`sin4theta` can be written as (a) `4sintheta(1-2sin^2 theta)sqrt(1-sin^2 theta)` (b) `2 sin theta cos thetasin^2 theta` (c) `4sin theta-6 sin^3 theta` (d) None of the above

sin4theta can be written as................... A) 4 sin theta (1-2 sin ^(2) theta ) sqrt(1- sin ^(2) theta ) B) 2 sin theta cos theta sin ^(2) theta C) 4 sin theta - 6 sin ^(3) theta D) 4 sin theta + 6 sin ^(2) theta

( sin 2 theta + 2 sin 4 theta + sin 6 theta )/(sin 4 theta + 2 sin 6 theta + sin 8 theta ) =

(sin^(2) theta)/(1-cos theta)-(cos^(2) theta)/(1-sin theta)=cos theta-sin theta

If sin 6theta = sin 4theta - sin 2theta , then theta =

int(2sin2 theta-cos theta)/(6-cos^(2)theta-4sin theta)d theta

" 4."(sin theta)/(1+cos theta)+(1+cos theta)/(sin theta)=(2)/(sin theta)

Prove that : sin theta cos^3 theta - cos theta sin^3 theta = 1/4 sin4theta .

4 sin theta * cos^(3) theta-4 cos theta * sin ^(3)theta= A) 4 cos theta B) cos 4 theta C) 4 sin theta D) sin 4 theta

If cos^2 theta-sin^2 theta= 1/2 then cos^4 theta-sin^4 theta=?

(sin 3 theta)/(sin theta)-(cos 3 theta)/(cos theta)= A) sin 2 theta B) cos 4 theta C)1D)2