Question

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

Answer 1

`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)))`

Explanation:

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.