cos theta*cos2 theta*cos2^(2)theta*cos2^(3)theta....cos2^(n-1)theta=(sin2^(n)theta)/(2^(n)sin theta)

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

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

Prove that cos theta cos 2theta cos2^2 theta … cos 2^(n-1) theta = (sin2^n theta)/(2^n.sin theta) , for all n in N

Prove that cos theta cos 2 theta cos 4 theta…. Cos 2 ^(n-1) theta = ( sin ( 2 ^(n) theta))/( 2 ^(n) (sin theta)) .

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

If cos theta+cos^(2)theta=1 then sin^(2)theta+2sin^(2)theta+sin^(2)theta=

Let f_(n)(theta)=((cos theta)/(2)+cos2 theta+(cos(7 theta))/(2)+...+cos(3n-2)((theta)/(2 ))/((sin theta)/(2)+sin2 theta+(sin(7 theta))/(2)+...+sin(3n-2)((theta)/(2))) then f_ (3)((3 pi)/(16))

(sin2 theta)/(1-cos2 theta)

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.

cos theta+sin theta=cos2 theta+sin2 theta