Complex sum of sine and cosine functions 
By using the complex representations of sine and cosine, show that $$\sum_{m=0}^n\sin m\theta =\frac{\sin\frac{n}{2}\theta\sin\frac{n+1}{2}\theta}{\sin\frac{1}{2}\theta}$$

So I am not too sure how to go about this proof. I tried to substitute sine for its complex representation but I can't see how to evaluate the sum?
 A: The sum is equal to the imaginary part of $\displaystyle \sum_{m=0}^n e^{im\theta}$. Can you sum this series up instead?
A: When you multiply the sum by the complex form of $sin \frac{\theta}{2}$ every term will cancel except four particular ones. Find those 4.
A: 
Based upon the finite geometric series formula and 
  $\sin \theta=\frac{1}{2i}\left(\exp(i\theta)-\exp(-i\theta)\right)$
we obtain:
\begin{align*}
&\color{blue}{\frac{\sin\frac{n}{2}\theta\sin\frac{n+1}{2}\theta}{\sin\frac{1}{2}\theta}}\\
&\qquad=\frac{\frac{1}{2i}\left(\exp\left(i\frac{n}{2}\theta\right)-\exp\left(-i\frac{n}{2}\theta\right)\right)
\frac{1}{2i}\left(\exp\left(i\frac{(n+1)}{2}\theta\right)-\exp\left(-i\frac{(n+1)}{2}\theta\right)\right)}
{\frac{1}{2i}\left(\exp\left(i\frac{\theta}{2}\right)-\exp\left(-i\frac{\theta}{2}\right)\right)}\\
&\qquad=\frac{\exp\left(i\left(n+\frac{1}{2}\right)\theta\right)-\exp\left(i\frac{\theta}{2}\right)
-\exp\left(-i\frac{\theta}{2}\right)+\exp\left(-i\left(n+\frac{1}{2}\right)\theta\right)}
{\frac{1}{2i}\left(\exp\left(i\frac{\theta}{2}\right)-\exp\left(-i\frac{\theta}{2}\right)\right)}\\
&\qquad=\frac{\exp\left(-i\frac{\theta}{2}\right)\left(\exp\left(i(n+1)\theta\right)-1\right)}{2i\exp\left(-i\frac{\theta}{2}\right)\left(\exp\left(i\theta\right)-1\right)}
-\frac{\exp\left(i\frac{\theta}{2}\right)\left(1-\exp\left(-i(n+1)\theta\right)\right)}{2i\exp\left(i\frac{\theta}{2}\right)\left(1-\exp\left(-i\theta\right)\right)}\\
&\qquad=\frac{1}{2i}\sum_{m=0}^n\exp\left(im\theta\right)-\frac{1}{2i}\sum_{m=0}^n\exp\left(-im\theta\right)\\
&\qquad=\sum_{m=0}^n\frac{1}{2i}\left(\exp\left(im\theta\right)-\exp\left(-im\theta\right)\right)\\
&\qquad\,\,\color{blue}{=\sum_{m=0}^n\sin m\theta}
\end{align*}
and the claim follows.

