TITLE: How was Euler able to create an infinite product for sinc by using its roots?
QUESTION [13 upvotes]: 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?

REPLY [18 votes]: 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
$$
\begin{align}
\frac{\sin x}{x} & = \frac{(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots}{(-\pi^2)(-4\pi^2)(-9\pi^2)\cdots} \\[8pt]
& = \left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\cdots.
\end{align}
$$