To take that a bit further, the expression in (2) can be simplified as $$\tan^{-1} \Bigl(\frac{\sin x}{1-\cos x}\Bigr) = \tan^{-1} \biggl(\frac{2\sin\frac x2\cos\frac x2}{2\sin^2\frac x2}\biggr) = \tan^{-1}\bigl(\cot\tfrac x2\bigr) = \tfrac\pi2 - \tfrac x2.$$ Hence $$\sum _{n=1}^{\infty} \frac{\sin n x}{n} = \tfrac12(\pi-x).$$ But that only works provided that $0<x< 2\pi$. At the endpoints of the interval, when $x=0$ or $2\pi$, the sum $\sum _{n=1}^{\infty} \frac{\sin n x}{n}$ is obviously $0$ (since each term vanishes).

As ZaidAlyafey points out, this sum is a Fourier series, namely for the function $\tfrac12(\pi-x)$ over the interval $[0,2\pi].$