you want. This can be given in pt's, px's (which i did) etc. //edit Hm I notice the \\hspace{} is actually not needed, but can be used in case of pmatrix. What happens is that the pmatrix brackets give a biased image of the matrices. When using vmatrix like: \\begin{align*} &\\begin{vmatrix} 1 & 2 & 3 \\\\ 3 & 4 & 5 \\end{vmatrix} \\\\[6px] &\\begin{vmatrix} 2 & 3 & 4 \\end{vmatrix} \\end{align*} It all goes well :). So basically, probably the easiest way to fix it is either use other brackets to make it look good or use the \\hspace to align as you like it. - If all else fails, PGF/TikZ can do this. See this example. - Wrap the thing in \\begin{align*} ... \\end{align*} and use & as the alignment marker in your formulas. Example: \\begin{align*} \\begin{pmatrix} ... vector here \\end{pmatrix} &\\begin{pmatrix} ... first matrix here \\end{pmatrix}\\\\ &\\begin{pmatrix} ... second matrix here","\\usepackage{amsmath} % for 'align*' and 'pmatrix' environments \\begin{document} \\begin{align*} H &= \\begin{pmatrix} 1 & x_{11} & \\dots & x_{1p} \\\\ 1 & x_{21} & \\dots & x_{2p} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ 1 & x_{n1} & \\dots & x_{np} \\end{pmatrix}%_{\\!\\!\\!n\\times(p+1)} \\begin{pmatrix} q_{11} & q_{12} & \\dots & q_{1,p+1} \\\\ q_{21} & q_{22} & \\dots & q_{2,p+1} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ q_{p+1,1} & q_{p+1,2} & \\dots & q_{p+1,p+1} \\end{pmatrix}%_{\\!\\!\\!(p+1)\\times(p+1)} \\begin{pmatrix} 1 & 1 & \\dots & 1 \\\\ x_{11} & x_{21} & \\dots & x_{n1} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ x_{1p} & x_{2p} & \\dots & x_{np} \\end{pmatrix}%_{\\!\\!\\!(p+1)\\times n} \\\\ &= \\begin{pmatrix} x_{1}{}'q_{1} & x_{1}{}'q_{2} & \\dots & x_{1}{}'q_{p+1} \\\\ x_{2}{}'q_{1} & x_{2}{}'q_{2} & \\dots & x_{2}{}'q_{p+1} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ x_{n}{}'q_{1} & x_{n}{}'q_{2} & \\dots & x_{n}{}'q_{p+1} \\end{pmatrix} \\begin{pmatrix} 1 & 1 & \\dots & 1 \\\\ x_{11} & x_{21} & \\dots & x_{n1} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ x_{1p} & x_{2p} & \\dots & x_{np} \\end{pmatrix} \\end{align*} \\end{document} • This positive vote was necessary :-) – Sebastiano Nov 23 '19 at 20:30 • @Sebastiano - Thanks so much! :-) – Mico Nov 23 '19 at 20:34 • @Sebastiano - In fact, your upvote raised my","\\end{pmatrix}F \\begin{pmatrix} u_r \\\\ v \\\\ 1 \\end{pmatrix}=0 ,\\end{align} and $$u_l$$ and $$u_r$$ become whatever the depth of the point $$P$$, so it must be \\begin{align} \\begin{pmatrix} u_l & v & 1 \\end{pmatrix} \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & s \\\\ 0 & -s & 0 \\end{pmatrix} \\begin{pmatrix} u_r \\\\ v \\\\ 1 \\end{pmatrix}=0 .\\end{align} That is, for $$\\tilde{\\ddot{m_l}}$$ and $$\\tilde{\\ddot{m_l}}$$, its base matrix $$f$$ is \\begin{align} f = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & s \\\\ 0 & -s & 0 \\end{pmatrix} .\\end{align} When this is returned to the original $$\\tilde{m_l}$$ and $$\\tilde{m_r}$$, it is \\begin{align} (A_rLRA_l^{-1}\\tilde{m_l})^T \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & s \\\\ 0 & -s & 0 \\end{pmatrix} A_rLA_r^{-1}\\tilde{m_r}= (\\tilde{m_l})^T(A_l^{-1})^TR^TL^TA_r^T \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & s \\\\ 0 & -s & 0 \\end{pmatrix} A_rLA_r^{-1}\\tilde{m_r}=0 ,\\end{align} so its base matrix $$F$$ is \\begin{align} F = (A_l^{-1})^TR^TL^TA_r^T \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & s \\\\ 0 & -s & 0 \\end{pmatrix} A_rLA_r^{-1} \\end{align} or \\begin{align} F = (A_l^{-1})^TR^TL^TA_r^TfA_rLA_r^{-1} \\end{align} for $$\\tilde{m_l}$$ and $$\\tilde{m_r}$$. ### Program distribution Then, the program which calculates a basic matrix by stereo processing is explained. graph_cut.cu , rectify.cpp , makefile , 2_R.png , 2_L.png Please download.","e_{12} \\\\ e_{13} \\end{pmatrix}, \\vec{e_2}=\\begin{pmatrix} e_{21} \\\\ e_{22} \\\\ e_{23} \\end{pmatrix}, \\vec{e_3}=\\begin{pmatrix} e_{31} \\\\ e_{32} \\\\ e_{33} \\end{pmatrix} \\end{align*} is an orthonormal vector, it is \\begin{align*} \\begin{pmatrix} (\\vec{e_1})^T \\\\ (\\vec{e_2})^T \\\\ (\\vec{e_3})^T \\end{pmatrix} \\begin{pmatrix} \\vec{e_1} & \\vec{e_2} & \\vec{e_3} \\end{pmatrix} =\\begin{pmatrix} e_{11} & e_{12} & e_{13} \\\\ e_{21} & e_{22} & e_{23} \\\\ e_{31} & e_{32} & e_{33} \\end{pmatrix} \\begin{pmatrix} e_{11} & e_{21} & e_{31} \\\\ e_{12} & e_{22} & e_{32} \\\\ e_{13} & e_{23} & e_{33} \\end{pmatrix} =\\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} .\\end{align*} That is, \\begin{align*} R=\\begin{pmatrix} e_{11} & e_{12} & e_{13} \\\\ e_{21} & e_{22} & e_{23} \\\\ e_{31} & e_{32} & e_{33} \\end{pmatrix} =(R^{-1})^T \\end{align*}\\begin{align*} R^{-1}=\\begin{pmatrix} e_{11} & e_{21} & e_{31} \\\\ e_{12} & e_{22} & e_{32} \\\\ e_{13} & e_{23} & e_{33} \\end{pmatrix} =R^T .\\end{align*} Camera and perspective transformation Consider a pinhole camera in which light from an object always passes through the pinhole. Align the origin $$o$$ of the 3D coordinate $$oxyz$$ with this pinhole. Next, consider the plane that intersects with the $$oz$$ axis at $$f$$ point $$(0, 0, f)$$ and calls it the imaging surface. However, $$f$$ is positive and the imaging surface is between the pinhole and the object. $$f$$ is also called the focus. In","\\end{pmatrix} \\end{align*} \\begin{align*} 3) \\hspace{.5cm} \\begin{pmatrix} 2 & 0 \\\\ 4 & 3 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} & \\mbox{answers may vary; one correct answer is} \\\\ & \\begin{pmatrix} 1 & 0 \\\\ 2 & 1 \\end{pmatrix} \\begin{pmatrix} 2 & 0 \\\\ 0 & 3 \\end{pmatrix} \\end{align*} \\begin{align*} 4) \\hspace{.5cm} \\begin{pmatrix} 2 & 1 \\\\ 6 & 4 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} & \\mbox{answers may vary; one correct answer is} \\\\ & \\begin{pmatrix} 1 & 0 \\\\ 3 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 2 & 0 \\\\ 0 & 1 \\end{pmatrix} \\end{align*} \\begin{align*} 5) \\hspace{.5cm} \\begin{pmatrix} 4 & 4 \\\\ 6 & 3 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} & \\mbox{answers may vary; one correct answer is} \\\\ & \\begin{pmatrix} 1 & 0 \\\\ \\frac{3}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 1 & −\\frac{4}{3} \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 0 \\\\ 0 & −3 \\end{pmatrix} \\end{align*} \\begin{align*} 6) \\hspace{.5cm} \\begin{pmatrix} 4 & 2 \\\\ 2 & 0 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} & \\mbox{answers may vary; one correct answer is} \\\\ & \\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 1 & −2 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 0 \\\\ 0 & −1 \\end{pmatrix} \\end{align*} \\begin{align*} 7) \\hspace{.5cm} \\begin{pmatrix} 1 & 0 & 0 \\\\ 4 & 1","& 1 \\\\ 0 & 2 & 4 \\\\ −1 & 1 & −1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\frac{1}{24} \\begin{pmatrix} 6 & −3 & −6 \\\\ 4 & 2 & 12 \\\\ −2 & 5 & −6 \\end{pmatrix} \\end{align*} \\begin{align*} 9) \\hspace{.5cm} \\begin{pmatrix} 1 & 2 & 1 & −2 \\\\ 1 & 1 & 0 & 0 \\\\ 0 & 1 & 0 & −1 \\\\ 2 & 1 & 0 & 2 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\begin{pmatrix} 0 & 3 & −2 & −1 \\\\ 0 & −2 & 2 & 1 \\\\ 1 & −3 & 0 & 1 \\\\ 0 & −2 & 1 & 1 \\end{pmatrix} \\end{align*} \\begin{align*} 10) \\hspace{.5cm} \\begin{pmatrix} 2 & 3 & 1 & 0 \\\\ −1 & 0 & 3 & 1 \\\\ 2 & 2 & 4 & 1 \\\\ 2 & −1 & 2 & −1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\frac{1}{25} \\begin{pmatrix} −11 & −17 & 16 & −1 \\\\ 14 & 8 & −9 & −1 \\\\ 5 & 10 & −5 & 5 \\\\ −26 & −22 & 31 & −16 \\end{pmatrix}\\end{align*} For each equation $Ax=b$, tell whether $A^{-1}$ exists. If it does, then compute the solution $x=A^{-1}b$. (You can view the solution by clicking on the problem.) \\begin{align*} 11) \\hspace{.5cm} \\begin{pmatrix} 1 & 4","2 & -2.2 + 2 \\\\ 0 & 0 & 1 \\end{pmatrix} . \\begin{pmatrix} 2 & 4 & 4 & 2\\\\ 1 & 1 & 3 & 3\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\\\\\\\ &= \\begin{pmatrix} 2 & 0 & -3 \\\\ 0 & 2 & -2 \\\\ 0 & 0 & 1 \\end{pmatrix} . \\begin{pmatrix} 2 & 4 & 4 & 2\\\\ 1 & 1 & 3 & 3\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\\\\\\\ \\begin{pmatrix} x_{1}^{\\prime} & x_{2}^{\\prime} & x_{3}^{\\prime} & x_{4}^{\\prime}\\\\ y_{1}^{\\prime} & y_{2}^{\\prime} & y_{3}^{\\prime} & y_{4}^{\\prime}\\\\ 1 & 1 & 1 & 1 \\end{pmatrix} &= \\begin{pmatrix} 1 & 5 & 5 & 1\\\\ 0 & 0 & 4 & 4\\\\ 1 & 1 & 1 & 1 \\end{pmatrix} \\end{aligned} ### Combination use-case: rotate image Now imagine you have an image in a view, the origin is not a the center of the view, it is probably at the top-left corner (implementations may vary), but you want to rotate the image at the center of the view :upside_down: This transformation is composed of the following sequence: • move anchor point to origin: $\\left( -t_{x}, -t_{y} \\right)$ • rotate by $\\theta$ • move back anchor point: $\\left( t_{x}, t_{y} \\right)$ Where $t$ is the anchor point of our rotation transformation. Our","-1 & 0 \\\\ 0 & 0 & 0 & -1 \\end{pmatrix} \\\\ \\quad + \\sum_{n=3}^{\\infty} \\sum_{i=0}^{n} \\frac{1}{i! (n-i)!} \\begin{pmatrix} 0 & \\rho_x & \\rho_y & \\rho_z \\\\ \\rho_x & 0 & -\\theta_z & \\theta_y \\\\ \\rho_y & \\theta_z & 0 & -\\theta_x \\\\ \\rho_z & -\\theta_y & \\theta_x & 0 \\end{pmatrix}^i \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -1 \\end{pmatrix} \\begin{pmatrix} 0 & \\rho_x & \\rho_y & \\rho_z \\\\ \\rho_x & 0 & \\theta_z & - \\theta_y \\\\ \\rho_y & -\\theta_z & 0 & \\theta_x \\\\ \\rho_z & \\theta_y & -\\theta_x & 0 \\end{pmatrix}^{n-i} \\\\ = \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -1 \\end{pmatrix}$ So we have: $c^2 \\, \\Delta t'_{AB} \\, \\Delta t'_{CD} \\; - \\; \\vec{x}'_{AB} \\, \\cdot \\, \\vec{x}'_{CD} \\\\ = \\left( \\Lambda \\; \\begin{pmatrix} c \\Delta t_{AB} \\\\ \\Delta x_{AB} \\\\ \\Delta y_{AB} \\\\ \\Delta z_{AB} \\end{pmatrix} \\right) ^T \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\","this, let $$A^{-1}\\equiv \\begin{pmatrix} a & b \\\\ c & d \\\\ \\end{pmatrix} , \\tag{10} \\label{10}$$ whereby \\begin{align} \\varepsilon (A^{-1})^T \\varepsilon^{-1} &= \\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} a & b \\\\ c & d \\\\ \\end{pmatrix} \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\\\ &= \\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} b & -a \\\\ d & -c \\\\ \\end{pmatrix} \\\\ &= \\begin{pmatrix} d & -c \\\\ -b & a \\\\ \\end{pmatrix}, \\tag{11} \\label{11} \\end{align} while \\begin{align} \\varepsilon^{-1} (A^{-1})^T \\varepsilon &= \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} a & b \\\\ c & d \\\\ \\end{pmatrix} \\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\\\ \\end{pmatrix} \\\\ &= \\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} -b & a \\\\ -d & c \\\\ \\end{pmatrix} \\\\ &= \\begin{pmatrix} d & -c \\\\ -b & a \\\\ \\end{pmatrix} = \\varepsilon (A^{-1})^T \\varepsilon^{-1}. \\tag{12} \\label{12} \\end{align} And yes, this does imply that $$\\varepsilon^T = \\varepsilon^{-1}. \\tag{13} \\label{13}$$ As to the indices, the wikipedia article on the transpose of a linear map states that if $$f : V \\to W$$ is a linear map, then the transpose of $$f$$ is defined to be $${}^t f : W^* \\to V^*","1 \\\\ \\end{pmatrix} & q_1&\\colon \\mathrm{ss\\_fff} \\\\ ^0T_2 &= \\begin{pmatrix} 333 & & & q_3 \\\\ & 333333 & & \\\\ & & 3333 & \\\\ & & & 1 \\\\ \\end{pmatrix} &q_2&\\colon \\mathrm{tt\\_bbb} \\end{align} \\end{document} • Yes that works, thank you! – Fee Jun 24 '19 at 15:10 The alignat environment enables you to control the spacing between the two columns of the environment: \\documentclass{report} \\usepackage{amsmath} \\begin{document} \\begin{alignat}{2} ^0T_1 &= \\begin{pmatrix} 1 & & & q_1 \\\\ & 1 & & \\\\ & & 1 & \\\\ & & & 1 \\\\ \\end{pmatrix} & q_1&\\colon \\mathrm{ss\\_fff} \\\\[2ex] ^0T_2 &= \\begin{pmatrix} 333 & & & q_3 \\\\ & 333333 & & \\\\ & & 3333 & \\\\ & & & 1 \\\\ \\end{pmatrix} &\\quad q_2&\\colon \\mathrm{tt\\_bbb} \\end{alignat} \\end{document}","b \\\\ c & d \\\\ \\end{pmatrix} , \\tag{10} \\label{10}$$ whereby \\begin{align} \\varepsilon (A^{-1})^T \\varepsilon^{-1} &= \\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} a & b \\\\ c & d \\\\ \\end{pmatrix} \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\\\ &= \\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} b & -a \\\\ d & -c \\\\ \\end{pmatrix} \\\\ &= \\begin{pmatrix} d & -c \\\\ -b & a \\\\ \\end{pmatrix}, \\tag{11} \\label{11} \\end{align} while \\begin{align} \\varepsilon^{-1} (A^{-1})^T \\varepsilon &= \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} a & b \\\\ c & d \\\\ \\end{pmatrix} \\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\\\ \\end{pmatrix} \\\\ &= \\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} -b & a \\\\ -d & c \\\\ \\end{pmatrix} \\\\ &= \\begin{pmatrix} d & -c \\\\ -b & a \\\\ \\end{pmatrix} = \\varepsilon (A^{-1})^T \\varepsilon^{-1}. \\tag{12} \\label{12} \\end{align} And yes, this does imply that $$\\varepsilon^T = \\varepsilon^{-1}. \\tag{13} \\label{13}$$ As to the indices, the wikipedia article on the transpose of a linear map states that if $$f : V \\to W$$ is a linear map, then the transpose of $$f$$ is defined to be $${}^t f : W^* \\to V^* , \\tag{14} \\label{14}$$ where $$V^*, W^*$$","that is a lot of work. I think I will just stick with $\\displaystyle \\text{diam}\\left(E\\right)$ Thanks though! 8. $\\displaystyle \\begin{pmatrix} a_{1,1} & a_{2,1} & \\ldots & a_{n,1}\\\\ a_{2,1} & a_{2,2} & \\ldots & a_{n,2}\\\\ \\hdotsfor[1.7]{4}\\\\ a_{n,1} & a_{2,n} & \\ldots & a_{n,n} \\end{pmatrix}$ $\\displaystyle \\begin{pmatrix} a_{1,1} & a_{2,1} & \\ldots & a_{n,1}\\\\ a_{2,1} & a_{2,2} & \\ldots & a_{n,2}\\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a_{n,1} & a_{2,n} & \\ldots & a_{n,n} \\end{pmatrix}$ 9. $\\displaystyle \\begin{matrix}0&1\\\\1&0\\end{matrix} \\hspace{1.5pc} \\begin{pmatrix}0&1\\\\1&0\\end{pmatrix} \\hspace{1.5pc} \\begin{bmatrix}0&1\\\\1&0\\end{bmatrix} \\hspace{1.5pc} \\begin{Bmatrix}0&1\\\\1&0\\end{Bmatrix} \\hspace{1.5pc} \\begin{vmatrix}0&1\\\\1&0\\end{vmatrix} \\hspace{1.5pc} \\begin{Vmatrix}0&1\\\\1&0\\end{Vmatrix}$ 10. $\\displaystyle \\sum_{k=0}^n$ $\\displaystyle \\sum_{k=0}^n a^k$ $\\displaystyle \\sum_{r=0}^n a^r$ $\\displaystyle \\sum_{r=0}^n\\ ^nC_r a^r$ $\\displaystyle \\sum_{r=0}^n ^nC_r a^rb^{n-r}$ $\\displaystyle (x+y)^n$ $\\displaystyle $$x+y$$^n$ 11. $\\displaystyle -e^{-x^2}$ equals 0 12. $\\displaystyle \\int\\limits_-\\cdots \\int\\limits_\\text{ }^-\\cdots\\int\\limits_a^b$ Math $\\displaystyle \\int\\limits_- f~d\\alpha=\\sup_{P}L\\left(P,f,\\alpha\\right)$ 13. . . . $\\displaystyle \\frac{1{\\color{red}\\rlap{/}}6}{{\\color{red}\\rlap{/}}64} \\;=\\;\\frac{1}{4} \\qquad\\qquad \\frac{1{\\color{red}\\rlap{/}}9}{{\\color{red}\\rlap{/}}95} \\;=\\;\\frac{1}{5}$ . . . . . $\\displaystyle \\frac{2{\\color{red}\\rlap{/}}6}{{\\color{red}\\rlap{/}}65} \\;=\\;\\frac{2}{5}\\qquad\\qquad \\frac{4{\\color{red}\\rlap{/}}9}{{\\color{red}\\rlap{/}}98} \\;=\\;\\frac{4}{8}\\;=\\;\\frac{1}{2}$ . . . . . $\\displaystyle \\frac{1-x^{{\\color{red}\\rlap{/}}2}}{(1+x)^{{\\color{red}\\rlap{/}} 2} }\\;=\\;\\frac{1-x}{1+x}$ All that work for a bad joke . . . (Good luck trying to read my LaTeX code.) . 14. A fairly comprehensive guide to LaTeX here: http://tobi.oetiker.ch/lshort/lshort.pdf 15. Just trying something fancy in LaTex. This came from that tutorial. $\\displaystyle \\text{Heron's formula}\\setlength{\\unitlength}{1cm} \\begin{picture}(6,5) \\thicklines \\put(1,0.5){\\line(2,1){3}} \\put(4,2){\\line(-2,1){2}} \\put(2,3){\\line(-2,-5){1}} \\put(0.7,0.3){$A$} \\put(4.05,1.9){$B$} \\put(1.7,2.95){$C$} \\put(3.1,2.5){$a$} \\put(1.3,1.7){$b$} \\put(2.5,1.05){$c$} \\put(0.3,4){$F= \\sqrt{s(s-a)(s-b)(s-c)}$} \\put(3.5,0.4){$\\displaystyle s:=\\frac{a+b+c}{2}$} \\end{picture}$","1 \\\\ 2 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{19} \\begin{pmatrix} 12 \\\\ 16 \\\\ −5 \\end{pmatrix} \\end{align*} \\begin{align*} 18) \\hspace{.5cm} \\begin{pmatrix} 3 & 1 & 1 \\\\ −1 & 1 & 2 \\\\ 4 & 3 & 5 \\end{pmatrix} x = \\begin{pmatrix} −3 \\\\ 3 \\\\ 4 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{3} \\begin{pmatrix} 1 \\\\ −34 \\\\ 22 \\end{pmatrix} \\end{align*} \\begin{align*} 19) \\hspace{.5cm} \\begin{pmatrix} 1 & 2 & −1 & −2 \\\\ 3 & 0 & 4 & 1 \\\\ 1 & 5 & 0 & 0 \\\\ 0 & 1 & 0 & 1 \\end{pmatrix} x = \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\\\ 2 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{10} \\begin{pmatrix} 45 \\\\ −9 \\\\ −41 \\\\ 29 \\end{pmatrix} \\end{align*} \\begin{align*} 20) \\hspace{.5cm} \\begin{pmatrix} 0 & 7 & 1 & 0 \\\\ 1 & 1 & 1 & 0 \\\\ 1 & −1 & 1 & −1 \\\\ 0 & 1 & 2 & 3 \\end{pmatrix} x = \\begin{pmatrix} 1 \\\\ 0 \\\\ 2 \\\\ −2 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x &=\\frac{1}{19} \\begin{pmatrix} −31 \\\\ −2 \\\\ 33 \\\\ −34 \\end{pmatrix} \\end{align*} Tags:","\\\\ 2 & 3 \\end{pmatrix} x = \\begin{pmatrix} 5 \\\\ 6 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{5} \\begin{pmatrix} 9 \\\\ 4 \\end{pmatrix} \\end{align*} \\begin{align*} 12) \\hspace{.5cm} \\begin{pmatrix} −1 & 5 \\\\ 1 & 1 \\end{pmatrix} x = \\begin{pmatrix} 2 \\\\ −1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{6} \\begin{pmatrix} −7 \\\\ 1 \\end{pmatrix} \\end{align*} \\begin{align*} 13) \\hspace{.5cm} \\begin{pmatrix} 3 & −7 \\\\ −9 & 21 \\end{pmatrix} x = \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\text{no inverse} \\end{align*} \\begin{align*} 14) \\hspace{.5cm} \\begin{pmatrix} 2 & 7 \\\\ 5 & −1 \\end{pmatrix} x = \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{37} \\begin{pmatrix} 31 \\\\ 7 \\end{pmatrix} \\end{align*} \\begin{align*} 15) \\hspace{.5cm} \\begin{pmatrix} 1 & 0 & 1 \\\\ 0 & −1 & 1 \\\\ 1 & 2 & 2 \\end{pmatrix} x = \\begin{pmatrix} 1 \\\\ −2 \\\\ −1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\begin{pmatrix} 3 \\\\ 0 \\\\ −2 \\end{pmatrix}\\end{align*} \\begin{align*} 16) \\hspace{.5cm} \\begin{pmatrix} 7 & 3 & 4 \\\\ 1 & 2 & 3 \\\\ 4 & −3 & −5 \\end{pmatrix} x = \\begin{pmatrix} 3 \\\\ 2 \\\\ 1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\text{no inverse} \\end{align*} \\begin{align*} 17) \\hspace{.5cm} \\begin{pmatrix} 3 & 4 & 1 \\\\ −2 & 3 & 1 \\\\ 0 & 3 & 2 \\end{pmatrix} x = \\begin{pmatrix} 5 \\\\"],"string":"[\n \"by clicking on the problem.) \\\\begin{align*} 1) \\\\hspace{.5cm} \\\\begin{pmatrix} 3 & 1 \\\\\\\\ 2 & 1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\begin{pmatrix} 1 & −1 \\\\\\\\ −2 & 3 \\\\end{pmatrix}\\\\end{align*} \\\\begin{align*} 