`64 cos^3thetasin^4theta=cos7theta-cos5theta-3cos3theta+3costheta`

64cos^(3)theta sin^(4)theta=cos7 theta-cos5 theta-3cos3 theta+3cos theta

cos theta+cos7 theta+cos3 theta+cos5 theta=0

4costheta cos 2theta cos3theta - 2cos 4theta=1

costheta.cos2theta.cos3theta=1/4

cos 4theta = cos 3theta - cos2theta

Prove that, If " "15 theta = pi, "show that," cos theta cos2 theta cos3theta cos 4theta cos5 theta cos6theta cos7theta =1/2^(7)

cos 3theta+cos 5theta+cos 7theta+cos 9theta=4 cos theta cos 2theta cos 6theta .

Prove that, 64cos^(3) theta sin^(4) theta = cos 7 theta - cos 5 theta - 3 cos 3theta+3 cos theta

Prove that: (sin2theta)/(costheta cos3theta)+(sin4theta)/(cos3theta.cos5theta)+(sin6theta)/(cos5theta.cos7theta)+… to terms = 1/(2sintheta) [sec(2n+1)theta-sectheta]

Prove that: (sin2theta)/(costheta cos3theta)+(sin4theta)/(cos3theta.cos5theta)+(sin6theta)/(cos5theta.cos7theta)+… to terms = 1/(2sintheta) [sec(2n+1)theta-sectheta]