# Proving $\sin 2x = 2\sin x \cos x$ using Taylor Series and Cauchy products

I have that the Taylor series of $\sin x$ and $\cos x$ are \begin{equation*} \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1} \\ \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} \end{equation*} which I understand yields the product series $\sum_{n=0}^{\infty} c_{n}x^n$ where \begin{equation*} c_n = \begin{cases} \sum\limits_{k=0}^{m} \frac{(-1)^{k}(-1)^{m-k}}{(2k+1)!(2m-2k)!} & n = 2m + 1 \\ \hspace{32 pt} 0 & n = 2m \end{cases} \end{equation*} $\forall \hspace{3 pt} m \in \mathbb{N}$. I then know by simple substitution that \begin{equation*} \frac{1}{2} \sin 2x = \sum_{n=0}^{\infty} \frac{(-1)^{n}2^{2n}}{(2n+1)!}x^{2n+1} \end{equation*} I know I need to show that the odd $c_n$'s and the terms of the above series equal (the even ones are irrelevant as they are all $0$ and so have no bearing on sum) but I am having trouble doing so. Can someone please show how to reduce said equality? Thanks in advance.

Note that \begin{align} c_{2n+1}&=\sum_{k=0}^{n} \frac{(-1)^{k}(-1)^{n-k}}{(2k+1)!(2n-2k)!}\\ &=\sum_{k=0}^{n} \frac{(-1)^{n}(2n+1)!}{(2n+1)!(2k+1)!(2n-2k)!}\\ &=\sum_{k=0}^{n} \frac{(-1)^{n}}{(2n+1)!}{2n+1 \choose 2k+1} \end{align} and that \begin{align} \sum_{k=0}^n{2n+1 \choose 2k+1}&=\sum_{k=0}^n({2n \choose 2k}+{2n \choose 2k+1})\\ &=\sum_{k=0}^{2n}{2n \choose k}\\ &=2^{2n} \end{align}