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. 
 A: The finite sum can be found in closed form exaclty using the geometric series and Euler's theorem. 
First Euler's theorem using imaginary numbers to relate trigonometric functions to the exponential function.
$$ e^{i\theta} = \cos\theta + i \sin(\theta) \qquad \textbf{(1)}$$
Second the finite geometric series tells us how to add a sum of power of a number,
$$ \sum_{k=0}^N x^k = \frac{x^{N+1}-1}{x-1} \qquad \textbf{(2)}$$
Since you seem to be interested in measuring the angle in degrees we need to include the conversion factor on our angle. I'll just represent this factor by the variable $\lambda=\pi/180$. We can write the series you are interested in as,
$$ \sum_{k=0}^N \sin( k^\circ) = \sum_{k=0}^N \sin( \lambda k) = Im\left(\sum_{k=0}^N e^{i\lambda k} \right) \qquad \textbf{(3)}$$
So if we find the imaginary part of the sum of exponential functions we will have succeeded in evaluating the sum.  
$$ \sum_{k=0}^N e^{i\lambda k} = \sum_{k=0}^N \left( e^{i \lambda} \right)^k = \frac{(e^{i\lambda})^{N+1} - 1 }{e^{i\lambda}-1}$$
To get the imaginary part of the sum we use the identity that the imaginary part of a complex number is the number minus its complex conjugate divided by twice $i$. 
$$ Im\left(\sum_{k=0}^N e^{i\lambda k} \right)=  \frac{1}{2i} \left( \frac{(e^{i\lambda})^{N+1} - 1 }{e^{i\lambda}-1} - \frac{(e^{-i\lambda})^{N+1} - 1 }{e^{-i\lambda}-1}\right)$$
$$  =  \frac{1}{2i} \left( \frac{(e^{i\lambda})^{N+1} - 1 }{e^{i\lambda}-1} - \frac{(e^{-i\lambda})^{N+1} - 1 }{e^{-i\lambda}-1}\right) $$
$$ = \frac{\sin(N\lambda) + \sin(\lambda) - \sin((N+1)\lambda)}{2(1-\cos(\lambda))}$$
If we substitute $N=180$ we get,
$$\sum_{k=0}^{180} \sin(k^\circ) = \frac{\sin(\pi) + \sin(\pi/180) - \sin(181 \pi/180)}{2(1-\cos(\pi/180))} \approx 114.6$$
Dividing this by $180$ we get an answer of,
$$\frac{1}{180 } \sum_{k=0}^{180} \sin(k^\circ) = \frac{114.6}{180} \approx .637 $$
A: I'll assume you meant the average,
$$
\frac{\sin0^\circ+\cdots+\sin180^\circ}{181}.
$$
This approximates the average value of the sine-in-radians function on the interval from $0$ to $\pi$, which is
$$
\frac 1 \pi \int_0^\pi \sin x\, dx = \frac 2 \pi.
$$
A: $$
\begin{align}
\sum_{i=0}^{180}\sin(i^{\circ})&=\sum_{i=0}^{180}\sin\left(i\frac{\pi}{180}\right)\\
&=\frac{180}{\pi}\sum_{i=0}^{180}\sin\left(i\frac{\pi}{180}\right)\frac{\pi}{180}\\
&=\frac{180}{\pi}\sum_{i=0}^{180}\sin\left(i\cdot\Delta{i}\right)\Delta{i}\\
&\approx\frac{180}{\pi}\int_{i=0}^{\pi}\sin(x)\,dx\\
&=\frac{180}{\pi}\left[-\cos(x)\right]_{x=0}^{x=\pi}\\
&=\frac{360}{\pi}\\
\end{align}
$$
If we average this out over the $181$ terms, $\frac{360}{181\pi}\approx0.633\ldots$.
A: Wikipedia says
$$\sum_{k=0}^n \sin{(\gamma + k\alpha)} = \frac{\sin\tfrac{(n+1)\alpha}{2} + \sin{(\gamma + \tfrac{n\alpha}{2})}}{\sin\tfrac\alpha2}$$So we may write $\gamma = 0$, $n=180$ and $\alpha=\frac{\pi}{180}$ to get $$\sum_{k=0}^{180} \sin{\tfrac{k\pi}{180}} = \frac{\sin\tfrac{181\pi}{360} \sin\tfrac{\pi}{2}}{\sin\tfrac{\pi}{720}} = \frac{\sin\tfrac{181\pi}{360} }{\sin\tfrac{\pi}{720}}$$
A proof of the formula can be found here.
A: 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
