How to prove that $$ \left(\frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^{\infty}{\frac{\cos(2 n x)}{4 n^2 - 1} }\right)\cdot\left(\frac{4}{\pi}\sum_{n=1}^{\infty}{\frac{\sin((2 n - 1) x)}{2 n - 1} }\right) = \sin(x)\;? $$
We know that $$\DeclareMathOperator{\sgn}{sgn} \lvert\sin(x)\rvert = \frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^{\infty}{\frac{\cos(2 n x)}{4 n^2 - 1} }\\ \sgn(\sin(x)) = \frac{4}{\pi}\sum_{n=1}^{\infty}{\frac{\sin((2 n - 1) x)}{2 n - 1} }$$ and the fact that $$ \lvert\sin(x)\rvert \cdot \sgn(\sin(x)) = \sin(x) $$