Find the sum of $\sum\limits_{k=1}^{\infty}q^k\sin k\alpha$

Question

Find the following sum $$\sum\limits_{k=1}^{\infty}q^k\sin k\alpha = q\sin\alpha + q^2\sin 2\alpha+\ldots+q^n\sin n\alpha+\ldots$$ I just really don't know any way to simplify that sum because of $q^k$ at $k$ term.
If it was just $$\sum\limits_{k=1}^{\infty}\sin k\alpha$$ there is a hint to time that sum by $2\sin\frac{\alpha}{2}$ and use the formula $2\sin\alpha \sin\beta = \cos(\alpha-\beta)-\cos(\alpha+\beta)$, hence $$\sum\limits_{k=1}^{\infty}2\sin k\alpha\sin\frac{\alpha}{2} = (\cos\frac{\alpha}{2}-\cos\frac{3\alpha}{2})+(\cos\frac{3\alpha}{2}-\cos\frac{5\alpha}{2})+\ldots+$$ and so on. But it doesn't work here and I don't know what to do with it

Answer 1

As $$\sin k\alpha=\frac{e^{ik\alpha}-e^{-ik\alpha}}{2i}$$ your sum is $$\frac1{2i}\sum_{k=1}^\infty q^ke^{ik\alpha} -\frac1{2i}\sum_{k=1}^\infty q^ke^{-ik\alpha}.$$ Each of these sums is a geometric series.