$\sum \cos$ when angles are in arithmetic progression 
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How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? 

Prove $$\cos(\alpha) + \cos(\alpha + \beta) + \cos(\alpha + 2\beta) + \dots + \cos[\alpha + (n-1)\beta] = \frac{\cos(\alpha + \frac{n-1}{2}\beta) \cdot \sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}} $$
 A: There is a solution. But I assume and hope that you have already tried to solve the problem by yourself. (I am french, so it could explain my strange written english. If you have any gramatical advice don't hesitate. :))
$\cos(\alpha)+\cos(\alpha+\beta)+...+\cos(\alpha+(n-1)\beta)={\displaystyle \dfrac{1}{2}\sum_{k=0}^{n-1}e^{i(\alpha+k\beta)}+e^{-i(\alpha+k\beta)}=\dfrac{1}{2}}\left(e^{i\alpha}\dfrac{e^{in\beta}-1}{e^{i\beta}-1}+e^{-i\alpha}\dfrac{e^{-in\beta}-1}{e^{-i\beta}-1}\right)=\dfrac{1}{2}\left(e^{i(\alpha+\dfrac{n-1}{2}\beta)}\dfrac{e^{i\dfrac{n}{2}\beta}-e^{-i\dfrac{n}{2}\beta}}{e^{i\dfrac{1}{2}\beta}-e^{-i\dfrac{1}{2}\beta}}+e^{-i(\alpha+\dfrac{n-1}{2}\beta)}\dfrac{e^{-i\dfrac{n}{2}\beta}-e^{i\dfrac{n}{2}\beta}}{e^{-i\dfrac{1}{2}\beta}-e^{i\dfrac{1}{2}\beta}}\right)=$
$\dfrac{e^{i(\alpha+\dfrac{n-1}{2}\beta)}+e^{-i(\alpha+\dfrac{n-1}{2}\beta)}}{2}\left(\dfrac{e^{i\dfrac{n}{2}\beta}-e^{-i\dfrac{n}{2}\beta}}{e^{i\dfrac{1}{2}\beta}-e^{-i\dfrac{1}{2}\beta}}\right)=\dfrac{\cos\left(\alpha+\dfrac{n-1}{2}\beta\right)\sin\left(\dfrac{n\beta}{2}\right)}{\sin\left(\dfrac{\beta}{2}\right)}$
