# How to prove that $2^k \cos(\theta) \cos(2\theta)\cdots \cos(2^{k-1} \theta)=1$, $\theta=\pi/m$, $m=2^k+1$?

I have verified many examples and I think that $2^k \cos(\theta) \cos(2\theta)\cdots \cos(2^{k-1} \theta)=1$, where $\theta=\frac{\pi}{2^k+1}$ is true.

For example, $8\,\cos \left( \pi/9 \right) \cos \left( 2\pi/9 \right) \cos \left( 4\pi/9 \right) =1$.

Are there some references about the proof of this identity? Thank you very much.

• Your title is missing the word "if". – DanielV Jul 11 '17 at 15:25

Multiply and divide LHS by $\sin \theta$ and use $2\sin \theta cos\theta = \sin 2\theta$.

You will get,

$$\sin(2^k\theta) = \sin\theta = \sin (\pi - \theta) \implies 2^k\theta = \pi - \theta \implies \theta = \frac{\pi}{2^k + 1}$$

Hint

$$\cos(\theta) \cos(2\theta)\cdots \cos(2^{k-1} \theta)=P$$

multiply both sides by by $\sin (\theta)$

$$\color{red}{\sin(\theta)\cos(\theta)} \cos(2\theta)\cdots \cos(2^{k-1} \theta)=P\sin (\theta)$$

but $\sin(\theta)\cos(\theta)=\sin(2\theta)/2$, so

$$\color{red}{\sin(2\theta)\cos(2\theta)}\cdots \cos(2^{k-1} \theta)=2P\sin (\theta)$$

can you go on?