complex numbers - How do I prove the following statement about a summation of a series?

I have not been able to completely solve this problem and it's driving me crazy. Could you please help.
The question is to show that,

$$\sum_{n=1}^N \frac{\sin n\theta}{2^n} =\frac{2^{N+1}\sin\theta+\sin N\theta-2\sin(N+1)\theta }{2^N(5-4\cos\theta)}$$
Where do I start? I tried solving this using de Moivre's Theorem but I don't know where I am going wrong. Could you please help me or if possible show other ways to tackle this particular problem.

Any Help is much appreciated!

Thanks in Advance!

Answer

If you follow one of the suggestions the summation is the imaginary part of

$$\begin{align*} \sum_{n = 1}^{N}\frac{e^{in\theta}}{2^{n}} &= \sum_{n = 1}^{N}(e^{i\theta}/2)^{n}\\ &= \frac{e^{i\theta}}{2} \frac{\left(1-\frac{e^{Ni\theta}}{2^N}\right)}{\left(1-\frac{e^{i\theta}}{2}\right)}\\ &= \frac{e^{i\theta}(2^N-e^{Ni\theta})}{2^N(2-e^{i\theta})}\\ &= \frac{e^{i\theta}(2^N-e^{Ni\theta})(2-e^{-i\theta})}{2^N(2-e^{i\theta})(2-e^{-i\theta})}\\ &= \frac{(2^Ne^{i\theta}-e^{(N+1)i\theta})(2-e^{-i\theta})}{2^N(4-2(e^{i\theta}+e^{-i\theta})+1)}\\ &= \frac{2^{(N+1)}e^{i\theta}-2e^{(N+1)i\theta}-2^N+e^{Ni\theta}}{2^N(4-2(e^{i\theta}+e^{-i\theta})+1)} \end{align*}$$

The imaginary part of this is

$$\frac{2^{(N+1)}\sin \theta - 2\sin (N+1)\theta + \sin N\theta}{2^N (5-4\cos \theta)}$$