2) \\\\hspace{.5cm} \\\\begin{pmatrix} 2 & 1 \\\\\\\\ 4 & 3 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*}\\\\begin{pmatrix} \\\\frac{3}{2} & −\\\\frac{1}{2} \\\\\\\\ −2 & 1 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 3) \\\\hspace{.5cm} \\\\begin{pmatrix} 0 & 1 \\\\\\\\ 5 & 7 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\begin{pmatrix} −\\\\frac{7}{5} & \\\\frac{1}{5} \\\\\\\\ 1 & 0 \\\\end{pmatrix}\\\\end{align*} \\\\begin{align*} 4) \\\\hspace{.5cm} \\\\begin{pmatrix} 2 & 1 \\\\\\\\ 6 & 3 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\text{not invertible} \\\\end{align*} \\\\begin{align*} 5) \\\\hspace{.5cm} \\\\begin{pmatrix} 1 & −2 & 0 \\\\\\\\ 0 & 1 & 0 \\\\\\\\ 0 & −1 & 1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\begin{pmatrix} 1 & 2 & 0 \\\\\\\\ 0 & 1 & 0 \\\\\\\\ 0 & 1 & 1 \\\\end{pmatrix}\\\\end{align*} \\\\begin{align*} 6) \\\\hspace{.5cm} \\\\begin{pmatrix} 1 & 0 & 1 \\\\\\\\ 2 & 3 & 4 \\\\\\\\ −1 & 1 & 0 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\begin{pmatrix} −4 & 1 & −3 \\\\\\\\ −4 & 1 & −2 \\\\\\\\ 5 & −1 & 3 \\\\end{pmatrix}\\\\end{align*} \\\\begin{align*} 7) \\\\hspace{.5cm} \\\\begin{pmatrix} 2 & 1 & 1 \\\\\\\\ 1 & 1 & 7 \\\\\\\\ 5 & 2 & −4 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\text{not invertible} \\\\end{align*} \\\\begin{align*} 8) \\\\hspace{.5cm} \\\\begin{pmatrix} 3 & 2\",\n \"be: \\\\begin{align}\\\\textbf{x}(t)=(18\\\\ln(e^t+1)+c_{1})e^{-t}\\\\begin{pmatrix}1\\\\\\\\0\\\\end{pmatrix}+(-3e^{2t}+6e^t-6\\\\ln(e^t+1)+c_{2})e^{-3t}\\\\begin{pmatrix}1\\\\\\\\-2\\\\end{pmatrix}\\\\end{align} Subbing in the initial condition $\\\\textbf{x}(0)=\\\\begin{pmatrix}1-3\\\\ln2\\\\\\\\-3-3\\\\ln2\\\\end{pmatrix}$: \\\\begin{align}1-3\\\\ln2=c_{1}+3+c_{2}\\\\end{align} \\\\begin{align}-3-3\\\\ln2=-6-2c_{2}\\\\end{align} Solving these equations for the constants gives: \\\\begin{align}c_{1}=\\\\frac{-19-9\\\\ln2}{2}\\\\end{align} \\\\begin{align}c_{2}=\\\\frac{-3+3\\\\ln2}{2}\\\\end{align} Therefore the general solution for the IVP is: \\\\begin{align}\\\\textbf{x}(t)=\\\\frac{-19-9\\\\ln2}{2}e^{-t}\\\\begin{pmatrix}1\\\\\\\\0\\\\end{pmatrix}+\\\\frac{-3+3\\\\ln2}{2}e^{-3t}\\\\begin{pmatrix}1\\\\\\\\-2\\\\end{pmatrix}+18\\\\ln(e^t+1)e^{-t}\\\\begin{pmatrix}1\\\\\\\\0\\\\end{pmatrix}-3e^{-t}\\\\begin{pmatrix}1\\\\\\\\-2\\\\end{pmatrix}+6e^{-2t}\\\\begin{pmatrix}1\\\\\\\\-2\\\\end{pmatrix}-6\\\\ln(e^t+1)e^{-3t}\\\\begin{pmatrix}1\\\\\\\\-2\\\\end{pmatrix} \\\\end{align}\",\n \"k & \\\\ell\\\\end{pmatrix}\\\\\\\\ & = \\\\begin{pmatrix}(ae+bg)i+(af+bh)k & (ae+bg)j+(af+bh)\\\\ell \\\\\\\\ (ce+dg)i+(cf+dh)k & (ce+dg)j+(cf+dh)\\\\ell\\\\end{pmatrix}\\\\\\\\ & = \\\\begin{pmatrix}aei+bgi+afk+bhk & aej+bgj+af\\\\ell+bh\\\\ell \\\\\\\\ cei+dgi+cfk+dhk & cej+dgj+cf\\\\ell+dh\\\\ell\\\\end{pmatrix}\\\\end{align*} \\\\begin{align*}A\\\\cdot (B\\\\cdot C)&=\\\\begin{pmatrix}a & b \\\\\\\\ c & d\\\\end{pmatrix}\\\\cdot\\\\left ( \\\\begin{pmatrix}e & f \\\\\\\\ g & h\\\\end{pmatrix}\\\\cdot \\\\begin{pmatrix}i & j \\\\\\\\ k & \\\\ell\\\\end{pmatrix}\\\\right )= \\\\begin{pmatrix}a & b \\\\\\\\ c & d\\\\end{pmatrix}\\\\cdot\\\\begin{pmatrix}ei+fk & ej+f\\\\ell \\\\\\\\ gi+hk & gj+h\\\\ell\\\\end{pmatrix} \\\\\\\\ & = \\\\begin{pmatrix}a(ei+fk)+b(gi+hk) & a(ej+f\\\\ell)+b(gj+h\\\\ell) \\\\\\\\ c(ei+fk)+d(gi+hk) & c(ej+f\\\\ell)+d(gj+h\\\\ell)\\\\end{pmatrix} \\\\\\\\ & = \\\\begin{pmatrix}aei+afk+bgi+bhk & aej+af\\\\ell+bgj+bh\\\\ell \\\\\\\\ cei+cfk+dgi+dhk & cej+cf\\\\ell+dgj+dh\\\\ell\\\\end{pmatrix}\\\\\\\\ & = \\\\begin{pmatrix}aei+bgi+afk+bhk & aej+bgj+af\\\\ell+bh\\\\ell \\\\\\\\ cei+dgi+cfk+dhk & cej+dgj+cf\\\\ell+dh\\\\ell\\\\end{pmatrix}\\\\end{align*} The results are the same. Therefore it holds that $(A\\\\cdot B)\\\\cdot C=A\\\\cdot (B\\\\cdot C)$ which means the multiplication over $M_2(\\\\mathbb{R})$ is associative. 2. Let $A=\\\\begin{pmatrix}a & b \\\\\\\\ c & d\\\\end{pmatrix}, \\\\ B=\\\\begin{pmatrix}e & f \\\\\\\\ g & h\\\\end{pmatrix}$. Then we have the following: \\\\begin{equation*}A\\\\cdot B=\\\\begin{pmatrix}a & b \\\\\\\\ c & d\\\\end{pmatrix}\\\\cdot \\\\begin{pmatrix}e & f \\\\\\\\ g & h\\\\end{pmatrix}=\\\\begin{pmatrix}ae+bg & af+bh \\\\\\\\ ce+dg & cf+dh\\\\end{pmatrix} \\\\end{equation*} Then we have the following: \\\\begin{equation*}B\\\\cdot A=\\\\begin{pmatrix}e & f \\\\\\\\ g & h\\\\end{pmatrix}\\\\cdot \\\\begin{pmatrix}a & b \\\\\\\\ c & d\\\\end{pmatrix}=\\\\begin{pmatrix}ea+fc & eb+fd \\\\\\\\ ga+hc & gb+hd\\\\end{pmatrix} \\\\end{equation*} We see that $A\\\\cdot B\\\\neq B\\\\cdot A$, which means that the multiplication over $M_2(\\\\mathbb{R})$ is not commutative. 3. The neutral element in respect of the multiplication over $M_2(\\\\mathbb{R})$ is the identity matrix \\\\begin{equation*}I_2=\\\\begin{pmatrix}1 & 0 \\\\\\\\ 0\",\n \"&=\\\\begin{pmatrix}0&0&1&-5\\\\\\\\ 0&0&0&-7\\\\\\\\ 0&0&0&7\\\\end{pmatrix}\\\\tag{R_3=R_3-R_1}\\\\\\\\\\\\\\\\ &=\\\\begin{pmatrix}0&0&1&-5\\\\\\\\ 0&0&0&1\\\\\\\\ 0&0&0&7\\\\tag{R_2=-\\\\frac17 R_2}\\\\end{pmatrix}\\\\\\\\\\\\\\\\ &=\\\\begin{pmatrix}0&0&1&-5\\\\\\\\ 0&0&0&1\\\\\\\\ 0&0&0&0\\\\tag{R_3=R_3 -7R_2}\\\\end{pmatrix} \\\\end{align}\",\n \"the observer). \\\\begin{aligned} R_x(\\\\theta)&=\\\\begin{pmatrix}1 &0 &0\\\\\\\\ 0 &\\\\cos\\\\theta &-\\\\sin\\\\theta\\\\\\\\ 0 &\\\\sin\\\\theta &\\\\cos\\\\theta \\\\end{pmatrix}\\\\\\\\ R_y(\\\\theta)&=\\\\begin{pmatrix}\\\\cos\\\\theta &0 &\\\\sin\\\\theta\\\\\\\\ 0 &1 &0\\\\\\\\ -\\\\sin\\\\theta &0 &\\\\cos\\\\theta \\\\end{pmatrix}\\\\\\\\ R_z(\\\\theta)&=\\\\begin{pmatrix} \\\\cos\\\\theta &-\\\\sin\\\\theta &0\\\\\\\\ \\\\sin\\\\theta &\\\\cos\\\\theta &0\\\\\\\\ 0 &0 &1 \\\\end{pmatrix} \\\\end{aligned} ## Example 1 Rotating $\\\\begin{pmatrix}1\\\\\\\\ 0\\\\\\\\ 0\\\\end{pmatrix}$ $45^\\\\circ$ anticlockwise about $z$-axis: \\\\begin{aligned} R_z(45^\\\\circ)\\\\begin{pmatrix}1\\\\\\\\ 0\\\\\\\\ 0\\\\end{pmatrix}&=\\\\begin{pmatrix} \\\\cos 45^\\\\circ &-\\\\sin 45^\\\\circ &0\\\\\\\\ \\\\sin 45^\\\\circ &\\\\cos 45^\\\\circ &0\\\\\\\\ 0 &0 &1 \\\\end{pmatrix} \\\\begin{pmatrix}1\\\\\\\\ 0\\\\\\\\ 0\\\\end{pmatrix}\\\\\\\\ &=\\\\begin{pmatrix}\\\\sqrt{2}/2\\\\\\\\\\\\sqrt{2}/2\\\\\\\\0\\\\end{pmatrix}. \\\\end{aligned} ## Example 2 Rotating $\\\\begin{pmatrix}0\\\\\\\\ 1\\\\\\\\ 0\\\\end{pmatrix}$ $45^\\\\circ$ anticlockwise about $z$-axis: \\\\begin{aligned} R_z(45^\\\\circ)\\\\begin{pmatrix}0\\\\\\\\ 1\\\\\\\\ 0\\\\end{pmatrix}&=\\\\begin{pmatrix} \\\\cos 45^\\\\circ &-\\\\sin 45^\\\\circ &0\\\\\\\\ \\\\sin 45^\\\\circ &\\\\cos 45^\\\\circ &0\\\\\\\\ 0 &0 &1 \\\\end{pmatrix} \\\\begin{pmatrix}0\\\\\\\\ 1\\\\\\\\ 0\\\\end{pmatrix}\\\\\\\\ &=\\\\begin{pmatrix}-\\\\sqrt{2}/2\\\\\\\\\\\\sqrt{2}/2\\\\\\\\0\\\\end{pmatrix}. \\\\end{aligned} Posted in Uncategorized | 1 Comment ## 十二平均律 12-tone Equal Temperament 《十二平均律》 是明朝 "布衣王子" 朱载堉 (1536年-1610年) 发明的,由当时在中国传教的意大利人 利马窦 (Matteo Ricci, 传给欧洲的法国数学家 Marin Mersenne (Mersenne Prime ) 。现代音乐之父 巴哈 Bach 第一个采用,制作世界第一架钢琴有12黑白键,并作曲 《Bach 12-tone Equal Temperament》。 Side Note: 1977年法国大学数学教授在课堂好奇地问我,你们祖先如何解代数?是用算盘吗?当时计算机还不流行,复杂的算法只能用Log Table 或 计算尺 (Slide Rule) 。 Bach 12-tone Equal Temperament https://zh.m.wikipedia.org/wiki/%E5%8D%81%E4%BA%8C%E5%B9%B3%E5%9D%87%E5%BE%8B View original post ## Richard Dedekind Julius Wilhelmina Richard Dedekind (6 Oct 1831 – 12 Feb 1916) – Last student of Gauss at Göttingen – Student and closed friend of Dirichlet who influenced his Mathematical education – Introduced the word Field (Körper) – Gave the first university course on Galois Theory – Developed Real Number ‘Dedekind Cut‘ in 1872 – Accomplished musician – Never married, lived with\",\n \"to align the \\\\Rightarrows – daleif Oct 6 '16 at 7:59 This is an automatic solution, based on the eqparbox package. I defined an \\\\eqmathbox command, which accepts $2$ optional arguments (the end of name of the box, and the alignment: l, r or c) and one mandatory argument: the contents of the math box. \\\\documentclass{article} %\\\\usepackage{enumitem} \\\\usepackage{mathtools} \\\\usepackage{xparse} \\\\usepackage{eqparbox} \\\\NewDocumentCommand\\\\Eqmathbox{O{0}O{l}m}{\\\\eqmakebox[Form#1][#2]{\\\\ensuremath{ \\\\displaystyle#3}}} \\\\usepackage{siunitx} \\\\sisetup{group-separator={,},group-minimum-digits = 4} \\\\begin{document} \\\\begin{enumerate} \\\\item $$\\\\Eqmathbox{\\\\begin{pmatrix*}[r] 20 & 25 & 20 \\\\\\\\ 15 & 20 & 18 \\\\\\\\ 18 & 23 & 22 \\\\end{pmatrix*} \\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ z \\\\end{pmatrix*} = \\\\begin{pmatrix*}[r] 975 \\\\\\\\ 810 \\\\\\\\ 965 \\\\end{pmatrix*}} \\\\Rightarrow\\\\Eqmathbox[1][r]{\\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ z \\\\end{pmatrix*} =} \\\\begin{pmatrix*}[r] 10 \\\\\\\\ 15 \\\\\\\\ 20 \\\\end{pmatrix*}$$ \\\\item $$\\\\Eqmathbox{\\\\begin{pmatrix*}[r] 90 & 95 & 80 \\\\\\\\ 80 & 85 & 90 \\\\\\\\ 75 & 90 & 95 \\\\end{pmatrix*} \\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ z \\\\end{pmatrix*} = \\\\begin{pmatrix*}[r] 86 \\\\\\\\ 86 \\\\\\\\ 88 \\\\end{pmatrix*}} \\\\Rightarrow\\\\Eqmathbox[1][r]{\\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ z \\\\end{pmatrix*} =} \\\\begin{pmatrix*}[r] 0.3 \\\\\\\\ 0.2 \\\\\\\\ 0.5 \\\\end{pmatrix*}$$ \\\\item $$\\\\Eqmathbox{\\\\begin{pmatrix*}[r] \\\\num{5000} & \\\\num{400}0 \\\\\\\\ \\\\num{6000} & \\\\num{5000} \\\\end{pmatrix*} \\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ \\\\end{pmatrix*} = \\\\begin{pmatrix*}[r] \\\\num{37000000} \\\\\\\\ \\\\num{45000000 }\\\\\\\\ \\\\end{pmatrix*}} \\\\Rightarrow\\\\Eqmathbox[1][r]{\\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ \\\\end{pmatrix*} =} \\\\begin{pmatrix*}[r] \\\\num{5000} \\\\\\\\ \\\\num{3000} \\\\end{pmatrix*}$$ \\\\end{enumerate} \\\\end{document} Here is one way. Note that you are adding a lot of unneeded\",\n \"\\\\end{pmatrix} \\\\[6px] &\\\\hspace{2px}\\\\begin{pmatrix} 2 & 3 & 4 \\\\end{pmatrix} \\\\end{align*} would do the job. For some strange reason the align gave errors when leaving out the first \\\"&\\\" symbol and it gave a 2px offset. I figured you wanted some space between the two if not leave the [6px]. You can always use \\\\hspace{amount of whitespace} to place your second matrix in the place you want. This can be given in pt's, px's (which i did) etc. //edit Hm I notice the \\\\hspace{} is actually not needed, but can be used in case of pmatrix. What happens is that the pmatrix brackets give a biased image of the matrices. When using vmatrix like: \\\\begin{align*} &\\\\begin{vmatrix} 1 & 2 & 3 \\\\\\\\ 3 & 4 & 5 \\\\end{vmatrix} \\\\\\\\[6px] &\\\\begin{vmatrix} 2 & 3 & 4 \\\\end{vmatrix} \\\\end{align*} It all goes well :). So basically, probably the easiest way to fix it is either use other brackets to make it look good or use the \\\\hspace to align as you like it. - If all else fails, PGF/TikZ can do this. See this example. - Wrap the thing in \\\\begin{align*} ... \\\\end{align*} and use & as the alignment marker in your formulas. Example: \\\\begin{align*} \\\\begin{pmatrix} ... vector here \\\\end{pmatrix} &\\\\begin{pmatrix} ... first matrix here \\\\end{pmatrix}\\\\\\\\ &\\\\begin{pmatrix} ... second matrix here\",\n \"\\\\usepackage{amsmath} % for 'align*' and 'pmatrix' environments \\\\begin{document} \\\\begin{align*} H &= \\\\begin{pmatrix} 1 & x_{11} & \\\\dots & x_{1p} \\\\\\\\ 1 & x_{21} & \\\\dots & x_{2p} \\\\\\\\ \\\\vdots & \\\\vdots & \\\\ddots & \\\\vdots \\\\\\\\ 1 & x_{n1} & \\\\dots & x_{np} \\\\end{pmatrix}%_{\\\\!\\\\!\\\\!n\\\\times(p+1)} \\\\begin{pmatrix} q_{11} & q_{12} & \\\\dots & q_{1,p+1} \\\\\\\\ q_{21} & q_{22} & \\\\dots & q_{2,p+1} \\\\\\\\ \\\\vdots & \\\\vdots & \\\\ddots & \\\\vdots \\\\\\\\ q_{p+1,1} & q_{p+1,2} & \\\\dots & q_{p+1,p+1} \\\\end{pmatrix}%_{\\\\!\\\\!\\\\!(p+1)\\\\times(p+1)} \\\\begin{pmatrix} 1 & 1 & \\\\dots & 1 \\\\\\\\ x_{11} & x_{21} & \\\\dots & x_{n1} \\\\\\\\ \\\\vdots & \\\\vdots & \\\\ddots & \\\\vdots \\\\\\\\ x_{1p} & x_{2p} & \\\\dots & x_{np} \\\\end{pmatrix}%_{\\\\!\\\\!\\\\!(p+1)\\\\times n} \\\\\\\\ &= \\\\begin{pmatrix} x_{1}{}'q_{1} & x_{1}{}'q_{2} & \\\\dots & x_{1}{}'q_{p+1} \\\\\\\\ x_{2}{}'q_{1} & x_{2}{}'q_{2} & \\\\dots & x_{2}{}'q_{p+1} \\\\\\\\ \\\\vdots & \\\\vdots & \\\\ddots & \\\\vdots \\\\\\\\ x_{n}{}'q_{1} & x_{n}{}'q_{2} & \\\\dots & x_{n}{}'q_{p+1} \\\\end{pmatrix} \\\\begin{pmatrix} 1 & 1 & \\\\dots & 1 \\\\\\\\ x_{11} & x_{21} & \\\\dots & x_{n1} \\\\\\\\ \\\\vdots & \\\\vdots & \\\\ddots & \\\\vdots \\\\\\\\ x_{1p} & x_{2p} & \\\\dots & x_{np} \\\\end{pmatrix} \\\\end{align*} \\\\end{document} • This positive vote was necessary :-) – Sebastiano Nov 23 '19 at 20:30 • @Sebastiano - Thanks so much! :-) – Mico Nov 23 '19 at 20:34 • @Sebastiano - In fact, your upvote raised my\",\n \"\\\\end{pmatrix}F \\\\begin{pmatrix} u_r \\\\\\\\ v \\\\\\\\ 1 \\\\end{pmatrix}=0 ,\\\\end{align} and $$u_l$$ and $$u_r$$ become whatever the depth of the point $$P$$, so it must be \\\\begin{align} \\\\begin{pmatrix} u_l & v & 1 \\\\end{pmatrix} \\\\begin{pmatrix} 0 & 0 & 0 \\\\\\\\ 0 & 0 & s \\\\\\\\ 0 & -s & 0 \\\\end{pmatrix} \\\\begin{pmatrix} u_r \\\\\\\\ v \\\\\\\\ 1 \\\\end{pmatrix}=0 .\\\\end{align} That is, for $$\\\\tilde{\\\\ddot{m_l}}$$ and $$\\\\tilde{\\\\ddot{m_l}}$$, its base matrix $$f$$ is \\\\begin{align} f = \\\\begin{pmatrix} 0 & 0 & 0 \\\\\\\\ 0 & 0 & s \\\\\\\\ 0 & -s & 0 \\\\end{pmatrix} .\\\\end{align} When this is returned to the original $$\\\\tilde{m_l}$$ and $$\\\\tilde{m_r}$$, it is \\\\begin{align} (A_rLRA_l^{-1}\\\\tilde{m_l})^T \\\\begin{pmatrix} 0 & 0 & 0 \\\\\\\\ 0 & 0 & s \\\\\\\\ 0 & -s & 0 \\\\end{pmatrix} A_rLA_r^{-1}\\\\tilde{m_r}= (\\\\tilde{m_l})^T(A_l^{-1})^TR^TL^TA_r^T \\\\begin{pmatrix} 0 & 0 & 0 \\\\\\\\ 0 & 0 & s \\\\\\\\ 0 & -s & 0 \\\\end{pmatrix} A_rLA_r^{-1}\\\\tilde{m_r}=0 ,\\\\end{align} so its base matrix $$F$$ is \\\\begin{align} F = (A_l^{-1})^TR^TL^TA_r^T \\\\begin{pmatrix} 0 & 0 & 0 \\\\\\\\ 0 & 0 & s \\\\\\\\ 0 & -s & 0 \\\\end{pmatrix} A_rLA_r^{-1} \\\\end{align} or \\\\begin{align} F = (A_l^{-1})^TR^TL^TA_r^TfA_rLA_r^{-1} \\\\end{align} for $$\\\\tilde{m_l}$$ and $$\\\\tilde{m_r}$$. ### Program distribution Then, the program which calculates a basic matrix by stereo processing is explained. graph_cut.cu , rectify.cpp , makefile , 2_R.png , 2_L.png Please download.\",\n \"e_{12} \\\\\\\\ e_{13} \\\\end{pmatrix}, \\\\vec{e_2}=\\\\begin{pmatrix} e_{21} \\\\\\\\ e_{22} \\\\\\\\ e_{23} \\\\end{pmatrix}, \\\\vec{e_3}=\\\\begin{pmatrix} e_{31} \\\\\\\\ e_{32} \\\\\\\\ e_{33} \\\\end{pmatrix} \\\\end{align*} is an orthonormal vector, it is \\\\begin{align*} \\\\begin{pmatrix} (\\\\vec{e_1})^T \\\\\\\\ (\\\\vec{e_2})^T \\\\\\\\ (\\\\vec{e_3})^T \\\\end{pmatrix} \\\\begin{pmatrix} \\\\vec{e_1} & \\\\vec{e_2} & \\\\vec{e_3} \\\\end{pmatrix} =\\\\begin{pmatrix} e_{11} & e_{12} & e_{13} \\\\\\\\ e_{21} & e_{22} & e_{23} \\\\\\\\ e_{31} & e_{32} & e_{33} \\\\end{pmatrix} \\\\begin{pmatrix} e_{11} & e_{21} & e_{31} \\\\\\\\ e_{12} & e_{22} & e_{32} \\\\\\\\ e_{13} & e_{23} & e_{33} \\\\end{pmatrix} =\\\\begin{pmatrix} 1 & 0 & 0 \\\\\\\\ 0 & 1 & 0 \\\\\\\\ 0 & 0 & 1 \\\\end{pmatrix} .\\\\end{align*} That is, \\\\begin{align*} R=\\\\begin{pmatrix} e_{11} & e_{12} & e_{13} \\\\\\\\ e_{21} & e_{22} & e_{23} \\\\\\\\ e_{31} & e_{32} & e_{33} \\\\end{pmatrix} =(R^{-1})^T \\\\end{align*}\\\\begin{align*} R^{-1}=\\\\begin{pmatrix} e_{11} & e_{21} & e_{31} \\\\\\\\ e_{12} & e_{22} & e_{32} \\\\\\\\ e_{13} & e_{23} & e_{33} \\\\end{pmatrix} =R^T .\\\\end{align*} Camera and perspective transformation Consider a pinhole camera in which light from an object always passes through the pinhole. Align the origin $$o$$ of the 3D coordinate $$oxyz$$ with this pinhole. Next, consider the plane that intersects with the $$oz$$ axis at $$f$$ point $$(0, 0, f)$$ and calls it the imaging surface. However, $$f$$ is positive and the imaging surface is between the pinhole and the object. $$f$$ is also called the focus. In\",\n \"\\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 3) \\\\hspace{.5cm} \\\\begin{pmatrix} 2 & 0 \\\\\\\\ 4 & 3 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} & \\\\mbox{answers may vary; one correct answer is} \\\\\\\\ & \\\\begin{pmatrix} 1 & 0 \\\\\\\\ 2 & 1 \\\\end{pmatrix} \\\\begin{pmatrix} 2 & 0 \\\\\\\\ 0 & 3 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 4) \\\\hspace{.5cm} \\\\begin{pmatrix} 2 & 1 \\\\\\\\ 6 & 4 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} & \\\\mbox{answers may vary; one correct answer is} \\\\\\\\ & \\\\begin{pmatrix} 1 & 0 \\\\\\\\ 3 & 1 \\\\end{pmatrix} \\\\begin{pmatrix} 1 & 1 \\\\\\\\ 0 & 1 \\\\end{pmatrix} \\\\begin{pmatrix} 2 & 0 \\\\\\\\ 0 & 1 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 5) \\\\hspace{.5cm} \\\\begin{pmatrix} 4 & 4 \\\\\\\\ 6 & 3 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} & \\\\mbox{answers may vary; one correct answer is} \\\\\\\\ & \\\\begin{pmatrix} 1 & 0 \\\\\\\\ \\\\frac{3}{2} & 1 \\\\end{pmatrix} \\\\begin{pmatrix} 1 & −\\\\frac{4}{3} \\\\\\\\ 0 & 1 \\\\end{pmatrix} \\\\begin{pmatrix} 4 & 0 \\\\\\\\ 0 & −3 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 6) \\\\hspace{.5cm} \\\\begin{pmatrix} 4 & 2 \\\\\\\\ 2 & 0 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} & \\\\mbox{answers may vary; one correct answer is} \\\\\\\\ & \\\\begin{pmatrix} 1 & 0 \\\\\\\\ \\\\frac{1}{2} & 1 \\\\end{pmatrix} \\\\begin{pmatrix} 1 & −2 \\\\\\\\ 0 & 1 \\\\end{pmatrix} \\\\begin{pmatrix} 4 & 0 \\\\\\\\ 0 & −1 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 7) \\\\hspace{.5cm} \\\\begin{pmatrix} 1 & 0 & 0 \\\\\\\\ 4 & 1\",\n \"& 1 \\\\\\\\ 0 & 2 & 4 \\\\\\\\ −1 & 1 & −1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\frac{1}{24} \\\\begin{pmatrix} 6 & −3 & −6 \\\\\\\\ 4 & 2 & 12 \\\\\\\\ −2 & 5 & −6 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 9) \\\\hspace{.5cm} \\\\begin{pmatrix} 1 & 2 & 1 & −2 \\\\\\\\ 1 & 1 & 0 & 0 \\\\\\\\ 0 & 1 & 0 & −1 \\\\\\\\ 2 & 1 & 0 & 2 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\begin{pmatrix} 0 & 3 & −2 & −1 \\\\\\\\ 0 & −2 & 2 & 1 \\\\\\\\ 1 & −3 & 0 & 1 \\\\\\\\ 0 & −2 & 1 & 1 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 10) \\\\hspace{.5cm} \\\\begin{pmatrix} 2 & 3 & 1 & 0 \\\\\\\\ −1 & 0 & 3 & 1 \\\\\\\\ 2 & 2 & 4 & 1 \\\\\\\\ 2 & −1 & 2 & −1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\frac{1}{25} \\\\begin{pmatrix} −11 & −17 & 16 & −1 \\\\\\\\ 14 & 8 & −9 & −1 \\\\\\\\ 5 & 10 & −5 & 5 \\\\\\\\ −26 & −22 & 31 & −16 \\\\end{pmatrix}\\\\end{align*} For each equation $Ax=b$, tell whether $A^{-1}$ exists. If it does, then compute the solution $x=A^{-1}b$. (You can view the solution by clicking on the problem.) \\\\begin{align*} 11) \\\\hspace{.5cm} \\\\begin{pmatrix} 1 & 4\",\n \"2 & -2.2 + 2 \\\\\\\\ 0 & 0 & 1 \\\\end{pmatrix} . \\\\begin{pmatrix} 2 & 4 & 4 & 2\\\\\\\\ 1 & 1 & 3 & 3\\\\\\\\ 1 & 1 & 1 & 1 \\\\end{pmatrix}\\\\\\\\\\\\\\\\ &= \\\\begin{pmatrix} 2 & 0 & -3 \\\\\\\\ 0 & 2 & -2 \\\\\\\\ 0 & 0 & 1 \\\\end{pmatrix} . \\\\begin{pmatrix} 2 & 4 & 4 & 2\\\\\\\\ 1 & 1 & 3 & 3\\\\\\\\ 1 & 1 & 1 & 1 \\\\end{pmatrix}\\\\\\\\\\\\\\\\ \\\\begin{pmatrix} x_{1}^{\\\\prime} & x_{2}^{\\\\prime} & x_{3}^{\\\\prime} & x_{4}^{\\\\prime}\\\\\\\\ y_{1}^{\\\\prime} & y_{2}^{\\\\prime} & y_{3}^{\\\\prime} & y_{4}^{\\\\prime}\\\\\\\\ 1 & 1 & 1 & 1 \\\\end{pmatrix} &= \\\\begin{pmatrix} 1 & 5 & 5 & 1\\\\\\\\ 0 & 0 & 4 & 4\\\\\\\\ 1 & 1 & 1 & 1 \\\\end{pmatrix} \\\\end{aligned} ### Combination use-case: rotate image Now imagine you have an image in a view, the origin is not a the center of the view, it is probably at the top-left corner (implementations may vary), but you want to rotate the image at the center of the view :upside_down: This transformation is composed of the following sequence: • move anchor point to origin: $\\\\left( -t_{x}, -t_{y} \\\\right)$ • rotate by $\\\\theta$ • move back anchor point: $\\\\left( t_{x}, t_{y} \\\\right)$ Where $t$ is the anchor point of our rotation transformation. Our\",\n \"-1 & 0 \\\\\\\\ 0 & 0 & 0 & -1 \\\\end{pmatrix} \\\\\\\\ \\\\quad + \\\\sum_{n=3}^{\\\\infty} \\\\sum_{i=0}^{n} \\\\frac{1}{i! (n-i)!