Proving $\frac{\sin x}{x} =\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right) \left(1-\frac{x^2}{3^2\pi^2}\right)\cdots$ How to prove the following product?
$$\frac{\sin(x)}{x}=
\left(1+\frac{x}{\pi}\right)
\left(1-\frac{x}{\pi}\right)
\left(1+\frac{x}{2\pi}\right)
\left(1-\frac{x}{2\pi}\right)
\left(1+\frac{x}{3\pi}\right)
\left(1-\frac{x}{3\pi}\right)\cdots$$
 A: 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 $[-\pi,\pi]$:
$$
f(x)=\frac{2\alpha\sin\pi\alpha}{\pi}\left(\frac{1}{2\alpha^2}+\sum\limits_{n=1}^\infty\frac{(-1)^n}{\alpha^2-n^2}\cos nx\right),
\quad x\in[-\pi,\pi]\tag{1}
$$
Now take $x=\pi$, then we get
$$
\cot\pi\alpha-\frac{1}{\pi\alpha}=\frac{2\alpha}{\pi}\sum\limits_{n=1}^\infty\frac{1}{\alpha^2-n^2},
\quad\alpha\in(-1,1)\tag{2}
$$
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
$$
\ln\frac{\sin \pi t}{\pi t}=\sum\limits_{n=1}^\infty\ln\left(1-\frac{t^2}{n^2}\right),
\quad t\in(0,1)
$$
Finally, substitute $x=\pi t$, to obtain 
$$
\frac{\sin x}{x}=\prod\limits_{n=1}^\infty\left(1-\frac{x^2}{\pi^2 n^2}\right),
\quad x\in(0,\pi)
$$
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 
  $$
f(z)=
z^r\exp(h(z))\prod\limits_{n=1}^\infty\left(1-\frac{z}{a_n}\right)
\exp\left(\sum\limits_{k=1}^{p_n}\frac{1}{k}\left(\frac{z}{a_n}\right)^{k}\right)
$$
  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 
  $$
\sum\limits_{n=1}^\infty\left(\frac{z}{a_n}\right)^{p_n+1} 
$$ 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 
$$
\sum\limits_{n=1}^\infty\left(\frac{z}{\pi n}\right)^2
$$
is uniformly convergent on each compact $K\subset \mathbb{C}$, then we may choose $p_n=1$. In this case we have
$$
\sin z=z\exp(h(z))\prod\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)
$$
Let $K\subset\mathbb{C}$ be a compact which doesn't contain zeros of $\sin z$. For all $z\in K$ we have
$$
\ln\sin z=h(z)+\ln(z)+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\ln\left(1-\frac{z}{\pi  n}\right)+\frac{z}{\pi n}\right)
$$
$$
\cot z=\frac{d}{dz}\ln\sin z=h'(z)+\frac{1}{z}+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\frac{1}{z-\pi n}+\frac{1}{\pi n}\right)
$$
It is known that (here you can find the proof)
$$
\cot z=\frac{1}{z}+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\frac{1}{z-\pi n}+\frac{1}{\pi n}\right).
$$
hence $h'(z)=0$ for all $z\in K$. Since $K$ is arbitrary then $h(z)=\mathrm{const}$. This means that
$$
\sin z=Cz\prod\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)
$$
Since $\lim\limits_{z\to 0}z^{-1}\sin z=1$, then $C=1$. Finally,
$$
\frac{\sin z}{z}=\prod\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)=
\lim\limits_{N\to\infty}\prod\limits_{n=-N,n\neq 0}^N\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)=
$$
$$
\lim\limits_{N\to\infty}\prod\limits_{n=1}^N\left(1-\frac{z^2}{\pi^2 n^2}\right)=
\prod\limits_{n=1}^\infty\left(1-\frac{z^2}{\pi^2 n^2}\right)
$$
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.
A: We will use Hardamard Factoriztation theorem to prove it (see https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem#Hadamard_factorization_theorem).
Observe $|\sin(\pi z)|\le e^{\pi |z|}$,  hence it has order of growth less than equal to one. Further it has simple zero at every integer $n \in Z$. Hence by Hardamard Factorization theorem we get
$$ \sin (\pi z)= z e^{az+b}\displaystyle \prod_{0 \ne n \in Z}\left(1-\frac zn\right)e^{\frac z n}, \text { for some } a, b \in  C.$$ Pairing $n$ and $-n$ together in the product  we get
\begin{equation}
 \sin (\pi z)= z e^{az+b}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).
\label{on}
\end{equation}
From above equation we will get
$$\frac {\sin (\pi z)}{\pi z}= \frac{e^{az+b}}{\pi}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
Taking $z \to 0$, in both side we get $\frac{e^{b}}{\pi}=1$.
From above equation we will get
$$\frac {\sin (\pi z)}{\pi z}= {e^{az}}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$ Again using the fact that
$\frac {\sin (\pi z)}{\pi z}$ is an even function, it can be easily seen that $a=0$.
Hence we get $$\frac {\sin (\pi z)}{\pi z}= \displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
Thus  $${\sin (\pi z)}= \pi z\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
{\cal P}_{N}\pars{x} & \equiv \prod_{n = 1}^{N}\pars{1 + {x \over n\pi}} =
\prod_{n = 1}^{N}\pars{n + x/\pi \over n} =
{1 \over N!}{\Gamma\pars{1 + x/\pi + N} \over \Gamma\pars{1 + x/\pi}}
\\[5mm] & =
{1 \over N!}{\pars{N + x/\pi}! \over \Gamma\pars{1 + x/\pi}}
\sr{{\rm as}\ N\ \to\ \infty}{\sim}
{1 \over \Gamma\pars{1 + x/\pi}}
{\root{2\pi}\pars{N + x/\pi}^{N + x/\pi + 1/2}\,\,\expo{-N - x/\pi} \over \root{2\pi}N^{N + 1/2}\,\,\expo{-N}}
\\[5mm] & \sr{{\rm as}\ N\ \to\ \infty}{\sim}
{1 \over \Gamma\pars{1 + x/\pi}}
{N^{N + x/\pi +1/2}\,\,\,\,\bracks{1 + \pars{x/\pi}/N}^{N}\,\,\expo{- x/\pi} \over N^{N + 1/2}}
\\[5mm] & \sr{{\rm as}\ N\ \to\ \infty}{\sim}
{N^{x/\pi} \over \Gamma\pars{1 + x/\pi}}
\\[1cm] & \mbox{Therefore,}
\\ &\color{#44f}{\lim_{N \to \infty}\bracks{{\cal P}_{N}\pars{x}{\cal P}_{N}\pars{-x}}} =
{1 \over \Gamma\pars{1 + x/\pi}\Gamma\pars{1 - x/\pi}} =
{1 \over \pars{x/\pi}\Gamma\pars{x/\pi}\Gamma\pars{1 - x/\pi}}
\\[5mm] & =
{\pi \over x}{1 \over \pi/\sin\pars{\pi\bracks{x/\pi}}} =
\bbx{\color{#44f}{\sin\pars{x} \over x}} \\ &
\end{align}
