$\sum_{n=1}^x\sin n$ is never greater than 2?

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

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

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