# Show that A_n=\sum_{k=1}^n\sin k is bounded?

Show that ${A}_{n}=\sum _{k=1}^{n}\mathrm{sin}k$ is bounded?
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Note that
$2\left(\mathrm{sin}k\right)\left(\mathrm{sin}\left(0.5\right)\right)=\mathrm{cos}\left(k-0.5\right)-\mathrm{cos}\left(k+0.5\right)$
This is obtained by using the ordinary expression for the cosine of a sum.
Add up, $k=1$ to n. On the right, there is mass cancellation. We get
$\mathrm{cos}\left(0.5\right)-\mathrm{cos}\left(n+0.5\right)$
Thus our sum of sines is
$\frac{\mathrm{cos}\left(0.5\right)-\mathrm{cos}\left(n+0.5\right)}{2\mathrm{sin}\left(0.5\right)}$
We can now obtain the desired bound for $|{A}_{n}|$. For example, 2 works, but not by much.
We could modify the appearance of the above formula by using the fact that $\mathrm{cos}\left(0.5\right)-\mathrm{cos}\left(n+0.5\right)=2\mathrm{sin}\left(\frac{n}{2}\right)\mathrm{sin}\left(\frac{n}{2}+0.5\right)$
###### Not exactly what you’re looking for?
dikgetse3u
Hint: Since
$\mathrm{sin}x=\frac{{e}^{ix}-{e}^{-ix}}{2i}$
we can rewrite
$\sum _{n=1}^{K}\mathrm{sin}n=\frac{1}{2i}\sum _{n=1}^{K}{\left({e}^{i}\right)}^{n}-\frac{1}{2i}\sum _{n=1}^{K}{\left({e}^{-i}\right)}^{n}$
and both of these are geometric series.
###### Not exactly what you’re looking for?
RizerMix
A more general formula would be: ${A}_{n}=\sum _{k=1}^{n}\mathrm{sin}k\theta$ $=\frac{\mathrm{sin}\theta +\mathrm{sin}n\theta -\mathrm{sin}\left(n+1\right)\theta }{2\left(1-\mathrm{cos}\theta \right)}$ So ${A}_{n}$ is clearly bounded (just simply check the case where $\theta =1$). The formula can be proved by induction using the trig identity: $\mathrm{sin}\alpha +\mathrm{sin}\beta =2\mathrm{sin}\left(\frac{\alpha +\beta }{2}\right)\mathrm{cos}\left(\frac{\alpha -\beta }{2}\right)$.