<p>A natural application of trigonometric functions is in the analysis of spatial data.  An example is the <a href="https://en.wikipedia.org/wiki/Weber_problem" rel="noreferrer">Weber problem</a> in location theory - finding the point which minimises the sum of transport costs to $n$ destinations.  There is more than one way to solve the problem but <a href="https://en.wikipedia.org/wiki/Weber_problem#Tellier.E2.80.99s_trigonometric_solution_of_the_Fermat_and_Weber_triangle_problems" rel="noreferrer">Tellier's solution</a> uses trigonometry.</p>

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2017-11-08T09:58:33.977Z
2017-11-08T17:10:46.277Z
2017-11-08T17:10:46.277Z
Adam Bailey
<p>Ignoring the intertemporal budget constraint, mergers and bankruptcies the distribution of returns for equity securities traded in a double auction is $$\Pr(\tilde{r}_t)=\left[\frac{\pi}{2}+\tan^{-1}\left(\frac{\mu}{\gamma}\right)\right]^{-1}\frac{\gamma}{\gamma^2+(\tilde{r}_t-\mu)^2}.$$</p>

<p>For this see: Harris, D.E. (2017) The Distribution of Returns. Journal
of Mathematical Finance , 7, 769-804.</p>

<p>For returns calculated as the difference of logs, the returns are:
$$\Pr(log(r_t))=\frac{1}{2\sigma}\text{sech}\left(\frac{\pi(\tilde{r}_t-\mu)}{2\sigma}\right)$$</p>

CC BY-SA 3.0
2017-11-10T19:53:30.507Z
2017-11-10T20:36:48.207Z
2017-11-10T20:36:48.207Z
Dave Harris
<p>For a concrete example of how trig (and inverse trig) functions may have financial or economic applications, here's one from &quot;Analysis of Financial Time Series&quot; by Ruey S. Tsay. Consider the AR(2) model:</p>
<p><span class="math-container">$$r_t = \phi_0 + \phi_1 r_{t-1} + \phi_2 r_{t-2} +a_t$$</span></p>
<p>Its autocorrelation function (ACF) <span class="math-container">$\rho_\ell = \operatorname{Corr}(r_t, r_{t-\ell})$</span> satisfies the difference equation <span class="math-container">$(1 - \phi_1 B - \phi_2 B^2) \rho _ \ell = 0$</span>, where <span class="math-container">$B$</span> is the back-shift operator, i.e. <span class="math-container">$B \rho_\ell = \rho_{\ell-1}$</span> and <span class="math-container">$B^2 \rho_\ell = \rho_{\ell-2}$</span>. (Some people prefer to write <span class="math-container">$L$</span> for lag operator instead.)</p>
<p>The second-order characteristic equation <span class="math-container">$1 - \phi_1 \omega - \phi_2 \omega^2 = 0$</span> has characteristic roots <span class="math-container">$\omega_1$</span> and <span class="math-container">$\omega_2$</span> given by:</p>
<p><span class="math-container">$$\omega = \frac{\phi_1 \pm \sqrt{\phi_1^2 + 4\phi_2}}{-2\phi_2}$$</span></p>
<p>If the characteristic roots are real, the behaviour is a mixture of two exponential decays. But if instead the discriminant <span class="math-container">$\phi_1^2 + 4\phi_2 &lt; 0$</span>, then the characteristic roots <span class="math-container">$\omega_1$</span> and <span class="math-container">$\omega_2$</span> form a complex-conjugate pair, and the plot of the ACF will exhibit damped sinusoidal waves. To quote Tsay:</p>
<blockquote>
<p>In business and economic applications, complex characteristic roots are important. They give rise to the behavior of business cycles. It is then common for economic time series models to have complex-valued characteristic roots. For an AR(2) model ... with a pair of complex characteristics roots, the <em>average</em> length of the stochastic cycles is</p>
<p><span class="math-container">$$ k = \frac{2 \pi}{\cos^{-1}[\phi_1 / (2\sqrt{-\phi_2})]} $$</span></p>
<p>where the cosine inverse is stated in radians. If one writes the complex solutions as <span class="math-container">$a \pm bi$</span>, where <span class="math-container">$i=\sqrt{-1}$</span>, then we have <span class="math-container">$\phi_1 = 2a$</span>, <span class="math-container">$\phi_2 = -(a^2 + b^2)$</span>, and</p>
<p><span class="math-container">$$ k = \frac{2 \pi}{\cos^{-1}(a / \sqrt{a^2+b^2})} $$</span></p>
</blockquote>
<p>Note that this second way of writing <span class="math-container">$k$</span> has a much more geometrically intuitive way of thinking about the inverse cosine.</p>

