This problem seems to send me down an endless path of failure. I could do with some hints.
Many thanks, Bobak
We will show that,
$\displaystyle \sum_{k=0}^{16}\cos (k\theta) = 0$ for $\displaystyle \theta = \frac{2\pi}{17}$Note that $\displaystyle \zeta = \cos \theta + i \sin \theta$ is a primitive root of unity of $\displaystyle 17$ which means all the roots are given by $\displaystyle 1,\zeta, ... ,\zeta^{16}$.
But the sum of the roots of unity is zero, thus (by de Moivre),
$\displaystyle \sum_{k=0}^{16}\zeta^k = 0\implies \sum_{k=0}^{16}\cos (k\theta) = \sum_{k=0}^{16}\sin (k\theta) = 0$.