Trigonometry (2)

We can write,

$S= \displaystyle\sum_{p=1}^{n-1} \cos^2 (p\pi/n) =\frac{1}{2}\displaystyle\sum_{p=1}^{n-1} [\cos (2p\pi/n)+1]$

$=\frac{1}{2} \left[\frac{\cos \left(\frac{2\pi}{n} \times \frac{1+n-1}{2} \right). \sin \left(\frac{(n-1)\pi}{n} \right)}{\sin \left(\frac{\pi}{n} \right)} +n-1 \right]$

$=\frac{1}{2} \left[\frac{-sin \left(\pi-\frac{\pi}{n} \right)}{\sin (\pi/n)} +n-1 \right]=\boxed{\frac{n-2}{2}}$

$\begin{aligned} S & = \cos^2 \frac \pi n + \cos^2 \frac {2\pi}n + \cos^2 \frac {3\pi}n + \cdots + \cos^2 \frac {(n-1)\pi}n \\ & = \sum_{k=1}^{n-1} \cos^2 \frac {k \pi}n \\ & = \sum_{k=1}^{n-1} \frac 12 \left(1+\cos \frac {2k \pi}n \right) \\ & = \frac 12 \sum_{k=1}^{n-1} 1 + \frac 12 \sum_{k=1}^{n-1} \cos \frac {2k \pi}n \\ & = \frac {n-1}2 + \frac 12 \sum_{k=1}^{n-1} \Re \left(e^{\frac {2k \pi}ni} \right) & \small \color{#3D99F6} \Re(z) \text{ is the real part of }z. \\ & = \frac {n-1}2 + \frac 12 \cdot \Re \sum_{k=1}^{n-1} e^{\frac {2k \pi}ni} \\ & = \frac {n-1}2 + \frac 12 \cdot \Re \left(e^{\frac {2\pi}ni} \cdot \frac {e^{\frac {2(n-1)\pi}ni}-1}{e^{\frac {2\pi}ni}-1} \right) \\ & = \frac {n-1}2 + \frac 12 \cdot \Re \left(\frac {1-e^{\frac {2\pi}ni}}{e^{\frac {2\pi}ni}-1} \right) \\ & = \frac {n-1}2 + \frac 12 \cdot \Re \left(-1\right) \\ & = \frac {n-1}2 - \frac 12 \\ & = \boxed{\dfrac {n-2}2} \end{aligned}$

Put n=2 ( Classic JEE technique :P)