We are considering a $2\pi$ periodic function defined on $x\in(-\pi,\pi)$ by $$f(x) = \pi - x, 0<x<\pi $$ and 0 otherwise.
I already computed the full Fourier series is equal to: $$f(x) = {\pi\over4}+ \sum_{p=0}^\infty {2\over{\pi(2p+1)^2}}\cos[(2p+1)x] + \sum_{n=1}^\infty {1\over n}\sin(nx) $$
The next piece is to show that for any $ x\epsilon[0, \pi]$ the following equality holds: $$\sum_{n=1}^\infty {1\over n}\sin(nx)= {\pi\over4}+ \sum_{p=0}^\infty {2\over{\pi(2p+1)^2}}\cos[(2p+1)x]$$
I emailed my professor and he told me to deduce this from a result of the series pointwise convergence (limit). The pointwise limit is:
$$\begin{cases} \pi-x, & \text{if $x \in (0,\pi)$} \\ 0, & \text{if $x\in(-\pi,0)$} \\ {\pi\over2}, & \text{if $x=0$} \\ \end{cases}$$
So how do I show the equality holds?
Let $0 < x \leqslant \pi$. By the known pointwise convergence of the Fourier series, we have
$$0 = f(-x) = \frac{\pi}{4} + \sum_{p=0}^\infty \frac{2}{\pi(2p+1)^2}\cos \left[(2p+1)(-x)\right] + \sum_{n=1}^\infty \frac{\sin \left(n(-x)\right)}{n}.$$
Since $\cos$ is an even function and $\sin$ odd, this becomes
$$0 = \frac{\pi}{4} + \sum_{p=0}^\infty \frac{2}{\pi(2p+1)^2}\cos \left[(2p+1)x\right] - \sum_{n=1}^\infty \frac{\sin (nx)}{n},$$
and that is evidently equivalent to
$$\sum_{n=1}^\infty \frac{\sin (nx)}{n} = \frac{\pi}{4} + \sum_{p=0}^\infty \frac{2}{\pi(2p+1)^2}\cos \left[(2p+1)x\right].\tag{1}$$
Note that $(1)$ does not hold for $x = 0$, only for $0 < x \leqslant \pi$ (by peridicity, for $2k \pi < x \leqslant (2k+1)\pi$).