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

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

If theta=(pi)/(7), prove that cos theta cos2 theta cos3 theta=(1)/(2^(3))

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

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

Using induction,prove that cos theta cos2 theta cos2^(2)theta...cos2^(n-1)theta=(sin2^(n)theta)/(2^(n)sin theta)

If f(theta)=lim_(n rarr oo)cos theta cos2 theta cos2^(2)theta...*cos2^(n-1)theta, then |(1)/(pi)f((3 pi)/(2))|

Prove that: (cos2 theta)/(1+s in2 theta-cos2 theta)=cot theta

Prove that cos theta* cos 2theta* cos2^(2)theta * cos2^(3)theta x... cos 2^(n-1) theta = 1/2^(n) , when theta = pi/(2^(n)+1) .

Prove that sin^(6)theta+cos^(6)theta=1-3sin^(2)theta.cos^(2)theta.