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.
Hint
Use the identity
$$\cos\frac x{2^n}=\frac12\frac{\sin\frac x{2^{n-1}}}{\sin\frac{x}{2^n}}$$ and telescope.
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 $$ x\prod_{k=1}^\infty\left(1-\frac{x^2}{k^2\pi^2}\right)=\sin(x)\tag{5} $$
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 $$\lim_{n\rightarrow\infty}2^{n}\sin\left(\frac{x}{2^{n}}\right)=x.$$