} \\\\begin{pmatrix} 0 & \\\\rho_x & \\\\rho_y & \\\\rho_z \\\\\\\\ \\\\rho_x & 0 & -\\\\theta_z & \\\\theta_y \\\\\\\\ \\\\rho_y & \\\\theta_z & 0 & -\\\\theta_x \\\\\\\\ \\\\rho_z & -\\\\theta_y & \\\\theta_x & 0 \\\\end{pmatrix}^i \\\\begin{pmatrix} 1 & 0 & 0 & 0 \\\\\\\\ 0 & -1 & 0 & 0 \\\\\\\\ 0 & 0 & -1 & 0 \\\\\\\\ 0 & 0 & 0 & -1 \\\\end{pmatrix} \\\\begin{pmatrix} 0 & \\\\rho_x & \\\\rho_y & \\\\rho_z \\\\\\\\ \\\\rho_x & 0 & \\\\theta_z & - \\\\theta_y \\\\\\\\ \\\\rho_y & -\\\\theta_z & 0 & \\\\theta_x \\\\\\\\ \\\\rho_z & \\\\theta_y & -\\\\theta_x & 0 \\\\end{pmatrix}^{n-i} \\\\\\\\ = \\\\begin{pmatrix} 1 & 0 & 0 & 0 \\\\\\\\ 0 & -1 & 0 & 0 \\\\\\\\ 0 & 0 & -1 & 0 \\\\\\\\ 0 & 0 & 0 & -1 \\\\end{pmatrix}$ So we have: $c^2 \\\\, \\\\Delta t'_{AB} \\\\, \\\\Delta t'_{CD} \\\\; - \\\\; \\\\vec{x}'_{AB} \\\\, \\\\cdot \\\\, \\\\vec{x}'_{CD} \\\\\\\\ = \\\\left( \\\\Lambda \\\\; \\\\begin{pmatrix} c \\\\Delta t_{AB} \\\\\\\\ \\\\Delta x_{AB} \\\\\\\\ \\\\Delta y_{AB} \\\\\\\\ \\\\Delta z_{AB} \\\\end{pmatrix} \\\\right) ^T \\\\begin{pmatrix} 1 & 0 & 0 & 0 \\\\\\\\ 0 & -1 & 0 & 0 \\\\\\\\ 0 & 0 & -1 & 0 \\\\\\\\\",\n \"this, let $$A^{-1}\\\\equiv \\\\begin{pmatrix} a & b \\\\\\\\ c & d \\\\\\\\ \\\\end{pmatrix} , \\\\tag{10} \\\\label{10}$$ whereby \\\\begin{align} \\\\varepsilon (A^{-1})^T \\\\varepsilon^{-1} &= \\\\begin{pmatrix} 0 & 1 \\\\\\\\ -1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} a & b \\\\\\\\ c & d \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} 0 & -1 \\\\\\\\ 1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\\\\\ &= \\\\begin{pmatrix} 0 & 1 \\\\\\\\ -1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} b & -a \\\\\\\\ d & -c \\\\\\\\ \\\\end{pmatrix} \\\\\\\\ &= \\\\begin{pmatrix} d & -c \\\\\\\\ -b & a \\\\\\\\ \\\\end{pmatrix}, \\\\tag{11} \\\\label{11} \\\\end{align} while \\\\begin{align} \\\\varepsilon^{-1} (A^{-1})^T \\\\varepsilon &= \\\\begin{pmatrix} 0 & -1 \\\\\\\\ 1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} a & b \\\\\\\\ c & d \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} 0 & 1 \\\\\\\\ -1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\\\\\ &= \\\\begin{pmatrix} 0 & 1 \\\\\\\\ -1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} -b & a \\\\\\\\ -d & c \\\\\\\\ \\\\end{pmatrix} \\\\\\\\ &= \\\\begin{pmatrix} d & -c \\\\\\\\ -b & a \\\\\\\\ \\\\end{pmatrix} = \\\\varepsilon (A^{-1})^T \\\\varepsilon^{-1}. \\\\tag{12} \\\\label{12} \\\\end{align} And yes, this does imply that $$\\\\varepsilon^T = \\\\varepsilon^{-1}. \\\\tag{13} \\\\label{13}$$ As to the indices, the wikipedia article on the transpose of a linear map states that if $$f : V \\\\to W$$ is a linear map, then the transpose of $$f$$ is defined to be $${}^t f : W^* \\\\to V^*\",\n \"1 \\\\\\\\ \\\\end{pmatrix} & q_1&\\\\colon \\\\mathrm{ss\\\\_fff} \\\\\\\\ ^0T_2 &= \\\\begin{pmatrix} 333 & & & q_3 \\\\\\\\ & 333333 & & \\\\\\\\ & & 3333 & \\\\\\\\ & & & 1 \\\\\\\\ \\\\end{pmatrix} &q_2&\\\\colon \\\\mathrm{tt\\\\_bbb} \\\\end{align} \\\\end{document} • Yes that works, thank you! – Fee Jun 24 '19 at 15:10 The alignat environment enables you to control the spacing between the two columns of the environment: \\\\documentclass{report} \\\\usepackage{amsmath} \\\\begin{document} \\\\begin{alignat}{2} ^0T_1 &= \\\\begin{pmatrix} 1 & & & q_1 \\\\\\\\ & 1 & & \\\\\\\\ & & 1 & \\\\\\\\ & & & 1 \\\\\\\\ \\\\end{pmatrix} & q_1&\\\\colon \\\\mathrm{ss\\\\_fff} \\\\\\\\[2ex] ^0T_2 &= \\\\begin{pmatrix} 333 & & & q_3 \\\\\\\\ & 333333 & & \\\\\\\\ & & 3333 & \\\\\\\\ & & & 1 \\\\\\\\ \\\\end{pmatrix} &\\\\quad q_2&\\\\colon \\\\mathrm{tt\\\\_bbb} \\\\end{alignat} \\\\end{document}\",\n \"b \\\\\\\\ c & d \\\\\\\\ \\\\end{pmatrix} , \\\\tag{10} \\\\label{10}$$ whereby \\\\begin{align} \\\\varepsilon (A^{-1})^T \\\\varepsilon^{-1} &= \\\\begin{pmatrix} 0 & 1 \\\\\\\\ -1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} a & b \\\\\\\\ c & d \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} 0 & -1 \\\\\\\\ 1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\\\\\ &= \\\\begin{pmatrix} 0 & 1 \\\\\\\\ -1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} b & -a \\\\\\\\ d & -c \\\\\\\\ \\\\end{pmatrix} \\\\\\\\ &= \\\\begin{pmatrix} d & -c \\\\\\\\ -b & a \\\\\\\\ \\\\end{pmatrix}, \\\\tag{11} \\\\label{11} \\\\end{align} while \\\\begin{align} \\\\varepsilon^{-1} (A^{-1})^T \\\\varepsilon &= \\\\begin{pmatrix} 0 & -1 \\\\\\\\ 1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} a & b \\\\\\\\ c & d \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} 0 & 1 \\\\\\\\ -1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\\\\\ &= \\\\begin{pmatrix} 0 & 1 \\\\\\\\ -1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} -b & a \\\\\\\\ -d & c \\\\\\\\ \\\\end{pmatrix} \\\\\\\\ &= \\\\begin{pmatrix} d & -c \\\\\\\\ -b & a \\\\\\\\ \\\\end{pmatrix} = \\\\varepsilon (A^{-1})^T \\\\varepsilon^{-1}. \\\\tag{12} \\\\label{12} \\\\end{align} And yes, this does imply that $$\\\\varepsilon^T = \\\\varepsilon^{-1}. \\\\tag{13} \\\\label{13}$$ As to the indices, the wikipedia article on the transpose of a linear map states that if $$f : V \\\\to W$$ is a linear map, then the transpose of $$f$$ is defined to be $${}^t f : W^* \\\\to V^* , \\\\tag{14} \\\\label{14}$$ where $$V^*, W^*$$\",\n \"that is a lot of work. I think I will just stick with $\\\\displaystyle \\\\text{diam}\\\\left(E\\\\right)$ Thanks though! 8. $\\\\displaystyle \\\\begin{pmatrix} a_{1,1} & a_{2,1} & \\\\ldots & a_{n,1}\\\\\\\\ a_{2,1} & a_{2,2} & \\\\ldots & a_{n,2}\\\\\\\\ \\\\hdotsfor[1.7]{4}\\\\\\\\ a_{n,1} & a_{2,n} & \\\\ldots & a_{n,n} \\\\end{pmatrix}$ $\\\\displaystyle \\\\begin{pmatrix} a_{1,1} & a_{2,1} & \\\\ldots & a_{n,1}\\\\\\\\ a_{2,1} & a_{2,2} & \\\\ldots & a_{n,2}\\\\\\\\ \\\\vdots & \\\\vdots & \\\\ddots & \\\\vdots \\\\\\\\ a_{n,1} & a_{2,n} & \\\\ldots & a_{n,n} \\\\end{pmatrix}$ 9. $\\\\displaystyle \\\\begin{matrix}0&1\\\\\\\\1&0\\\\end{matrix} \\\\hspace{1.5pc} \\\\begin{pmatrix}0&1\\\\\\\\1&0\\\\end{pmatrix} \\\\hspace{1.5pc} \\\\begin{bmatrix}0&1\\\\\\\\1&0\\\\end{bmatrix} \\\\hspace{1.5pc} \\\\begin{Bmatrix}0&1\\\\\\\\1&0\\\\end{Bmatrix} \\\\hspace{1.5pc} \\\\begin{vmatrix}0&1\\\\\\\\1&0\\\\end{vmatrix} \\\\hspace{1.5pc} \\\\begin{Vmatrix}0&1\\\\\\\\1&0\\\\end{Vmatrix}$ 10. $\\\\displaystyle \\\\sum_{k=0}^n$ $\\\\displaystyle \\\\sum_{k=0}^n a^k$ $\\\\displaystyle \\\\sum_{r=0}^n a^r$ $\\\\displaystyle \\\\sum_{r=0}^n\\\\ ^nC_r a^r$ $\\\\displaystyle \\\\sum_{r=0}^n ^nC_r a^rb^{n-r}$ $\\\\displaystyle (x+y)^n$ $\\\\displaystyle $$x+y$$^n$ 11. $\\\\displaystyle -e^{-x^2}$ equals 0 12. $\\\\displaystyle \\\\int\\\\limits_-\\\\cdots \\\\int\\\\limits_\\\\text{ }^-\\\\cdots\\\\int\\\\limits_a^b$ Math $\\\\displaystyle \\\\int\\\\limits_- f~d\\\\alpha=\\\\sup_{P}L\\\\left(P,f,\\\\alpha\\\\right)$ 13. . . . $\\\\displaystyle \\\\frac{1{\\\\color{red}\\\\rlap{/}}6}{{\\\\color{red}\\\\rlap{/}}64} \\\\;=\\\\;\\\\frac{1}{4} \\\\qquad\\\\qquad \\\\frac{1{\\\\color{red}\\\\rlap{/}}9}{{\\\\color{red}\\\\rlap{/}}95} \\\\;=\\\\;\\\\frac{1}{5}$ . . . . . $\\\\displaystyle \\\\frac{2{\\\\color{red}\\\\rlap{/}}6}{{\\\\color{red}\\\\rlap{/}}65} \\\\;=\\\\;\\\\frac{2}{5}\\\\qquad\\\\qquad \\\\frac{4{\\\\color{red}\\\\rlap{/}}9}{{\\\\color{red}\\\\rlap{/}}98} \\\\;=\\\\;\\\\frac{4}{8}\\\\;=\\\\;\\\\frac{1}{2}$ . . . . . $\\\\displaystyle \\\\frac{1-x^{{\\\\color{red}\\\\rlap{/}}2}}{(1+x)^{{\\\\color{red}\\\\rlap{/}} 2} }\\\\;=\\\\;\\\\frac{1-x}{1+x}$ All that work for a bad joke . . . (Good luck trying to read my LaTeX code.) . 14. A fairly comprehensive guide to LaTeX here: http://tobi.oetiker.ch/lshort/lshort.pdf 15. Just trying something fancy in LaTex. This came from that tutorial. $\\\\displaystyle \\\\text{Heron's formula}\\\\setlength{\\\\unitlength}{1cm} \\\\begin{picture}(6,5) \\\\thicklines \\\\put(1,0.5){\\\\line(2,1){3}} \\\\put(4,2){\\\\line(-2,1){2}} \\\\put(2,3){\\\\line(-2,-5){1}} \\\\put(0.7,0.3){$A$} \\\\put(4.05,1.9){$B$} \\\\put(1.7,2.95){$C$} \\\\put(3.1,2.5){$a$} \\\\put(1.3,1.7){$b$} \\\\put(2.5,1.05){$c$} \\\\put(0.3,4){$F= \\\\sqrt{s(s-a)(s-b)(s-c)}$} \\\\put(3.5,0.4){$\\\\displaystyle s:=\\\\frac{a+b+c}{2}$} \\\\end{picture}$\",\n \"1 \\\\\\\\ 2 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} x = \\\\frac{1}{19} \\\\begin{pmatrix} 12 \\\\\\\\ 16 \\\\\\\\ −5 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 18) \\\\hspace{.5cm} \\\\begin{pmatrix} 3 & 1 & 1 \\\\\\\\ −1 & 1 & 2 \\\\\\\\ 4 & 3 & 5 \\\\end{pmatrix} x = \\\\begin{pmatrix} −3 \\\\\\\\ 3 \\\\\\\\ 4 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} x = \\\\frac{1}{3} \\\\begin{pmatrix} 1 \\\\\\\\ −34 \\\\\\\\ 22 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 19) \\\\hspace{.5cm} \\\\begin{pmatrix} 1 & 2 & −1 & −2 \\\\\\\\ 3 & 0 & 4 & 1 \\\\\\\\ 1 & 5 & 0 & 0 \\\\\\\\ 0 & 1 & 0 & 1 \\\\end{pmatrix} x = \\\\begin{pmatrix} 1 \\\\\\\\ 0 \\\\\\\\ 0 \\\\\\\\ 2 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} x = \\\\frac{1}{10} \\\\begin{pmatrix} 45 \\\\\\\\ −9 \\\\\\\\ −41 \\\\\\\\ 29 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 20) \\\\hspace{.5cm} \\\\begin{pmatrix} 0 & 7 & 1 & 0 \\\\\\\\ 1 & 1 & 1 & 0 \\\\\\\\ 1 & −1 & 1 & −1 \\\\\\\\ 0 & 1 & 2 & 3 \\\\end{pmatrix} x = \\\\begin{pmatrix} 1 \\\\\\\\ 0 \\\\\\\\ 2 \\\\\\\\ −2 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} x &=\\\\frac{1}{19} \\\\begin{pmatrix} −31 \\\\\\\\ −2 \\\\\\\\ 33 \\\\\\\\ −34 \\\\end{pmatrix} \\\\end{align*} Tags:\",\n \"\\\\\\\\ 2 & 3 \\\\end{pmatrix} x = \\\\begin{pmatrix} 5 \\\\\\\\ 6 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} x = \\\\frac{1}{5} \\\\begin{pmatrix} 9 \\\\\\\\ 4 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 12) \\\\hspace{.5cm} \\\\begin{pmatrix} −1 & 5 \\\\\\\\ 1 & 1 \\\\end{pmatrix} x = \\\\begin{pmatrix} 2 \\\\\\\\ −1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} x = \\\\frac{1}{6} \\\\begin{pmatrix} −7 \\\\\\\\ 1 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 13) \\\\hspace{.5cm} \\\\begin{pmatrix} 3 & −7 \\\\\\\\ −9 & 21 \\\\end{pmatrix} x = \\\\begin{pmatrix} 0 \\\\\\\\ 1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\text{no inverse} \\\\end{align*} \\\\begin{align*} 14) \\\\hspace{.5cm} \\\\begin{pmatrix} 2 & 7 \\\\\\\\ 5 & −1 \\\\end{pmatrix} x = \\\\begin{pmatrix} 3 \\\\\\\\ 4 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} x = \\\\frac{1}{37} \\\\begin{pmatrix} 31 \\\\\\\\ 7 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 15) \\\\hspace{.5cm} \\\\begin{pmatrix} 1 & 0 & 1 \\\\\\\\ 0 & −1 & 1 \\\\\\\\ 1 & 2 & 2 \\\\end{pmatrix} x = \\\\begin{pmatrix} 1 \\\\\\\\ −2 \\\\\\\\ −1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} x = \\\\begin{pmatrix} 3 \\\\\\\\ 0 \\\\\\\\ −2 \\\\end{pmatrix}\\\\end{align*} \\\\begin{align*} 16) \\\\hspace{.5cm} \\\\begin{pmatrix} 7 & 3 & 4 \\\\\\\\ 1 & 2 & 3 \\\\\\\\ 4 & −3 & −5 \\\\end{pmatrix} x = \\\\begin{pmatrix} 3 \\\\\\\\ 2 \\\\\\\\ 1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\text{no inverse} \\\\end{align*} \\\\begin{align*} 17) \\\\hspace{.5cm} \\\\begin{pmatrix} 3 & 4 & 1 \\\\\\\\ −2 & 3 & 1 \\\\\\\\ 0 & 3 & 2 \\\\end{pmatrix} x = \\\\begin{pmatrix} 5 \\\\\\\\\"\n]"},"ps_retrievals":{"kind":"list like","value":["be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}","-1 & 3\\\\ 0 & 0 & 1 & 1 & 0 & -1 \\end{array}\\right) \\\\[4mm]&\\xrightarrow{R1 := R1 - R2} \\left(\\begin{array}{lll|lll} 1 & 0 & 0 & 0 & 1 & -1\\\\ 0 & 1 & 0 & -1 & -1 & 3\\\\ 0 & 0 & 1 & 1 & 0 & -1 \\end{array}\\right) \\end{align*} So the solution is: \\begin{align*} \\begin{pmatrix} x_1\\\\x_2\\\\x_3 \\end{pmatrix}= \\begin{pmatrix} 0 & 1 & -1\\\\ -1 & -1 & 3\\\\ 1 & 0 & -1 \\end{pmatrix} \\begin{pmatrix} 1\\\\2\\\\3 \\end{pmatrix} =\\begin{pmatrix} -1\\\\6\\\\-2 \\end{pmatrix}. \\end{align*} ​To see the full video page and find related videos, click the following link. Linear Algebra for Math 308: L6E1inv 2. Solve the system:\\begin{align*} x_1 - x_2 + x_3 &=1\\\\ x_1 + 3x_2 + 5x_3 &=0\\\\ 3x_1 + 2x_2 + x_3 &=1. \\end{align*} Note: There is no video solution for this exercise. We are going to find the inverse matrix: \\begin{align*} \\left(\\begin{array}{ccc|ccc} 1 & -1 & 1 & 1 & 0 & 0\\\\ 1 & 3 & 5 & 0 & 1 & 0\\\\ 3 & 2 & 1 & 0 & 0 & 1 \\end{array}\\right) &\\xrightarrow[{R2 := \\frac{1}{4}(R2 - R1)}]{R3 := R3 - 3R1} \\left(\\begin{array}{ccc|ccc} 1 & -1 & 1 & 1 & 0 & 0\\\\ 0 & 1 & 1 & -1/4 & 1/4 & 0\\\\ 0 &","column: \\begin{align*} \\left(\\begin{array}{llll|l} 1 & 1 & 2 & 0 & 3\\\\ 0 & 1 & 1 & 0 & 1\\\\ 0 & 0 & 0 & 1 & 0 \\end{array}\\right) \\xrightarrow{R1 : R1 - R2} \\left(\\begin{array}{llll|l} 1 & 0 & 1 & 0 & 2\\\\ 0 & 1 & 1 & 0 & 1\\\\ 0 & 0 & 0 & 1 & 0 \\end{array}\\right) \\end{align*} This is the same as this system: $$\\hspace{.5in}\\begin{array}{ll} x_1 + x_3 &= 2\\\\ x_2 + x_3 &= 1\\\\ x_4 &= 0 \\end{array}$$ Which is the same as $$\\begin{array}{ll} x_1 &= 2 - x_3\\\\ x_2 &= 1 - x_3\\\\ x_3 &= \\\\ x_4 &= 0\\end{array}$$ As we have seen above, this means that the equations place no restrictions on $$x_3$$. That is, for every value of $$x_3$$, there is a solution and so we call it a \"free variable''. (This is discussed more below). The solution can be written in vector form: \\begin{align} \\begin{pmatrix}x_1\\\\x_2\\\\x_3\\\\x_4\\end{pmatrix} =\\begin{pmatrix}2\\\\1\\\\0\\\\0\\end{pmatrix} + t\\begin{pmatrix}-1\\\\-1\\\\1\\\\0\\end{pmatrix} \\end{align} Note that we have use \"$$t$$'' instead of $$x_3$$''. This is because \"$$x_3$$'' is really only a place holder and also appears on the left hand side of the previous equation. Also, it is usually customary to write the set of solutions in \"set builder notation'': \\begin{align*} \\left\\{\\begin{pmatrix}2\\\\1\\\\0\\\\0\\end{pmatrix} + t\\begin{pmatrix}-1\\\\-1\\\\1\\\\0\\end{pmatrix} : t\\in\\mathbb{R}\\right\\} \\end{align*} ​To see","-1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix}$$. Therefore, $$U_2 =\\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0\\\\ 0 & 1 & 0 & 0 \\end{pmatrix}$$ Thus, $$U_1 = U_2$$.","{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}t={\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}(6-3t)}$ and so any vector in the form for ${\\displaystyle S_{1}}$ can be stated in the form needed for inclusion in ${\\displaystyle S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we look for ${\\displaystyle t}$ so that these equations hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}1&+&0t&=&1+0s\\\\2&+&3t&=&8-1s\\end{array}}}$ Rewrite that as ${\\displaystyle {\\begin{array}{*{1}{rc}r}1&=&1\\\\t&=&2-(1/3)s\\end{array}}}$ and so ${\\displaystyle {\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}s={\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}(2-(1/3)s).}$ 2. These two are equal. To show that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we check that for any ${\\displaystyle t,s}$ we can find an appropriate ${\\displaystyle m,n}$ so that these hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}4m&-&4n&=&1t+2s\\\\7m&-&2n&=&3t+1s\\\\7m&-&10n&=&1t+5s\\end{array}}}$ Use Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\7&-2&3t+1s\\\\7&-10&1t+5s\\end{array}}\\right)&{\\xrightarrow[{(-7/4)\\rho _{1}+\\rho _{3}}]{(-7/4)\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&-3&-(3/4)t+(6/4)s\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to conclude that ${\\displaystyle {\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}t+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}s={\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}((1/4)t-(1/2)s)}$ and so ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&+&2s&=&4m-4n\\\\3t&+&1s&=&7m-2n\\\\1t&+&5s&=&7m-10n\\end{array}}}$ with Gaussian reduction ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\3&1&7m-2n\\\\1&5&7m-10n\\end{array}}\\right)&{\\xrightarrow[{-\\rho _{1}+\\rho _{3}}]{-3\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&3&3m-6n\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to get ${\\displaystyle {\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}m+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}n={\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}(2m)+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}(m-2n)}$ and so any member of ${\\displaystyle S_{2}}$ can be expressed in the form needed for ${\\displaystyle S_{1}}$. 3. These sets are equal. To prove that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we must be able to solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}2m&+&4n&=&1t\\\\4m&+&8n&=&2t\\end{array}}}$ for ${\\displaystyle m}$ and ${\\displaystyle n}$ in terms of ${\\displaystyle t}$. Apply Gaussian reduction ${\\displaystyle \\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\4&8&2t\\end{array}}\\right){\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}\\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\0&0&0\\end{array}}\\right)}$ to conclude that any pair ${\\displaystyle m,n}$ where ${\\displaystyle 2m+4n=t}$ will do. For instance, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(0)}$ or ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((-3/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(t).