Sum of Cosines is $-1$

Question

Using this formula, $$z^k-1=(z-1)(1+z+z^2...+z^{k-1})$$ where $z \in \mathbb{C}$. How can we prove the following, $$ cos(\frac{2\pi}{k}) + cos(\frac{4\pi}{k}) + ... + cos(\frac{2(k-1)\pi}{k}) = -1 $$

Answer 1

Note $$ I= 1+e^{i \frac{2\pi}{k}} + e^{i \frac{4\pi}{k}}+ ... + e^{i \frac{2(k-1)\pi}{k}}= \frac {(e^{i \frac{2\pi}{k}})^k -1}{e^{i \frac{2\pi}{k}} -1} =0 $$

Thus, with $\cos x=Re(e^{ix})$,

$$ \cos\frac{2\pi}{k}+ \cos\frac{4\pi}{k}+ ... + \cos\frac{2(k-1)\pi}{k}= Re(I-1)=-1 $$