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

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

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

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

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

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

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 : cos ec theta+cos ec2 theta+cos ec2^(2)theta+.........+cos ec2^(n-1)theta=cot((theta)/(2))-cot2^(n-)