}$ Thus ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&=&2m+4n\\\\2t&=&4m+8n\\end{array}}}$ with Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\2&4m+8n\\end{array}}\\right)&{\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\0&0\\end{array}}\\right)\\end{array}}}$ to deduce that any","& 1 \\\\ 0 & 2 & 4 \\\\ −1 & 1 & −1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\frac{1}{24} \\begin{pmatrix} 6 & −3 & −6 \\\\ 4 & 2 & 12 \\\\ −2 & 5 & −6 \\end{pmatrix} \\end{align*} \\begin{align*} 9) \\hspace{.5cm} \\begin{pmatrix} 1 & 2 & 1 & −2 \\\\ 1 & 1 & 0 & 0 \\\\ 0 & 1 & 0 & −1 \\\\ 2 & 1 & 0 & 2 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\begin{pmatrix} 0 & 3 & −2 & −1 \\\\ 0 & −2 & 2 & 1 \\\\ 1 & −3 & 0 & 1 \\\\ 0 & −2 & 1 & 1 \\end{pmatrix} \\end{align*} \\begin{align*} 10) \\hspace{.5cm} \\begin{pmatrix} 2 & 3 & 1 & 0 \\\\ −1 & 0 & 3 & 1 \\\\ 2 & 2 & 4 & 1 \\\\ 2 & −1 & 2 & −1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\frac{1}{25} \\begin{pmatrix} −11 & −17 & 16 & −1 \\\\ 14 & 8 & −9 & −1 \\\\ 5 & 10 & −5 & 5 \\\\ −26 & −22 & 31 & −16 \\end{pmatrix}\\end{align*} For each equation $Ax=b$, tell whether $A^{-1}$ exists. If it does, then compute the solution $x=A^{-1}b$. (You can view the solution by clicking on the problem.) \\begin{align*} 11) \\hspace{.5cm} \\begin{pmatrix} 1 & 4","the observer). \\begin{aligned} R_x(\\theta)&=\\begin{pmatrix}1 &0 &0\\\\ 0 &\\cos\\theta &-\\sin\\theta\\\\ 0 &\\sin\\theta &\\cos\\theta \\end{pmatrix}\\\\ R_y(\\theta)&=\\begin{pmatrix}\\cos\\theta &0 &\\sin\\theta\\\\ 0 &1 &0\\\\ -\\sin\\theta &0 &\\cos\\theta \\end{pmatrix}\\\\ R_z(\\theta)&=\\begin{pmatrix} \\cos\\theta &-\\sin\\theta &0\\\\ \\sin\\theta &\\cos\\theta &0\\\\ 0 &0 &1 \\end{pmatrix} \\end{aligned} ## Example 1 Rotating $\\begin{pmatrix}1\\\\ 0\\\\ 0\\end{pmatrix}$ $45^\\circ$ anticlockwise about $z$-axis: \\begin{aligned} R_z(45^\\circ)\\begin{pmatrix}1\\\\ 0\\\\ 0\\end{pmatrix}&=\\begin{pmatrix} \\cos 45^\\circ &-\\sin 45^\\circ &0\\\\ \\sin 45^\\circ &\\cos 45^\\circ &0\\\\ 0 &0 &1 \\end{pmatrix} \\begin{pmatrix}1\\\\ 0\\\\ 0\\end{pmatrix}\\\\ &=\\begin{pmatrix}\\sqrt{2}/2\\\\\\sqrt{2}/2\\\\0\\end{pmatrix}. \\end{aligned} ## Example 2 Rotating $\\begin{pmatrix}0\\\\ 1\\\\ 0\\end{pmatrix}$ $45^\\circ$ anticlockwise about $z$-axis: \\begin{aligned} R_z(45^\\circ)\\begin{pmatrix}0\\\\ 1\\\\ 0\\end{pmatrix}&=\\begin{pmatrix} \\cos 45^\\circ &-\\sin 45^\\circ &0\\\\ \\sin 45^\\circ &\\cos 45^\\circ &0\\\\ 0 &0 &1 \\end{pmatrix} \\begin{pmatrix}0\\\\ 1\\\\ 0\\end{pmatrix}\\\\ &=\\begin{pmatrix}-\\sqrt{2}/2\\\\\\sqrt{2}/2\\\\0\\end{pmatrix}. \\end{aligned} Posted in Uncategorized | 1 Comment ## 十二平均律 12-tone Equal Temperament 《十二平均律》 是明朝 "布衣王子" 朱载堉 (1536年-1610年) 发明的,由当时在中国传教的意大利人 利马窦 (Matteo Ricci, 传给欧洲的法国数学家 Marin Mersenne (Mersenne Prime ) 。现代音乐之父 巴哈 Bach 第一个采用,制作世界第一架钢琴有12黑白键,并作曲 《Bach 12-tone Equal Temperament》。 Side Note: 1977年法国大学数学教授在课堂好奇地问我,你们祖先如何解代数?是用算盘吗?当时计算机还不流行,复杂的算法只能用Log Table 或 计算尺 (Slide Rule) 。 Bach 12-tone Equal Temperament https://zh.m.wikipedia.org/wiki/%E5%8D%81%E4%BA%8C%E5%B9%B3%E5%9D%87%E5%BE%8B View original post ## Richard Dedekind Julius Wilhelmina Richard Dedekind (6 Oct 1831 – 12 Feb 1916) – Last student of Gauss at Göttingen – Student and closed friend of Dirichlet who influenced his Mathematical education – Introduced the word Field (Körper) – Gave the first university course on Galois Theory – Developed Real Number ‘Dedekind Cut‘ in 1872 – Accomplished musician – Never married, lived with","by clicking on the problem.) \\begin{align*} 1) \\hspace{.5cm} \\begin{pmatrix} 3 & 1 \\\\ 2 & 1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\begin{pmatrix} 1 & −1 \\\\ −2 & 3 \\end{pmatrix}\\end{align*} \\begin{align*} 2) \\hspace{.5cm} \\begin{pmatrix} 2 & 1 \\\\ 4 & 3 \\end{pmatrix} \\end{align*} Solution: \\begin{align*}\\begin{pmatrix} \\frac{3}{2} & −\\frac{1}{2} \\\\ −2 & 1 \\end{pmatrix} \\end{align*} \\begin{align*} 3) \\hspace{.5cm} \\begin{pmatrix} 0 & 1 \\\\ 5 & 7 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\begin{pmatrix} −\\frac{7}{5} & \\frac{1}{5} \\\\ 1 & 0 \\end{pmatrix}\\end{align*} \\begin{align*} 4) \\hspace{.5cm} \\begin{pmatrix} 2 & 1 \\\\ 6 & 3 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\text{not invertible} \\end{align*} \\begin{align*} 5) \\hspace{.5cm} \\begin{pmatrix} 1 & −2 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & −1 & 1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 1 & 1 \\end{pmatrix}\\end{align*} \\begin{align*} 6) \\hspace{.5cm} \\begin{pmatrix} 1 & 0 & 1 \\\\ 2 & 3 & 4 \\\\ −1 & 1 & 0 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\begin{pmatrix} −4 & 1 & −3 \\\\ −4 & 1 & −2 \\\\ 5 & −1 & 3 \\end{pmatrix}\\end{align*} \\begin{align*} 7) \\hspace{.5cm} \\begin{pmatrix} 2 & 1 & 1 \\\\ 1 & 1 & 7 \\\\ 5 & 2 & −4 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\text{not invertible} \\end{align*} \\begin{align*} 8) \\hspace{.5cm} \\begin{pmatrix} 3 & 2","to align the \\Rightarrows – daleif Oct 6 '16 at 7:59 This is an automatic solution, based on the eqparbox package. I defined an \\eqmathbox command, which accepts $2$ optional arguments (the end of name of the box, and the alignment: l, r or c) and one mandatory argument: the contents of the math box. \\documentclass{article} %\\usepackage{enumitem} \\usepackage{mathtools} \\usepackage{xparse} \\usepackage{eqparbox} \\NewDocumentCommand\\Eqmathbox{O{0}O{l}m}{\\eqmakebox[Form#1][#2]{\\ensuremath{ \\displaystyle#3}}} \\usepackage{siunitx} \\sisetup{group-separator={,},group-minimum-digits = 4} \\begin{document} \\begin{enumerate} \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 20 & 25 & 20 \\\\ 15 & 20 & 18 \\\\ 18 & 23 & 22 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 975 \\\\ 810 \\\\ 965 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 10 \\\\ 15 \\\\ 20 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 90 & 95 & 80 \\\\ 80 & 85 & 90 \\\\ 75 & 90 & 95 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 86 \\\\ 86 \\\\ 88 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 0.3 \\\\ 0.2 \\\\ 0.5 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] \\num{5000} & \\num{400}0 \\\\ \\num{6000} & \\num{5000} \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} = \\begin{pmatrix*}[r] \\num{37000000} \\\\ \\num{45000000 }\\\\ \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} =} \\begin{pmatrix*}[r] \\num{5000} \\\\ \\num{3000} \\end{pmatrix*}$$ \\end{enumerate} \\end{document} Here is one way. Note that you are adding a lot of unneeded","\\\\ 0\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (1 \\cdot 0 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 0 + 1 \\cdot 0) \\\\ 1^2 + 0^2 - 0^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\1 \\end{pmatrix} \\\\ \\phi(|1\\rangle) = \\phi\\left(\\begin{pmatrix}0 \\\\ 1\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (0 \\cdot 1 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 1 + 0 \\cdot 0) \\\\ 0^2 + 0^2 - 1^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\-1 \\end{pmatrix} \\\\ \\phi(|+\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}} + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\frac{1}{\\sqrt{2}} + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|-\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(-\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} -1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|{+i}\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{i}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot 0 + 0 \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ 2 \\cdot \\left(-0 \\cdot 0 + \\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - 0^2 - \\left(\\frac{1}{\\sqrt{2}}\\right)^2 \\end{pmatrix} = \\begin{pmatrix}0\\\\1\\\\0 \\end{pmatrix} \\\\ \\phi(|{-i}\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{i}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot 0 + 0 \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right)\\right) \\\\ 2 \\cdot \\left(-0 \\cdot 0","\\stackrel{E_7}{\\sim} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} \\stackrel{E_6E_5E_4E_3E_2E_1}{\\sim} L \\end{eqnarray*} \\begin{eqnarray*} E_7= \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} =E_7^{-1} \\end{eqnarray*} \\begin{eqnarray*} M=(E_1^{-1}E_2^{-1}E_3^{-1})(E_4^{-1}E_5^{-1}E_6^{-1}) (E_7^{-1}) U=LDPU\\\\ \\end{eqnarray*} \\begin{eqnarray*} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} = \\begin{pmatrix} 1&0&0&0\\\\ 0&1&0&0\\\\ -2&0&1&0\\\\ 0&-1&1&1\\\\ \\end{pmatrix} \\begin{pmatrix} 2&0&0&0\\\\ 0&1&0&0\\\\ 0&0&3&0\\\\ 0&0&1&-3\\\\ \\end{pmatrix} \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\begin{pmatrix} 1&0&\\frac{-3}{2}&\\frac{1}{2}\\\\ 0&1&2&2\\\\ 0&0&1&\\frac43\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\end{eqnarray*}","F_{n}={\\cfrac {1}{\\sqrt {5}}}\\left({\\cfrac {1+{\\sqrt {5}}}{2}}\\right)^{n}-{\\cfrac {1}{\\sqrt {5}}}\\left({\\cfrac {1-{\\sqrt {5}}}{2}}\\right)^{n}.}$ Equivalently, the same computation may performed by diagonalization of A through use of its eigendecomposition: {\\displaystyle {\\begin{aligned}A&=S\\Lambda S^{-1},\\\\A^{n}&=S\\Lambda ^{n}S^{-1},\\end{aligned}}} where ${\\displaystyle \\Lambda ={\\begin{pmatrix}\\varphi &0\\\\0&-\\varphi ^{-1}\\end{pmatrix}}}$ and ${\\displaystyle S={\\begin{pmatrix}\\varphi &-\\varphi ^{-1}\\\\1&1\\end{pmatrix}}.}$ The closed-form expression for the nth element in the Fibonacci series is therefore given by {\\displaystyle {\\begin{aligned}{F_{n+1} \\choose F_{n}}&=A^{n}{F_{1} \\choose F_{0}}\\\\&=S\\Lambda ^{n}S^{-1}{F_{1} \\choose F_{0}}\\\\&=S{\\begin{pmatrix}\\varphi ^{n}&0\\\\0&(-\\varphi )^{-n}\\end{pmatrix}}S^{-1}{F_{1} \\choose F_{0}}\\\\&={\\begin{pmatrix}\\varphi &-\\varphi ^{-1}\\\\1&1\\end{pmatrix}}{\\begin{pmatrix}\\varphi ^{n}&0\\\\0&(-\\varphi )^{-n}\\end{pmatrix}}{\\frac {1}{\\sqrt {5}}}{\\begin{pmatrix}1&\\varphi ^{-1}\\\\-1&\\varphi \\end{pmatrix}}{1 \\choose 0},\\end{aligned}}} which again yields ${\\displaystyle F_{n}={\\cfrac {\\varphi ^{n}-(-\\varphi )^{-n}}{\\sqrt {5}}}.}$ The matrix A has a determinant of −1, and thus it is a 2×2 unimodular matrix. This property can be understood in terms of the continued fraction representation for the golden ratio: ${\\displaystyle \\varphi =1+{\\cfrac {1}{1+{\\cfrac {1}{1+{\\cfrac {1}{1+\\ddots }}}}}}.}$ The Fibonacci numbers occur as the ratio of successive convergents of the continued fraction for φ, and the matrix formed from successive convergents of any continued fraction has a determinant of +1 or −1. The matrix representation gives the following closed-form expression for the Fibonacci numbers: ${\\displaystyle {\\begin{pmatrix}1&1\\\\1&0\\end{pmatrix}}^{n}={\\begin{pmatrix}F_{n+1}&F_{n}\\\\F_{n}&F_{n-1}\\end{pmatrix}}.}$ For a given n, this matrix can be computed in O(log(n)) arithmetic operations, using the exponentiation by squaring method. Taking the determinant of both sides of this equation yields Cassini's identity, ${\\displaystyle (-1)^{n}=F_{n+1}F_{n-1}-{F_{n}}^{2}.}$ Moreover, since An Am = An+m for any square matrix A, the","& 0 \\\\0 & 0 & 0 & -1 \\\\0 & 1 & 0 & 0 \\\\1 & 0 & 0 & 0 \\end{pmatrix} = \\begin{pmatrix} 0 & 0 & 1 & 0 \\\\0 & 0 & 0 & -1 \\\\0 & -1 & 0 & 0 \\\\-1 & 0 & 0 & 0 \\end{pmatrix}$$","/ a^* = e^{2 i \\phi}$ for some $\\phi$ real – reuns Oct 7 '16 at 14:24 ## 1 Answer \\begin{align*} \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} &= \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\begin{pmatrix} \\bar{d} & -\\bar{b} \\\\ -\\bar{c} & \\bar{a} \\end{pmatrix} \\\\ &= \\begin{pmatrix} a\\bar{d}-b\\bar{c} & b\\bar{a}-a\\bar{b} \\\\ c\\bar{d}-d\\bar{c} & d\\bar{a}-c\\bar{b} \\end{pmatrix} \\end{align*} Equating \"$0$\": \\left \\{ \\begin{align*} \\frac{a}{\\bar{a}} &= \\frac{b}{\\bar{b}} \\\\ \\frac{c}{\\bar{c}} &= \\frac{d}{\\bar{d}} \\\\ \\end{align*} \\right. Let \\left \\{ \\begin{align*} a &= Ae^{i\\theta} \\\\ b &= Be^{i\\theta} \\end{align*} \\right. and \\left \\{ \\begin{align*} c &= Ce^{i\\phi} \\\\ d &= De^{i\\phi}\\ \\end{align*} \\right. where $A,B,C,D\\in \\mathbb{R}$ Equating \"$1$\": \\left \\{ \\begin{align*} (AD-BC)e^{i(\\theta-\\phi)} &= 1 \\\\ (AD-BC)e^{i(\\phi-\\theta)} &= 1 \\\\ \\end{align*} \\right. \\implies \\left \\{ \\begin{align*} AD-BC &= 1 \\\\ \\theta &= \\phi \\\\ \\end{align*} \\right.","& -3/50 \\end{array} \\right) \\left( \\begin{array}{cc} 1 & -2 \\\\ 1 & 3 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.1 \\\\ 0 \\end{pmatrix}$ Multiply throughout by $\\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right)^{-1} = 1/5 \\left( \\begin{array}{cc} 2 & 3 \\\\ -1 & 1 \\end{array} \\right)$ to get $\\frac{d}{dt} \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\left( \\begin{array}{cc} -1/50 & 0 \\\\ 0 & -6/50 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.04 \\\\ -0.02 \\end{pmatrix}$ this is equivalent to the differential equations $\\frac{du_1}{dt}=-1/50u_1+0.04$ $\\frac{du_2}{dt}=-6/50u_1-0.02$ These have the solutions $u_1=2+Ae^{-t/50}, \\: u_2= -1/6+Be^{-6t/50}$ We can write this as $\\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix}$ Now transform back to $x_1, \\: x_2$ . \\begin{aligned} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} &= P \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} \\\\ &= \\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right) \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix} \\\\ &= \\begin{pmatrix} 5/2+Ae^{-t/50}-3Be^{-6t/50} \\\\ 5/3+Ae^{-t/50}+2Be^{-6t/50} \\end{pmatrix} \\end{aligned} . These solutions can be fitted to any initial conditions to find the constants.","&=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&-7\\\\ 0&0&0&7\\end{pmatrix}\\tag{R_3=R_3-R_1}\\\\\\\\ &=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&1\\\\ 0&0&0&7\\tag{R_2=-\\frac17 R_2}\\end{pmatrix}\\\\\\\\ &=\\begin{pmatrix}0&0&1&-5\\\\ 0&0&0&1\\\\ 0&0&0&0\\tag{R_3=R_3 -7R_2}\\end{pmatrix} \\end{align}","1 \\\\ -1 & 1 & 3 & -2 \\end{pmatrix}\\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}.$$ Our normal equation becomes $$\\begin{pmatrix} 6 & -1 & 7 \\\\ -1 & 7 & -8 \\\\ 7 & -8 & 15 \\end{pmatrix}\\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix},$$ or as an augmented matrix, \\begin{align*} \\left(\\begin{array}{ccc|c} 6 & -1 & 7 & 2 \\\\ -1 & 7 & -8 & 1 \\\\ 7 & -8 & 15 & 1 \\end{array}\\right) &\\sim \\left(\\begin{array}{ccc|c} 1 & -7 & 8 & -1 \\\\ 6 & -1 & 7 & 2 \\\\ 7 & -8 & 15 & 1 \\end{array}\\right) \\\\ &\\sim \\left(\\begin{array}{ccc|c} 1 & -7 & 8 & -1 \\\\ 0 & 41 & -41 & 8 \\\\ 0 & 41 & -41 & 8 \\end{array}\\right) \\\\ &\\sim \\left(\\begin{array}{ccc|c} 1 & -7 & 8 & -1 \\\\ 0 & 1 & -1 & \\frac{8}{41} \\\\ 0 & 0 & 0 & 0 \\end{array}\\right) \\\\ &\\sim \\left(\\begin{array}{ccc|c} 1 & 0 & \\color{red}1 & \\frac{15}{41} \\\\ 0 & 1 & \\color{red}{-1} & \\frac{8}{41} \\\\ 0 & 0 & \\color{red}0 & 0 \\end{array}\\right) \\end{align*} Note that, as we always expect from the normal equation, we have consistency; there is no","\\vdots \\\\ a & a & \\ldots & 1+a\\end{pmatrix} =\\begin{pmatrix}1 & 0 & \\ldots & 0 \\\\ 0 & 1 & \\ldots & 0 \\\\ \\vdots & \\vdots & \\vdots & \\vdots \\\\ 0 & 0 & \\ldots & 1\\end{pmatrix}+\\begin{pmatrix}a & a & \\ldots & a \\\\ a & a & \\ldots & a \\\\ \\vdots & \\vdots & \\vdots & \\vdots \\\\ a & a & \\ldots & a\\end{pmatrix} =I+a\\begin{pmatrix}1 & 1 & \\ldots & 1 \\\\ 1 & 1 & \\ldots & 1 \\\\ \\vdots & \\vdots & \\vdots & \\vdots \\\\ 1 & 1 & \\ldots & 1\\end{pmatrix} \\\\ & =I+a\\begin{pmatrix}1 \\\\ 1 \\\\ \\vdots \\\\ 1 \\end{pmatrix}\\begin{pmatrix}1 & 1 & \\ldots & 1\\end{pmatrix} =I-a\\begin{pmatrix}-1 \\\\ -1 \\\\ \\vdots \\\\ -1 \\end{pmatrix}\\begin{pmatrix}-1 & -1 & \\ldots & -1\\end{pmatrix}\\end{align*} So we have in this case $$\\vec{v}=\\begin{pmatrix}-1 \\\\ -1 \\\\ \\vdots \\\\ -1 \\end{pmatrix}$$. Therefore the inverse matrix is $$A^{-1}=I+b\\vec{v}\\vec{v}^T$$ with $$b=a\\left (I-a\\vec{v}\\vec{v}^T\\right )^{-1}$$. Is this part correct? • Mar 27, 2020 at 10:55 The linear operator $$A = I - a v v^T$$ is quite simple. You can check that $$Av = (1-a)v$$ and so $$v$$ is an eigenvector, whilst for any vector $$w$$ perpendicular to $$v$$ we have $$Aw = w$$. Therefore there is a one-dimensional eigenspace with eigenvalue $$(1-a)$$ spanned by $$v$$, and an","3}, \\phi_{1, 3}$ but by common or differential rotation of both two flat mirrors $\\theta_\\pm \\equiv \\theta_1 \\pm \\theta_3, \\phi_\\pm \\equiv \\phi_1 \\pm \\phi_3$. Therefore, we divide $A$ into two parts $A_1, A_2$ (relation $A = A_1 A_2$): • $A_1$: relation between the beam parameters and the PIT and YAW rotation with $\\theta_\\pm, \\phi_\\pm$ $\\begin{pmatrix} x_\\mathrm{c} \\\\ y_\\mathrm{c} \\\\ \\theta_\\mathrm{c} \\\\ \\phi_\\mathrm{c} \\end{pmatrix} = A_1 \\begin{pmatrix} \\theta_+ \\\\ \\theta_- \\\\ \\theta_2 \\\\ \\phi_+ \\\\ \\phi_- \\\\ \\phi_2 \\end{pmatrix}.$ • $A_2$: relation between $\\theta_\\pm, \\phi_\\pm$ and $\\theta_{1, 3}, \\phi_{1, 3}$ $\\begin{pmatrix} \\theta_+ \\\\ \\theta_- \\\\ \\theta_2 \\\\ \\phi_+ \\\\ \\phi_- \\\\ \\phi_2 \\end{pmatrix} = A_2 \\begin{pmatrix} \\theta_1 \\\\ \\theta_2 \\\\ \\theta_3 \\\\ \\phi_1 \\\\ \\phi_2 \\\\ \\phi_3 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 1 & 0 & 0 & 0 \\\\ 1 & 0 & -1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & -1 \\\\ 0 & 0 & 0 & 0 & 1 & 0 \\\\ \\end{pmatrix} \\begin{pmatrix} \\theta_1 \\\\ \\theta_2 \\\\ \\theta_3 \\\\ \\phi_1 \\\\ \\phi_2 \\\\ \\phi_3 \\end{pmatrix}.$ $A_1$ is represented by revewing F. Kawazoe+ (2011): $A_1 = \\begin{pmatrix} 0 &","$\\newcommand{\\FirstMat}{\\begin{pmatrix*}[r] 1 & -3 & 4 \\\\ 4 & -7 & 8 \\\\ 6 & -7 & 7 \\end{pmatrix*}} \\begin{array}{c} (1) \\\\ \\vphantom{\\FirstMat}\\\\ \\end{array} \\begin{array}{c} A =\\FirstMat\\;,\\\\ \\vphantom{X}\\\\ \\end{array} \\quad \\begin{array}{c} (2) \\\\ \\vphantom{\\FirstMat}\\\\ \\end{array} \\begin{array}{c} A = \\begin{pmatrix*}[r] 0 & -2 & 3 & 2 \\\\ 1 & 1 & -1 & -1 \\\\ 0 & 0 & 2 & 0 \\\\ 1 & -1 & 0 & 1 \\end{pmatrix*}\\;.