Sum of Sines of Fractions of Pi

Theorem
Let $n \in \Z$ such that $n > 1$.

Then:
 * $\displaystyle \sum_{k \mathop = 1}^{n - 1} \sin \frac {2 \pi} n = 0$

Proof
Consider the equation:
 * $z^n - 1 = 0$

whose solutions are the complex roots of unity:
 * $1, e^{2 \pi i / n}, e^{4 \pi i / n}, e^{6 \pi i / n}, \ldots, e^{2 \left({n - 1}\right) \pi i / n}$

By Sum of Roots of Polynomial:
 * $1 + e^{2 \pi i / n} + e^{4 \pi i / n} + e^{6 \pi i / n} + \cdots + e^{2 \left({n - 1}\right) \pi i / n} = 0$

From Euler's Formula:
 * $e^{i \theta} = \cos \theta + i \sin \theta$

from which comes:
 * $\left({1 + \cos \dfrac {2 \pi} n + \cos \dfrac {4 \pi} n + \cdots + \cos \dfrac {2 \left({n - 1}\right) \pi} n}\right) + i \left({\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \left({n - 1}\right) \pi} n}\right) = 0$

Equating imaginary parts:
 * $\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \left({n - 1}\right) \pi} n = 0$