Let $S = \sin\,1+\sin\, 2+\sin\, 3+\cdots+\sin\, n$
hence $2S \sin \, 1 = 2\sin\,1 \sin \, 1 + 2\sin\, 2 \sin \, 1+2 \sin\, 3 \sin \, 1+\cdots+2\sin\, n \sin \, 1$
$= \cos\,0-\cos\, 2+\cos\, 1-\cos\, 3+\cos\, 2-\cos\, 4+\cdots+\cos(n-2)-\cos\, n+\cos(n-1)-\cos(n+1)$
$= \cos\,0 + \cos \, 1 - (\cos(n+1) + \cos\, n)$
So $S= \frac{2\cos \frac{1}{2} \cos \frac{1}{2}- 2 \cos (n+\frac{1}{2})\cos \frac{1}{2}}{4 \sin \frac{1}{2}\cos\frac{1}{2}}$
$= \frac{\cos \frac{1}{2} - \cos (n+\frac{1}{2})}{2 \sin \frac{1}{2}}$
$ < \frac{\cos \frac{1}{2} +1}{2 \sin \frac{1}{2}}$
$ < \frac{\cos^2 \frac{1}{4}}{2 \cos \frac{1}{4}\sin \frac{1}{4}}$
or $S < \frac{1}{2} \cot \frac {1}{4}\cdots(1)$
we have for $0 < x < \frac{\pi}{2}$ $ x < \tan x $
Hence $\tan \frac{1}{4} > \frac{1}{4}$ or $ \cot \frac{1}{4} < 4 $
From above and (1) we get
$ S < \frac{1}{2} \cot \frac {1}{4} < \frac{1}{2} * 4$ or $S < 2$
