Sum of Sines of Fractions of Pi

Theorem

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

Then:

$\ds \sum_{k \mathop = 1}^{n - 1} \sin \frac {2 k \pi} n = 0$

Proof

Consider the equation:

$z^n - 1 = 0$

whose solutions are the complex roots of unity:

$1, e^{2 \pi i / n}, e^{4 \pi i / n}, e^{6 \pi i / n}, \ldots, e^{2 \paren {n - 1} \pi i / n}$

By Sum of Roots of Polynomial:

$1 + e^{2 \pi i / n} + e^{4 \pi i / n} + e^{6 \pi i / n} + \cdots + e^{2 \paren {n - 1} \pi i / n} = 0$

From Euler's Formula:

$e^{i \theta} = \cos \theta + i \sin \theta$

from which comes:

$\paren {1 + \cos \dfrac {2 \pi} n + \cos \dfrac {4 \pi} n + \cdots + \cos \dfrac {2 \paren {n - 1} \pi} n} + i \paren {\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \paren {n - 1} \pi} n} = 0$

Equating imaginary parts:

$\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \paren {n - 1} \pi} n = 0$

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: The $n$th Roots of Unity: $38 \ \text{(b)}$