<s>[INST] How was Euler able to create an infinite product for sinc by using its roots? In the Wikipedia page for the Basel problem, it says that Euler, in his proof, found that
$$\begin{align*}
\frac{\sin(x)}{x} &=
\left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots
\end{align*}$$
because the roots are at $\pm\pi, \pm2\pi, \pm3\pi, \cdots$ and finite polynomials are in this form (i.e. $(x-\text{root}_1)(x-\text{root}_2)\cdots$).
How was he able to do this?  Why does this not simply make a polynomial function that has the roots same roots of $(\sin x)/x$?  Can this method be used to make other trigonometric functions?
 [/INST] Nearly the same question was posted here recently.  I hope this will add a little that is not in the other answers to this present question.
We know that $\dfrac{\sin x}{x}=0$ when $\sin x= 0$ and $x\neq0$, and we know that $\dfrac{\sin x}{x}$ "$=$" $1$ when $x=0$ (I think Euler's way of saying this is that $\sin x = x$ when $x$ is infinitely small).  So this function should be $0$ when $x=\pm\pi$ or $\pm2\pi$ or $\pm3\pi$, etc., so it is
$$
\begin{align}
& \text{constant}\cdot(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\cdots \\[8pt]
& = \text{constant}\cdot(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots.
\end{align}
$$
When $x=0$, this is $(-\pi^2)(-4\pi^2)(-9\pi^2)\cdots$.  But we saw above that when $x=0$, this is $1$.  Hence we have
$$
\begin{align}
\frac{\sin x}{x} & = \frac{(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots}{(-\pi^2)(-4\pi^2)(-9\pi^2)\cdots} \\[8pt]
& = \left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\cdots.
\end{align}
$$
</s>