# $\sum \cos$ when angles are in arithmetic progression [duplicate]

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

• 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.) Mar 6, 2012 at 15:48

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