Using trigonometric identities to prove following series...

Question

I'm stuck on this problem:

If $(2^n + 1)\theta = \pi$,

then find the value of:

$$2^n \prod_{k=0}^{n-1}\cos(2^k\theta)$$

I have no idea how to start. Help would be appreciated!

Answer 1

$$2^n \cos\theta \cos2\theta\cos2^2 \theta... \cos 2^{n-1} \theta =\frac{2^n\sin\theta \cos\theta \cos2\theta\cos2^2 \theta... \cos 2^{n-1} \theta}{\sin\theta}=$$ $$=\frac{\sin2^n\theta}{\sin\theta}=\frac{\sin(\pi-\theta)}{\sin\theta}=1.$$

Answer 2

HINT

Use Viète's infinite product

$$\cos\frac{\alpha}{2} \cdot \cos \frac{\alpha}{4} \cdot \cos \frac{\alpha}{8} \cdots = \prod_{n=1}^\infty \cos \frac{\alpha}{2^n} = \frac{\sin \alpha}{\alpha}$$

and plug in $\alpha=2^n \theta$