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Find expression of: $$ \sum_{r=1}^n \dbinom{n}{r} \sin {r\theta} = ??? $$

Namas Thapa
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    Similar: https://math.stackexchange.com/questions/2371738/tricky-sum-involving-binomial-coefficients-and-sine – Matti P. Feb 20 '20 at 11:45
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    Write as two sums involving complex exponentials and consider the Binomial expansion – Paul Feb 20 '20 at 11:58
  • Perhaps the recursion $$ \sin r\theta = 2\cos\theta\sin((r-1)\theta)-\sin((r-2)\theta) $$ may help. – Math1000 Feb 20 '20 at 12:17

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$$X=\sum_{r=1}^n \dbinom{n}{r} \sin {r\theta} = ?$$

Consider $$\begin{align} S:=\sum_{r=1}^n \dbinom{n}{r} e^{ir\theta} &= \sum_{r=1}^n \dbinom{n}{r} (\underbrace{e^{i\theta}}_{:=a})^r \\ &= \sum_{r=1}^n \dbinom{n}{r} a^r \\ &= -1+\sum_{r=0}^n \dbinom{n}{r} a^r1^{n-r} \\ &= -1+(1+a)^n \\ &= (1+e^{i\theta})^n-1\\ \end{align}$$ Because $e^{ix}=\cos x+i\sin x$, we also have: $$\begin{align} S&= \sum_{r=1}^n \dbinom{n}{r} \big(\cos r\theta+i\sin r\theta\big) \\ &= \sum_{r=1}^n \dbinom{n}{r} \cos r\theta + i\underbrace{\sum_{r=1}^n \dbinom{n}{r} \sin r\theta}_{=X} \\ \end{align}$$

Hence: $\def\Im{\operatorname{Imag}}$ $$\begin{align} X&= \Im(S) \\ &= \Im\big((1+e^{i\theta})^n-1\big) \\ &= \Im\big((1+e^{i\theta})^n\big) \end{align}$$ where $\Im$ denotes the imaginary part of a complex number.

emacs drives me nuts
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