Question

Prove that ${\sum }_{k=1}^{n-1}(n-k)cos\frac{2k\pi }{n}=-\frac{n}{2}$
Where $n\ge 3$ is an integer.

Answer 1

Expanding the sigma on putting k = 1,2,3,....,n
$S=(n-1)cos\frac{2\pi }{n}+(n-2)cos\frac{4\pi }{n}+.........+1cos(n-1)\frac{2\pi }{n}...\left(1\right)$
Replacing each angle by $\theta by2\pi -\theta$ in (1), we get
$S=(n-1)cos(n-1)\frac{2\pi }{n}+(n-2)cos(n-2)\frac{2\pi }{n}+.........+1cos(n-1)\frac{2\pi }{n}...by\left(1\right)$
Add terms having the same angle and take n common.
$\therefore 2S=n[cos\frac{2\pi }{n}+cos\frac{4\pi }{n}+cos\frac{6\pi }{n}+......+cos(n-1\left)\frac{2\pi }{n}\right]$
Angles are in A.P. of $d=\frac{2\pi }{n}$
$=\left[\frac{sin(n-1)\frac{\pi }{n}}{sin\frac{\pi }{n}}cos\frac{\frac{2\pi }{n}+(n-1)\frac{2\pi }{n}}{2}\right]$
$=n\cdot 1cos\pi =-n\because sin(\pi -\theta )=sin\theta$
$\therefore S=-n/2$