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