jisu61hbke

2022-03-31

$\frac{1}{{\mathrm{sin}8}^{\circ }}+\frac{1}{{\mathrm{sin}16}^{\circ }}+\dots +\frac{1}{{\mathrm{sin}4090}^{\circ }}+\frac{1}{{\mathrm{sin}8192}^{\circ }}=\frac{1}{\mathrm{sin}\alpha }$, find $\alpha$

Riya Erickson

Using the identity from the Weierstrass Substitution
$\mathrm{tan}\left(\frac{x}{2}\right)=\frac{\mathrm{sin}\left(x\right)}{1+\mathrm{cos}\left(x\right)}$
we get
$\frac{1}{\mathrm{tan}\left(\frac{x}{2}\right)}-\frac{1}{\mathrm{tan}\left(x\right)}=\frac{1+\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}-\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$
$=\frac{1}{\mathrm{sin}\left(x\right)}$
The rest is the same telescoping series
$\sum _{k=0}^{n}\frac{1}{\mathrm{sin}\left({2}^{k}x\right)}=\sum _{k=0}^{n}\left[\frac{1}{\mathrm{tan}\left({2}^{k-1}x\right)}-\frac{1}{\mathrm{tan}\left({2}^{k}x\right)}\right]$
$=\frac{1}{\mathrm{tan}\left(\frac{x}{2}\right)}-\frac{1}{\mathrm{tan}\left({2}^{n}x\right)}$
The question has $x={8}^{\circ }$ and $n=10$, so we get
$\sum _{k=0}^{10}\frac{1}{\mathrm{sin}\left({2}^{k}{8}^{\circ }\right)}=\frac{1}{\mathrm{tan}\left({4}^{\circ }\right)}-\frac{1}{\mathrm{tan}\left({8192}^{\circ }\right)}$
$=\frac{1}{\mathrm{tan}\left({4}^{\circ }\right)}+\frac{1}{\mathrm{tan}\left({88}^{\circ }\right)}$
$=\frac{1}{\mathrm{tan}\left({4}^{\circ }\right)}+\mathrm{tan}\left({2}^{\circ }\right)$
$=\frac{\mathrm{cos}\left({4}^{\circ }\right)}{\mathrm{sin}\left({4}^{\circ }\right)}+\frac{\mathrm{sin}\left({4}^{\circ }\right)}{1+\mathrm{cos}\left({4}^{\circ }\right)}$
$=\frac{1}{\mathrm{sin}\left({4}^{\circ }\right)}$

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