Use \(n\)-th roots of unity (i.e. solutions of \(z^{n}-1=0\) ) to show that
\[ \cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\cos \frac{6 \pi}{n}+\cdots+\cos \frac{2(n-1) \pi}{n}+=-1 \text { and } \\ \sin \frac{2 \pi}{n}+\sin \frac{4 \pi}{n}+\sin \frac{6 \pi}{n}+\cdots \frac{2(n-1) \pi}{n}=0 \]
Hint: If \(z^{n}+c_{1} z^{n-1}+\cdots+c_{n-1} z+c_{n}=0\) has roots \(z_{1}, z_{2}, \ldots, z_{n}\), then \[ z_{1}+z_{2}+\cdots+z_{n}=-c_{1} \\ z_{1} z_{2} \cdots z_{n}=(-1)^{n} c_{n} \text { (not used) } \]
Solution
The final summand in the displayed sine identity is missing its \(\sin\) in the source. With that typographical correction, both identities follow at once from the \(n\)-th roots of unity.
Let \(\omega=e^{2\pi i/n}\). Since \[ 1+\omega+\omega^2+\cdots+\omega^{n-1}=0, \] we have \[ \sum_{k=1}^{n-1}\omega^k=-1. \] Taking real and imaginary parts gives \[ \boxed{\sum_{k=1}^{n-1}\cos\frac{2\pi k}{n}=-1}, \qquad \boxed{\sum_{k=1}^{n-1}\sin\frac{2\pi k}{n}=0}. \]