Finding the sum of $\sin(0^\circ) + \sin(1^\circ) + \sin(2^\circ) + \cdots +\sin(180^\circ)$ I need help understanding the sum of $\sin(0^\circ) + \sin(1^\circ) + \sin(2^\circ) + \cdots +\sin(180^\circ)$ or $\displaystyle \sum_{i=0}^{180} \sin(i)$ 
This might be related to a formula to find the average voltage from a generator used to gauge waves: $V_\text{avg} = 0.637 \times V_\text{peak}$. I am currently learning about AC circuits in the military. 

For some positive integer $k | 180$,
\begin{align}
\sum_{i=0}^{180\over k} \sin(i\cdot k^\circ) = \text{ctan}(\frac{k\pi}{360})
\end{align}
edit:
this sum of $\sin$ functions is the imaginary part of $\sum e^{ik^\circ}$, sum of vectors. We connect each of these vector (aka summing up), from $e^0$ to $e^{i180^\circ}$, it actually forms the right-half (${180\over k}+1$ pieces of tangential $360\over k$-regular-polygon. The distance from bottom and top of this half of regular $360\over k$-regular polygon, is the sum of imaginary part those connecting vectors, aka $\sum_{i=0}^{180\over k} \sin(i\cdot k^\circ)$. It also happens to be diameter $d$ of the inscribing circle of this polygon. Noting that side length of this polygon is 1, thus we have
\begin{align}
\frac{1\over2}{d\over2}=\tan({k\over2}^\circ) = \tan({k\pi\over 360})
\end{align}
An illustrating example, let $k=60$. Then connecting $e^0 + e^{i60^\circ} + e^{i120^\circ} + e^{i180^\circ}$ gives

$d$ is the sum we want to solve. The plot shows 4 pieces of a regular hexagon, apparently
\begin{align}
\frac{0.5}{0.5d}= \tan(30^\circ) 
\end{align}
Therefore $\sum_{i=0}^3 \sin(i\cdot 60^\circ)=\text{ctan}(30^\circ)$
credit to wewill2009@mitbbs
