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.
 A: Hint
Use the identity
$$\cos\frac x{2^n}=\frac12\frac{\sin\frac x{2^{n-1}}}{\sin\frac{x}{2^n}}$$
and telescope.
A: In this answer, it is shown that using induction and the identity
$$
\cos(x2^{-k})=\frac{\sin(x2^{1-k})}{2\sin(x2^{-k})}\tag{1}
$$
we get
$$
\prod_{k=1}^\infty\cos(x2^{-k})=\frac{\sin(x)}x\tag{2}
$$

In this answer, it is shown that
$$
\frac1x+\sum_{k=1}^\infty\frac{2x}{x^2-k^2}=\pi\cot(\pi x)\tag{3}
$$
Integrating $(3)$ gives
$$
\log(\pi x)+\sum_{k=1}^\infty\log\left(1-\frac{x^2}{k^2}\right)=\log(\sin(\pi x))\tag{4}
$$
Substituting $x\mapsto x/\pi$, and exponentiating yields
$$
x\prod_{k=1}^\infty\left(1-\frac{x^2}{k^2\pi^2}\right)=\sin(x)\tag{5}
$$
A: 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 $$\lim_{n\rightarrow\infty}2^{n}\sin\left(\frac{x}{2^{n}}\right)=x.$$
