Sum over k from 1 to n of n Choose k by Sine of n Theta
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Theorem
- $\ds \sum_{k \mathop = 1}^n \dbinom n k \sin k \theta = \paren {2 \cos \dfrac \theta 2}^n \sin \dfrac {n \theta} 2$
Proof
\(\ds \paren {1 + e^{i \theta} }^n\) | \(=\) | \(\ds \sum_{k \mathop = 0}^n \dbinom n k e^{i k \theta}\) | Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^n \dbinom n k \paren {\cos k \theta + i \sin k \theta}\) | Euler's Formula | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \Im {\paren {1 + e^{i \theta} }^n}\) | \(=\) | \(\ds \map \Im {\sum_{k \mathop = 0}^n \dbinom n k \paren {\cos k \theta + i \sin k \theta} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^n \dbinom n k \sin k \theta\) | taking imaginary parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \dbinom n k \sin k \theta\) | as the zeroth term vanishes: $\sin 0 = 0$ |
At the same time:
\(\ds \map \Im {\paren {1 + e^{i \theta} }^n}\) | \(=\) | \(\ds \map \Im {e^{i n \theta / 2} \paren {e^{i \theta / 2} + e^{-i \theta / 2} }^n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \Im {e^{i n \theta / 2} \paren {2 \cos \dfrac \theta 2}^n}\) | Euler's Cosine Identity | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \Im {\paren {\cos \dfrac {n \theta} 2 + i \sin \dfrac {n \theta} 2} \paren {2 \cos \dfrac \theta 2}^n}\) | Euler's Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin \dfrac {n \theta} 2 \paren {2 \cos \dfrac \theta 2}^n\) |
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: Example $2$.