How do you find the exact values of the sine, cosine, and tangent of the angle $(13pi)/12$ ?

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Answer 1 · 2017-04-13T20:50:40

$sin((13pi)/12)=-sqrt(2-sqrt(3))/2$

$cos((13pi)/12)=-sqrt(2+sqrt(3))/2$

$tan((13pi)/12)=sqrt((2-sqrt(3))/(2+sqrt(3)))$

Recall:

$sin theta=pm sqrt((1-cos 2theta)/2)$

$cos theta=pm sqrt((1+cos 2theta)/2)$

Since $(13pi)/12$ is in the third quadrant,

$sin((13pi)/12)<0$ , $cos((13pi)/12)<0$ , $tan((13pi)/12)>0$

$sin((13pi)/12)=-sqrt((1-cos((13pi)/6))/2) =-sqrt((1-sqrt(3)/2)/2) =-sqrt(2-sqrt(3))/2$

$cos((13pi)/12)=-sqrt((1+cos((13pi)/6))/2) =-sqrt((1+sqrt(3)/2)/2) =-sqrt(2+sqrt(3))/2$

$tan((13pi)/12)=(sin((13pi)/12))/(cos((13pi)/12)) =(-sqrt(2-sqrt(3))/2)/(-sqrt(2+sqrt(3))/2) =sqrt((2-sqrt(3))/(2+sqrt(3)))$

I hope that this was clear.

How do you find the exact values of the sine, cosine, and tangent of the angle (13pi)/12(13pi)/12?