I'd like to prove $sin(x) = x\prod\limits_{k=1}^{+\infty}(1-\frac{x^{2}}{\pi^{2}k^{2}})$ using Dominated Converge theorem which states :
Let $\{a_{k,n}\}_{n \in \mathbb{N},k \in I}$ be a family of real positive numbers. If $\begin{cases}n \to +\infty & a_{k,n} \longmapsto a_{k}* \hspace{0.1cm} \forall k \\ \forall k \in I & \lvert {a_{k,n}} \rvert \leq b_{k} \\ \sum\limits_{k \in I}b_{k} & <+\infty\end{cases} \rightarrow \lim\limits_{n \to +\infty} \prod\limits_{k=0}^{+\infty}(1+a_{k,n}) = \prod\limits_{k=0}^{+\infty}(1+a_{k}*)$
I was able,making other considerations to state that $sin(x) = \lim\limits_{n \to +\infty} x \prod\limits_{0<k<\frac{n}{2}}(1-\frac{x^{2}}{n^{2}tan^{2}(\frac{\pi k}{n})})$
Knowing that $\lim\limits_{n \to +\infty} {n \cdot tan(\frac{\pi k}{n})} = \pi k$ ,we can easily see this with Taylor, and knowing actually that $n^{2}tan^{2}(\frac{\pi k}{n}) \geq \pi^{2}k^{2}$
I'd like to say that $\lim\limits_{n \to +\infty} x \prod\limits_{0<k<\frac{n}{2}}(1-\frac{x^{2}}{n^{2}tan^{2}(\frac{\pi k}{n})}) \longmapsto x \prod\limits_{k=1}^{+\infty}(1-\frac{x^{2}}{\pi^{2}k^{2}})$.
To do so I'd like to compare and find inequalities between $(1-\frac{x^{2}}{n^{2}tan^{2}(\frac{\pi k}{n})})$ ,$(1-\frac{x^{2}}{\pi^{2}k^{2}})$ and finish thanks to the theorem,
But I'm really stuck on this last part,specially into finding the bound and the convergent series.
Any solution,tip or hint that relies on the cited theorem would be appreacited.
Using $\sin z \geqslant 2z/\pi$ for $0 \leqslant z \leqslant \pi/2$, we have
$$\left|\frac{x^2}{n^2\tan^2 \frac{\pi k}{n}} \right| \leqslant \frac{x^2}{n^2\sin^2 \frac{\pi k}{n}} \leqslant \frac{x^2}{n^2\left(\frac{2\pi k}{\pi n}\right)^2} = \frac{x^2}{4 k^2},$$
and thus the infinite product is uniformly convergent for all $n$. This justifies the interchange of limit and product.
Here we use the theorem that uniform convergence of the series $\sum_k |a_k(n)|$ implies uniform convergence of the infinite product $\prod_k(1 + a_k(n))$.