Reducing Radian AngleHow would I reduce a large radian angle into a smaller one?For example, if you're asked to find a related angle for cos ⁡ 3177 π 12 . I usually try to break it down to π 2 , π 3 , π 4 , or π 6 to visualize it in a unit circle, but I wouldn't know where the terminal arm would lie in such a big angle.I know that a period for one revolution in a unit circle is equal to 2 π . So I could divide the big angle 3177 π 12 by 2 π .So that's 3177 π 12 ⋅ 1 2 π = 132.375. So I found out the number of rotations. But I still don't know how to interpret that in a unit circle. If I were to draw the angle 3177 π 12 , which quadrant would the terminal arm lie?

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Reducing Radian AngleHow would I reduce a large radian angle into a smaller one?For example, if you&#039;re asked to find a related angle for   cos  ⁡            3177      π        12    . I usually try to break it down to       π    2    ,      π    3    ,      π    4    , or       π    6   to visualize it in a unit circle, but I wouldn&#039;t know where the terminal arm would lie in such a big angle.I know that a period for one revolution in a unit circle is equal to   2  π  . So I could divide the big angle             3177      π        12   by   2  π  .So that&#039;s             3177      π        12    ⋅      1          2      π        =  132.375. So I found out the number of rotations. But I still don&#039;t know how to interpret that in a unit circle. If I were to draw the angle             3177      π        12    , which quadrant would the terminal arm lie?

Dana Simmons · Accepted Answer

You are almost there. You have            3177      π              /            12              2      π        =  132.375So you can subtract 132 complete rotations  3177  π      /    12  −  132  ⋅  2  π  =  2  π      9    24    =  2  π      3    8  Less than   1      /    4 of a circle would be in the first quadrant,   3      /    8 is in the second quadrant. Second quadrant are angles between   1      /    4 and   1      /    2 of a full circle (  2  π)

Raiden Barr · Accepted Answer

You want to find the unique integer k such that  0  ≤            3177      π        12    −  2  k  π  &amp;lt;  2  πThis inequality becomes  0  ≤  3177  −  24  k  &amp;lt;  24so you want the quotient of dividing   3177 by   24, but also the remainder is important.                    3                    1                    7                    7                    24                                  7                    7                          132                                        5                    7                                              9            or   3177  =  132  ⋅  24  +  9.Hence            3177      π        12    =  132  ⋅  2  π  +            9      π        12    =  132  ⋅  2  π  +            3      π        4

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