CC BY-SA 3.0
2017-11-10T10:30:43.070Z
I have quoted Tsay verbatim re "complex characteristic roots are important. They give rise to the behavior of business cycles" because I think that claim should be treated with skepticism - see the answer by Alecos but also e.g. Stephan Kolassa's comments [here](https://stats.stackexchange.com/a/312906/22228).  I wonder if the book is being oversimplified for its audience (although a graduate-level text, the emphasis is for practitioners). If the lengths of cycles are non-stochastic, however, the formula for $k$ holds true.
Silverfish
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2017-11-10T00:28:08.687Z
2017-11-10T00:28:08.687Z
2020-06-17T09:40:54.643Z
Silverfish
<p>The main property of trig functions is their cyclicality. Then one would think that they could be ideal in time series analysis, to model "fluctuations around a trend". I believe that the reasons they are not actually used in such a setting are </p>

<p>1) They are <em>deterministic</em> functions, so they do not allow for fluctuations to be stochastic</p>

<p>2) If the researcher wants to create a model that <em>produces</em> up and down fluctuations (oscillations) around a trend, he would want to <em>obtain</em> that property from the behavioral and other assumptions of the model. If he were to use a trig function, he would <em>a priori</em> impose on the model the sought theoretical outcome.</p>

<p>Instead, one opts for difference-differential equations. There we obtain oscillations (damped or not) if some characteristic roots are complex -and then the trig functions appear, but as an alternative representation, not as buidling blocks.</p>

CC BY-SA 3.0
2017-11-08T17:21:52.207Z
I'm not sure I would agree with you. There's an area called spectral analysis in Time Series which is mainly the use of trig functions, Fourier transform, etc. You learn that you can decompose a stationary time series in a sum of sinusoidal components with uncorrelated random coefficients.
An old man in the sea.
CC BY-SA 3.0
2017-11-08T17:24:31.823Z
@Anoldmaninthesea. Certainly and good that you pointed that out (I would suggest to make an answer out of it). But spectral analysis is mainly used for atheoretical forecasting purposes, not for structural economic modelling.
Alecos Papadopoulos
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2017-11-09T12:05:21.913Z
Alecos, unfortunately I would need to study it in detail to provide a good answer. Maybe during the weekend. :D
An old man in the sea.
CC BY-SA 3.0
2017-11-21T18:50:25.937Z
Just to say that I read up on the subject and it involves stochastic integration (the decomposition into a series of sinusoidal components), which is something I have no clue about... The remaining of the reading was simply stating that spectral analysis is equivalent to the usual time-domain analysis, but without entering into much detail. I'm adding this comment so that you know I didn't forget, and tried, but I simply don't know enough. ;)
An old man in the sea.
CC BY-SA 3.0
2017-11-21T18:58:08.647Z
@Anoldmaninthesea. Try chapter 2 of Granger and Newbold "Forecasting Economic Time Series" (2nd ed). Is is an old book but full of wisdom, realism, and expositional power (and not just for spectral analysis).
Alecos Papadopoulos
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2017-11-08T17:12:55.657Z
2017-11-08T17:12:55.657Z
Alecos Papadopoulos
<p>Are there any applications of trig functions (ie $\sin(x)$, $\cos(x)$,$\tan(x)$) in economics? </p>

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2017-11-08T11:18:48.097Z
Why do you care?
Michael Greinecker
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2017-11-08T13:04:45.960Z
@MichaelGreinecker general interest.
EconJohn
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2017-11-09T19:19:00.543Z
Relevant https://stats.stackexchange.com/questions/312883/when-would-i-use-a-trig-function-in-a-regression-time-model-series
EconJohn
<mathematical-economics>
Applications of Trig functions in Economics?
EconJohn
# Applications of Trig functions in Economics?

(Asked by: username_0 on 2017-11-08)

Are there any applications of trig functions (ie $\sin(x)$, $\cos(x)$,$\tan(x)$) in economics?

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: Why do you care?  
> **username_0**: @MichaelGreinecker general interest.  
> **username_0**: Relevant https://stats.stackexchange.com/questions/312883/when-would-i-use-a-trig-function-in-a-regression-time-model-series  

---

## 4 answers

### Answer by username_5 on 2017-11-08. Score: 16 | Accepted answer

The main property of trig functions is their cyclicality. Then one would think that they could be ideal in time series analysis, to model "fluctuations around a trend". I believe that the reasons they are not actually used in such a setting are

1) They are *deterministic* functions, so they do not allow for fluctuations to be stochastic

2) If the researcher wants to create a model that *produces* up and down fluctuations (oscillations) around a trend, he would want to *obtain* that property from the behavioral and other assumptions of the model. If he were to use a trig function, he would *a priori* impose on the model the sought theoretical outcome.

Instead, one opts for difference-differential equations. There we obtain oscillations (damped or not) if some characteristic roots are complex -and then the trig functions appear, but as an alternative representation, not as buidling blocks.

