Show that $\sum_{n=1}^{\infty}(-1)^n\frac{\sin(nx)}{n} = \frac{-x}{2}$ for $-\pi < x < \pi$.

Question

Show that $\sum_{n=1}^{\infty}(-1)^n\frac{\sin(nx)}{n} = \frac{-x}{2}$ for $-\pi < x < \pi$.

I rewrote to $\sum_{n=1}^{\infty}\frac{\sin(n(x+\pi))}{n}$, but then I'm stuck.