\\\\ \\end{array}$"],"string":"[\n \"be: \\\\begin{align}\\\\textbf{x}(t)=(18\\\\ln(e^t+1)+c_{1})e^{-t}\\\\begin{pmatrix}1\\\\\\\\0\\\\end{pmatrix}+(-3e^{2t}+6e^t-6\\\\ln(e^t+1)+c_{2})e^{-3t}\\\\begin{pmatrix}1\\\\\\\\-2\\\\end{pmatrix}\\\\end{align} Subbing in the initial condition $\\\\textbf{x}(0)=\\\\begin{pmatrix}1-3\\\\ln2\\\\\\\\-3-3\\\\ln2\\\\end{pmatrix}$: \\\\begin{align}1-3\\\\ln2=c_{1}+3+c_{2}\\\\end{align} \\\\begin{align}-3-3\\\\ln2=-6-2c_{2}\\\\end{align} Solving these equations for the constants gives: \\\\begin{align}c_{1}=\\\\frac{-19-9\\\\ln2}{2}\\\\end{align} \\\\begin{align}c_{2}=\\\\frac{-3+3\\\\ln2}{2}\\\\end{align} Therefore the general solution for the IVP is: \\\\begin{align}\\\\textbf{x}(t)=\\\\frac{-19-9\\\\ln2}{2}e^{-t}\\\\begin{pmatrix}1\\\\\\\\0\\\\end{pmatrix}+\\\\frac{-3+3\\\\ln2}{2}e^{-3t}\\\\begin{pmatrix}1\\\\\\\\-2\\\\end{pmatrix}+18\\\\ln(e^t+1)e^{-t}\\\\begin{pmatrix}1\\\\\\\\0\\\\end{pmatrix}-3e^{-t}\\\\begin{pmatrix}1\\\\\\\\-2\\\\end{pmatrix}+6e^{-2t}\\\\begin{pmatrix}1\\\\\\\\-2\\\\end{pmatrix}-6\\\\ln(e^t+1)e^{-3t}\\\\begin{pmatrix}1\\\\\\\\-2\\\\end{pmatrix} \\\\end{align}\",\n \"-1 & 3\\\\\\\\ 0 & 0 & 1 & 1 & 0 & -1 \\\\end{array}\\\\right) \\\\\\\\[4mm]&\\\\xrightarrow{R1 := R1 - R2} \\\\left(\\\\begin{array}{lll|lll} 1 & 0 & 0 & 0 & 1 & -1\\\\\\\\ 0 & 1 & 0 & -1 & -1 & 3\\\\\\\\ 0 & 0 & 1 & 1 & 0 & -1 \\\\end{array}\\\\right) \\\\end{align*} So the solution is: \\\\begin{align*} \\\\begin{pmatrix} x_1\\\\\\\\x_2\\\\\\\\x_3 \\\\end{pmatrix}= \\\\begin{pmatrix} 0 & 1 & -1\\\\\\\\ -1 & -1 & 3\\\\\\\\ 1 & 0 & -1 \\\\end{pmatrix} \\\\begin{pmatrix} 1\\\\\\\\2\\\\\\\\3 \\\\end{pmatrix} =\\\\begin{pmatrix} -1\\\\\\\\6\\\\\\\\-2 \\\\end{pmatrix}. \\\\end{align*} ​To see the full video page and find related videos, click the following link. Linear Algebra for Math 308: L6E1inv 2. Solve the system:\\\\begin{align*} x_1 - x_2 + x_3 &=1\\\\\\\\ x_1 + 3x_2 + 5x_3 &=0\\\\\\\\ 3x_1 + 2x_2 + x_3 &=1. \\\\end{align*} Note: There is no video solution for this exercise. We are going to find the inverse matrix: \\\\begin{align*} \\\\left(\\\\begin{array}{ccc|ccc} 1 & -1 & 1 & 1 & 0 & 0\\\\\\\\ 1 & 3 & 5 & 0 & 1 & 0\\\\\\\\ 3 & 2 & 1 & 0 & 0 & 1 \\\\end{array}\\\\right) &\\\\xrightarrow[{R2 := \\\\frac{1}{4}(R2 - R1)}]{R3 := R3 - 3R1} \\\\left(\\\\begin{array}{ccc|ccc} 1 & -1 & 1 & 1 & 0 & 0\\\\\\\\ 0 & 1 & 1 & -1/4 & 1/4 & 0\\\\\\\\ 0 &\",\n \"column: \\\\begin{align*} \\\\left(\\\\begin{array}{llll|l} 1 & 1 & 2 & 0 & 3\\\\\\\\ 0 & 1 & 1 & 0 & 1\\\\\\\\ 0 & 0 & 0 & 1 & 0 \\\\end{array}\\\\right) \\\\xrightarrow{R1 : R1 - R2} \\\\left(\\\\begin{array}{llll|l} 1 & 0 & 1 & 0 & 2\\\\\\\\ 0 & 1 & 1 & 0 & 1\\\\\\\\ 0 & 0 & 0 & 1 & 0 \\\\end{array}\\\\right) \\\\end{align*} This is the same as this system: $$\\\\hspace{.5in}\\\\begin{array}{ll} x_1 + x_3 &= 2\\\\\\\\ x_2 + x_3 &= 1\\\\\\\\ x_4 &= 0 \\\\end{array}$$ Which is the same as $$\\\\begin{array}{ll} x_1 &= 2 - x_3\\\\\\\\ x_2 &= 1 - x_3\\\\\\\\ x_3 &= \\\\\\\\ x_4 &= 0\\\\end{array}$$ As we have seen above, this means that the equations place no restrictions on $$x_3$$. That is, for every value of $$x_3$$, there is a solution and so we call it a \\\"free variable''. (This is discussed more below). The solution can be written in vector form: \\\\begin{align} \\\\begin{pmatrix}x_1\\\\\\\\x_2\\\\\\\\x_3\\\\\\\\x_4\\\\end{pmatrix} =\\\\begin{pmatrix}2\\\\\\\\1\\\\\\\\0\\\\\\\\0\\\\end{pmatrix} + t\\\\begin{pmatrix}-1\\\\\\\\-1\\\\\\\\1\\\\\\\\0\\\\end{pmatrix} \\\\end{align} Note that we have use \\\"$$t$$'' instead of $$x_3$$''. This is because \\\"$$x_3$$'' is really only a place holder and also appears on the left hand side of the previous equation. Also, it is usually customary to write the set of solutions in \\\"set builder notation'': \\\\begin{align*} \\\\left\\\\{\\\\begin{pmatrix}2\\\\\\\\1\\\\\\\\0\\\\\\\\0\\\\end{pmatrix} + t\\\\begin{pmatrix}-1\\\\\\\\-1\\\\\\\\1\\\\\\\\0\\\\end{pmatrix} : t\\\\in\\\\mathbb{R}\\\\right\\\\} \\\\end{align*} ​To see\",\n \"-1 & 1 & -1\\\\\\\\ 1 & 1 & -1 & -1\\\\\\\\1 & -1 & -1 & 1 \\\\end{pmatrix}$$. Therefore, $$U_2 =\\\\begin{pmatrix} 1 & 1 & 1 & 1\\\\\\\\ 1 & -1 & 1 & -1\\\\\\\\ 1 & 1 & -1 & -1\\\\\\\\1 & -1 & -1 & 1 \\\\end{pmatrix} \\\\begin{pmatrix} 1 & 0 & 0 & 0\\\\\\\\ 0 & 1 & 0 & 0\\\\\\\\ 0 & 0 & 0 & 1\\\\\\\\ 0 & 0 & 1 & 0 \\\\end{pmatrix} \\\\begin{pmatrix} 1 & 1 & 1 & 1\\\\\\\\ 1 & -1 & 1 & -1\\\\\\\\ 1 & 1 & -1 & -1\\\\\\\\1 & -1 & -1 & 1 \\\\end{pmatrix} = \\\\begin{pmatrix} 1 & 0 & 0 & 0\\\\\\\\ 0 & 0 & 0 & 1\\\\\\\\ 0 & 0 & 1 & 0\\\\\\\\ 0 & 1 & 0 & 0 \\\\end{pmatrix}$$ Thus, $$U_1 = U_2$$.\",\n \"{\\\\begin{pmatrix}1\\\\\\\\2\\\\end{pmatrix}}+{\\\\begin{pmatrix}0\\\\\\\\3\\\\end{pmatrix}}t={\\\\begin{pmatrix}1\\\\\\\\8\\\\end{pmatrix}}+{\\\\begin{pmatrix}0\\\\\\\\-1\\\\end{pmatrix}}(6-3t)}$ and so any vector in the form for ${\\\\displaystyle S_{1}}$ can be stated in the form needed for inclusion in ${\\\\displaystyle S_{2}}$. For ${\\\\displaystyle S_{2}\\\\subseteq S_{1}}$, we look for ${\\\\displaystyle t}$ so that these equations hold. ${\\\\displaystyle {\\\\begin{array}{*{2}{rc}r}1&+&0t&=&1+0s\\\\\\\\2&+&3t&=&8-1s\\\\end{array}}}$ Rewrite that as ${\\\\displaystyle {\\\\begin{array}{*{1}{rc}r}1&=&1\\\\\\\\t&=&2-(1/3)s\\\\end{array}}}$ and so ${\\\\displaystyle {\\\\begin{pmatrix}1\\\\\\\\8\\\\end{pmatrix}}+{\\\\begin{pmatrix}0\\\\\\\\-1\\\\end{pmatrix}}s={\\\\begin{pmatrix}1\\\\\\\\2\\\\end{pmatrix}}+{\\\\begin{pmatrix}0\\\\\\\\3\\\\end{pmatrix}}(2-(1/3)s).}$ 2. These two are equal. To show that ${\\\\displaystyle S_{1}\\\\subseteq S_{2}}$, we check that for any ${\\\\displaystyle t,s}$ we can find an appropriate ${\\\\displaystyle m,n}$ so that these hold. ${\\\\displaystyle {\\\\begin{array}{*{2}{rc}r}4m&-&4n&=&1t+2s\\\\\\\\7m&-&2n&=&3t+1s\\\\\\\\7m&-&10n&=&1t+5s\\\\end{array}}}$ Use Gauss' method ${\\\\displaystyle {\\\\begin{array}{rcl}\\\\left({\\\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\\\\\7&-2&3t+1s\\\\\\\\7&-10&1t+5s\\\\end{array}}\\\\right)&{\\\\xrightarrow[{(-7/4)\\\\rho _{1}+\\\\rho _{3}}]{(-7/4)\\\\rho _{1}+\\\\rho _{2}}}&\\\\left({\\\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\\\\\0&5&(5/4)t-(10/4)s\\\\\\\\0&-3&-(3/4)t+(6/4)s\\\\end{array}}\\\\right)\\\\\\\\&{\\\\xrightarrow[{}]{(3/5)\\\\rho _{2}+\\\\rho _{3}}}&\\\\left({\\\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\\\\\0&5&(5/4)t-(10/4)s\\\\\\\\0&0&0\\\\end{array}}\\\\right)\\\\end{array}}}$ to conclude that ${\\\\displaystyle {\\\\begin{pmatrix}1\\\\\\\\3\\\\\\\\1\\\\end{pmatrix}}t+{\\\\begin{pmatrix}2\\\\\\\\1\\\\\\\\5\\\\end{pmatrix}}s={\\\\begin{pmatrix}4\\\\\\\\7\\\\\\\\7\\\\end{pmatrix}}((1/2)t)+{\\\\begin{pmatrix}-4\\\\\\\\-2\\\\\\\\-10\\\\end{pmatrix}}((1/4)t-(1/2)s)}$ and so ${\\\\displaystyle S_{1}\\\\subseteq S_{2}}$. For ${\\\\displaystyle S_{2}\\\\subseteq S_{1}}$, solve ${\\\\displaystyle {\\\\begin{array}{*{2}{rc}r}1t&+&2s&=&4m-4n\\\\\\\\3t&+&1s&=&7m-2n\\\\\\\\1t&+&5s&=&7m-10n\\\\end{array}}}$ with Gaussian reduction ${\\\\displaystyle {\\\\begin{array}{rcl}\\\\left({\\\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\\\\\3&1&7m-2n\\\\\\\\1&5&7m-10n\\\\end{array}}\\\\right)&{\\\\xrightarrow[{-\\\\rho _{1}+\\\\rho _{3}}]{-3\\\\rho _{1}+\\\\rho _{2}}}&\\\\left({\\\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\\\\\0&-5&-5m+10n\\\\\\\\0&3&3m-6n\\\\end{array}}\\\\right)\\\\\\\\&{\\\\xrightarrow[{}]{(3/5)\\\\rho _{2}+\\\\rho _{3}}}&\\\\left({\\\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\\\\\0&-5&-5m+10n\\\\\\\\0&0&0\\\\end{array}}\\\\right)\\\\end{array}}}$ to get ${\\\\displaystyle {\\\\begin{pmatrix}4\\\\\\\\7\\\\\\\\7\\\\end{pmatrix}}m+{\\\\begin{pmatrix}-4\\\\\\\\-2\\\\\\\\-10\\\\end{pmatrix}}n={\\\\begin{pmatrix}1\\\\\\\\3\\\\\\\\1\\\\end{pmatrix}}(2m)+{\\\\begin{pmatrix}2\\\\\\\\1\\\\\\\\5\\\\end{pmatrix}}(m-2n)}$ and so any member of ${\\\\displaystyle S_{2}}$ can be expressed in the form needed for ${\\\\displaystyle S_{1}}$. 3. These sets are equal. To prove that ${\\\\displaystyle S_{1}\\\\subseteq S_{2}}$, we must be able to solve ${\\\\displaystyle {\\\\begin{array}{*{2}{rc}r}2m&+&4n&=&1t\\\\\\\\4m&+&8n&=&2t\\\\end{array}}}$ for ${\\\\displaystyle m}$ and ${\\\\displaystyle n}$ in terms of ${\\\\displaystyle t}$. Apply Gaussian reduction ${\\\\displaystyle \\\\left({\\\\begin{array}{*{2}{c}|c}2&4&1t\\\\\\\\4&8&2t\\\\end{array}}\\\\right){\\\\xrightarrow[{}]{-2\\\\rho _{1}+\\\\rho _{2}}}\\\\left({\\\\begin{array}{*{2}{c}|c}2&4&1t\\\\\\\\0&0&0\\\\end{array}}\\\\right)}$ to conclude that any pair ${\\\\displaystyle m,n}$ where ${\\\\displaystyle 2m+4n=t}$ will do. For instance, ${\\\\displaystyle {\\\\begin{pmatrix}1\\\\\\\\2\\\\end{pmatrix}}t={\\\\begin{pmatrix}2\\\\\\\\4\\\\end{pmatrix}}((1/2)t)+{\\\\begin{pmatrix}4\\\\\\\\8\\\\end{pmatrix}}(0)}$ or ${\\\\displaystyle {\\\\begin{pmatrix}1\\\\\\\\2\\\\end{pmatrix}}t={\\\\begin{pmatrix}2\\\\\\\\4\\\\end{pmatrix}}((-3/2)t)+{\\\\begin{pmatrix}4\\\\\\\\8\\\\end{pmatrix}}(t).}$ Thus ${\\\\displaystyle S_{1}\\\\subseteq S_{2}}$. For ${\\\\displaystyle S_{2}\\\\subseteq S_{1}}$, we solve ${\\\\displaystyle {\\\\begin{array}{*{2}{rc}r}1t&=&2m+4n\\\\\\\\2t&=&4m+8n\\\\end{array}}}$ with Gauss' method ${\\\\displaystyle {\\\\begin{array}{rcl}\\\\left({\\\\begin{array}{*{1}{c}|c}1&2m+4n\\\\\\\\2&4m+8n\\\\end{array}}\\\\right)&{\\\\xrightarrow[{}]{-2\\\\rho _{1}+\\\\rho _{2}}}&\\\\left({\\\\begin{array}{*{1}{c}|c}1&2m+4n\\\\\\\\0&0\\\\end{array}}\\\\right)\\\\end{array}}}$ to deduce that any\",\n \"& 1 \\\\\\\\ 0 & 2 & 4 \\\\\\\\ −1 & 1 & −1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\frac{1}{24} \\\\begin{pmatrix} 6 & −3 & −6 \\\\\\\\ 4 & 2 & 12 \\\\\\\\ −2 & 5 & −6 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 9) \\\\hspace{.5cm} \\\\begin{pmatrix} 1 & 2 & 1 & −2 \\\\\\\\ 1 & 1 & 0 & 0 \\\\\\\\ 0 & 1 & 0 & −1 \\\\\\\\ 2 & 1 & 0 & 2 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\begin{pmatrix} 0 & 3 & −2 & −1 \\\\\\\\ 0 & −2 & 2 & 1 \\\\\\\\ 1 & −3 & 0 & 1 \\\\\\\\ 0 & −2 & 1 & 1 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 10) \\\\hspace{.5cm} \\\\begin{pmatrix} 2 & 3 & 1 & 0 \\\\\\\\ −1 & 0 & 3 & 1 \\\\\\\\ 2 & 2 & 4 & 1 \\\\\\\\ 2 & −1 & 2 & −1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\frac{1}{25} \\\\begin{pmatrix} −11 & −17 & 16 & −1 \\\\\\\\ 14 & 8 & −9 & −1 \\\\\\\\ 5 & 10 & −5 & 5 \\\\\\\\ −26 & −22 & 31 & −16 \\\\end{pmatrix}\\\\end{align*} For each equation $Ax=b$, tell whether $A^{-1}$ exists. If it does, then compute the solution $x=A^{-1}b$. (You can view the solution by clicking on the problem.) \\\\begin{align*} 11) \\\\hspace{.5cm} \\\\begin{pmatrix} 1 & 4\",\n \"the observer). \\\\begin{aligned} R_x(\\\\theta)&=\\\\begin{pmatrix}1 &0 &0\\\\\\\\ 0 &\\\\cos\\\\theta &-\\\\sin\\\\theta\\\\\\\\ 0 &\\\\sin\\\\theta &\\\\cos\\\\theta \\\\end{pmatrix}\\\\\\\\ R_y(\\\\theta)&=\\\\begin{pmatrix}\\\\cos\\\\theta &0 &\\\\sin\\\\theta\\\\\\\\ 0 &1 &0\\\\\\\\ -\\\\sin\\\\theta &0 &\\\\cos\\\\theta \\\\end{pmatrix}\\\\\\\\ R_z(\\\\theta)&=\\\\begin{pmatrix} \\\\cos\\\\theta &-\\\\sin\\\\theta &0\\\\\\\\ \\\\sin\\\\theta &\\\\cos\\\\theta &0\\\\\\\\ 0 &0 &1 \\\\end{pmatrix} \\\\end{aligned} ## Example 1 Rotating $\\\\begin{pmatrix}1\\\\\\\\ 0\\\\\\\\ 0\\\\end{pmatrix}$ $45^\\\\circ$ anticlockwise about $z$-axis: \\\\begin{aligned} R_z(45^\\\\circ)\\\\begin{pmatrix}1\\\\\\\\ 0\\\\\\\\ 0\\\\end{pmatrix}&=\\\\begin{pmatrix} \\\\cos 45^\\\\circ &-\\\\sin 45^\\\\circ &0\\\\\\\\ \\\\sin 45^\\\\circ &\\\\cos 45^\\\\circ &0\\\\\\\\ 0 &0 &1 \\\\end{pmatrix} \\\\begin{pmatrix}1\\\\\\\\ 0\\\\\\\\ 0\\\\end{pmatrix}\\\\\\\\ &=\\\\begin{pmatrix}\\\\sqrt{2}/2\\\\\\\\\\\\sqrt{2}/2\\\\\\\\0\\\\end{pmatrix}. \\\\end{aligned} ## Example 2 Rotating $\\\\begin{pmatrix}0\\\\\\\\ 1\\\\\\\\ 0\\\\end{pmatrix}$ $45^\\\\circ$ anticlockwise about $z$-axis: \\\\begin{aligned} R_z(45^\\\\circ)\\\\begin{pmatrix}0\\\\\\\\ 1\\\\\\\\ 0\\\\end{pmatrix}&=\\\\begin{pmatrix} \\\\cos 45^\\\\circ &-\\\\sin 45^\\\\circ &0\\\\\\\\ \\\\sin 45^\\\\circ &\\\\cos 45^\\\\circ &0\\\\\\\\ 0 &0 &1 \\\\end{pmatrix} \\\\begin{pmatrix}0\\\\\\\\ 1\\\\\\\\ 0\\\\end{pmatrix}\\\\\\\\ &=\\\\begin{pmatrix}-\\\\sqrt{2}/2\\\\\\\\\\\\sqrt{2}/2\\\\\\\\0\\\\end{pmatrix}. \\\\end{aligned} Posted in Uncategorized | 1 Comment ## 十二平均律 12-tone Equal Temperament 《十二平均律》 是明朝 "布衣王子" 朱载堉 (1536年-1610年) 发明的,由当时在中国传教的意大利人 利马窦 (Matteo Ricci, 传给欧洲的法国数学家 Marin Mersenne (Mersenne Prime ) 。现代音乐之父 巴哈 Bach 第一个采用,制作世界第一架钢琴有12黑白键,并作曲 《Bach 12-tone Equal Temperament》。 Side Note: 1977年法国大学数学教授在课堂好奇地问我,你们祖先如何解代数?是用算盘吗?当时计算机还不流行,复杂的算法只能用Log Table 或 计算尺 (Slide Rule) 。 Bach 12-tone Equal Temperament https://zh.m.wikipedia.org/wiki/%E5%8D%81%E4%BA%8C%E5%B9%B3%E5%9D%87%E5%BE%8B View original post ## Richard Dedekind Julius Wilhelmina Richard Dedekind (6 Oct 1831 – 12 Feb 1916) – Last student of Gauss at Göttingen – Student and closed friend of Dirichlet who influenced his Mathematical education – Introduced the word Field (Körper) – Gave the first university course on Galois Theory – Developed Real Number ‘Dedekind Cut‘ in 1872 – Accomplished musician – Never married, lived with\",\n \"by clicking on the problem.) \\\\begin{align*} 1) \\\\hspace{.5cm} \\\\begin{pmatrix} 3 & 1 \\\\\\\\ 2 & 1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\begin{pmatrix} 1 & −1 \\\\\\\\ −2 & 3 \\\\end{pmatrix}\\\\end{align*} \\\\begin{align*} 2) \\\\hspace{.5cm} \\\\begin{pmatrix} 2 & 1 \\\\\\\\ 4 & 3 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*}\\\\begin{pmatrix} \\\\frac{3}{2} & −\\\\frac{1}{2} \\\\\\\\ −2 & 1 \\\\end{pmatrix} \\\\end{align*} \\\\begin{align*} 3) \\\\hspace{.5cm} \\\\begin{pmatrix} 0 & 1 \\\\\\\\ 5 & 7 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\begin{pmatrix} −\\\\frac{7}{5} & \\\\frac{1}{5} \\\\\\\\ 1 & 0 \\\\end{pmatrix}\\\\end{align*} \\\\begin{align*} 4) \\\\hspace{.5cm} \\\\begin{pmatrix} 2 & 1 \\\\\\\\ 6 & 3 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\text{not invertible} \\\\end{align*} \\\\begin{align*} 5) \\\\hspace{.5cm} \\\\begin{pmatrix} 1 & −2 & 0 \\\\\\\\ 0 & 1 & 0 \\\\\\\\ 0 & −1 & 1 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\begin{pmatrix} 1 & 2 & 0 \\\\\\\\ 0 & 1 & 0 \\\\\\\\ 0 & 1 & 1 \\\\end{pmatrix}\\\\end{align*} \\\\begin{align*} 6) \\\\hspace{.5cm} \\\\begin{pmatrix} 1 & 0 & 1 \\\\\\\\ 2 & 3 & 4 \\\\\\\\ −1 & 1 & 0 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\begin{pmatrix} −4 & 1 & −3 \\\\\\\\ −4 & 1 & −2 \\\\\\\\ 5 & −1 & 3 \\\\end{pmatrix}\\\\end{align*} \\\\begin{align*} 7) \\\\hspace{.5cm} \\\\begin{pmatrix} 2 & 1 & 1 \\\\\\\\ 1 & 1 & 7 \\\\\\\\ 5 & 2 & −4 \\\\end{pmatrix} \\\\end{align*} Solution: \\\\begin{align*} \\\\text{not invertible} \\\\end{align*} \\\\begin{align*} 8) \\\\hspace{.5cm} \\\\begin{pmatrix} 3 & 2\",\n \"to align the \\\\Rightarrows – daleif Oct 6 '16 at 7:59 This is an automatic solution, based on the eqparbox package. I defined an \\\\eqmathbox command, which accepts $2$ optional arguments (the end of name of the box, and the alignment: l, r or c) and one mandatory argument: the contents of the math box. \\\\documentclass{article} %\\\\usepackage{enumitem} \\\\usepackage{mathtools} \\\\usepackage{xparse} \\\\usepackage{eqparbox} \\\\NewDocumentCommand\\\\Eqmathbox{O{0}O{l}m}{\\\\eqmakebox[Form#1][#2]{\\\\ensuremath{ \\\\displaystyle#3}}} \\\\usepackage{siunitx} \\\\sisetup{group-separator={,},group-minimum-digits = 4} \\\\begin{document} \\\\begin{enumerate} \\\\item $$\\\\Eqmathbox{\\\\begin{pmatrix*}[r] 20 & 25 & 20 \\\\\\\\ 15 & 20 & 18 \\\\\\\\ 18 & 23 & 22 \\\\end{pmatrix*} \\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ z \\\\end{pmatrix*} = \\\\begin{pmatrix*}[r] 975 \\\\\\\\ 810 \\\\\\\\ 965 \\\\end{pmatrix*}} \\\\Rightarrow\\\\Eqmathbox[1][r]{\\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ z \\\\end{pmatrix*} =} \\\\begin{pmatrix*}[r] 10 \\\\\\\\ 15 \\\\\\\\ 20 \\\\end{pmatrix*}$$ \\\\item $$\\\\Eqmathbox{\\\\begin{pmatrix*}[r] 90 & 95 & 80 \\\\\\\\ 80 & 85 & 90 \\\\\\\\ 75 & 90 & 95 \\\\end{pmatrix*} \\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ z \\\\end{pmatrix*} = \\\\begin{pmatrix*}[r] 86 \\\\\\\\ 86 \\\\\\\\ 88 \\\\end{pmatrix*}} \\\\Rightarrow\\\\Eqmathbox[1][r]{\\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ z \\\\end{pmatrix*} =} \\\\begin{pmatrix*}[r] 0.3 \\\\\\\\ 0.2 \\\\\\\\ 0.5 \\\\end{pmatrix*}$$ \\\\item $$\\\\Eqmathbox{\\\\begin{pmatrix*}[r] \\\\num{5000} & \\\\num{400}0 \\\\\\\\ \\\\num{6000} & \\\\num{5000} \\\\end{pmatrix*} \\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ \\\\end{pmatrix*} = \\\\begin{pmatrix*}[r] \\\\num{37000000} \\\\\\\\ \\\\num{45000000 }\\\\\\\\ \\\\end{pmatrix*}} \\\\Rightarrow\\\\Eqmathbox[1][r]{\\\\begin{pmatrix*}[r] x \\\\\\\\ y \\\\\\\\ \\\\end{pmatrix*} =} \\\\begin{pmatrix*}[r] \\\\num{5000} \\\\\\\\ \\\\num{3000} \\\\end{pmatrix*}$$ \\\\end{enumerate} \\\\end{document} Here is one way. Note that you are adding a lot of unneeded\",\n \"\\\\\\\\ 0\\\\end{pmatrix}\\\\right) = \\\\begin{pmatrix} 2 \\\\cdot (1 \\\\cdot 0 + 0 \\\\cdot 0) \\\\\\\\ 2 \\\\cdot (-0 \\\\cdot 0 + 1 \\\\cdot 0) \\\\\\\\ 1^2 + 0^2 - 0^2 - 0^2 \\\\end{pmatrix} = \\\\begin{pmatrix} 0\\\\\\\\0\\\\\\\\1 \\\\end{pmatrix} \\\\\\\\ \\\\phi(|1\\\\rangle) = \\\\phi\\\\left(\\\\begin{pmatrix}0 \\\\\\\\ 1\\\\end{pmatrix}\\\\right) = \\\\begin{pmatrix} 2 \\\\cdot (0 \\\\cdot 1 + 0 \\\\cdot 0) \\\\\\\\ 2 \\\\cdot (-0 \\\\cdot 1 + 0 \\\\cdot 0) \\\\\\\\ 0^2 + 0^2 - 1^2 - 0^2 \\\\end{pmatrix} = \\\\begin{pmatrix} 0\\\\\\\\0\\\\\\\\-1 \\\\end{pmatrix} \\\\\\\\ \\\\phi(|+\\\\rangle) = \\\\phi\\\\left(\\\\begin{pmatrix}\\\\frac{1}{\\\\sqrt{2}} \\\\\\\\ \\\\frac{1}{\\\\sqrt{2}}\\\\end{pmatrix}\\\\right) = \\\\begin{pmatrix} 2 \\\\cdot \\\\left(\\\\frac{1}{\\\\sqrt{2}} \\\\cdot \\\\frac{1}{\\\\sqrt{2}} + 0 \\\\cdot 0\\\\right) \\\\\\\\ 2 \\\\cdot \\\\left(-0 \\\\cdot \\\\frac{1}{\\\\sqrt{2}} + \\\\frac{1}{\\\\sqrt{2}} \\\\cdot 0\\\\right) \\\\\\\\ \\\\left(\\\\frac{1}{\\\\sqrt{2}}\\\\right)^2 + 0^2 - \\\\left(\\\\frac{1}{\\\\sqrt{2}}\\\\right)^2 - 0^2 \\\\end{pmatrix} = \\\\begin{pmatrix} 1\\\\\\\\0\\\\\\\\0 \\\\end{pmatrix} \\\\\\\\ \\\\phi(|-\\\\rangle) = \\\\phi\\\\left(\\\\begin{pmatrix}\\\\frac{1}{\\\\sqrt{2}} \\\\\\\\ -\\\\frac{1}{\\\\sqrt{2}}\\\\end{pmatrix}\\\\right) = \\\\begin{pmatrix} 2 \\\\cdot \\\\left(\\\\frac{1}{\\\\sqrt{2}} \\\\cdot \\\\left(-\\\\frac{1}{\\\\sqrt{2}}\\\\right) + 0 \\\\cdot 0\\\\right) \\\\\\\\ 2 \\\\cdot \\\\left(-0 \\\\cdot \\\\left(-\\\\frac{1}{\\\\sqrt{2}}\\\\right) + \\\\frac{1}{\\\\sqrt{2}} \\\\cdot 0\\\\right) \\\\\\\\ \\\\left(\\\\frac{1}{\\\\sqrt{2}}\\\\right)^2 + 0^2 - \\\\left(-\\\\frac{1}{\\\\sqrt{2}}\\\\right)^2 - 0^2 \\\\end{pmatrix} = \\\\begin{pmatrix} -1\\\\\\\\0\\\\\\\\0 \\\\end{pmatrix} \\\\\\\\ \\\\phi(|{+i}\\\\rangle) = \\\\phi\\\\left(\\\\begin{pmatrix}\\\\frac{1}{\\\\sqrt{2}} \\\\\\\\ \\\\frac{i}{\\\\sqrt{2}}\\\\end{pmatrix}\\\\right) = \\\\begin{pmatrix} 2 \\\\cdot \\\\left(\\\\frac{1}{\\\\sqrt{2}} \\\\cdot 0 + 0 \\\\cdot \\\\frac{1}{\\\\sqrt{2}}\\\\right) \\\\\\\\ 2 \\\\cdot \\\\left(-0 \\\\cdot 0 + \\\\frac{1}{\\\\sqrt{2}} \\\\cdot \\\\frac{1}{\\\\sqrt{2}}\\\\right) \\\\\\\\ \\\\left(\\\\frac{1}{\\\\sqrt{2}}\\\\right)^2 + 0^2 - 0^2 - \\\\left(\\\\frac{1}{\\\\sqrt{2}}\\\\right)^2 \\\\end{pmatrix} = \\\\begin{pmatrix}0\\\\\\\\1\\\\\\\\0 \\\\end{pmatrix} \\\\\\\\ \\\\phi(|{-i}\\\\rangle) = \\\\phi\\\\left(\\\\begin{pmatrix}\\\\frac{1}{\\\\sqrt{2}} \\\\\\\\ -\\\\frac{i}{\\\\sqrt{2}}\\\\end{pmatrix}\\\\right) = \\\\begin{pmatrix} 2 \\\\cdot \\\\left(\\\\frac{1}{\\\\sqrt{2}} \\\\cdot 0 + 0 \\\\cdot \\\\left(-\\\\frac{1}{\\\\sqrt{2}}\\\\right)\\\\right) \\\\\\\\ 2 \\\\cdot \\\\left(-0 \\\\cdot 0\",\n \"\\\\stackrel{E_7}{\\\\sim} \\\\begin{pmatrix} 0&1&2&2\\\\\\\\ 2&0&-3&1\\\\\\\\ -4&0&9&2\\\\\\\\ 0&-1&1&-1\\\\\\\\ \\\\end{pmatrix} \\\\stackrel{E_6E_5E_4E_3E_2E_1}{\\\\sim} L \\\\end{eqnarray*} \\\\begin{eqnarray*} E_7= \\\\begin{pmatrix} 0&1&0&0\\\\\\\\ 1&0&0&0\\\\\\\\ 0&0&1&0\\\\\\\\ 0&0&0&1\\\\\\\\ \\\\end{pmatrix} =E_7^{-1} \\\\end{eqnarray*} \\\\begin{eqnarray*} M=(E_1^{-1}E_2^{-1}E_3^{-1})(E_4^{-1}E_5^{-1}E_6^{-1}) (E_7^{-1}) U=LDPU\\\\\\\\ \\\\end{eqnarray*} \\\\begin{eqnarray*} \\\\begin{pmatrix} 0&1&2&2\\\\\\\\ 2&0&-3&1\\\\\\\\ -4&0&9&2\\\\\\\\ 0&-1&1&-1\\\\\\\\ \\\\end{pmatrix} = \\\\begin{pmatrix} 1&0&0&0\\\\\\\\ 0&1&0&0\\\\\\\\ -2&0&1&0\\\\\\\\ 0&-1&1&1\\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} 2&0&0&0\\\\\\\\ 0&1&0&0\\\\\\\\ 0&0&3&0\\\\\\\\ 0&0&1&-3\\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} 0&1&0&0\\\\\\\\ 1&0&0&0\\\\\\\\ 0&0&1&0\\\\\\\\ 0&0&0&1\\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} 1&0&\\\\frac{-3}{2}&\\\\frac{1}{2}\\\\\\\\ 0&1&2&2\\\\\\\\ 0&0&1&\\\\frac43\\\\\\\\ 0&0&0&1\\\\\\\\ \\\\end{pmatrix} \\\\end{eqnarray*}\",\n \"F_{n}={\\\\cfrac {1}{\\\\sqrt {5}}}\\\\left({\\\\cfrac {1+{\\\\sqrt {5}}}{2}}\\\\right)^{n}-{\\\\cfrac {1}{\\\\sqrt {5}}}\\\\left({\\\\cfrac {1-{\\\\sqrt {5}}}{2}}\\\\right)^{n}.