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

Prove $$\cos(\alpha) + \cos(\alpha + \beta) + \cos(\alpha + 2\beta) + \dots + \cos[\alpha + (n-1)\beta] = \frac{\cos(\alpha + \frac{n-1}{2}\beta) \cdot \sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}} $$

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marked as duplicate by Aryabhata, Did, t.b., Américo Tavares, Quixotic Mar 6 '12 at 20:38

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You first! (Please don't write your question in the imperative. If it's your assignment to prove the identity, please let us know what you've already tried.) –  Adam Saltz Mar 6 '12 at 15:48

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There is a solution. But I assume and hope that you have already try 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. :))

$\cos(\alpha)+\cos(\alpha+\beta)+...+\cos(\alpha+(n-1)\beta)={\displaystyle \dfrac{1}{2}\sum_{k=0}^{n-1}e^{i(\alpha+k\beta)}+e^{-i(\alpha+k\beta)}=\dfrac{1}{2}}\left(e^{i\alpha}\dfrac{e^{in\beta}-1}{e^{i\beta}-1}+e^{-i\alpha}\dfrac{e^{-in\beta}-1}{e^{-i\beta}-1}\right)=\dfrac{1}{2}\left(e^{i(\alpha+\dfrac{n-1}{2}\beta)}\dfrac{e^{i\dfrac{n}{2}\beta}-e^{-i\dfrac{n}{2}\beta}}{e^{i\dfrac{1}{2}\beta}-e^{-i\dfrac{1}{2}\beta}}+e^{-i(\alpha+\dfrac{n-1}{2}\beta)}\dfrac{e^{-i\dfrac{n}{2}\beta}-e^{i\dfrac{n}{2}\beta}}{e^{-i\dfrac{1}{2}\beta}-e^{i\dfrac{1}{2}\beta}}\right)=$

$\dfrac{e^{i(\alpha+\dfrac{n-1}{2}\beta)}+e^{-i(\alpha+\dfrac{n-1}{2}\beta)}}{2}\left(\dfrac{e^{i\dfrac{n}{2}\beta}-e^{-i\dfrac{n}{2}\beta}}{e^{i\dfrac{1}{2}\beta}-e^{-i\dfrac{1}{2}\beta}}\right)=\dfrac{\cos\left(\alpha+\dfrac{n-1}{2}\beta\right)\sin\left(\dfrac{n\beta}{2}\right)}{\sin\left(\dfrac{\beta}{2}\right)}$

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