`costheta+cos2theta+cos3theta+.......+cos(n-1)theta+cosntheta=`

costheta-cos2theta=sin3theta

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

costheta.cos2theta.cos3theta=1/4

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

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 (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)

cos theta cos2 theta cos3 theta=(1)/(4)

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