Sum of Cosines of Fractions of Pi
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Theorem
Let $n \in \Z$ such that $n > 1$.
Then:
- $\ds \sum_{k \mathop = 1}^{n - 1} \cos \frac {2 k \pi} n = -1$
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 real parts:
- $1 + \cos \dfrac {2 \pi} n + \cos \dfrac {4 \pi} n + \cdots + \cos \dfrac {2 \paren {n - 1} \pi} n = 0$
whence the result.
$\blacksquare$
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: $38 \ \text{(a)}$