Calculating $\sum_{k=0}^{n}\sin(k\theta)$ 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.

$$
2\sin \frac{\theta}2 \sin k\theta = \cos \frac{(2k-1)\theta}2 -\cos \frac{(2k+1)\theta}2
$$
so (telescoping sum)
$$
2 \sin \frac{\theta}2 \sum_{k=1}^n\sin k\theta = \cos (\frac{\theta}2) -\cos \frac{(2n+1)\theta}2 \\
= 2 \sin \frac{n\theta}2 \sin \frac{(n+1)\theta}2
$$
giving
$$
\sum_{k=1}^n\sin k\theta = \frac{\sin \frac{n\theta}2 \sin \frac{(n+1)\theta}2}{\sin \frac{\theta}2}
$$
