$\sum_{n=1}^x\sin n$ 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.

Solution 1:

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

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

Solution 2:

$\sin x + \sin 2x + \cdots + \sin nx = (\cos \frac x 2 - \cos (n + \frac 1 2) x)/ 2 \sin \frac x 2$. Now put $x = 1$. It's trigonometry.

Solution 3:

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 $$2\sin\frac{\beta}{2}\left[\sin \alpha +\sin (\alpha+\beta)+\sin (\alpha+2\beta) +\cdots +\sin(\alpha+(n-1)\beta) \right]=\cos\left(\frac{\alpha-\beta}{2}\right) - \cos\left(\alpha+\frac{(n-1)\beta}{2}\right) $$ $$\implies 2\sin\frac{\beta}{2}.S=2\sin\left(\alpha+\frac{(n-1)\beta}{2}\right).\sin\frac{n\beta}{2}$$ $$=\frac{\sin{\frac{n\beta}{2}}}{\sin{\frac{\beta}{2}}}{\sin\left[ {\alpha + \frac{\beta}{2}{(n-1)}}\right]}$$