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.
 A: You may write, for any $\theta \in \mathbb{R}$ such that $\sin(\theta/2) \neq 0$,
$$
\begin{align}
\sum_{k=0}^{n} \sin (k\theta)&=\Im \sum_{k=0}^{n} e^{ik\theta}\\\\
&=\Im\left(\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right)\\\\
&=\Im\left( \frac{e^{i(n+1)\theta/2}\left(e^{i(n+1)\theta/2}-e^{-i(n+1)\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\
&=\Im\left( \frac{e^{in\theta/2}\left(2i\sin((n+1)\theta/2)\right)}{\left(2i\sin(\theta/2)\right)}\right)\\\\
&=\Im\left( e^{in\theta/2}\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}\right)\\\\
&=\Im\left( \left(\cos (n\theta/2)+i\sin (n\theta/2)\right)\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}\right)\\\\
&=\frac{\sin(n\theta/2)\sin ((n+1)\theta/2)}{\sin(\theta/2)}.
\end{align}
$$
A: If you know complex variables, 
$$
e^{ik\theta}=\cos(k\theta)+i\sin(k\theta).
$$
Then 
$$
\sum_{k=0}^n\sin(k\theta)=\Im\left(\sum_{k=0}^ne^{ik\theta}\right)=\Im\left(\sum_{k=0}^n\left(e^{i\theta}\right)^k\right).
$$
This, however, is a geometric series (provided $\theta$ is not a multiple of $2\pi$), so we know its formula: 
$$
\Im\left(\frac{1-e^{i\theta(n+1)}}{1-e^{i\theta}}\right)=\Im\left(\frac{1-\cos((n+1)\theta)-i\sin((n+1)\theta)}{1-\cos(\theta)-i\sin(\theta)}\right)
$$
Now, rationalize the denominator (and take the imaginary part) to get
$$
\frac{(1-\cos((n+1)\theta))\sin(\theta)-\sin((n+1)\theta)(1-\cos(\theta))}{1-2\cos(\theta)+\cos^2(\theta)+\sin^2(\theta)}
$$
A: $$
2\sin \frac{\theta}2 \sin k\theta = \cos \frac{(2k-1)\theta}2 -\cos \frac{(2k+1)\theta}2
$$
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
$$
\sum_{k=1}^n\sin k\theta = \frac{\sin \frac{n\theta}2 \sin \frac{(n+1)\theta}2}{\sin \frac{\theta}2}
$$
