# Proving sinxx=(1−x2π2)(1−x222π2)(1−x232π2)⋯\frac{\sin x}{x} =\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right) \left(1-\frac{x^2}{3^2\pi^2}\right)\cdots

How to prove the following product?

Real analysis approach.

Let $\alpha\in(0,1)$, then define on the interval $[-\pi,\pi]$ the function $f(x)=\cos(\alpha x)$ and $2\pi$-periodically extended it the real line. It is straightforward to compute its Fourier series. Since $f$ is $2\pi$-periodic and continuous on $[-\pi,\pi]$, then its Fourier series converges pointwise to $f$ on $[-\pi,\pi]$:

Now take $x=\pi$, then we get

Fix $t\in(0,1)$. Note that for each $\alpha\in(0,t)$ we have $|(\alpha^2-n^2)^{-1}|\leq(n^2-t^2)^{-1}$ and the series $\sum_{n=1}^\infty(n^2-t^2)^{-1}$ is convergent. By Weierstrass $M$-test the series in the right hand side of $(2)$ is uniformly convergent for $\alpha\in(0,t)$. Hence we can integrate $(2)$ over the interval $[0,t]$. And we get

Finally, substitute $x=\pi t$, to obtain

Complex analysis approach

We will need the following theorem (due to Weierstrass).

Let $f$ be an entire function with infinite number of zeros $\{a_n:n\in\mathbb{N}\}$. Assume that $a_0=0$ is zero of order $r$ and $\lim\limits_{n\to\infty}a_n=\infty$, then

for some entire function $h$ and sequence of positive integers $\{p_n:n\in\mathbb{N}\}$. The sequence $\{p_n:n\in\mathbb{N}\}$ can be chosen arbitrary with only one requirement $-$ the series
is uniformly convergent on each compact $K\subset\mathbb{C}$.

Now we apply this theorem to the entire function $\sin z$. In this case we have $a_n=\pi n$ and $r=1$. Since the series

is uniformly convergent on each compact $K\subset \mathbb{C}$, then we may choose $p_n=1$. In this case we have

Let $K\subset\mathbb{C}$ be a compact which doesn’t contain zeros of $\sin z$. For all $z\in K$ we have

It is known that (here you can find the proof)

hence $h'(z)=0$ for all $z\in K$. Since $K$ is arbitrary then $h(z)=\mathrm{const}$. This means that

Since $\lim\limits_{z\to 0}z^{-1}\sin z=1$, then $C=1$. Finally,

This result is much more stronger because it holds for all complex numbers. But in this proof I cheated because series representation for $\cot z$ given above require additional efforts and use of Mittag-Leffler’s theorem.