`cos A *cos 2A *cos 2^2 A *cos 2^3 A..... cos 2^n A = sin(2^n A) / (2^n Sin A)`

Prove that : cos A cos 2A cos 2^2 A cos 2^3 A........ cos 2^(n-1) A= (sin 2^n A)/(2^n sinA) .

Prove that cos A cos 2A cos 2^(2) A cos 2^(3)A....cos 2^(n-1) A=(sin 2^(n) A)/(2^(n) sin A)

Prove that: cos A cos 2A cos 2^(2)A cos 2^(3)A….....cos 2^(n-1)A=("sin" 2^(n)A)/(2^(n)"sin"A) .

cos A.cos2A.cos2^(^^)2A*cos2^(^^)3A cos2^(^^)nA=sin(2^(^^)nA)/2^(^^)n sin A

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

(cos h 2 + sin h 2)^(n) =

cos x*cos2x*cos4x......cos(2^(n-1)x)=(sin2^(n)x)/(2^(n)sin x)AA n in N