If `theta=(2pi)/(2009), then costhetacos 2thetacos3theta... cos1004theta` is

If theta=pi/(2^n+1) then costhetacos2thetacos2^2theta....cos2^(n-1)theta is equal to

If theta=(2 pi)/(4033), then cos theta*cos2 theta*cos3 theta*....cos2016 theta is equal

cos3thetacos5theta-cos7thetacos9theta=0

If 13theta=pi ,show that costhetacos2theta cos3theta cos4theta cos5theta cos6theta=2^-6

Solve : costhetacos2thetacos3theta=1/4 , 0lethetalepi

If theta=pi/(2^n+1) , prove that: 2^ncosthetacos2thetacos2^2theta....... cos2^(n-1)theta=1.

sin^4theta+2sin^2thetacos^2theta=1-cos^4 theta

Solve: costhetacos2thetacos3theta=1/4 , where 0 lethetalex .

Solve: costhetacos2thetacos3theta=1/4 , where 0 lethetalex .

sin5 theta+sin2 theta-sin thetacos5 theta+2cos3 theta+2cos^(2)theta+cos theta to a.tan theta b.cos theta c.cot theta d.none of these