}$ Equivalently, the same computation may performed by diagonalization of A through use of its eigendecomposition: {\\\\displaystyle {\\\\begin{aligned}A&=S\\\\Lambda S^{-1},\\\\\\\\A^{n}&=S\\\\Lambda ^{n}S^{-1},\\\\end{aligned}}} where ${\\\\displaystyle \\\\Lambda ={\\\\begin{pmatrix}\\\\varphi &0\\\\\\\\0&-\\\\varphi ^{-1}\\\\end{pmatrix}}}$ and ${\\\\displaystyle S={\\\\begin{pmatrix}\\\\varphi &-\\\\varphi ^{-1}\\\\\\\\1&1\\\\end{pmatrix}}.}$ The closed-form expression for the nth element in the Fibonacci series is therefore given by {\\\\displaystyle {\\\\begin{aligned}{F_{n+1} \\\\choose F_{n}}&=A^{n}{F_{1} \\\\choose F_{0}}\\\\\\\\&=S\\\\Lambda ^{n}S^{-1}{F_{1} \\\\choose F_{0}}\\\\\\\\&=S{\\\\begin{pmatrix}\\\\varphi ^{n}&0\\\\\\\\0&(-\\\\varphi )^{-n}\\\\end{pmatrix}}S^{-1}{F_{1} \\\\choose F_{0}}\\\\\\\\&={\\\\begin{pmatrix}\\\\varphi &-\\\\varphi ^{-1}\\\\\\\\1&1\\\\end{pmatrix}}{\\\\begin{pmatrix}\\\\varphi ^{n}&0\\\\\\\\0&(-\\\\varphi )^{-n}\\\\end{pmatrix}}{\\\\frac {1}{\\\\sqrt {5}}}{\\\\begin{pmatrix}1&\\\\varphi ^{-1}\\\\\\\\-1&\\\\varphi \\\\end{pmatrix}}{1 \\\\choose 0},\\\\end{aligned}}} which again yields ${\\\\displaystyle F_{n}={\\\\cfrac {\\\\varphi ^{n}-(-\\\\varphi )^{-n}}{\\\\sqrt {5}}}.}$ The matrix A has a determinant of −1, and thus it is a 2×2 unimodular matrix. This property can be understood in terms of the continued fraction representation for the golden ratio: ${\\\\displaystyle \\\\varphi =1+{\\\\cfrac {1}{1+{\\\\cfrac {1}{1+{\\\\cfrac {1}{1+\\\\ddots }}}}}}.}$ The Fibonacci numbers occur as the ratio of successive convergents of the continued fraction for φ, and the matrix formed from successive convergents of any continued fraction has a determinant of +1 or −1. The matrix representation gives the following closed-form expression for the Fibonacci numbers: ${\\\\displaystyle {\\\\begin{pmatrix}1&1\\\\\\\\1&0\\\\end{pmatrix}}^{n}={\\\\begin{pmatrix}F_{n+1}&F_{n}\\\\\\\\F_{n}&F_{n-1}\\\\end{pmatrix}}.}$ For a given n, this matrix can be computed in O(log(n)) arithmetic operations, using the exponentiation by squaring method. Taking the determinant of both sides of this equation yields Cassini's identity, ${\\\\displaystyle (-1)^{n}=F_{n+1}F_{n-1}-{F_{n}}^{2}.}$ Moreover, since An Am = An+m for any square matrix A, the\",\n \"& 0 \\\\\\\\0 & 0 & 0 & -1 \\\\\\\\0 & 1 & 0 & 0 \\\\\\\\1 & 0 & 0 & 0 \\\\end{pmatrix} = \\\\begin{pmatrix} 0 & 0 & 1 & 0 \\\\\\\\0 & 0 & 0 & -1 \\\\\\\\0 & -1 & 0 & 0 \\\\\\\\-1 & 0 & 0 & 0 \\\\end{pmatrix}$$\",\n \"/ a^* = e^{2 i \\\\phi}$ for some $\\\\phi$ real – reuns Oct 7 '16 at 14:24 ## 1 Answer \\\\begin{align*} \\\\begin{pmatrix} 1 & 0 \\\\\\\\ 0 & 1 \\\\end{pmatrix} &= \\\\begin{pmatrix} a & b \\\\\\\\ c & d \\\\end{pmatrix} \\\\begin{pmatrix} \\\\bar{d} & -\\\\bar{b} \\\\\\\\ -\\\\bar{c} & \\\\bar{a} \\\\end{pmatrix} \\\\\\\\ &= \\\\begin{pmatrix} a\\\\bar{d}-b\\\\bar{c} & b\\\\bar{a}-a\\\\bar{b} \\\\\\\\ c\\\\bar{d}-d\\\\bar{c} & d\\\\bar{a}-c\\\\bar{b} \\\\end{pmatrix} \\\\end{align*} Equating \\\"$0$\\\": \\\\left \\\\{ \\\\begin{align*} \\\\frac{a}{\\\\bar{a}} &= \\\\frac{b}{\\\\bar{b}} \\\\\\\\ \\\\frac{c}{\\\\bar{c}} &= \\\\frac{d}{\\\\bar{d}} \\\\\\\\ \\\\end{align*} \\\\right. Let \\\\left \\\\{ \\\\begin{align*} a &= Ae^{i\\\\theta} \\\\\\\\ b &= Be^{i\\\\theta} \\\\end{align*} \\\\right. and \\\\left \\\\{ \\\\begin{align*} c &= Ce^{i\\\\phi} \\\\\\\\ d &= De^{i\\\\phi}\\\\ \\\\end{align*} \\\\right. where $A,B,C,D\\\\in \\\\mathbb{R}$ Equating \\\"$1$\\\": \\\\left \\\\{ \\\\begin{align*} (AD-BC)e^{i(\\\\theta-\\\\phi)} &= 1 \\\\\\\\ (AD-BC)e^{i(\\\\phi-\\\\theta)} &= 1 \\\\\\\\ \\\\end{align*} \\\\right. \\\\implies \\\\left \\\\{ \\\\begin{align*} AD-BC &= 1 \\\\\\\\ \\\\theta &= \\\\phi \\\\\\\\ \\\\end{align*} \\\\right.\",\n \"& -3/50 \\\\end{array} \\\\right) \\\\left( \\\\begin{array}{cc} 1 & -2 \\\\\\\\ 1 & 3 \\\\end{array} \\\\right) \\\\begin{pmatrix} u_1 \\\\\\\\ u_2 \\\\end{pmatrix} + \\\\begin{pmatrix} 0.1 \\\\\\\\ 0 \\\\end{pmatrix}$ Multiply throughout by $\\\\left( \\\\begin{array}{cc} 1 & -3 \\\\\\\\ 1 & 2 \\\\end{array} \\\\right)^{-1} = 1/5 \\\\left( \\\\begin{array}{cc} 2 & 3 \\\\\\\\ -1 & 1 \\\\end{array} \\\\right)$ to get $\\\\frac{d}{dt} \\\\begin{pmatrix} u_1 \\\\\\\\ u_2 \\\\end{pmatrix} = \\\\left( \\\\begin{array}{cc} -1/50 & 0 \\\\\\\\ 0 & -6/50 \\\\end{array} \\\\right) \\\\begin{pmatrix} u_1 \\\\\\\\ u_2 \\\\end{pmatrix} + \\\\begin{pmatrix} 0.04 \\\\\\\\ -0.02 \\\\end{pmatrix}$ this is equivalent to the differential equations $\\\\frac{du_1}{dt}=-1/50u_1+0.04$ $\\\\frac{du_2}{dt}=-6/50u_1-0.02$ These have the solutions $u_1=2+Ae^{-t/50}, \\\\: u_2= -1/6+Be^{-6t/50}$ We can write this as $\\\\begin{pmatrix} u_1 \\\\\\\\ u_2 \\\\end{pmatrix} = \\\\begin{pmatrix} 2+Ae^{-t/50} \\\\\\\\ -1/6+Be^{-6t/50} \\\\end{pmatrix}$ Now transform back to $x_1, \\\\: x_2$ . \\\\begin{aligned} \\\\begin{pmatrix} x_1 \\\\\\\\ x_2 \\\\end{pmatrix} &= P \\\\begin{pmatrix} u_1 \\\\\\\\ u_2 \\\\end{pmatrix} \\\\\\\\ &= \\\\left( \\\\begin{array}{cc} 1 & -3 \\\\\\\\ 1 & 2 \\\\end{array} \\\\right) \\\\begin{pmatrix} 2+Ae^{-t/50} \\\\\\\\ -1/6+Be^{-6t/50} \\\\end{pmatrix} \\\\\\\\ &= \\\\begin{pmatrix} 5/2+Ae^{-t/50}-3Be^{-6t/50} \\\\\\\\ 5/3+Ae^{-t/50}+2Be^{-6t/50} \\\\end{pmatrix} \\\\end{aligned} . These solutions can be fitted to any initial conditions to find the constants.\",\n \"&=\\\\begin{pmatrix}0&0&1&-5\\\\\\\\ 0&0&0&-7\\\\\\\\ 0&0&0&7\\\\end{pmatrix}\\\\tag{R_3=R_3-R_1}\\\\\\\\\\\\\\\\ &=\\\\begin{pmatrix}0&0&1&-5\\\\\\\\ 0&0&0&1\\\\\\\\ 0&0&0&7\\\\tag{R_2=-\\\\frac17 R_2}\\\\end{pmatrix}\\\\\\\\\\\\\\\\ &=\\\\begin{pmatrix}0&0&1&-5\\\\\\\\ 0&0&0&1\\\\\\\\ 0&0&0&0\\\\tag{R_3=R_3 -7R_2}\\\\end{pmatrix} \\\\end{align}\",\n \"1 \\\\\\\\ -1 & 1 & 3 & -2 \\\\end{pmatrix}\\\\begin{pmatrix} 1 \\\\\\\\ 1 \\\\\\\\ 1 \\\\\\\\ 1 \\\\end{pmatrix} = \\\\begin{pmatrix} 2 \\\\\\\\ 1 \\\\\\\\ 1 \\\\end{pmatrix}.$$ Our normal equation becomes $$\\\\begin{pmatrix} 6 & -1 & 7 \\\\\\\\ -1 & 7 & -8 \\\\\\\\ 7 & -8 & 15 \\\\end{pmatrix}\\\\begin{pmatrix} x_1 \\\\\\\\ x_2 \\\\\\\\ x_3 \\\\end{pmatrix} = \\\\begin{pmatrix} 2 \\\\\\\\ 1 \\\\\\\\ 1 \\\\end{pmatrix},$$ or as an augmented matrix, \\\\begin{align*} \\\\left(\\\\begin{array}{ccc|c} 6 & -1 & 7 & 2 \\\\\\\\ -1 & 7 & -8 & 1 \\\\\\\\ 7 & -8 & 15 & 1 \\\\end{array}\\\\right) &\\\\sim \\\\left(\\\\begin{array}{ccc|c} 1 & -7 & 8 & -1 \\\\\\\\ 6 & -1 & 7 & 2 \\\\\\\\ 7 & -8 & 15 & 1 \\\\end{array}\\\\right) \\\\\\\\ &\\\\sim \\\\left(\\\\begin{array}{ccc|c} 1 & -7 & 8 & -1 \\\\\\\\ 0 & 41 & -41 & 8 \\\\\\\\ 0 & 41 & -41 & 8 \\\\end{array}\\\\right) \\\\\\\\ &\\\\sim \\\\left(\\\\begin{array}{ccc|c} 1 & -7 & 8 & -1 \\\\\\\\ 0 & 1 & -1 & \\\\frac{8}{41} \\\\\\\\ 0 & 0 & 0 & 0 \\\\end{array}\\\\right) \\\\\\\\ &\\\\sim \\\\left(\\\\begin{array}{ccc|c} 1 & 0 & \\\\color{red}1 & \\\\frac{15}{41} \\\\\\\\ 0 & 1 & \\\\color{red}{-1} & \\\\frac{8}{41} \\\\\\\\ 0 & 0 & \\\\color{red}0 & 0 \\\\end{array}\\\\right) \\\\end{align*} Note that, as we always expect from the normal equation, we have consistency; there is no\",\n \"\\\\vdots \\\\\\\\ a & a & \\\\ldots & 1+a\\\\end{pmatrix} =\\\\begin{pmatrix}1 & 0 & \\\\ldots & 0 \\\\\\\\ 0 & 1 & \\\\ldots & 0 \\\\\\\\ \\\\vdots & \\\\vdots & \\\\vdots & \\\\vdots \\\\\\\\ 0 & 0 & \\\\ldots & 1\\\\end{pmatrix}+\\\\begin{pmatrix}a & a & \\\\ldots & a \\\\\\\\ a & a & \\\\ldots & a \\\\\\\\ \\\\vdots & \\\\vdots & \\\\vdots & \\\\vdots \\\\\\\\ a & a & \\\\ldots & a\\\\end{pmatrix} =I+a\\\\begin{pmatrix}1 & 1 & \\\\ldots & 1 \\\\\\\\ 1 & 1 & \\\\ldots & 1 \\\\\\\\ \\\\vdots & \\\\vdots & \\\\vdots & \\\\vdots \\\\\\\\ 1 & 1 & \\\\ldots & 1\\\\end{pmatrix} \\\\\\\\ & =I+a\\\\begin{pmatrix}1 \\\\\\\\ 1 \\\\\\\\ \\\\vdots \\\\\\\\ 1 \\\\end{pmatrix}\\\\begin{pmatrix}1 & 1 & \\\\ldots & 1\\\\end{pmatrix} =I-a\\\\begin{pmatrix}-1 \\\\\\\\ -1 \\\\\\\\ \\\\vdots \\\\\\\\ -1 \\\\end{pmatrix}\\\\begin{pmatrix}-1 & -1 & \\\\ldots & -1\\\\end{pmatrix}\\\\end{align*} So we have in this case $$\\\\vec{v}=\\\\begin{pmatrix}-1 \\\\\\\\ -1 \\\\\\\\ \\\\vdots \\\\\\\\ -1 \\\\end{pmatrix}$$. Therefore the inverse matrix is $$A^{-1}=I+b\\\\vec{v}\\\\vec{v}^T$$ with $$b=a\\\\left (I-a\\\\vec{v}\\\\vec{v}^T\\\\right )^{-1}$$. Is this part correct? • Mar 27, 2020 at 10:55 The linear operator $$A = I - a v v^T$$ is quite simple. You can check that $$Av = (1-a)v$$ and so $$v$$ is an eigenvector, whilst for any vector $$w$$ perpendicular to $$v$$ we have $$Aw = w$$. Therefore there is a one-dimensional eigenspace with eigenvalue $$(1-a)$$ spanned by $$v$$, and an\",\n \"3}, \\\\phi_{1, 3}$ but by common or differential rotation of both two flat mirrors $\\\\theta_\\\\pm \\\\equiv \\\\theta_1 \\\\pm \\\\theta_3, \\\\phi_\\\\pm \\\\equiv \\\\phi_1 \\\\pm \\\\phi_3$. Therefore, we divide $A$ into two parts $A_1, A_2$ (relation $A = A_1 A_2$): • $A_1$: relation between the beam parameters and the PIT and YAW rotation with $\\\\theta_\\\\pm, \\\\phi_\\\\pm$ $\\\\begin{pmatrix} x_\\\\mathrm{c} \\\\\\\\ y_\\\\mathrm{c} \\\\\\\\ \\\\theta_\\\\mathrm{c} \\\\\\\\ \\\\phi_\\\\mathrm{c} \\\\end{pmatrix} = A_1 \\\\begin{pmatrix} \\\\theta_+ \\\\\\\\ \\\\theta_- \\\\\\\\ \\\\theta_2 \\\\\\\\ \\\\phi_+ \\\\\\\\ \\\\phi_- \\\\\\\\ \\\\phi_2 \\\\end{pmatrix}.$ • $A_2$: relation between $\\\\theta_\\\\pm, \\\\phi_\\\\pm$ and $\\\\theta_{1, 3}, \\\\phi_{1, 3}$ $\\\\begin{pmatrix} \\\\theta_+ \\\\\\\\ \\\\theta_- \\\\\\\\ \\\\theta_2 \\\\\\\\ \\\\phi_+ \\\\\\\\ \\\\phi_- \\\\\\\\ \\\\phi_2 \\\\end{pmatrix} = A_2 \\\\begin{pmatrix} \\\\theta_1 \\\\\\\\ \\\\theta_2 \\\\\\\\ \\\\theta_3 \\\\\\\\ \\\\phi_1 \\\\\\\\ \\\\phi_2 \\\\\\\\ \\\\phi_3 \\\\end{pmatrix} = \\\\begin{pmatrix} 1 & 0 & 1 & 0 & 0 & 0 \\\\\\\\ 1 & 0 & -1 & 0 & 0 & 0 \\\\\\\\ 0 & 1 & 0 & 0 & 0 & 0 \\\\\\\\ 0 & 0 & 0 & 1 & 0 & 1 \\\\\\\\ 0 & 0 & 0 & 1 & 0 & -1 \\\\\\\\ 0 & 0 & 0 & 0 & 1 & 0 \\\\\\\\ \\\\end{pmatrix} \\\\begin{pmatrix} \\\\theta_1 \\\\\\\\ \\\\theta_2 \\\\\\\\ \\\\theta_3 \\\\\\\\ \\\\phi_1 \\\\\\\\ \\\\phi_2 \\\\\\\\ \\\\phi_3 \\\\end{pmatrix}.$ $A_1$ is represented by revewing F. Kawazoe+ (2011): $A_1 = \\\\begin{pmatrix} 0 &\",\n \"$\\\\newcommand{\\\\FirstMat}{\\\\begin{pmatrix*}[r] 1 & -3 & 4 \\\\\\\\ 4 & -7 & 8 \\\\\\\\ 6 & -7 & 7 \\\\end{pmatrix*}} \\\\begin{array}{c} (1) \\\\\\\\ \\\\vphantom{\\\\FirstMat}\\\\\\\\ \\\\end{array} \\\\begin{array}{c} A =\\\\FirstMat\\\\;,\\\\\\\\ \\\\vphantom{X}\\\\\\\\ \\\\end{array} \\\\quad \\\\begin{array}{c} (2) \\\\\\\\ \\\\vphantom{\\\\FirstMat}\\\\\\\\ \\\\end{array} \\\\begin{array}{c} A = \\\\begin{pmatrix*}[r] 0 & -2 & 3 & 2 \\\\\\\\ 1 & 1 & -1 & -1 \\\\\\\\ 0 & 0 & 2 & 0 \\\\\\\\ 1 & -1 & 0 & 1 \\\\end{pmatrix*}\\\\;.\\\\\\\\ \\\\end{array}$\"\n]"}}},{"rowIdx":6940,"cells":{"problem":{"kind":"string","value":"The sphere with radius 1 and center $(0,0,1)$ rests on the $xy$-plane. A light source is at $P = (0,-1,2).$ Then the boundary of the shadow of the sphere can be expressed in the form $y = f(x),$ for some function $f(x).$ Find the function $f(x).$"},"level":{"kind":"string","value":"Level 5"},"type":{"kind":"string","value":"Precalculus"},"solution":{"kind":"string","value":"Let $O = (0,0,1)$ be the center of the sphere, and let $X = (x,y,0)$ be a point on the boundary of the shadow. Since $X$ is on the boundary, $\\overline{PX}$ is tangent to the sphere; let $T$ be the point of tangency. Note that $\\angle PTO = 90^\\circ.$ Also, lengths $OP$ and $OT$ are fixed, so $\\angle OPT = \\angle OPX$ is constant for all points $X$ on the boundary.\n\n[asy]\nimport three;\nimport solids;\n\nsize(250);\ncurrentprojection = perspective(6,3,2);\n\ntriple O = (0,0,1), P = (0,-1,2), X = (3, 3^2/4 - 1, 0), T = P + dot(O - P, X - P)/dot(X - P,X - P)*(X - P);\nreal x;\n\npath3 shadow = (-1,1/4 - 1,0);\n\nfor (x = -1; x <= 3.1; x = x + 0.1) {\n shadow = shadow--(x,x^2/4 - 1,0);\n}\n\ndraw(surface(shadow--(3,9/4 - 1,0)--(3,3,0)--(-1,3,0)--(-1,1/4 - 1,0)--cycle),gray(0.8),nolight);\ndraw((3,0,0)--(-2,0,0));\ndraw((0,3,0)--(0,-1.5,0));\ndraw(shadow);\ndraw(shift((0,0,1))*surface(sphere(1)),gray(0.8));\ndraw(O--P,dashed + red);\ndraw(P--X,red);\ndraw(O--T,dashed + red);\n\ndot(\"$O$\", O, SE, white);\ndot(\"$P$\", P, NW);\ndot(\"$X$\", X, S);\ndot(T, red);\nlabel(\"$T$\", T, W);\n[/asy]\n\nIf we take $X = (0,-1,0)$ and $T = (0,-1,1),$ we see that $\\angle OPX = 45^\\circ.$ Hence, the angle between $\\overrightarrow{PX}$ and $\\overrightarrow{PO}$ is $45^\\circ.$ This means\n\\[\\frac{\\begin{pmatrix} x \\\\ y + 1 \\\\ -2 \\end{pmatrix} \\cdot \\begin{pmatrix} 0 \\\\ 1 \\\\ -1 \\end{pmatrix}}{\\left\\| \\begin{pmatrix} x \\\\ y + 1 \\\\ -2 \\end{pmatrix} \\right\\| \\left\\| \\begin{pmatrix} 0 \\\\ 1 \\\\ -1 \\end{pmatrix} \\right\\|} = \\cos 45^\\circ = \\frac{1}{\\sqrt{2}}.\\]Then\n\\[\\frac{(y + 1)(1) + (-2)(-1)}{\\sqrt{x^2 + (y + 1)^2 + (-2)^2} \\cdot \\sqrt{2}} = \\frac{1}{\\sqrt{2}},\\]or $y + 3 = \\sqrt{x^2 + (y + 1)^2 + 4}.$ Squaring both sides, we get\n\\[y^2 + 6y + 9 = x^2 + y^2 + 2y + 1 + 4.\\]Solving for $y,$ we find $y = \\frac{x^2}{4} - 1.$ Thus, $f(x) = \\boxed{\\frac{x^2}{4} - 1}.$"},"answer":{"kind":"string","value":"\\frac{x^2}{4} - 1"},"p_retrievals":{"kind":"list like","value":["An upside-down right cone of height $h$ and base radius $r$ with its axis perpendicular to a plane has its tip touching that plane. A point light source is placed infinitely far away at an angle of $\\frac{\\pi}{6}$ from the plane. Find the area of the shadow in terms of $h$ and $r$.","A circle of radius 10 units is centered at $$(0, 0, 50)$$. The plane it lies in, has a normal vector $$(1, 0, 0)$$. A uniform direction ray of light falls on the circle, generating a shadow on the floor (which is the XY plane). The direction of the shadow is given by $$d = (1, -0.5, -0.75)$$. The shadow is an ellipse. Find the sum of the semi-major and semi-minor axes of this ellipse.","A sphere of radius 3 m is placed on a flat ground. A light source is held 7 m above the ground vertically on top of the centre of the sphere. What is the area in m$$^2$$ of the shadow cast by the light source?","In the $xyz$-coordinate system, a thin solid square of side length $S$ lies in the $xy$-plane with its center at the origin. An isotropic light source is positioned at $(x,y,z) = \\left(0,0,\\frac{S}{2}\\right)$. What percentage of the light energy is incident on the square $($to one decimal place; i.e., $66\\frac{2}{3}\\%$ would be entered as 66.7$)?