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?
sin(x)x=(1+xπ)(1xπ)(1+x2π)(1x2π)(1+x3π)(1x3π)

Answer

Real analysis approach.

Let α(0,1), then define on the interval [π,π] the function f(x)=cos(αx) and 2π-periodically extended it the real line. It is straightforward to compute its Fourier series. Since f is 2π-periodic and continuous on [π,π], then its Fourier series converges pointwise to f on [π,π]:
f(x)=2αsinπαπ(12α2+n=1(1)nα2n2cosnx),x[π,π]
Now take x=π, then we get
cotπα1πα=2απn=11α2n2,α(1,1)
Fix t(0,1). Note that for each α(0,t) we have |(α2n2)1|(n2t2)1 and the series n=1(n2t2)1 is convergent. By Weierstrass M-test the series in the right hand side of (2) is uniformly convergent for α(0,t). Hence we can integrate (2) over the interval [0,t]. And we get
lnsinπtπt=n=1ln(1t2n2),t(0,1)
Finally, substitute x=πt, to obtain
sinxx=n=1(1x2π2n2),x(0,π)

Complex analysis approach

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

Let f be an entire function with infinite number of zeros {an:nN}. Assume that a0=0 is zero of order r and limnan=, then
f(z)=zrexp(h(z))n=1(1zan)exp(pnk=11k(zan)k)
for some entire function h and sequence of positive integers {pn:nN}. The sequence {pn:nN} can be chosen arbitrary with only one requirement the series
n=1(zan)pn+1 is uniformly convergent on each compact KC.

Now we apply this theorem to the entire function sinz. In this case we have an=πn and r=1. Since the series
n=1(zπn)2
is uniformly convergent on each compact KC, then we may choose pn=1. In this case we have
sinz=zexp(h(z))nZ{0}(1zπn)exp(zπn)
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.

Attribution
Source : Link , Question Author : Michael Li , Answer Author : Norbert

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