Geometric series and complex numbers

I'm new to this site, english is not my mother tongue, and I'm just learning LaTeX. I'm basically a noob, so please be indulgent if I break any rule or habits.

I'm stuck at proving the following equation. I suppose I should use the formula for geometric series ($\sum\limits_{k=0}^{n}q^k=\frac{1-q^{n+1}}{1-q}$), and also use somewhere that $e^{î\theta}=cos(\theta)+i\sin(\theta)$

So here is the equation I have to prove :
$$\frac{1}{2}+cos(\theta)+cos(2\theta)+...+cos(n\theta)=\frac{sin(n+\frac{1}{2}\theta)}{2sin(\frac{\theta}{2})}$$

Thanks for your help

Answer

$$\begin{aligned} \dfrac{1}{2} + \sum_{r=1}^{n} \cos(r\theta) & = \Re \left\{ \dfrac{1}{2} + \sum_{r=1}^{n} e^{ir\theta} \right\} \\ & = \Re \left\{ -\dfrac{1}{2} + \sum_{r=0}^{n} e^{ir\theta} \right\} \\ & = \Re \left\{ -\dfrac{1}{2} + \dfrac{1-e^{(n+1)\theta}}{1-e^{i\theta}} \right\} \\ & = \Re \left\{ \dfrac{1-2e^{(n+1)\theta}+e^{i\theta}}{2(1-e^{i\theta})} \right\} \\ & = \Re \left\{ \dfrac{e^{-i\theta/2}-2e^{(n+1/2)\theta}+e^{i\theta/2}}{2(e^{-i\theta/2}-e^{i\theta/2})} \right\} \\ & = \Re \left\{ \dfrac{2\cos\left(\frac{\theta}{2}\right)-2\cos(n+1/2)\theta-2i\sin(n+1/2)\theta}{-2i\sin\left(\frac{\theta}{2}\right)} \right\} \end{aligned}$$

Multiply through by $1=\frac{i}{i}$, take the real part and there you have it. ^_^