$","jump to navigation ## Problem of the Day #228: Shadow of the ConeNovember 2, 2011 Posted by Albert in : potd , trackback An upside-down right cone of height $h$ and base radius $r$ with its axis perpendicular to a plane has its tip touching that plane. A point light source is placed infinitely far away at an angle of $\\frac{\\pi}{6}$ from the plane. Find the area of the shadow in terms of $h$ and $r$. ## Comments» no comments yet - be the first?","both functions cut into each other, ea, share the same z value for a given x/y point. Take it from there. Greetz, Leo 5. Nov 16, 2004 NateTG I mentioned in my post that the 'shadow' that this region casts into the xy plane is a circle. It's also relatively easy to identify that the z range is $$(4,-4)$$.","points $A$, $B$, and $C$, where the line $L_1$ intersects the $xy$-plane, the $yz$-plane, and the $xz$-plane, respectively. Check back soon! Problem 44 Consider the points $P$ such that the distance from $P$ to $A (-1, 5, 3)$ is twice the distance from $P$ to $B (6, 2, -2)$. Show that the set of all such points is a sphere, and find its center and radius. Check back soon! Problem 45 Find an equation of the set of all points equidistant from the points $A (-1, 5, 3)$ and $B (6, 2, -2)$. Describe the set. Check back soon! Problem 46 Find the volume of the solid that lies inside both of the spheres $$x^2 + y^2 + z^2 + 4x -2y + 4z + 5 = 0$$ and $$x^2 + y^2 + z^2 = 4$$ Check back soon! Problem 47 Find the distance between the spheres $x^2 + y^2 + z^2 = 4$ and $x^2 + y^2 + z^2 = 4x + 4y + 4z - 11$. Check back soon! Problem 48 Describe and sketch a solid with the following properties. When illuminated by rays parallel to the z-axis, its shadow is a circular disk. If the rays are parallel to the y-axis, its shadow is a square. If the rays are parallel to the x-axis, its","intersects the big sphere with $$0 < t < 1$$, so the sphere casts a shadow over $$P_1$$. There’s a literal edge case we need to consider. Consider the ray $$P + t\\vec{L}$$. If we look for intersections starting from $$t_{min} = 0$$, we’ll find one at $$P$$ itself! We know $$P$$ is on a sphere, so for $$t = 0$$, $$P + 0\\vec{L} = P$$ ; in other words, every point would be casting a shadow over itself! The simplest workaround is to set $$t_{min}$$ to a very small value $$\\epsilon$$ instead of $$0$$. Geometrically, we’re saying we want the ray to start just a tiny bit off the surface where $$P$$ is, rather than exactly at $$P$$. So the range will be $$[\\epsilon, +\\infty]$$ for directional lights and $$[\\epsilon, 1]$$ for point lights. It might be tempting to fix this by just not computing intersections between the ray and the sphere $$P$$ belongs to. This would work for spheres, but it would fail for objects with more complex shapes. For example, when you use your hand to protect your eyes from the Sun, your hand is casting a shadow over your face, and both surfaces are part of the same object - your body. Let’s turn the above discussion into pseudocode. In its previous version, TraceRay","Problem 72 A light is to be placed above the center of a circular area of radius a. What height gives the best illumination on a circular walk surrounding the area? (When light from a point source strikes a surface obliquely, the intensity of illumination is $I = \\dfrac{k \\sin \\theta}{d^2}$ where θ is the angle of incidence and d the distance from the source.) Solution: Problem 73 It is shown in the theory of attraction that a wire bent in the form of a circle of radius a exerts upon a particle in the axis of the circle (i.e., in the line through the center of the circle perpendicular to the plane) an attraction proportional to $\\dfrac{h}{(a^2 + h^2)^{3/2}}$ where h is the height of the particle above the plane of the circle. Find h, for maximum attraction. (Compare with Problem 72 above) Solution: Problem 74 In Problem 73 above, if the wire has instead the form of a square of side $2l$, the attraction is proportional to $\\dfrac{h}{(h^2 + l^2)\\sqrt{h^2 + 2l^2}}$ Find h for maximum attraction. Solution: 0 likes","out, the problem was because the sphere was placed very far away and the radius is too big. This produces a normal with a high value z. A solution for me was to place the sphere very close to the origin—about z=0.25, and set the radius of the sphere less than z— about r=0.125. light_source (-1,0,0) light_source (0,0,-1)","# A light and a shadow Geometry Level 2 A rectangular billboard $ABCD$ is illuminated by a lantern $E$ and casts a shadow in the $xy$-plane. The positions $(x,y,z)$ of the lantern and the billboard's vertices are $A = (0,0,0), \\quad B = (0,3,0), \\quad C = (0,3,2), \\quad D = (0, 0 ,2), \\quad E = (3, -1, 4).$ What is the area of the shadow? Assumptions: The light source is a point and the billboard has zero thickness. The ground is the $xy$-plane. ×","such as a light bulb. Objects which are closer to the source are brighter. The contribution of the light somehow seems to decrease as the distance from the light source to the objects increases. The question now, is to find out what the rule of that falloff is. In fact, the amount of light energy arriving on a point in the scene from a point light source, depends on the area of the \"sphere\" which itself depends on the sphere radius. Note that this sphere doesn't really exist. We just use this image to help you visualising the process. When we speak of sphere what we actually mean is some sort of virtual sphere centred around the point light source origin and whose radius is the distance from the point light source to $P$ a point in the scene that we wish to shade. What we are trying to find is how much light arrives at $P$ from that point light source. As suggested, the contribution of the point light source depends on the area of the sphere of radius $r$: $$A = 4\\pi r^2.$$ The intensity of the light arriving at $P$ is in fact inversely proportional to the sphere area. In other words: $$L_i = \\dfrac{\\text{light intensity * light color}}{4 \\pi r^2}.$$ Where $r$ is equal","lying in the plane. The entire process is illustrated in Interactive Illustration 3.20, and after the reader has explored it, the math will be derived. $L$ $V_1$ $V_2$ $V_3$ $S$ $\\vc{n}$ $\\vc{d}_1$ $\\vc{d}_2$ $\\vc{d}_3$ Interactive Illustration 3.20: This illustration shows how a shadow, projected onto a plane, can be calculated. Initially, there is a light source (located at $L$), illustrated as a yellow circle, a triangle with three vertices, $V_i$, and a ground plane. The plane is described as: $\\vc{n}\\cdot(P-S)=0$. Click/touch Forward to commence the illustration. Interactive Illustration 3.20: Finally, a shadow triangle is shown. Note that the light source position can be moved around in this illustration. Be careful, however, in certain situations, unpredictable illustrations will occur. Still, it is interesting to move the light source so that it is located under the triangle. The current calculations still produce a shadow (sometimes called an anti-shadow), even though this would not be physically correct. To calculate where the shadow \"lands\" on the plane, we need to create one ray per vertex. All three rays will start at the light source position, $L$, and the direction per vertex will be: $\\vc{d}_i = V_i - L$, i.e., the ray direction is formed by the line segment created from the $L$ and $V_i$. Hence, the line equations for the rays will","sources lie in the same half-space divided by the plane of this area, :$\\text \\mathbf\\text dA\\text E_v\\left(\\mathbf\\right) = \\sum _.$ In the case of a single point light source of intensity ''Iv'', at a distance ''r'' and normally incident, this reduces to :$E_v\\left(r\\right) = \\frac.$","Shadow boundary on convex body in $\\mathbb{R}^3$ Let $S$ be the surface of a compact, convex, smooth ($C^\\infty$) body in $\\mathbb{R}^3$, with strictly positive Gaussian curvature at every point of $S$. Fix a direction $z$ in a Cartesian coordinate system, and consider all the lines parallel to $z$ and tangent to $S$, which form a topological cylinder enclosing $S$, touching $S$ on the shadow boundary resulting from a light source at $z=+\\infty$ (yellow in the figure below). Parametrize these lines from $s=0$ to $s=1$ around the cylinder, and let $h(s)$ be the height of the point of tangency to $S$ above the $xy$-plane, orthogonal to $z$. My question is: Can $h(s)$ have an arbitrarily large number of local maxima and minima? I am interested to learn if this shadow-boundary curve is \"well-behaved\" in some sense, for smooth convex bodies. Thanks for pointers/suggestions/counterexmaples! - The shadow boundary can be any $C^\\infty$ curve with (quadratically) strictly convex projection to the $xy$-plane. For simplicity, let me stick to the case when the projection is a circle. So consider a smooth function $h:S^1\\to\\mathbb R$. I am going to construct a smooth, compact, quadratically convex surface whose shadow boundary equals the set $\\{(s,h(s)):s\\in S^1\\}\\subset S^1\\times\\mathbb R\\subset\\mathbb R^2\\times\\mathbb R$. Here $S^1$ is the standard circle $x^2+y^2=1$ in the $xy$-plane. For an $\\varepsilon>0$, define a","ray of light along $x+\\sqrt 3 \\, y=\\sqrt 3$ gets reflected upon reaching X-axis, the equation of the reflected ray is JEE Main 2013 Straight Lines ## 7. The centre of the circle passing through the point (0, 1)and touching the curve $y = x^2 at (2,4)$ is IIT JEE 1983 Conic Sections ## 8. Let AB be a chord of the circle $x^2 + y^2 = r^2$ subtending a right angle at the centre. Then, the locus of the centroid of the $\\Delta PAB$ as P moves on the circle, is IIT JEE 2001 Conic Sections","represent our light ray and our spheres. The light ray is simple - it can be composed of a starting position, and a parameterized direction vector, like so: $\\Large L = p_0+ t Dir$ Where L is the light ray vector, $p_0$ is the starting point of the ray, t is a variable, and Dir is a directional vector. A sphere can also easily represented in vector form like so: $\\Large (p - c)\\cdot(p - c)= Radius^2$ Where p is any point on the sphere and c is the center of the sphere (a vector relative to the origin, or 0,0,0). This equation is effectively saying that the distance from any point on the sphere to the center of the sphere squared is equal to the radius squared - which makes sense! (A dot product of something with itself gives the magnitude of that thing squared). Now finding whether or not the line intersects the sphere is as easy as substituting the line equation in to the sphere equation. After some manipulation, this gives a quadratic equation of the form $ax^2 + bx +c = 0$, where: $a = Dir \\cdot Dir$ $b = 2(p_0 - c)\\cdot Dir$ $c =(p_0-c)\\cdot(p_0-c)-Radius^2$ As expected with all quadratic equations, this leaves us with three possible results. The determinant, $b^2 - 4ac$,","r1 ... xN yN zN rN The first line of a dataset contains a positive integer N which is the number of spheres. The next line contains three integers u, v and w separated by single spaces, where (u, v,w) is the direction of the laser ray initially emitted from the origin. Each of the following N lines contains four integers separated by single spaces. The i-th line corresponds to the i-th sphere, and the numbers represent the center position (xi, yi, zi) and the radius ri. N, u, v,w,xi, yi, zi and ri satisfy the following conditions. 1 ≤ N ≤ 100 -100 ≤ u,v,w ≤ 100 -100 ≤ xi,yi,zi ≤ 100 5 ≤ ri ≤ 30 u2 + v2 + w2 > 0 You can assume that the distance between the surfaces of any two spheres is no less than 0.1. You can also assume that the origin (0; 0; 0) is located outside of any sphere, and is at least 0.1 distant from the surface of any sphere. The ray is known to be reflected by the sphere surfaces at least once, and at most five times. You can assume that the angle between the ray and the line connecting the sphere center and the reflection point, which is known as the angle of reflection","line contains three integers u, v and w separated by single spaces, where (u, v, w) is the direction of the laser ray initially emitted from the origin. Each of the following N lines contains four integers separated by single spaces. The i-th line corresponds to the i-th sphere, and the numbers represent the center position (xi, yi, zi ) and the radius ri . N, u, v, w, xi, yi, zi and ri satisfy the following conditions. 1 ≤ N ≤ 100 −100 ≤ u, v, w ≤ 100 −100 ≤ xi, yi, zi ≤ 100 5 ≤ ri ≤ 30 u2 + v2 + w2 > 0 You can assume that the distance between the surfaces of any two spheres is no less than 0.1. You can also assume that the origin (0, 0, 0) is located outside of any sphere, and is at least 0.1 distant from the surface of any sphere. The ray is known to be reflected by the sphere surfaces at least once, and at most five times. You can assume that the angle between the ray and the line connecting the sphere center and the reflection point, which is known as the angle of reflection (i.e. θ in Figure 1), is less than 85 degrees for each point of reflection. The","# Shadow of a rod AB is a rod which is held such that $$A=(1,-2,3)$$ and $$B=(2,3,-4)$$ . A source of light is at the origin. Find the length of the shadow of the rod on a plane screen whose equation is $$x+y+2z=1$$ I figured out that origin and point B are on one side of given plane and point A is on other side. I found projection points of A and B and calculated distance between them but that is not matching with answer. Please help. ## 3 Answers If what you say is correct, that the points A and B are on opposite sides of the plane, then the you answer will be wrong indeed, what you are looking for is not the projection of a vector representing the rod on the plane, but the shadow of the rod. Shadows are used to explain projections, but its a very particular analogy requiring specific circumstances. Draw it out for yourself and you'll see that actually the lenght you're looking for is the distance between the projection point of B on the plane, and the point where the the rod actually intersects the plane, as the part on the opposite side of the plane (opposite of the light source) won't cast a shadow on it The rod can"],"string":"[\n \"An upside-down right cone of height $h$ and base radius $r$ with its axis perpendicular to a plane has its tip touching that plane. A point light source is placed infinitely far away at an angle of $\\\\frac{\\\\pi}{6}$ from the plane. Find the area of the shadow in terms of $h$ and $r$.\",\n \"A circle of radius 10 units is centered at $$(0, 0, 50)$$. The plane it lies in, has a normal vector $$(1, 0, 0)$$. A uniform direction ray of light falls on the circle, generating a shadow on the floor (which is the XY plane). The direction of the shadow is given by $$d = (1, -0.5, -0.75)$$. The shadow is an ellipse. Find the sum of the semi-major and semi-minor axes of this ellipse.\",\n \"A sphere of radius 3 m is placed on a flat ground. A light source is held 7 m above the ground vertically on top of the centre of the sphere. What is the area in m$$^2$$ of the shadow cast by the light source?\",\n \"In the $xyz$-coordinate system, a thin solid square of side length $S$ lies in the $xy$-plane with its center at the origin. An isotropic light source is positioned at $(x,y,z) = \\\\left(0,0,\\\\frac{S}{2}\\\\right)$. What percentage of the light energy is incident on the square $($to one decimal place; i.e., $66\\\\frac{2}{3}\\\\%$ would be entered as 66.7$)?$\",\n \"jump to navigation ## Problem of the Day #228: Shadow of the ConeNovember 2, 2011 Posted by Albert in : potd , trackback An upside-down right cone of height $h$ and base radius $r$ with its axis perpendicular to a plane has its tip touching that plane. A point light source is placed infinitely far away at an angle of $\\\\frac{\\\\pi}{6}$ from the plane. Find the area of the shadow in terms of $h$ and $r$. ## Comments» no comments yet - be the first?\",\n \"both functions cut into each other, ea, share the same z value for a given x/y point. Take it from there. Greetz, Leo 5. Nov 16, 2004 NateTG I mentioned in my post that the 'shadow' that this region casts into the xy plane is a circle. It's also relatively easy to identify that the z range is $$(4,-4)$$.\",\n \"points $A$, $B$, and $C$, where the line $L_1$ intersects the $xy$-plane, the $yz$-plane, and the $xz$-plane, respectively. Check back soon! Problem 44 Consider the points $P$ such that the distance from $P$ to $A (-1, 5, 3)$ is twice the distance from $P$ to $B (6, 2, -2)$. Show that the set of all such points is a sphere, and find its center and radius. Check back soon! Problem 45 Find an equation of the set of all points equidistant from the points $A (-1, 5, 3)$ and $B (6, 2, -2)$. Describe the set. Check back soon! Problem 46 Find the volume of the solid that lies inside both of the spheres $$x^2 + y^2 + z^2 + 4x -2y + 4z + 5 = 0$$ and $$x^2 + y^2 + z^2 = 4$$ Check back soon! Problem 47 Find the distance between the spheres $x^2 + y^2 + z^2 = 4$ and $x^2 + y^2 + z^2 = 4x + 4y + 4z - 11$. Check back soon! Problem 48 Describe and sketch a solid with the following properties. When illuminated by rays parallel to the z-axis, its shadow is a circular disk. If the rays are parallel to the y-axis, its shadow is a square. If the rays are parallel to the x-axis, its\",\n \"intersects the big sphere with $$0 < t < 1$$, so the sphere casts a shadow over $$P_1$$. There’s a literal edge case we need to consider. Consider the ray $$P + t\\\\vec{L}$$. If we look for intersections starting from $$t_{min} = 0$$, we’ll find one at $$P$$ itself! We know $$P$$ is on a sphere, so for $$t = 0$$, $$P + 0\\\\vec{L} = P$$ ; in other words, every point would be casting a shadow over itself! The simplest workaround is to set $$t_{min}$$ to a very small value $$\\\\epsilon$$ instead of $$0$$. Geometrically, we’re saying we want the ray to start just a tiny bit off the surface where $$P$$ is, rather than exactly at $$P$$. So the range will be $$[\\\\epsilon, +\\\\infty]$$ for directional lights and $$[\\\\epsilon, 1]$$ for point lights. It might be tempting to fix this by just not computing intersections between the ray and the sphere $$P$$ belongs to. This would work for spheres, but it would fail for objects with more complex shapes. For example, when you use your hand to protect your eyes from the Sun, your hand is casting a shadow over your face, and both surfaces are part of the same object - your body. Let’s turn the above discussion into pseudocode. In its previous version, TraceRay\",\n \"Problem 72 A light is to be placed above the center of a circular area of radius a. What height gives the best illumination on a circular walk surrounding the area? (When light from a point source strikes a surface obliquely, the intensity of illumination is $I = \\\\dfrac{k \\\\sin \\\\theta}{d^2}$ where θ is the angle of incidence and d the distance from the source.) Solution: Problem 73 It is shown in the theory of attraction that a wire bent in the form of a circle of radius a exerts upon a particle in the axis of the circle (i.e., in the line through the center of the circle perpendicular to the plane) an attraction proportional to $\\\\dfrac{h}{(a^2 + h^2)^{3/2}}$ where h is the height of the particle above the plane of the circle. Find h, for maximum attraction. (Compare with Problem 72 above) Solution: Problem 74 In Problem 73 above, if the wire has instead the form of a square of side $2l$, the attraction is proportional to $\\\\dfrac{h}{(h^2 + l^2)\\\\sqrt{h^2 + 2l^2}}$ Find h for maximum attraction. Solution: 0 likes\",\n \"out, the problem was because the sphere was placed very far away and the radius is too big. This produces a normal with a high value z. A solution for me was to place the sphere very close to the origin—about z=0.25, and set the radius of the sphere less than z— about r=0.125. light_source (-1,0,0) light_source (0,0,-1)\",\n \"# A light and a shadow Geometry Level 2 A rectangular billboard $ABCD$ is illuminated by a lantern $E$ and casts a shadow in the $xy$-plane. The positions $(x,y,z)$ of the lantern and the billboard's vertices are $A = (0,0,0), \\\\quad B = (0,3,0), \\\\quad C = (0,3,2), \\\\quad D = (0, 0 ,2), \\\\quad E = (3, -1, 4).