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



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

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.

