TITLE: $\sum_{n=1}^x\sin n$ is never greater than 2?
QUESTION [11 upvotes]: 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.

REPLY [10 votes]: 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 
$$\begin{align}
\sum_{n=1}^N\sin(n)&=\frac{\cos(1/2)-\cos(N+1/2)}{2\sin(1/2)}\\\\
&\ge \frac{\cos(1/2)-1}{2\sin(1/2)}\\\\
&\approx -0.127670960610518
\end{align}$$