Finding an infinite trigonometric sum Find the following infinite sum : $$q\sin a+q^2\sin 2a+\ldots+q^n\sin na+\ldots$$  where $|q|<1$ .It would be good if you could find it without the help of any auxiliary sequences using only trigonometric formulas.
 A: Note that for $\vert q \vert < 1$
$$\sum_{k=1}^{\infty} (qe^{ia})^k = \dfrac{qe^{ia}}{1-qe^{ia}}$$
Now look at the imaginary part of the above expression.

EDIT
If complex numbers are not allowed, then proceed as follows. We have
$$S_n(q,a) = \sum_{k=1}^n q^k \sin(ka)$$
\begin{align}
\cos(a)S_n(q,a) & = \sum_{k=1}^n q^k \sin(ka) \cos(a)\\
& = \sum_{k=1}^n q^k \dfrac{\sin((k+1)a) + \sin((k-1)a)}2\\
& = \dfrac12 \left(\sum_{k=1}^n q^k \sin((k+1)a) + \sum_{k=1}^n q^k \sin((k-1)a)\right)\\
& = \dfrac12 \left(\dfrac{S_{n+1}(q,a) - q \sin(a)}{q} + q S_{n-1}(q,a)\right)
\end{align}
If $\lim_{n \to \infty} S_n(q,a) = S(q,a)$ exists, we then have
$$\cos(a) S(q,a) = \dfrac12 \left(\dfrac{S(q,a) - q \sin(a)}q + q S(q,a)\right)$$
This gives us
$$\left(\cos(a) - \dfrac12 \left(q+\dfrac1q\right)\right) S(q,a) = - \dfrac{\sin(a)}2$$
This gives us
$$S(q,a) = \dfrac{q \sin(a)}{q^2 + 1 - 2q \cos(a)}$$
A: Hint
We calculate the sum of geometric series
$$\sum_{n=1}^\infty q^ne^{ina}$$
then we take the imaginary part.
