Question

Prove that $cos\alpha +cos(\alpha +\beta )+cos(\alpha +2\beta )...cos(\alpha +(n-1\left)\beta \right)=\frac{cos\frac{n\beta }{2}}{cos\frac{\beta }{2}}\{\alpha +(n-1\left)\frac{\beta }{2}\right\}$

Answer 1

$\cos \alpha +\cos (\alpha +\beta )+\cos (\alpha +2\beta )..........\cos (\alpha +(n-1\left)\beta \right)$

$=\frac{1}{2}{\sum }_{k=0}^{n=1}{e}^{i(\alpha +k\beta )}+e^{-i(\alpha +k\beta )}$

$=\frac{1}{2}(e^{i\alpha }\cdot \frac{e^{in\beta -1}}{e^{i\beta }-1}+e^{-i\alpha }\frac{e^{-in\beta -1}}{e^{-i\beta }-1})$

$=\frac{1}{2}(e^{i(\alpha +\frac{n-1}{2}\beta )}\cdot \frac{e^{i\frac{n}{2}\cdot \beta }-e^{-i\frac{n}{2}\beta }}{e^{i\frac{1}{2}\cdot \beta }-e^{-i\frac{1}{2}\beta }}+e^{-i(\alpha +\frac{n-1}{2}\beta )}\cdot \frac{e^{-i\frac{n}{2}\cdot \beta }-e^{i\frac{n}{2}\beta }}{e^{-i\frac{1}{2}\cdot \beta }-e^{i\frac{1}{2}\beta }})$

$=\frac{e^{i(\alpha +\frac{n-1}{2}\beta )}+e^{-i(\alpha +\frac{n-1}{2}\beta )}}{2}\left(\frac{e^{i\frac{n}{2}\cdot \beta }-e^{-i\frac{n}{2}\beta }}{e^{i\frac{1}{2}\cdot \beta }-e^{-i\frac{1}{2}\beta }}\right)$

$=\frac{\cos (\alpha +\left(\frac{n-1}{2}\right)\beta )\sin \left(\frac{n\beta }{2}\right)}{\sin \left(\frac{\beta }{2}\right)}$

LHS=RHS

Hence proved.