" (iv) "2^(3)sin theta cos theta cos2 theta cos4 theta=sin8 theta

Prove the following: 2^(3)sin theta cos theta cos2 theta cos4 theta=sin8 theta

8sin theta cos theta cos2 theta cos4 theta=sin x the x=

8sin theta cos theta cos2 theta cos4 theta=sin x rArr x=

8sin theta cos theta*cos2 theta cos4 theta=sin x rArr x=

cos theta+sin theta=cos2 theta+sin2 theta

8 sin theta * cos theta * cos 2 theta * cos 4 theta = A) sin 4 theta B) sin 8 theta C) sin 16 theta D) cos 8 theta

(sin3 theta+cos3 theta)/(cos theta-sin theta)=1+2xx sin2 theta

2(cos theta+cos2 theta)+(1+2cos theta)sin2 theta=2sin theta

Simplify: (sin8 theta cos theta-sin6 theta cos3 theta)/(cos2 theta cos theta-sin3 theta sin4 theta)

Prove that ( cos theta + cos 2 theta + cos 3 theta + cos 4 theta)/(sin theta + sin 2 theta + sin 3 theta + sin 4 theta)= cot ""(5theta)/(2).