calculus - Calculating $sum_{k=0}^{n}sin(ktheta)$

I'm given the task of calculating the sum $\sum_{i=0}^{n}\sin(i\theta)$.

So far, I've tried converting each $\sin(i\theta)$ in the sum into its taylor series form to get:
$\sin(\theta)=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}...$
$\sin(2\theta)=2\theta-\frac{(2\theta)^3}{3!}+\frac{(2\theta)^5}{5!}-\frac{(2\theta)^7}{7!}...$
$\sin(3\theta)=3\theta-\frac{(3\theta)^3}{3!}+\frac{(3\theta)^5}{5!}-\frac{(3\theta)^7}{7!}...$
...
$\sin(n\theta)=n\theta-\frac{(n\theta)^3}{3!}+\frac{(n\theta)^5}{5!}-\frac{(n\theta)^7}{7!}...$

Therefore the sum becomes,
$\theta(1+...+n)-\frac{\theta^3}{3!}(1^3+...+n^3)+\frac{\theta^5}{5!}(1^5+...+n^5)-\frac{\theta^7}{7!}(1^7+...+n^7)...$

But it's not immediately obvious what the next step should be.

I also considered expanding each $\sin(i\theta)$ using the trigonemetry identity $\sin(A+B)$, however I don't see a general form for $\sin(i\theta)$ to work with.

Answer

You may write, for any $\theta \in \mathbb{R}$ such that $\sin(\theta/2) \neq 0$,
$$\begin{align} \sum_{k=0}^{n} \sin (k\theta)&=\Im \sum_{k=0}^{n} e^{ik\theta}\\\\ &=\Im\left(\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right)\\\\ &=\Im\left( \frac{e^{i(n+1)\theta/2}\left(e^{i(n+1)\theta/2}-e^{-i(n+1)\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\ &=\Im\left( \frac{e^{in\theta/2}\left(2i\sin((n+1)\theta/2)\right)}{\left(2i\sin(\theta/2)\right)}\right)\\\\ &=\Im\left( e^{in\theta/2}\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\Im\left( \left(\cos (n\theta/2)+i\sin (n\theta/2)\right)\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\frac{\sin(n\theta/2)\sin ((n+1)\theta/2)}{\sin(\theta/2)}. \end{align}$$