As we know, we can define that:
$$\sin x= \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!},\quad x \in \boldsymbol{R}$$ and also
$$\cos x= \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n}}{(2n)!},\quad x \in \boldsymbol{R}$$
Furthermore, we know this equation is true:
$$\sin x \cos x =\dfrac{1}{2} \sin 2x \tag{1}$$
I have tried to prove this as follows: using the Cauchy product,
\begin{eqnarray} \sin x \cos x& =&\left(\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}\right)\left(\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n}}{(2n)!}\right) \\&=&\sum_{n=0}^{\infty}\left(\sum_{i=0}^{n}(-1)^i\dfrac{x^{2i+1}}{(2i+1)!}(-1)^{n-i}\dfrac{x^{2n-2i}}{(2n-2i)!}\right)\\ &=& \sum_{n=0}^{\infty}\left(\sum_{i=0}^{n}(-1)^n \dfrac{x^{2n+1}}{(2i+1)!(2n-2i)!}\right)\\ &=&\sum_{n=0}^{\infty}(-1)^n x^{2n+1}\left(\sum_{i=0}^{n}\dfrac{1}{(2i+1)!(2n-2i)!}\right) \end{eqnarray}
But the right side of equation (1) is $\dfrac{1}{2} \sum_{n=0}^{\infty} (-1)^n \dfrac{2^{2n+1} x^{2n+1}}{(2n+1)!} $. It's so weird! I don't know how to make them equal!
