Question

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

Answer 1

Answer: 4

$T_1=\frac{1}{\sin 8^{\circ }}=\frac{\sin \left(8^{\circ }-4^{\circ }\right)}{\sin 4^{\circ }\sin 8^{\circ }}=\cot 4^{\circ }-\cot 8^{\circ }$ and so on
Given $S=\frac{1}{\sin 8^{\circ }}+\frac{1}{\sin 16^{\circ }}+\frac{1}{\sin 32^{\circ }}+\dots \dots \dots \dots +\frac{1}{\sin 4096^{\circ }}+\frac{1}{\sin 8192^{\circ }}$
$=\left(\cot 4^{\circ }-\cot 8^{\circ }\right)+\left(\cot 8^{\circ }-\cot 16^{\circ }\right)+\left(\cot 16^{\circ }-\cot 32^{\circ }\right)+\dots \dots \dots \dots +\left(\cot 2048^{\circ }-\cot 4096^{\circ }\right)+\left(\cot 4096^{\circ }-\cot 8192^{\circ }\right)$
$=\left(\cot 4^{\circ }-\cot 8192^{\circ }\right)]$
$\text{ Also }\cot 8192^{\circ }=\cot \left(45\times 180^{\circ }+92^{\circ }\right)=\cot 92^{\circ }\left(\text{ As }45\times 180^{\circ }=8100^{\circ }\right)$
$\therefore S=\cot 4^{\circ }-\cot 92^{\circ }=\frac{1-{\tan }^2{2}^{\circ }}{2\tan 2^{\circ }}+\tan 2^{\circ }=\frac{1+{\tan }^2{2}^{\circ }}{2\tan 2^{\circ }}=\frac{{\sec }^2{2}^{\circ }}{2\tan 2^{\circ }}=\frac{1}{\sin 4^{\circ }}$
$\Rightarrow \alpha =4^{\circ }$