Find expression for following summation

Question

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

Answer 1

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

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.