Question

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

Answer 1

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