Using induction, prove that `cos theta · cos 2theta · cos 2^2 theta. ... cos 2^(n-1) theta =(sin2^n theta)/(2^n sin theta)`

Prove that cos theta cos 2theta cos2^2 theta … cos 2^(n-1) theta = (sin2^n theta)/(2^n.sin theta) , for all n in N

Prove that cos theta cos 2 theta cos 4 theta…. Cos 2 ^(n-1) theta = ( sin ( 2 ^(n) theta))/( 2 ^(n) (sin theta)) .

Using induction, prove that, costhetacos2thetacos2^(2)theta . . .cos2^(n)theta=(sin2^(n+1)theta)/(2^(n+1)sintheta) .

cos theta + cos 2 theta + cos 3 theta + …. + cos { (n-1) theta}+ cos n theta=

Prove that sin^(4) theta + cos^(4) theta = 1 - 2 sin^2 theta cos^(2) theta

cos theta+cos2 theta+cos3 theta+......+cos(n-1)theta+cos n theta=

If (2^n+1)theta=pi then 2^n costheta cos2theta cos2^2 theta .......cos 2^(n-1) theta=

If theta=pi/(2^n+1) then cos theta cos 2 theta cos 2^(2) theta...... Cos2^(n-1) theta is equal to

Using mathematical induction prove that, costheta+cos2theta+ . . .+cosntheta=("cos"(n+1)/(2)theta"sin"(ntheta)/(2))/("sin"(theta)/(2)) .

Prove that : sin^(4)theta + cos^(4)theta = 1 - 2 cos^(2) theta + 2 cos^(4)theta