$\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+…+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$,find $\alpha$

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.

Solutions Collecting From Web of "$\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+…+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$,find $\alpha$"

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?

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
$$
\begin{align}
\sum_{k=0}^{10}\frac1{\sin\left(2^k8^\circ\right)}
&=\frac1{\tan(4^\circ)}-\frac1{\tan(8192^\circ)}\\
&=\frac1{\tan(4^\circ)}+\frac1{\tan(88^\circ)}\\
&=\frac1{\tan(4^\circ)}+\tan(2^\circ)\\
&=\frac{\cos(4^\circ)}{\sin(4^\circ)}+\frac{\sin(4^\circ)}{1+\cos(4^\circ)}\\
&=\frac{\cos(4^\circ)}{\sin(4^\circ)}+\frac{1-\cos(4^\circ)}{\sin(4^\circ)}\\
&=\frac1{\sin(4^\circ)}
\end{align}
$$