> **Comments:**
> **username_6**: I'm not sure I would agree with you. There's an area called spectral analysis in Time Series which is mainly the use of trig functions, Fourier transform, etc. You learn that you can decompose a stationary time series in a sum of sinusoidal components with uncorrelated random coefficients.  
> **username_5**: @Anoldmaninthesea. Certainly and good that you pointed that out (I would suggest to make an answer out of it). But spectral analysis is mainly used for atheoretical forecasting purposes, not for structural economic modelling.  
> **username_6**: Alecos, unfortunately I would need to study it in detail to provide a good answer. Maybe during the weekend. :D  
> **username_6**: Just to say that I read up on the subject and it involves stochastic integration (the decomposition into a series of sinusoidal components), which is something I have no clue about... The remaining of the reading was simply stating that spectral analysis is equivalent to the usual time-domain analysis, but without entering into much detail. I'm adding this comment so that you know I didn't forget, and tried, but I simply don't know enough. ;)  
> **username_5**: @Anoldmaninthesea. Try chapter 2 of Granger and Newbold "Forecasting Economic Time Series" (2nd ed). Is is an old book but full of wisdom, realism, and expositional power (and not just for spectral analysis).  

---

### Answer by username_2 on 2017-11-08. Score: 13

A natural application of trigonometric functions is in the analysis of spatial data. An example is the [Weber problem](https://en.wikipedia.org/wiki/Weber_problem) in location theory - finding the point which minimises the sum of transport costs to $n$ destinations. There is more than one way to solve the problem but [Tellier's solution](https://en.wikipedia.org/wiki/Weber_problem#Tellier.E2.80.99s_trigonometric_solution_of_the_Fermat_and_Weber_triangle_problems) uses trigonometry.


---

### Answer by username_3 on 2017-11-10. Score: 5

Ignoring the intertemporal budget constraint, mergers and bankruptcies the distribution of returns for equity securities traded in a double auction is $$\Pr(\tilde{r}\_t)=\left[\frac{\pi}{2}+\tan^{-1}\left(\frac{\mu}{\gamma}\right)\right]^{-1}\frac{\gamma}{\gamma^2+(\tilde{r}\_t-\mu)^2}.$$

For this see: Harris, D.E. (2017) The Distribution of Returns. Journal
of Mathematical Finance , 7, 769-804.

For returns calculated as the difference of logs, the returns are:
$$\Pr(log(r\_t))=\frac{1}{2\sigma}\text{sech}\left(\frac{\pi(\tilde{r}\_t-\mu)}{2\sigma}\right)$$


---

### Answer by username_4 on 2017-11-10. Score: 4

For a concrete example of how trig (and inverse trig) functions may have financial or economic applications, here's one from "Analysis of Financial Time Series" by Ruey S. Tsay. Consider the AR(2) model:

$$r\_t = \phi\_0 + \phi\_1 r\_{t-1} + \phi\_2 r\_{t-2} +a\_t$$

Its autocorrelation function (ACF) $\rho\_\ell = \operatorname{Corr}(r\_t, r\_{t-\ell})$ satisfies the difference equation $(1 - \phi\_1 B - \phi\_2 B^2) \rho \_ \ell = 0$, where $B$ is the back-shift operator, i.e. $B \rho\_\ell = \rho\_{\ell-1}$ and $B^2 \rho\_\ell = \rho\_{\ell-2}$. (Some people prefer to write $L$ for lag operator instead.)

The second-order characteristic equation $1 - \phi\_1 \omega - \phi\_2 \omega^2 = 0$ has characteristic roots $\omega\_1$ and $\omega\_2$ given by:

$$\omega = \frac{\phi\_1 \pm \sqrt{\phi\_1^2 + 4\phi\_2}}{-2\phi\_2}$$

If the characteristic roots are real, the behaviour is a mixture of two exponential decays. But if instead the discriminant $\phi\_1^2 + 4\phi\_2 < 0$, then the characteristic roots $\omega\_1$ and $\omega\_2$ form a complex-conjugate pair, and the plot of the ACF will exhibit damped sinusoidal waves. To quote Tsay:

> In business and economic applications, complex characteristic roots are important. They give rise to the behavior of business cycles. It is then common for economic time series models to have complex-valued characteristic roots. For an AR(2) model ... with a pair of complex characteristics roots, the *average* length of the stochastic cycles is
>
> $$ k = \frac{2 \pi}{\cos^{-1}[\phi\_1 / (2\sqrt{-\phi\_2})]} $$
>
> where the cosine inverse is stated in radians. If one writes the complex solutions as $a \pm bi$, where $i=\sqrt{-1}$, then we have $\phi\_1 = 2a$, $\phi\_2 = -(a^2 + b^2)$, and
>
> $$ k = \frac{2 \pi}{\cos^{-1}(a / \sqrt{a^2+b^2})} $$

Note that this second way of writing $k$ has a much more geometrically intuitive way of thinking about the inverse cosine.

> **Comments:**
> **username_4**: I have quoted Tsay verbatim re "complex characteristic roots are important. They give rise to the behavior of business cycles" because I think that claim should be treated with skepticism - see the answer by Alecos but also e.g. Stephan Kolassa's comments [here](https://stats.stackexchange.com/a/312906/22228).  I wonder if the book is being oversimplified for its audience (although a graduate-level text, the emphasis is for practitioners). If the lengths of cycles are non-stochastic, however, the formula for $k$ holds true.  

---

