Alex Cervantes

2022-01-23

Show that ${A}_{n}=\sum _{k=1}^{n}\mathrm{sin}k$ is bounded?

Addisyn Thompson

Beginner2022-01-24Added 16 answers

Note that

$2\left(\mathrm{sin}k\right)\left(\mathrm{sin}\left(0.5\right)\right)=\mathrm{cos}(k-0.5)-\mathrm{cos}(k+0.5)$

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

$\mathrm{cos}\left(0.5\right)-\mathrm{cos}(n+0.5)$

Thus our sum of sines is

$\frac{\mathrm{cos}\left(0.5\right)-\mathrm{cos}(n+0.5)}{2\mathrm{sin}\left(0.5\right)}$

We can now obtain the desired bound for$\left|{A}_{n}\right|$ . For example, 2 works, but not by much.

We could modify the appearance of the above formula by using the fact that$\mathrm{cos}\left(0.5\right)-\mathrm{cos}(n+0.5)=2\mathrm{sin}\left(\frac{n}{2}\right)\mathrm{sin}(\frac{n}{2}+0.5)$

This is obtained by using the ordinary expression for the cosine of a sum.

Add up,

Thus our sum of sines is

We can now obtain the desired bound for

We could modify the appearance of the above formula by using the fact that

dikgetse3u

Beginner2022-01-25Added 10 answers

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.

we can rewrite

and both of these are geometric series.

RizerMix

Expert2022-01-27Added 656 answers

A more general formula would be:
${A}_{n}=\sum _{k=1}^{n}\mathrm{sin}k\theta $
$=\frac{\mathrm{sin}\theta +\mathrm{sin}n\theta -\mathrm{sin}(n+1)\theta}{2(1-\mathrm{cos}\theta )}$
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: $\mathrm{sin}\alpha +\mathrm{sin}\beta =2\mathrm{sin}(\frac{\alpha +\beta}{2})\mathrm{cos}(\frac{\alpha -\beta}{2})$ .

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