Proving a trigonometric identity by infinite series

Question

We know that: $$\cos(x) = \sum_{k=0}^\infty \dfrac{(-1)^kx^{2k}}{(2k)!}.$$ And $$\sin(x) = \sum_{k=0}^\infty \dfrac{(-1)^kx^{2k+1}}{(2k+1)!}.$$ Using this prove that $$\cos(x) \cdot \sin(x) = \dfrac{1}{2} \cdot \sum_{k=0}^\infty \dfrac{(-1)^k(2x)^{2k+1}}{(2k+1)!} =\dfrac{\sin(2x)}{2}.$$

Answer 1

Obviously $\sin(x)\cdot\cos(x)$ is an entire odd function, hence $$[x^{2k}]\sin(x)\cos(x)=0,$$ while: $$\begin{eqnarray*}[x^{2k+1}]\sin(x)\cos(x) &=& \sum_{j=0}^{k}\frac{(-1)^{k-j}}{(2(k-j))!}\cdot\frac{(-1)^j}{(2j+1)!}\\&=&\frac{(-1)^k}{(2k+1)!}\sum_{j=0}^{k}\binom{2k+1}{2j}\\&=&\frac{(-4)^k}{(2k+1)!}\end{eqnarray*}$$ is exactly $[x^{2k+1}]\frac{\sin(2x)}{2}$. This is more or less the same as writing $$\sin x=\frac{e^{ix}-e^{-ix}}{2},\qquad \cos x=\frac{e^{ix}+e^{-ix}}{2}$$ and multiplying the two expressions.

Answer 2

Here is a hint: If $$f(x) = \sum_{k=0}^\infty a_k x^k$$ and $$g(x) = \sum_{k=0}^\infty b_k x^k,$$ then $$f(x) g(x) = \sum_{k=0}^\infty c_k x^k,$$ where $$c_k = \sum_{j=0}^k a_j b_{k-j}.$$ This is known as the Cauchy product or convolution of ordinary generating functions.