On $\sin x=x \cdot \prod_{n=1}^{\infty}(1-(\frac{x}{\pi n})^2)$ I found this product representation of $\sin x$
 online and I want to know the intuition behind this. I know that there are other questions about this product in this site, but I haven't gotten what I wanted from them.$$\sin x=x\prod_{n=1}^{\infty}\left(1-\left(\frac{x}{\pi n}\right)^2\right)$$
I recognize that the product has roots $x=\pm\pi n$ for all $n\in \Bbb N\cup\{0\}$, which are the same roots as $\sin x$. Is that all to the product, or are there any deeper-level meanings to this product? What is the use of this product (other than the fact that $\Pi$ notation is the coolest)? Is there any geometric intuition to this product? How would one integrate this product (assuming that they didn't know that it represented $\sin x$)? Is there any connection to this product and the Taylor series for $\sin x$? 
 A: Euler wanted to compute $\zeta(2):=\sum_{n\ge 1}n^{-2}$. He noted that, while $\sin x$ isn't a polynomial, if it were you could factorise it based on its roots. He non-rigorously concluded the title equation from such a factorisation, by noting $\frac{\sin x}{x}$ (i) has the non-zero integer multiples of $\pi$ as roots, none of them "repeated" by calculus, and (ii) $\to 1$ as $x\to 0$. (If $p$ is a polynomial with non-repeated root $a$, $p'(a)\ne 0$; and $\frac{\cos x}{x}-\frac{\sin x}{x^2}\ne 0$ for such $x$.) Thus $$\frac{\sin x}{x}=\prod_{n=1}^\infty (1-\frac{x}{n\pi})\prod_{n=1}^\infty (1-\frac{x}{-n\pi})=\prod_{n=1}^\infty (1-\frac{x^2}{n^2\pi^2}).$$Satisfied of his expression, Euler equated $x^2$ coefficients viz. $-\frac{1}{6}=-\pi^{-2}\zeta(2)$. If you want to make this proof of $\zeta(2)=\frac{\pi^2}{6}$ rigorous, you need to come up with a better reason for why this factorisation should be valid. This was eventually provided via the Weierstrass factorisation theorem, but that goes beyond intuition.
A: Just as a polynomial is the product of its linear factors times a constant, Euler wrote $\sin x $ as $x$ times products of $(1\pm x/(n \pi ))$ to get the results that he was looking for.
Note that $\sin x=0$ at $x=0$, and $x=\pm n\pi$
