How to prove that $\frac{\sin \pi x}{\pi x}=\prod_{n=1}^{\infty}(1-\frac{x^2}{n^2})$ How to prove that $$\frac{\sin \pi x}{\pi x}=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)$$
I tried it with the Taylor series of $\sin(x)$ but I failed. 
Is there any help? 

You can start from the identity

$$ \frac{\pi \cot(\pi x)}{x} = \sum_{n=0}^{\infty}\frac{1}{x^2-n^2} \quad x\neq \mathbb{Z}$$

$$ \implies \frac{\pi \cot(\pi x)}{x}= \frac{1}{x^2}+2\sum_{n=1}^{\infty}\frac{1}{x^2-n^2}$$
$$ \implies {\pi \cot(\pi x)}= \frac{1}{x}+2\sum_{n=1}^{\infty}\frac{x}{x^2-n^2}. $$
Now, you can advance by integrating both sides w.r.t. $x$.
