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.