In the equation: $$\sin(x) = x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right),$$
I know that if I set $$0 = x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right),$$
all of the roots are really going to be the roots of $\sin(x)$. But why is this generally equal to $\sin(x)$? How can I know that $\sin(2.285672364)$ will be $\prod_{n=1}^\infty \left(1-\frac{(2.285672364)^2}{n^2\pi^2}\right)$, for example?
(I mean, how do I know that the graphs of both functions will be the same?).
How can I assume that the coefficients of $\frac{\sin(x)}{x}$ in the product expansion are going to be the same of the sum expansion, i.e. in the Taylor series? $\left(\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots\right)$ like Euler showed in the 'Basel problem', by saying: $$\frac{x^2}{3!} = \frac{x^2}{\pi^2}\left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots\right)$$ So: $$\zeta(2) = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + ... = \frac{\pi^2}{6}$$
