Is there a mistake in an old question on this site: "Where does the sum of sin(n) formula come from?" 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

hint
Yes, Abel's result seems to be false.
$$2S\sin(1)=\sum_{k=1}^n\Big((\cos(k-1)-\cos(k))+(\cos(k)-\cos(k+1))\Bigr)=$$
$$1-\cos(n)+\cos(1)-\cos(n+1)=$$
$$2\sin^2(\frac{n+1}{2})+\cos(1)-\cos(n)=$$
$$2\sin(\frac{n+1}{2})\Bigl(\sin(\frac{n+1}{2})+\sin(\frac{n-1}{2})\Bigr)=$$
$$4\sin(\frac{n+1}{2})\sin(\frac n2)\cos(\frac 12)$$
Thus
$$\boxed{S=\frac{\sin(\frac{n+1}{2})\sin(\frac n2)}{\sin(\frac 12)}}$$
