By considering the geometric series $1+z+z^{2}+...+z^{n-1}$ where $z=\cos(\theta)+i\sin(\theta)$, show that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+...+\cos(n-1)\theta$ = ${1-\cos(\theta)+\cos(n-1)\theta-\cos(n\theta)}\over {2-2\cos(\theta)}$  I've tried expressing $\cos(n\theta)$ as ${e^{in\theta}+e^{-in\theta}} \over {2}$ but I don't think that will lead anywhere. Does it help that $1+z+z^{2}+z^{3}+...+z^{n-1}$=$e^{0i\theta}+e^{i\theta}+e^{2i\theta}+e^{3i\theta}+...+e^{(n-1)i\theta}$?  So the sum $\sum_{r=0}^{n-1} e^{ir\theta}$=${e^{ni\theta}-1} \over {e^{i\theta}-1}$  Thank you in advance :)
Possible Duplicate:          Prove $$\cos(\alpha) + \cos(\alpha + \beta) + \cos(\alpha + 2\beta) + \dots + \cos[\alpha + (n-1)\beta] = \frac{\cos(\alpha + \frac{n-1}{2}\beta) \cdot \sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}} $$
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