Subtracting terms from a Fourier series

Question

It is known that $\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}=\frac{\pi-x}{2}$ in $]0,\pi]$, mostly because this is a way of evaluating $\zeta(2)$. Knowing this, is there a way to evaluate $\sum_{n=1}^{\infty}\frac{\sin(2nx)}{2n}$? It's an exercise in one of my professor past papers, and he talks about "antisimmetric part" of a function. I suspect that it is possible to do something similar to what can be done with matrices, but I have no idea why or how. Anyone?

Answer 1

On the surface, all that happened was plugging $2x$ instead of $x$, and dividing the sum by $2$. Which results in $$\sum_{n=1}^{\infty}\frac{\sin(2nx)}{2n} = \frac12 \frac{\pi - 2x}{2},\quad x\in (0,\pi) \tag1$$ However, this only works because the original identity $$\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}=\frac{\pi-x}{2} \tag2$$ was true on the larger interval $x\in (0,2\pi)$. (This latter fact does involve some symmetry consideration: replacing $x$ by $2\pi -x$ changes the sign of both sides.)

In general: if something like (2) is true in some interval $I$, and we have an invertible transformation $x=f(t)$, $t=f^{-1}(x)$, then $$\sum_{n=1}^{\infty}\frac{\sin(n f(t))}{n}=\frac{\pi-f(t)}{2} $$ holds in $f^{-1}(I)$. And we can rename $t$ as $x$ if we want. It's not something about Fourier series, just a manipulation of symbols.