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Let $\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+\frac{1}{\sin 32^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$ where $\alpha\in(0,90^\circ)$,then find $\alpha$(in degree.)

$\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+\frac{1}{\sin 32^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$

$\frac{2\cos8^\circ}{\sin 16^\circ}+\frac{2\cos16^\circ}{\sin 32^\circ}+\frac{2\cos32^\circ}{\sin 64^\circ}+....+\frac{2\cos4096^\circ}{\sin 8192^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$

$\frac{2^2\cos8^\circ\cos16^\circ}{\sin 32^\circ}+\frac{2^2\cos16^\circ\cos32^\circ}{\sin 64^\circ}+\frac{2^2\cos32^\circ\cos64^\circ}{\sin 128^\circ}+....+\frac{2\cos4096^\circ}{\sin 8192^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$

In this way this series is getting complicated at each stage,is there any way to simplify it?Please help me.Thanks.

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HINT:

For $\sin A\ne0\iff A\ne m\pi$ where $m$ is any integer,

$$\cot A-\cot2A=\dfrac{\sin(2A-A)}{\sin2A\sin A}=\csc2A$$

Do you recognize the Telescoping Series?

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    $\begingroup$ Why 'DO' and not 'Do'? $\endgroup$ – G-man Dec 28 '15 at 5:28
  • $\begingroup$ @G-man, Please ignore the case difference. Btw could you follow my method? $\endgroup$ – lab bhattacharjee Dec 28 '15 at 5:30
  • $\begingroup$ Ummm..yes. But why did you feel the need to ask me that? $\endgroup$ – G-man Dec 28 '15 at 5:32
  • $\begingroup$ @G-man, I want to be sure that : my method is understandable enough $\endgroup$ – lab bhattacharjee Dec 28 '15 at 5:33
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    $\begingroup$ @diya, See cemc.uwaterloo.ca/contests/past_contests/2010/… $$9 a(ii)$$ $\endgroup$ – lab bhattacharjee Dec 28 '15 at 6:25
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This is just the same idea as in lab bhattacharjee's answer, but using the identity from the Weierstrass Substitution $$ \tan(x/2)=\frac{\sin(x)}{1+\cos(x)} $$ we get $$ \begin{align} \frac1{\tan(x/2)}-\frac1{\tan(x)} &=\frac{1+\cos(x)}{\sin(x)}-\frac{\cos(x)}{\sin(x)}\\ &=\frac1{\sin(x)} \end{align} $$ The rest is the same telescoping series $$ \begin{align} \sum_{k=0}^n\frac1{\sin\left(2^kx\right)} &=\sum_{k=0}^n\left[\frac1{\tan\left(2^{k-1}x\right)}-\frac1{\tan\left(2^kx\right)}\right]\\ &=\frac1{\tan(x/2)}-\frac1{\tan\left(2^nx\right)} \end{align} $$ The question has $x=8^\circ$ and $n=10$, so we get $$ \begin{align} \sum_{k=0}^{10}\frac1{\sin\left(2^k8^\circ\right)} &=\frac1{\tan(4^\circ)}-\frac1{\tan(8192^\circ)}\\ &=\frac1{\tan(4^\circ)}+\frac1{\tan(88^\circ)}\\ &=\frac1{\tan(4^\circ)}+\tan(2^\circ)\\ &=\frac{\cos(4^\circ)}{\sin(4^\circ)}+\frac{\sin(4^\circ)}{1+\cos(4^\circ)}\\ &=\frac{\cos(4^\circ)}{\sin(4^\circ)}+\frac{1-\cos(4^\circ)}{\sin(4^\circ)}\\ &=\frac1{\sin(4^\circ)} \end{align} $$

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