Summation of sine by considering the imaginary part of exp(ik*seta) 
prove that
  $$
\sum\limits_{k=o}^{n} {\sin k\theta} = \frac{\cos{\frac {1} {2}}\theta - \cos(n + {\frac 1 2}\theta)} {2\sin \frac{1} {2}\theta} 
$$

I would like to solve this by considering the imaginary part of $\sum {\exp(ik\theta)}$.
This is an alternative method suggested by the paper.
$$
S_n = \sum\limits_{k=0}^n {e^{ik\theta}} = 1+e^{i\theta} + e^i\theta + \cdots +e^{in\theta}  
$$
$$
= \frac{1 - e^{in\theta}} {1-e^{i\theta}}    
$$
$$
=\frac{1-(\cos n\theta + i\sin n\theta)}{1-(\cos\theta + i\sin \theta)}
$$
Considering only the imaginary parts:
$$
\sum\limits_{k=o}^{n} {\sin k\theta} = \frac {\sin n\theta}{\sin \theta} = \frac {\sin ((n+{\frac 1 2})-\frac {1} {2})\theta}{{2\sin {\frac{1} {2}}\theta cos {\frac{1} {2}}\theta}} 
$$
$$
= \frac {\sin (n + {\frac {1} {2}}) \theta \cos {\frac{1} {2}}\theta - \sin {\frac {1} {2}}\theta \cos (n+{\frac{1} {2}})\theta} {2\sin {\frac{1} {2}}\theta cos {\frac{1} {2}}\theta}
$$
Now I have all the parts required + some extra terms that I need to get rid of.
How do i do that?
Many thanks in advance.
 A: Since
$$\sum_{k = 0}^n e^{ik\theta} = \frac{1 - e^{i(n+1)\theta}}{1 - e^{i\theta}} = \frac{e^{i(n+1)\theta/2}}{e^{i\theta/2}}\frac{e^{-i(n+1)\theta/2} - e^{i(n+1)\theta/2}}{e^{-i\theta/2} - e^{i\theta/2}} = e^{in\theta/2} \frac{\sin [(n+1)\theta/2]}{\sin \theta/2},$$
Taking imaginary parts yields
\begin{align}\sum_{k = 0}^n \sin k\theta &= \frac{\sin(n\theta/2)\sin [(n+1)\theta/2]}{\sin(\theta/2)}\\
&= \frac{\cos\left(\frac{n\theta}{2} - \frac{(n+1)\theta}{2}\right) - \cos\left(\frac{n\theta}{2} + \frac{(n+1)\theta}{2}\right)}{2\sin(\theta/2)}\\
&= \frac{\cos(\theta/2)- \cos[(n + 1/2)\theta]}{2\sin(\theta/2)}.
\end{align}
A: HINT: 
Try using 
$$\begin{align}
S_n&=\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\\\\
&=\frac{(1-e^{i(n+1)\theta})(1-e^{-i\theta})}{2(1-\cos(\theta))}
\end{align}$$
The imaginary part is
$$\frac{(\sin(n\theta)+\sin(\theta)-\sin((n+1)\theta)}{2(1-\cos(\theta))}$$
Now, using $1-\cos(\theta)=2\sin^2(\frac{\theta}{2})$, $\sin(\theta)=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$, and the addition angle formulae for the sine and cosine functions show
$$\begin{align}
\text{Im}(S_n)&=\frac{\sin(n\theta)(1-\cos(\theta))+\sin(\theta)(1-\cos(n\theta))}{4\sin^2(\theta/2)}\\\\
&=\frac{\sin(n\theta)\sin(\theta/2)+\cos(\theta/2)(1-\cos(n\theta))}{2\sin(\theta/2)}\\\\
&=\frac{\cos(\theta/2)-\cos((n+1/2)\theta)}{2\sin(\theta/2)}
\end{align}$$
