$\sum_{n=1}^x\sin n$ is never greater than 2? I was playing around with Desmos, and I put in the following equation:
$$y=\sum_{n=1}^{100x}\sin n$$
I'm not quite sure what I was expecting, but I noticed that the seemingly random dots it produced were never greater than y=2 (and never less than y=-0.25).
I'm wondering if there is any proof and/or explanation that that is the case.
Also, what branch of math would this problem fall under? It really interests me and I would like to know.
Thanks.
 A: $\sin x + \sin 2x + \cdots + \sin nx = (\cos \frac x 2 - \cos (n + \frac 1 2) x)/ 2 \sin \frac x 2$. Now put $x = 1$. It's trigonometry.
A: Note that 
$$\begin{align}
\sum_{n=1}^N\sin(n)&=\text{Im}\left(\sum_{n=1}^Ne^{in}\right)\\\\
&=\text{Im}\left(\frac{e^i-e^{i(N+1)}}{1-e^i}\right)\\\\
&=\text{Im}\left(e^{i(N+1)/2}\frac{\sin(N/2)}{\sin(1/2)}\right)\\\\
&=\frac{\sin\left(\frac{N+1}{2}\right)\sin(N/2)}{\sin(1/2)}\\\\
&\le \csc(1/2)\\\\
&\approx 2.08582964293349 
\end{align}$$
As pointed out by ClementC., we can write
$$\sin\left(\frac{N+1}{2}\right)\sin(N/2)=\frac12\left(\cos(1/2)-\cos(N+1/2)\right)$$
to reveal that
$$\begin{align}
\sum_{n=1}^N\sin(n)&=\frac{\cos(1/2)-\cos(N+1/2)}{2\sin(1/2)}\\\\
&\le \frac{\cos(1/2)+1}{2\sin(1/2)}\\\\
&\approx 1.95815868232297
\end{align}$$
Moreover, we find that 
$$\begin{align}
\sum_{n=1}^N\sin(n)&=\frac{\cos(1/2)-\cos(N+1/2)}{2\sin(1/2)}\\\\
&\ge \frac{\cos(1/2)-1}{2\sin(1/2)}\\\\
&\approx -0.127670960610518
\end{align}$$
A: Sum of $sines $ series when angles are in Arithmetical progression
$S=\sin \alpha +\sin (\alpha+\beta)+\sin (\alpha+2\beta) +\cdots n$ terms
We know that 
$$2\sin A.\sin B=\cos(A-B) -\cos(A+B) $$
$$\implies 2\sin\alpha.\sin\frac{\beta}{2}=\cos\left(\alpha-\frac{\beta}{2}\right) -\cos(\alpha+\frac{\beta}{2})$$
$$\implies 2\sin(\alpha + \beta).\sin\frac{\beta}{2}=\cos\left(\alpha +\frac{\beta}{2}\right) -\cos\left(\alpha+\frac{3\beta}{2}\right)$$
$$\implies 2\sin\left(\alpha + 2\beta\right).\sin\frac{\beta}{2}=\cos\left(\alpha +\frac{3\beta}{2}\right) -\cos\left(\alpha+\frac{5\beta}{2}\right)$$
$\cdots$
$\cdots$
$$\implies 2\sin\left(\alpha + (n-1)\beta\right).\sin\frac{\beta}{2}=\cos\left(\alpha +( n-1)\beta -\frac{\beta}{2}\right) -\cos\left(\alpha+(n-1)\beta+\frac{\beta}{2}\right)$$
By adding
$$2\sin\frac{\beta}{2}\left[\sin \alpha +\sin (\alpha+\beta)+\sin (\alpha+2\beta) +\cdots +\sin(\alpha+(n-1)\beta) \right]=\cos\left(\frac{\alpha-\beta}{2}\right) - \cos\left(\alpha+\frac{(n-1)\beta}{2}\right) $$
$$\implies 2\sin\frac{\beta}{2}.S=2\sin\left(\alpha+\frac{(n-1)\beta}{2}\right).\sin\frac{n\beta}{2}$$
$$=\frac{\sin{\frac{n\beta}{2}}}{\sin{\frac{\beta}{2}}}{\sin\left[ {\alpha + \frac{\beta}{2}{(n-1)}}\right]}$$
