Q:

Showing that $\sin(2x)=2\sin x\cos x$ by multiplying power series

okay so 
$$\sin(x)= \sum_0^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}\\
\cos(x)= 1+\sum_0^\infty (-1)^n \frac{x^{2n}}{(2n)!}$$
I am asked to show directly by multiplying power series. Tried finding the Cauchy product by setting $a_n=(-1)^n\frac{x^{2n+1}}{(2n+1)!}$ and $b_n=(-1)^n \frac{x^{2n}}{(2n)!}$
$$\sum_0^\infty c_n ~\text{ where }~ c_n=a_0b_n+a_1b_{n-1}+ \cdots +a_nb_0$$
But couldnt work it out any help?

A:

$$ 2\sin x \cos x =2 \left(\sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{(2k+1)!}\right) \left(\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}\right)$$
$$= 2\sum_{k=0}^\infty \sum_{i=0}^k (-1)^i \frac{x^{2i+1}}{(2i+1)!} (-1)^{k-i} \frac{x^{2(k-i)}}{(2k-2i)!} = \sum_{k=0}^\infty (-1)^k x^{2k+1} \sum_{i=0}^k\frac{2}{(2i+1)! (2k-2i)!} $$
$$ = \sum_{k=0}^\infty (-1)^k \frac{(2x)^{2k+1}}{(2k+1)!} = \sin(2x),$$
since
$$\sum_{i=0}^k \frac{2}{(2i+1)!(2k-2i)!} = \frac{2^{2k+1}}{(2k+1)!}.$$

