How do i show this equality without using reccurence method:$\prod_{k=0}^{k=n}{\cos\frac{\theta}{2^k}}={\frac {\sin\theta}{2^n\sin(2^{-n}\theta)}}$?? I would like to show this without using reccurence method for all $n$ $\in $ $\mathbb{N}$ and $\theta \in \mathbb{R}$ :
$${\cos\frac{\theta}{2}}\cos\frac{\theta}{2^2}\cos\frac{\theta}{2^3}\cdots \cos\frac{\theta}{2^n} ={\frac {\sin\theta}{2^n\sin(2^{-n}\theta)}} $$
Note: I have used many trigonomitrics transformations but i can't succed !!
$$\prod_{k=0}^{k=n}{\cos\frac{\theta}{2^k}}={\frac {\sin\theta}{2^n\sin(2^{-n}\theta)}}$$
Thank you for any help 

First, notice that:
\begin{align}
2\cos x\sin x&=\sin 2x\\
\implies \cos x&=\frac{1}{2}\frac{\sin 2x}{\sin x}
\end{align}
Then, given an integer number $n>0$ we have
$$\cos \left(\frac{\theta}{2}\right)\cdot \cos \left(\frac{\theta}{2^2}\right)\cdot \ldots \cdot \cos \left(\frac{\theta}{2^n}\right)=\frac{1}{2^n}\frac{\sin\left(\theta\right)}{\sin \left(\frac{\theta}{2}\right)}\cdot \frac{\sin \left(\frac{\theta}{2}\right)}{\sin \left(\frac{\theta}{2^2}\right)}\cdot \ldots\cdot \frac{\sin \left(\frac{\theta}{2^{n-1}}\right)}{\sin \left(\frac{\theta}{2^n}\right)}$$
Which is a "telescopic product", and the inequality follows from here.
