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

If theta = (pi)/(2 ^(n) +1) prove that 2 ^(n) cos theta cos 2 theta cos 2 ^(2) theta…. Cos 2 ^(n -1) theta =1

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

For all n in N, cos theta cos 2theta cos 4theta ....cos2^(n-1)theta equals to

For all n in N, cos theta cos 2theta cos 4theta ....cos2^(n-1)theta equals to

If theta = pi/(2^(n) + 1), "show that",2^(n) cos theta cos 2 theta cos2^(2) theta cos2^(3) theta.... cos2^(n-1) theta = 1.

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