$\sum_{k=0}^{n-1}\sin(k\frac{\pi}{n}+\theta)$ I'm trying to find the closed form of the above formula. This link shows the solution of tan version. Sum of tangent functions where arguments are in specific arithmetic series
 Though I'm trying to find the way to apply the argument used in the link to this case, it becomes quite complicated. Could you show another method to get the formula or modify my method? Do you think your formula is similar to the closed formula of sum of cos's?
My method: 
Using  De Moivre's formula,
$$\sin rx=\binom r1\cos^{r-1}x\sin x-\binom r1\cos^{r-3}x\sin^3x+\binom r5\cos^{r-5}x\sin^5x-\cdots$$
$$=\cos^nx\left(\binom r1 \tan x-\binom r1\tan^3x+\binom r5\tan^5x-\cdots\right)$$  and
If $r$ is even, $=2m$(say) 
$$\sin2m\theta=\sin2mx=\cos^n{x}\left(\binom {2m}1\tan x-\binom {2m}3\tan^3x+\cdots+\binom {2m}{2m-1}(-1)^{m-1}\tan^{2m-1}x\right)$$
If $n$ is even, I can convert $$\cos^nx=\frac{1}{1+\tan^{n/2}x}$$
and if I investigate each case of the parity of $m$, I can get the formula of the case in which $r$ is even. But otherwise, I can't. Also, I have to investigate the case in which $n$ is odd. 
 A: For sums of sines (and cosines) where the arguments are in an arithmetic progression, we have a much simpler way using Euler's formula:
$$\begin{align}
\sum_{k=0}^{n-1} \sin \left(k\frac\pi n + \theta\right)
&= \operatorname{Im}\sum_{k=0}^{n-1} \exp \left(i\left(k\frac\pi n+\theta\right) \right)\\
&= \operatorname{Im} e^{i\theta} \sum_{k=0}^{n-1} \exp\left(ik\frac\pi n\right)\\
&= \operatorname{Im} e^{i\theta} \frac{1-e^{i\pi}}{1-e^{i\pi/n}}\\
&= \operatorname{Im} \frac{2e^{i\theta}e^{-i\pi/(2n)}}{e^{-i\pi/(2n)}-e^{i\pi/(2n)}}\\
&= \operatorname{Im} \frac{i\exp\left(i\left(\theta-\frac{\pi}{2n}\right)\right)}{\sin \left(\frac{\pi}{2n}\right)}\\
&= \frac{\cos\left(\theta-\frac{\pi}{2n}\right)}{\sin \left(\frac{\pi}{2n}\right)}.
\end{align}$$
A: Use
$$\sum_{k=0}^{n-1} \sin{\left ( k \frac{\pi}{n} + \theta\right)} = \Im{\left [e^{i \theta} \sum_{k=0}^{n-1} e^{i k \pi/n} \right ]}$$
Now, use a geometric series and a half-angle formula to show that
$$\sum_{k=0}^{n-1} e^{i k \pi/n} = \frac{1-e^{-i \pi/n}}{2 \sin^2{\left ( \frac{\pi}{2 n}\right)}}$$
Then you will come across several trig identities in showing that 
$$\sum_{k=0}^{n-1} \sin{\left ( k \frac{\pi}{n} + \theta\right)} = \frac{1}{2 \sin^2{\left ( \frac{\pi}{2 n}\right)}} \Im{\left [e^{i \theta} \left ( 1-e^{-i \pi/n}\right) \right ]} = \frac{\cos{\left ( \theta-\frac{\pi}{2 n}\right)}}{\sin{\frac{\pi}{2 n}}}$$
