import math


def count_sines_vs_cosines(m):
    """
    What comes in:  A non-negative integer m.
    What goes out:  Returns the number of integers from -m to m,
       inclusive, whose sine is greater than its cosine.
    Side effects:   None.
    Examples:
    Since:  sine(-5) is about  0.96  and cosine(-5) is about  0.28
            sine(-4) is about  0.76  and cosine(-4) is about -0.65
            sine(-3) is about -0.14  and cosine(-3) is about -0.99
            sine(-2) is about -0.91  and cosine(-2) is about -0.42
            sine(-1) is about -0.84  and cosine(-1) is about  0.54
            sine(0)  is about  0.00  and  cosine(0) is about  1.00
            sine(1)  is about  0.84  and  cosine(1) is about  0.54
            sine(2)  is about  0.91  and  cosine(2) is about -0.42
            sine(3)  is about  0.14  and  cosine(3) is about -0.99
            sine(4)  is about -0.76  and  cosine(4) is about -0.65
            sine(5)  is about -0.96  and  cosine(5) is about  0.28
      -- count_sines_vs_cosines(5) returns 6 because
           for -5, -4, -3, 1, 2, and 3, their sine is larger than their cosine
      -- count_sines_vs_cosines(3) returns 4
      -- count_sines_vs_cosines(0) returns 0
      -- count_sines_vs_cosines(1) returns 1
      -- Also:  count_sines_vs_cosines(101) returns 100 (trust me!)
    """
    # -------------------------------------------------------------------------
    # DONE: 7. Implement and test this function.
    #   Note that you should write its TEST function first (above).
    #
    # IMPORTANT: As in previous problems in this session,
    #   you must NOT use the 2 or 3-parameter versions
    #   of the RANGE expression, if you happen to know them.
    # -------------------------------------------------------------------------
    count = 0
    for k in range(2 * m):
        if math.sin(k - m) > math.cos(k - m):
            count = count + 1
    return count