Solution for $\sum_{i=0}^{n} \sin(\frac{i\pi}{2n})$? While I was trying to find the formula of something by my own means I came across this sum which I need to solve, however I don't know if there is a solution for it, maybe it doesn't mean anything and I made a mistake. However if there's an equation which can replace this sum I will appreciate it a lot if you show me which one and how did you find the answer!
 A: Let $S = \displaystyle\sum_{k = 1}^{n}\sin\dfrac{k\pi}{2n}$. (The $k = 0$ term is $0$, so we can ignore it). Then, by using the product to difference identity $\sin A \sin B = \dfrac{1}{2}\left(\cos(A-B)-\cos(A+B)\right)$, we have:
$S\sin \dfrac{\pi}{4n}$ $= \displaystyle\sum_{k = 1}^{n}\sin\dfrac{k\pi}{2n}\sin\dfrac{\pi}{4n}$ $= \dfrac{1}{2}\displaystyle\sum_{k = 1}^{n}\left(\cos\left(\dfrac{k\pi}{2n}-\dfrac{\pi}{4n}\right)-\cos\left(\dfrac{k\pi}{2n}-\dfrac{\pi}{4n}\right)\right)$ 
$= \dfrac{1}{2}\displaystyle\sum_{k = 1}^{n}\left(\cos\dfrac{(2k-1)\pi}{4n} - \cos\dfrac{(2k+1)\pi}{4n}\right)$ $= \dfrac{1}{2}\left(\cos\dfrac{\pi}{4n} - \cos\dfrac{(2n+1)\pi}{4n}\right)$. 
Now, divide both sides by $\sin\dfrac{\pi}{4n}$ to get $S = \dfrac{\cos\tfrac{\pi}{4n} - \cos\tfrac{(2n+1)\pi}{4n}}{2\sin\tfrac{\pi}{4n}}$. 
A: Multiply your sum by $\sin\left(\frac{\pi}{4n}\right)$ and use the formula
$$\sin\left(\frac{i \pi}{2n}\right)\sin\left(\frac{\pi}{4n}\right)=\frac{1}{2}\left[\cos\left(\frac{i \pi}{2n}-\frac{\pi}{4n}\right)-\cos\left(\frac{i \pi}{2n}+\frac{\pi}{4n}\right)\right]$$
A: Let $z=e^{i\theta}=\cos\theta+i\sin\theta$. Then $z^j=e^{ij\theta}=\cos j\theta+i\sin j\theta$ for all $j\in\mathbb{N}$.
Thus
$\begin{eqnarray}
1+\sin\theta+\sin2\theta+...+\sin n\theta&=&Im(z^0+z^1+z^2+...+z^n)\\
&=&Im\left(\frac{z^{m+1}-1}{z-1}\right)
\end{eqnarray}$
Putting $\theta=\frac{\pi}{2n}$ you get an implicit form for your sum (of course, you must continue simplifying the RHS)