$ What is the area of the shadow? Assumptions: The light source is a point and the billboard has zero thickness. The ground is the $xy$-plane. ×\",\n \"such as a light bulb. Objects which are closer to the source are brighter. The contribution of the light somehow seems to decrease as the distance from the light source to the objects increases. The question now, is to find out what the rule of that falloff is. In fact, the amount of light energy arriving on a point in the scene from a point light source, depends on the area of the \\\"sphere\\\" which itself depends on the sphere radius. Note that this sphere doesn't really exist. We just use this image to help you visualising the process. When we speak of sphere what we actually mean is some sort of virtual sphere centred around the point light source origin and whose radius is the distance from the point light source to $P$ a point in the scene that we wish to shade. What we are trying to find is how much light arrives at $P$ from that point light source. As suggested, the contribution of the point light source depends on the area of the sphere of radius $r$: $$A = 4\\\\pi r^2.$$ The intensity of the light arriving at $P$ is in fact inversely proportional to the sphere area. In other words: $$L_i = \\\\dfrac{\\\\text{light intensity * light color}}{4 \\\\pi r^2}.$$ Where $r$ is equal\",\n \"lying in the plane. The entire process is illustrated in Interactive Illustration 3.20, and after the reader has explored it, the math will be derived. $L$ $V_1$ $V_2$ $V_3$ $S$ $\\\\vc{n}$ $\\\\vc{d}_1$ $\\\\vc{d}_2$ $\\\\vc{d}_3$ Interactive Illustration 3.20: This illustration shows how a shadow, projected onto a plane, can be calculated. Initially, there is a light source (located at $L$), illustrated as a yellow circle, a triangle with three vertices, $V_i$, and a ground plane. The plane is described as: $\\\\vc{n}\\\\cdot(P-S)=0$. Click/touch Forward to commence the illustration. Interactive Illustration 3.20: Finally, a shadow triangle is shown. Note that the light source position can be moved around in this illustration. Be careful, however, in certain situations, unpredictable illustrations will occur. Still, it is interesting to move the light source so that it is located under the triangle. The current calculations still produce a shadow (sometimes called an anti-shadow), even though this would not be physically correct. To calculate where the shadow \\\"lands\\\" on the plane, we need to create one ray per vertex. All three rays will start at the light source position, $L$, and the direction per vertex will be: $\\\\vc{d}_i = V_i - L$, i.e., the ray direction is formed by the line segment created from the $L$ and $V_i$. Hence, the line equations for the rays will\",\n \"sources lie in the same half-space divided by the plane of this area, :$\\\\text \\\\mathbf\\\\text dA\\\\text E_v\\\\left(\\\\mathbf\\\\right) = \\\\sum _.$ In the case of a single point light source of intensity ''Iv'', at a distance ''r'' and normally incident, this reduces to :$E_v\\\\left(r\\\\right) = \\\\frac.$\",\n \"Shadow boundary on convex body in $\\\\mathbb{R}^3$ Let $S$ be the surface of a compact, convex, smooth ($C^\\\\infty$) body in $\\\\mathbb{R}^3$, with strictly positive Gaussian curvature at every point of $S$. Fix a direction $z$ in a Cartesian coordinate system, and consider all the lines parallel to $z$ and tangent to $S$, which form a topological cylinder enclosing $S$, touching $S$ on the shadow boundary resulting from a light source at $z=+\\\\infty$ (yellow in the figure below). Parametrize these lines from $s=0$ to $s=1$ around the cylinder, and let $h(s)$ be the height of the point of tangency to $S$ above the $xy$-plane, orthogonal to $z$. My question is: Can $h(s)$ have an arbitrarily large number of local maxima and minima? I am interested to learn if this shadow-boundary curve is \\\"well-behaved\\\" in some sense, for smooth convex bodies. Thanks for pointers/suggestions/counterexmaples! - The shadow boundary can be any $C^\\\\infty$ curve with (quadratically) strictly convex projection to the $xy$-plane. For simplicity, let me stick to the case when the projection is a circle. So consider a smooth function $h:S^1\\\\to\\\\mathbb R$. I am going to construct a smooth, compact, quadratically convex surface whose shadow boundary equals the set $\\\\{(s,h(s)):s\\\\in S^1\\\\}\\\\subset S^1\\\\times\\\\mathbb R\\\\subset\\\\mathbb R^2\\\\times\\\\mathbb R$. Here $S^1$ is the standard circle $x^2+y^2=1$ in the $xy$-plane. For an $\\\\varepsilon>0$, define a\",\n \"ray of light along $x+\\\\sqrt 3 \\\\, y=\\\\sqrt 3$ gets reflected upon reaching X-axis, the equation of the reflected ray is JEE Main 2013 Straight Lines ## 7. The centre of the circle passing through the point (0, 1)and touching the curve $y = x^2 at (2,4)$ is IIT JEE 1983 Conic Sections ## 8. Let AB be a chord of the circle $x^2 + y^2 = r^2$ subtending a right angle at the centre. Then, the locus of the centroid of the $\\\\Delta PAB$ as P moves on the circle, is IIT JEE 2001 Conic Sections\",\n \"represent our light ray and our spheres. The light ray is simple - it can be composed of a starting position, and a parameterized direction vector, like so: $\\\\Large L = p_0+ t Dir$ Where L is the light ray vector, $p_0$ is the starting point of the ray, t is a variable, and Dir is a directional vector. A sphere can also easily represented in vector form like so: $\\\\Large (p - c)\\\\cdot(p - c)= Radius^2$ Where p is any point on the sphere and c is the center of the sphere (a vector relative to the origin, or 0,0,0). This equation is effectively saying that the distance from any point on the sphere to the center of the sphere squared is equal to the radius squared - which makes sense! (A dot product of something with itself gives the magnitude of that thing squared). Now finding whether or not the line intersects the sphere is as easy as substituting the line equation in to the sphere equation. After some manipulation, this gives a quadratic equation of the form $ax^2 + bx +c = 0$, where: $a = Dir \\\\cdot Dir$ $b = 2(p_0 - c)\\\\cdot Dir$ $c =(p_0-c)\\\\cdot(p_0-c)-Radius^2$ As expected with all quadratic equations, this leaves us with three possible results. The determinant, $b^2 - 4ac$,\",\n \"r1 ... xN yN zN rN The first line of a dataset contains a positive integer N which is the number of spheres. The next line contains three integers u, v and w separated by single spaces, where (u, v,w) is the direction of the laser ray initially emitted from the origin. Each of the following N lines contains four integers separated by single spaces. The i-th line corresponds to the i-th sphere, and the numbers represent the center position (xi, yi, zi) and the radius ri. N, u, v,w,xi, yi, zi and ri satisfy the following conditions. 1 ≤ N ≤ 100 -100 ≤ u,v,w ≤ 100 -100 ≤ xi,yi,zi ≤ 100 5 ≤ ri ≤ 30 u2 + v2 + w2 > 0 You can assume that the distance between the surfaces of any two spheres is no less than 0.1. You can also assume that the origin (0; 0; 0) is located outside of any sphere, and is at least 0.1 distant from the surface of any sphere. The ray is known to be reflected by the sphere surfaces at least once, and at most five times. You can assume that the angle between the ray and the line connecting the sphere center and the reflection point, which is known as the angle of reflection\",\n \"line contains three integers u, v and w separated by single spaces, where (u, v, w) is the direction of the laser ray initially emitted from the origin. Each of the following N lines contains four integers separated by single spaces. The i-th line corresponds to the i-th sphere, and the numbers represent the center position (xi, yi, zi ) and the radius ri . N, u, v, w, xi, yi, zi and ri satisfy the following conditions. 1 ≤ N ≤ 100 −100 ≤ u, v, w ≤ 100 −100 ≤ xi, yi, zi ≤ 100 5 ≤ ri ≤ 30 u2 + v2 + w2 > 0 You can assume that the distance between the surfaces of any two spheres is no less than 0.1. You can also assume that the origin (0, 0, 0) is located outside of any sphere, and is at least 0.1 distant from the surface of any sphere. The ray is known to be reflected by the sphere surfaces at least once, and at most five times. You can assume that the angle between the ray and the line connecting the sphere center and the reflection point, which is known as the angle of reflection (i.e. θ in Figure 1), is less than 85 degrees for each point of reflection. The\",\n \"# Shadow of a rod AB is a rod which is held such that $$A=(1,-2,3)$$ and $$B=(2,3,-4)$$ . A source of light is at the origin. Find the length of the shadow of the rod on a plane screen whose equation is $$x+y+2z=1$$ I figured out that origin and point B are on one side of given plane and point A is on other side. I found projection points of A and B and calculated distance between them but that is not matching with answer. Please help. ## 3 Answers If what you say is correct, that the points A and B are on opposite sides of the plane, then the you answer will be wrong indeed, what you are looking for is not the projection of a vector representing the rod on the plane, but the shadow of the rod. Shadows are used to explain projections, but its a very particular analogy requiring specific circumstances. Draw it out for yourself and you'll see that actually the lenght you're looking for is the distance between the projection point of B on the plane, and the point where the the rod actually intersects the plane, as the part on the opposite side of the plane (opposite of the light source) won't cast a shadow on it The rod can\"\n]"},"s_retrievals":{"kind":"list like","value":["much as if we were in 2D TikZ. So, setting settings.render to some other value is required for the plane to 'cover' stuff layered below. These should also probably really be surfaces, I guess, perhaps with no light. \\documentclass{article} \\usepackage{asymptote} \\begin{document} \\begin{asy} settings.outformat = \"pdf\"; settings.prc = false; settings.render = 16; import three; import bsp; texpreamble(\"\\usepackage{euler,beton}\"); size(5cm, 0); currentprojection=orthographic((5,4,3)); path3 pl =plane((0,-2,0),(-2,0,0),(0,1,0)); path3 pl1=rotate(-28,X)*pl; path3 pl2=rotate(-56,X)*pl; path3 pl3=shift(-0.3*normal(pl1))*pl1; draw (pl); draw (pl1); draw (pl2); draw (pl3); triple[] asd=intersectionpoints(pl2,pl3); triple A=asd[1]; triple B=asd[0]; triple[] asf=intersectionpoints(pl,pl2); triple C=asf[1]; triple D=asf[0]; triple[] asg=intersectionpoints(pl,pl3); triple E=asg[1]; triple F=asg[0]; surface s1=surface(pl); draw(s1,white+opacity(.6),light=nolight); surface s2=surface(pl1); draw(s2,white+opacity(.6),light=nolight); surface s3=surface(pl2); draw(s3,white+opacity(.6),light=nolight); surface s4=surface(pl3); draw(s4,white+opacity(.6),light=nolight); draw(B--A); draw(C--D); draw(E--F); dot(A,red); dot(B,red); dot(C,red); dot(D,red); dot(E,red); dot(F,red); //From Charles Staats's tutorial //Direction of a point toward the camera. triple cameradirection(triple pt, projection P=currentprojection) { if (P.infinity) { return unit(P.camera); } else { return unit(P.camera - pt); } } //Move a point closer to the camera. triple towardcamera(triple pt, real distance=1, projection P=currentprojection) { return pt + distance * cameradirection(pt, P); } label(\"$A$\",align=NE,position=towardcamera((A))); label(\"$B$\",align=S,position=towardcamera((B))); label(\"$C$\",align=SE,position=towardcamera((C))); label(\"$D$\",align=SE,position=towardcamera((D))); label(\"$E$\",align=NE,position=towardcamera((E))); label(\"$F$\",align=S,position=towardcamera((F))); \\end{asy} \\end{document} # EDIT If you simply must have vector graphics - a restriction which was not mentioned in the question - then you need to take care of the drawing order yourself. This requires splitting each surface into pieces and reassembling them","did was to follow this great tutorial, which has the desired syntax on the bottom of p. 60. The striking difference to you code is that there the normal of the 2D arrow gets specified, too, and that it is actually +Z (and not -Z). \\documentclass[border=3.14mm]{standalone} \\usepackage{asypictureB} \\begin{document} \\begin{asypicture}{name=arc} settings.outformat = \"png\"; settings.render = 16; size(6cm,0); import solids; currentlight=Headlamp; // inutile car Headlamp est la valeur par defaut nslice=4*nslice; revolution boule=sphere(O,1); draw(surface(boule),lightgrey+white+opacity(.5)); draw((0,0,1)--(0,0,-1),dashed); dot(\"$P1$\",(0,0,1),N); dot(\"$P2$\",(0,0,-1),2.2S); dot(\"$O$\",(0,0,0),SE); draw(O--(1,0,0),red); draw(O--(Cos(60),Sin(60),0),red); triple v1= 0.4*(1,0,0), v2 = 0.4*(Cos(60),Sin(60),0); //draw(arc(c=O,v1,v2),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); for (real alpha = 90; alpha<=450; alpha+= 30) { if (alpha==150) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red); if (alpha==330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red + dashed); if (alpha != 150 && alpha != 330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),black); }; draw(arc(c=O, v2, v1, normal=-Z), red, arrow = Arrow3(TeXHead2(normal=Z), emissive(red)),L=Label(\"$\\varphi$\",align=S, position=MidPoint)); \\end{asypicture} \\end{document} • Tank you Marmot for your anwer – user273366 Nov 24 '18 at 7:45 • indeed, the normal parameter was missing in the arrow function – user273366 Nov 24 '18 at 7:47","// spherexplane.asy // // run // asy -f pdf -render=4 -noprc spherexplane.asy // to get a standalone raster spherexplane.pdf // import solids; size(8cm); size3(100,100); currentprojection=orthographic(camera=(66,40,-9),zoom=0.9); pen linePen=darkblue+1.3bp; pen dotPen= darkblue+3bp; pen dashPen=1bp+linetype(new real[]{4,3})+linecap(0); // Eqn of the sphere (x - 1)^2 + (y + 1)^ 2 + (z - 2)^ 2 - 25 = 0 triple O=(1,-1,2); real R=5; // Eqn of the plane 2 x - 2 y + z - 15 = 0 triple fp(real x, real y){return (x,y,- 2 x + 2 y + 15);} triple Np=unit((2,-2,1)); // plane normal triple A=fp(0,0); // any point on the plane triple C=O+dot(A-O,Np)*Np; // center of the circle cross section real d=abs(C-O); real r=sqrt(R^2-d^2); guide3 baseArc=Arc(O,O+Np*R,O-Np*R,normal=cross(Z,Np)); revolution b=revolution(O,baseArc,axis=Np); // spherical surface skeleton s; real t=acos(d/R)/pi; // fraction of the arc length at the cutting point b.transverse(s,reltime(b.g,t),P=currentprojection); guide3 circCut=s.transverse.back[0] & s.transverse.front[0] & cycle; triple D=relpoint(circCut,0.6); draw(surface(b),paleblue+opacity(0.3)); draw(surface(circCut),orange+opacity(0.3)); draw(s.transverse.front,linePen); draw(s.transverse.back, dashPen); draw(O--C--D, dashPen); dot(\"$O$\",O,dotPen); dot(Label(\"$C$\",unit(C-O)),C,dotPen); dot(\"$D$\",D,dotPen); xaxis3(xmin=0,xmax=1,red,above=true); yaxis3(ymin=0,ymax=1,deepgreen,above=true); zaxis3(zmin=0,zmax=1,blue,above=true); • +1, I am still learning your asymptote answers. – user213378 Oct 31 '20 at 11:33 • Thank you very much. – minhthien_2016 Oct 31 '20 at 12:30 • @g.kov Does asymptote always draw the hidden parts dashed automatically? – minhthien_2016 Dec 19 '20 at 14:59 • @minhthien_2016: I don't think it does. – g.kov Dec 19 '20 at 16:44 •","box are available. • SurfaceHex(N, B, L, orient) • this operator allows to build the surface mesh of a 3d box • int[int] N=[nx,ny,nz]; // the number of seg in the 3 direction • real [int,int] B=[[xmin,xmax],[ymin,ymax],[zmin,zmax]]; // bounding bax • int [int,int] L=[[1,2],[3,4],[5,6]]; // the label of the 6 face left,right, front, back, down, right • orient the global orientation of the surface 1 extern (-1 intern), • returns a meshS type • Ellipsoide (RX, RY, RZ, h, L, OX, OY, OZ, orient) • h is the mesh size • L is the label • orient the global orientation of the surface 1 extern (-1 intern) • OX, OY, OZ are real numbers to give the Ellipsoide center ( optinal, by default is (0,0,0) ) • where RX, RY, RZ are real numbers such as the parametric equations of the ellipsoid is: • returns a meshS type $\\forall u \\in [- \\frac{\\pi}{2},\\frac{\\pi}{2} [ \\text{ and } v \\in [0, 2 \\pi], \\vectthree{x=\\text{Rx } cos(u)cos(v) + \\text{Ox }}{y=\\text{Ry } cos(u)sin(v) + \\text{Oy }}{z = \\text{Rz } sin(v) + \\text{Oz } }$ • Sphere(R, h, L, OX, OY, OZ, orient) • where R is the raduis of the sphere, • OX, OY, OZ are real numbers to give the Ellipsoide center ( optinal, by default is (0,0,0) ) •","= circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,$$which rearranges to the desired. ## Solution 4 (Quick) Suppose we label the points as shown below. $[asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot(\"A\",A,dir(A-O2/2)); dot(\"B\",B,dir(B-O2/2)); dot(\"O_2\",O2,right+up); dot(\"O_1\",O1,left+up); dot(\"O\",O,dir(O-O2)); dot(\"D\",D,dir(170)); dot(\"E\",E,dir(E-O1)); dot(\"X\",X,dir(X-D)); dot(\"Y\",Y,dir(Y-D)); label(\"R\",O--E,right+up,p); label(\"R\",O--D,left+down,p); label(\"2R\",(X+O)/2-(150,0),down,p); label(\"961\",O1--D,2*(left+down),p); label(\"625\",O2--E,2*(right+up),p); MA(\"\",E,D,O1,100,fuchsia+linewidth(1)); MA(\"\",X,D,O,100,fuchsia+linewidth(1)); MA(\"\",Y,E,O2,100,orange+linewidth(1)); MA(\"\",D,E,O,100,orange+linewidth(1)); [/asy]$","# I can not draw arc in a clockwise direction with this program in asymptote I have to trace the meridians. settings.outformat = \"png\"; settings.render = 16; size(6cm,0); import solids; currentlight=Headlamp; // inutile car Headlamp est la valeur par défaut nslice=4*nslice; revolution boule=sphere(O,1); draw(surface(boule),lightgrey+white+opacity(.5)); draw((0,0,1)--(0,0,-1),dashed); dot(\"$P1$\",(0,0,1),N); dot(\"$P2$\",(0,0,-1),2.2S); dot(\"$O$\",(0,0,0),SE); draw(O--(1,0,0),red); draw(O--(Cos(60),Sin(60),0),red); triple v1= 0.4(1,0,0), v2 = 0.4(Cos(60),Sin(60),0); draw(arc(c=O,v1,v2,normal=Z),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); for (real alpha = 90; alpha<=450; alpha+= 30) { if (alpha==150) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red); if (alpha==330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red + dashed); if (alpha != 150 && alpha != 330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),black); } But I can not draw arc in a clockwise direction with this code (this is the critic point!): draw(arc(c=O,v1,v2,normal=Z),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); I tried: 1. to change v1 and v2 order with the same normal (Z), but in this way i have the angle greater than ninety degrees (i need the explementar angle) 2. to change v1 and v2 order and take normal = - Z, but in this way asymptote runs on a infinite loop. I don't know why. ## 1 Answer It took me a while to understand the question since IMHO it is not very clearly written. An additional complication arises from the fact that you did not present an MWE and seem to insist to have a special character in a comment: \"défaut\". All I","settings.outformat=\"pdf\"; settings.prc=false; settings.render=0; import x11colors; import graph3; size(8cm,0); /* real h = 5; triple[] axes = new triple[] {X, Y, Z}; for(int i=0; i