Sum of Sines of Arithmetic Sequence of Angles/Formulation 2
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Theorem
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
\(\ds \sum_{k \mathop = 1}^n \map \sin {\theta + k \alpha}\) | \(=\) | \(\ds \map \sin {\theta + \alpha} + \map \sin {\theta + 2 \alpha} + \map \sin {\theta + 3 \alpha} + \dotsb\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \sin {\theta + \frac {n + 1} 2 \alpha}\frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }\) |
Proof
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 2:
- $\ds \sum_{k \mathop = 1}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the right hand side is not defined.
From Euler's Formula, this can be expressed as:
- $\ds \sum_{k \mathop = 1}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
Equating imaginary parts:
- $\ds \sum_{k \mathop = 1}^n \map \sin {\theta + k \alpha} = \map \sin {\theta + \frac {n + 1} 2 \alpha} \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: Example $1$