I've been working on two proofs of my own on the factorisation of $\sin(x)$ and consequentially a sum for $\cot(x)$ (though I do not claim these to be my own at all as I haven't done any research yet on their authenticity), and I'd like to check with the mathematical community s to whether they are entirely rigorous.
Part 1: Factorising $\sin(x):$
Begin by assuming that there exists some constant $\gamma$ such that $\sin(x)$ can be factorised in terms of its roots: $$\sin(x)=\gamma x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)...$$ Regrouping yields: $$\sin(x)=\gamma x(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)...$$ (i)$$=\gamma x\prod_{n\in\mathbb{N}^*}(x^2-n^2\pi^2)$$ What remains is to discern what the value of $\gamma$ is. To do, we need to make sure that there are a few coherences between our factorised form and certain properties of $\sin(x)$ to be satisfied. Playing with many, we settle with the following condition that must be satisfied: $$\lim_{x\to 0}(\frac{\sin(x)}{x})=1\Leftrightarrow\lim_{x\to0}\gamma\prod_{n\in\mathbb{N}^*}(x^2-n^2\pi^2)=1\Leftrightarrow\gamma=\frac{1}{\prod_{n\in\mathbb{N}^*}(0^2-n^2\pi^2)}$$ $$=\frac{1}{(-\pi^2)(-4\pi^2)(-9\pi^2)...}$$ This may seem like a useless specification. After all, this means that $\gamma$ is undefined, as the negative products make it's sign alternate and converge to 0. Nevertheless, when substituting this expression back into (i) , we get the following: $$\sin(x)=x\frac{(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)...}{(-\pi^2)(-4\pi^2)(-9\pi^2)...}$$ If it doesn't strike you immediately, then look at how, because this is a product of fractions, we can regroup the terms in the following manner: $$\sin(x)=x(\frac{x^2-\pi^2}{-\pi^2}\frac{x^2-4\pi^2}{-4\pi^2}\frac{x^2-9\pi^2}{-9\pi^2}...)=x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})...$$ $$\Leftrightarrow\sin(x)=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)$$ Every term is well defined in this product expression! We have thus factorised $\sin(x)$ into a well defined series.
Part 2: Deriving a sum for $\cot(x)$:
We begin as before, assuming that there exists a $\gamma$ such that we can factorise $\sin(x)$ in terms of its roots: $$\sin(x)=\gamma\prod_{n\in\mathbb{Z}}(x-n\pi)=\gamma x\prod_{n\in\mathbb{N}^*}(x^2-n^2\pi^2)$$ Applying the natural logarithm on both sides yields: $$\ln(\sin(x))=\ln(\gamma)+\ln(x)+\sum_{n\in\mathbb{N}^*}\ln(x^2-n^2\pi^2)$$ And now taking the derivative with respect to $x$ on both sides yields: $$\cot(x)=\frac{1}{x}+\sum_{n\in\mathbb{N}^*}\frac{2x}{x^2-n^2\pi^2}$$
Motivation and problem:
These proofs are a lot simpler than having to use the Weierstrass factorisation theorem for proving part 1 and then having to use the Mittag-Leffler theorem for part 2. Despite these proofs checking out numerically and graphically, I question whether the use of this constant $\gamma$ in part 2 is rigorous at all, since it disappears later on in the proof without ever having defined it. I'm also unsure about it's existence and validity of use in part 1, considering it technically remains undefined due to the alternation of signs on the denominator. So, despite these proofs being nice and elegant and leading us to the right answer, are they really valid?
Let $$P_n=\prod_{k=1}^n (x-k \pi)(x+k\pi)=(-1)^n \pi ^{2 n} \left(1+\frac{x }{\pi }\right)_n \left(1-\frac{x}{\pi }\right)_n$$
Expand to $$\sin(x)-\gamma_n\, x\, P_n$$ around $x=0$ to obtain $$\gamma_n=\frac {(-1)^n}{\pi ^{2 n}\, (n!)^2}$$