To prove $\prod\limits_{n=1}^\infty\cos\frac{x}{2^n}=\frac{\sin x}{x},x\neq0$ Prove $$\prod_{n=1}^\infty\cos\frac{x}{2^n}=\frac{\sin x}{x},x\neq0$$ 
This equation may be famous, but I have no idea how to start. I suppose it is related to another eqution:
(Euler)And how can I prove the $follwing$ eqution?
$$\sin x=x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{2^2\pi^2})\cdots=x\prod_{n=1}^\infty (1-\frac{n^2}{2^2\pi^2})$$
I can't find the relation of the two. Maybe I am stuck in a wrong way,thanks for your help.

By double angle formula we have $$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)=4\sin\left(\frac{x}{4}\right)\cos\left(\frac{x}{4}\right)\cos\left(\frac{x}{2}\right)=\dots=2^{n}\sin\left(\frac{x}{2^{n}}\right)\prod_{k\leq n}\cos\left(\frac{x}{2^{k}}\right)$$ now remains to note that $$\lim_{n\rightarrow\infty}2^{n}\sin\left(\frac{x}{2^{n}}\right)=x.$$
