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?

Solutions Collecting From Web of "How was Euler able to create an infinite product for sinc by using its roots?"

To summarize,

How was he able to do this?

He lucked out, really. It was a coincidence that it worked and an intuitive guess. As Ragib Zaman said, “Euler most likely relied on his incredible intuition for the guess, then his great calculating ability to check it’s validity numerically.”

The derivation (though not accurate for most other functions) is:

\begin{align}
\frac{\sin(x)}{x}&=(x-\pm \pi)(x-\pm 2\pi)\cdots\\
&=\left(\frac{x}{\pi}-\pm 1\right)\left(\frac{x}{2\pi}-\pm1\right)\cdots\\
&=\left(\frac{x}{\pi}-+1\right)\left(\frac{x}{\pi}–1\right)\left(\frac{x}{2\pi}-+1\right)\left(\frac{x}{2\pi}–1\right)\cdots\\
&=\left(\frac{x}{\pi}-1\right)\left(\frac{x}{\pi}+1\right)\left(\frac{x}{2\pi}-1\right)\left(\frac{x}{2\pi}+1\right)\cdots\\
&=\prod_{k=1}^{\infty}\left(\frac{x}{k\pi}-1 \right)\left(\frac{x}{k\pi}+1 \right)
\end{align}

Why does this not simply make a polynomial function that has the roots same roots of sinx/x?

Note that it does make a function* with the same roots as $\frac{\sin(x)}{x}$. You can see this by equating the first equation with $0$.

Can this method be used to make other trigonometric functions?

I don’t think so. From what I understand, there are six basic trigonometric functions and that’s all there is. There is no specific limit on the number of trigonometric functions; rather, these six are the only ones that have caught on due to their specialness, power, and use. (You may find it intriguing that there are hyperbolic analogs of these functions: Hyperbolic functions.)

*I do not think this function could be considered a polynomial function.

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