Question

Prove by PMI:
$cos\theta +cos2\theta +.....+cosn\theta =sin\frac{n\theta }{2}.cos\theta c\frac{\theta }{2}.cos\frac{(n+1)\theta }{2}$

Answer 1

$\begin{array}{c}\cos \theta +\cos 2\theta +\cos 3\theta +...........\cos n\theta \\ e^{i\theta }=\cos \theta +i\sin \theta \\ {\sum }_{k=1}^n\cos k\theta =R{\sum }_{k=1}^n{e}^{ki\theta }\\ =R.e^{i\theta }\left(1+e^{i\theta }+e^{2i\theta }+.....e^{\left(n-1\right)i\theta }\right)\\ =R\left[e^{i\theta }.\left(\frac{e^{i\theta }-1}{e^{i\theta }-1}\right)\right]\\ =R\left[e^{i\theta }.\frac{e^{\frac{\theta }{2}}\left(2i\sin \frac{n\theta }{2}\right)}{e^{\frac{i\theta }{2}}.2i\sin \frac{\theta }{2}}\right]\\ =R\left[e^{\frac{i\theta \left(n+1\right)}{2}}.\frac{\sin n\frac{\theta }{2}}{\sin \frac{\theta }{2}}\right]\\ =R\left[\cos \left(n+1\right)\frac{\theta }{2}+i\sin \left(n+1\right)\frac{\theta }{2}.\frac{\sin n\frac{\theta }{2}}{\sin \frac{\theta }{2}}\right]\\ =\frac{\sin n\frac{\theta }{2}}{\sin \frac{\theta }{2}}.\cos \left(n+1\right)\frac{\theta }{2}\\ =\sin \frac{n\theta }{2}.\cos ec\frac{\theta }{2}.\cos \left(n+1\right)\frac{\theta }{2}\\ HenceProved.\end{array}$