Factorization of the sine I am working on the Basel problem for a project for my Mathematics study. I need to prove that one could write the sine as a factorization of its linear roots. I know the proofs is in general done using the Weierstrass Factorization theorem but are there any other proofs specifically for the sine that are less complicated?
P.S. I'm in my first year as a Mathematics student so my knowledge is limited.
 A: This is a proof I stumbled upon while taking Fourier analysis, it uses Fourier series to show that
$$\sin(z)=z\prod_{n=1}^{\infty}\left (  1-\frac{z^2}{n^2 \pi^2}\right )$$
As one would expect the derivation was already known. This derivation doesn't use the Weierstrass Factorization Theorem and if you have some experience with integrals and series then it might not be too complicated.

Before I delve into the derivation I will first say a little on Fourier series in case you are not familiar. I will try to not make this too dense as it is only to give a general idea of what we will be using.
A Fourier series will take some function $f(x)$ and represent it as a trigonometric series on an interval. So for us it will take a function
$$f:[-\pi, \pi]\rightarrow\mathbb{R}$$
and represent it as a series of the following form
$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left (  a_n\cos(nx)+b_n\sin(nx)\right)\;\;\;\;\;\;\;\;\;(1)$$ 
The coefficients $a_0, a_n, b_n$ are called the Fourier coefficients and they are calculated by the following integrals:
$$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx$$
$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx$$
$$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx$$

To begin, we need to choose a function. We will choose
$$f(x)=\cos(kx)$$
This choice seems fairly arbitrary at the moment but as we progress the reason behind the choice will become clear.
We now wish to calculate the Fourier coefficients. So for our $f(x)$ and the integrals for calculating the Fourier coefficients given above, we get the following:
$$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}\cos(kx)dx=\frac{2\sin(k\pi)}{k}$$
$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}\cos(kx)\cos(nx)dx=\frac{2k\sin(\pi k)\cos(\pi n)-2n\cos(\pi k)\sin(\pi n)}{k^2-n^2}$$
$$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}\cos(kx)\sin(nx)dx=0$$
Plugging these into the series $(1)$ we get:
$$\cos(kx)=\frac{2k\sin(\pi k)}{2\pi k}+\sum_{n=1}^{\infty}\frac{2k\sin(\pi k)(-1)^n}{k^2-n^2}\cos(nx)\;\;\;\;\;\;\;\;\;(2)$$
Which is the Fourier series for $\cos(kx)$ on $[-\pi,\pi]$
Note: since we have $k^2-n^2$ in the denominator the series will not be valid for $k\in\mathbb{Z}$. 

In our series $(2)$ we will let $x=\pi$. This will give us
$$\cos(k\pi)=\frac{2k\sin(\pi k)}{2\pi k}+\sum_{n=1}^{\infty}\frac{2k\sin(\pi k)(-1)^n}{k^2-n^2}\cos(n\pi)$$
Which can be rearranged to give
$$\pi\cot(\pi k)=\frac{1}{k}+2k\sum_{n=1}^{\infty}\frac{1}{k^2-n^2}\;\;\;\;\;\;\;\;\;(3)$$
We make a change of variables in series $(3)$, replace $k$ with $t/\pi$:
$$\begin{align*}
 \pi\cot(t)&=\frac{\pi}{t}+2\frac{t}{\pi}\sum_{n=1}^{\infty}\frac{1}{t^2 / \pi^2-n^2} \\ 
 &=\frac{\pi}{t}+2t\pi\sum_{n=1}^{\infty}\frac{1}{t^2-n^2 \pi^2} 
\end{align*}$$
$$\Rightarrow\cot(t)-\frac{1}{t}=2t\sum_{n=1}^{\infty}\frac{1}{t^2-n^2\pi^2}\;\;\;\;\;\;\;\;\;(4)$$

This next part is where the reason behind the choice $f(x)=\cos(kx)$ will become clear. In order to yield our product we are going to integrate sum $(4)$, the result will allow us to take advantage of logarithms to convert a sum to a product.
The sum $(4)$ converges uniformly on the domain of $\cot(t)-1/t$ so we can integrate it. I will not be proving it here but it is still worth noting.
The path of integration in $\mathbb{C}$ will be a rectifiable curve $\Gamma:[0,z]$ that doesn't intersect any points $t=q\pi$ where $q\in\mathbb{Z}\setminus\left \{  0\right \}$. The reason $\cot(t)-1/t$ is defined at $t=0$ is because it can be classified as a removable singularity as $\lim_{t \rightarrow 0}(\cot(t)-1/t)=0$. Using this path of integration we get
$$\int_\Gamma\left (  \cot(t)-\frac{1}{t}\right )dt=\int_{0}^{z}\left (  \cot(t)-\frac{1}{t}\right )dt$$
as the integral will be path independent. Upon evaluation of the rightmost integral and the equality in $(4)$:
$$\int_{0}^{z}\left (  \cot(t)-\frac{1}{t}\right )dt=\int_{0}^{z}\left (  2t\sum_{n=1}^{\infty}\frac{1}{t^2-n^2\pi^2}\right )dt$$
$$\Rightarrow \ln(\sin(z))-\ln(z)=\sum_{n=1}^{\infty}\ln\left (  \frac{n\pi-z}{n\pi}\cdot\frac{n\pi+z}{n\pi}\right )$$
$$\Rightarrow\frac{\sin(z)}{z}=\prod_{n=1}^{\infty}\left (  \frac{n\pi-z}{n\pi}\cdot\frac{n\pi+z}{n\pi}\right )$$
Rearranging:
$$\sin(z)=z\prod_{n=1}^{\infty}\left (  1-\frac{z^2}{n^2 \pi^2}\right )$$
A: Perhaps you will be interested in this: http://arxiv.org/pdf/math/0701039.pdf
It uses means and methods that you will likely don't have difficulty in understanding-with a bit of further studying perhaps-while being one of the best approaches to the problem I have seen.. 
