solution of a trigonometric identity

Question

We know that $\cos θ\cdot\cos 2θ\cdot\cos 4θ\cdot \cos 8θ\cdot...\cdot \cos (2^{n-1}θ)=\sin (2^nθ)/(2^n\cdot\sin θ))$

Can anyone show me how to derive the value of

$\sin θ\cdot\sin 2θ\cdot \sin 4θ\cdot\sin 8θ\cdot ...\cdot \sin (2^{n-1}θ)$ using this?

I have been unable to do it by myself. Please help.

Answer 1

Try to multiplicate the both sides of the first equation by $sin\theta ...sin(2^{n-1}\theta) $ and use $sin(2\theta)=2sin(\theta)cos(\theta)$

Answer 2

There may be a simpler way to compute the product other than using the other equation. If $x=\sin\theta\sin2\theta...\sin2^{n-1}\theta$, what happens if you multiply both sides by $\cos \theta$?