# Math Help - Trig question

1. ## Trig question

This problem seems to send me down an endless path of failure. I could do with some hints.

Many thanks, Bobak

2. We will show that,
$\sum_{k=0}^{16}\cos (k\theta) = 0$ for $\theta = \frac{2\pi}{17}$
Note that $\zeta = \cos \theta + i \sin \theta$ is a primitive root of unity of $17$ which means all the roots are given by $1,\zeta, ... ,\zeta^{16}$.

But the sum of the roots of unity is zero, thus (by de Moivre),
$\sum_{k=0}^{16}\zeta^k = 0\implies \sum_{k=0}^{16}\cos (k\theta) = \sum_{k=0}^{16}\sin (k\theta) = 0$
.

3. Yeah i actually got a bit lost on the first "show by direct substitution part" i kept going around in circles.