How do you find the exact values of the sine, cosine, and tangent of the angle 13π12?

1 Answer
Apr 13, 2017

sin(13π12)=232

cos(13π12)=2+32

tan(13π12)=232+3

Explanation:

Recall:

sinθ=±1cos2θ2

cosθ=±1+cos2θ2

Since 13π12 is in the third quadrant,

sin(13π12)<0, cos(13π12)<0, tan(13π12)>0

sin(13π12)= 1cos(13π6)2=1322=232

cos(13π12)= 1+cos(13π6)2=1+322=2+32

tan(13π12)=sin(13π12)cos(13π12)=2322+32=232+3

I hope that this was clear.