https://math.stackexchange.com/a/1119137/29156
Shouldn't Abel's answer be:
we will use the fact that $$2 \sin 1 \sin k = \cos(k-1) -\cos(k + 1)$$let $S = \sin 1 + \sin 2 \cdots + \sin n,$ then
$\begin{align} 2S \sin 1 &= 2\sin 1 \sin 1+ 2 \sin 1 \sin 2 + 2 \sin 1 \sin 3\cdots +2 \sin 1 \sin n \\ &=(1 - \cos 2) +(\color{red}{\cos 1} - \cos3) +(\color{red}{\cos 2} - \cos 4)+\cdots +(\cos (n- 1) - \cos ( n + 1) )\\ &=1 + \color{red}{\cos 1} - \color{red}{\cos (n)} - \color{red}{\cos(n+1)} \end{align}$
?
Obviously this is still bounded