A closed form of the series $ \sum_{n=1}^{\infty} q^n \sin(n\alpha) $ I am having problems with the following series:
$$
\sum_{n=1}^{\infty} q^n \sin(n\alpha), \quad|q| < 1.
$$
No restrictions on $\alpha$. I need to find out whether it converges and if yes, evaluate its sum.
I can see that it's convergent using the comparison test. But I fail to find its sum. So far I tried grouping subsequent terms and using trigonometric formulas, but it didn't help me much. 
Where should I start when I see trigonometric functions in a series? In general, I have no idea where to take off in such situations.
Thanks in advance.
 A: Hint. We assume $\alpha\in \mathbb{R}$ and $-1<q<1$. Then one may write
$$
\sum_{n=1}^{\infty} q^n \sin(n\alpha)=\Im \sum_{n=1}^{\infty} (qe^{i\alpha})^n =\Im\: \frac{qe^{i\alpha}}{1-qe^{i\alpha}}
$$ where we have used the standard evaluation of a geometric series.
A: I'd like to expand Olivier Oloa's hint:

$$ \sum _{ n=1 }^{ \infty  } q^{ n }\sin  (n\alpha )=q\sin { \alpha +{ q }^{ 2 }\sin { 2\alpha +...+{ q }^{ n }\sin { n\alpha +...\quad \quad \quad \quad \quad \quad \quad \quad \left( 1 \right)  }  }  } \quad \\ \sum _{ n=1 }^{ \infty  } q^{ n }\cos { \left( n\alpha  \right)  } =q\cos { \alpha +{ q }^{ 2 }\cos { 2\alpha  } +...+{ q }^{ n }\cos { n\alpha +... }  } ,\left| q \right| <1\quad \quad \left( 2 \right) $$

let denote partial sums of $(1)$ and $(2)$as follows:

$${ u }_{ n }=\sum _{ n=1 }^{ \infty  } q^{ n }\sin  (n\alpha )=q\sin { \alpha +{ q }^{ 2 }\sin { 2\alpha +...+{ q }^{ n }\sin { n\alpha  }  }  } \quad \quad \\ { v }_{ n }=\sum _{ n=1 }^{ \infty  } q^{ n }\cos { \left( n\alpha  \right)  } =q\cos { \alpha +{ q }^{ 2 }\cos { 2\alpha  } +...+{ q }^{ n }\cos { n\alpha  }  
} $$

by using Euler's formula ${ e }^{ i\varphi  }=\cos { \varphi +i\sin { \varphi  }  } $ we get

$${ u }_{ n }+i{ v }_{ n }=q\left( \sin { \alpha  } +i\cos { \alpha  }  \right) +{ q }^{ 2 }\left( \sin { 2\alpha  } +\cos { 2\alpha  }  \right) +...+{ q }^{ n }\left( \sin { n\alpha +i\cos { n\alpha  }  }  \right) =\\ =\frac { q{ e }^{ i\alpha  }-{ q }^{ n+1 }{ e }^{ i\left( n+1 \right) \alpha  } }{ 1-q{ e }^{ i\alpha  } } $$

since $\left| q \right| <1\Rightarrow \left| q{ e }^{ i\alpha  } \right| <1$
we have 

$$\\ \lim _{ n\rightarrow \infty  }{ \left( { q }^{ n+1 }{ e }^{ i\left( n+1 \right) \alpha  } \right) =0 } $$

finally we get 

$$u+iv=\lim _{ n\rightarrow \infty  }{ \left( { u }_{ n }+i{ v }_{ n } \right) =\frac { q{ e }^{ i\alpha  } }{ 1-q{ e }^{ i\alpha  } } =q\left( \frac { \cos { \alpha -q }  }{ 1-2q\cos { \alpha +{ q }^{ 2 } }  } +i\frac { \sin { \alpha  }  }{ 1-2q\cos { \alpha +{ q }^{ 2 } }  }  \right)  } $$
  where  $$u=q\frac { \cos { \alpha -q }  }{ 1-2q\cos { \alpha +{ q }^{ 2 } }  } ,v=\frac { q\sin { \alpha  }  }{ 1-2q\cos { \alpha +{ q }^{ 2 } }  } $$


To be more clearly understant the last part 
$$\frac { q{ e }^{ i\alpha  } }{ 1-q{ e }^{ i\alpha  } } =q\frac { \cos { \alpha +i\sin { \alpha  }  }  }{ 1-q\cos { \alpha -iq\sin { \alpha  }  }  } =q\frac { \cos { \alpha +i\sin { \alpha  }  }  }{ \left( 1-q\cos { \alpha  }  \right) -iq\sin { \alpha  }  } =q\frac { \left( \cos { \alpha +i\sin { \alpha  }  }  \right) \left( \left( 1-q\cos { \alpha  }  \right) +iq\sin { \alpha  }  \right)  }{ \left( \left( 1-q\cos { \alpha  }  \right) -iq\sin { \alpha  }  \right) \left( \left( 1-q\cos { \alpha  }  \right) +iq\sin { \alpha  }  \right)  } =\\ =q\frac { \cos { \alpha -q\cos ^{ 2 }{ \alpha +iq\cos { \alpha  } \sin { \alpha +i\sin { \alpha  } -iq\sin { \alpha  } \cos { \alpha -q\sin ^{ 2 }{ \alpha  }  }  }  }  }  }{ 1-2q\cos { \alpha +{ q }^{ 2 }\cos ^{ 2 }{ \alpha +{ q }^{ 2 }\sin ^{ 2 }{ \alpha  }  }  }  } =q\frac { \cos { \alpha -q+i\sin { \alpha  }  }  }{ 1-2q\cos { \alpha +{ q }^{ 2 } }  } =\\ =q\left( \frac { \cos { \alpha -q }  }{ 1-2q\cos { \alpha +{ q }^{ 2 } }  } +i\frac { \sin { \alpha  }  }{ 1-2q\cos { \alpha +{ q }^{ 2 } }  }  \right) $$
