Problem (Algebra):
Ross starts with an angle of measure $8^{\circ}$ and doubles it 10 times until he obtains $8192^{\circ}$. He then adds up the reciprocals of the sines of these 11 angles. That is, he calculates

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

Determine, without using a calculator, the measure of the acute angle $\alpha$ so that $S=\frac{1}{\sin \alpha}$.

Solution:
We first prove Lemma(i): If $\theta$ is an angle whose measure is not an integer multiple of $90^{\circ}$, then
$$
\cot \theta-\cot 2 \theta=\frac{1}{\sin 2 \theta}
$$

Proof. 
$$
\begin{aligned}
\mathrm{LS} & =\cot \theta-\cot 2 \theta \\
& =\frac{\cos \theta}{\sin \theta}-\frac{\cos 2 \theta}{\sin 2 \theta} \\
& =\frac{\cos \theta}{\sin \theta}-\frac{\cos 2 \theta}{2 \sin \theta \cos \theta} \\
& =\frac{2 \cos ^{2} \theta-\cos 2 \theta}{2 \sin \theta \cos \theta} \\
& =\frac{2 \cos ^{2} \theta-\left(2 \cos ^{2} \theta-1\right)}{\sin 2 \theta} \\
& =\frac{1}{\sin 2 \theta} \\
& =\mathrm{RS}
\end{aligned}
$$

as required.

We use (i) to note that $\frac{1}{\sin 8^{\circ}}=\cot 4^{\circ}-\cot 8^{\circ}$ and $\frac{1}{\sin 16^{\circ}}=\cot 8^{\circ}-\cot 16^{\circ}$ and so on. Thus,

$$
\begin{aligned}
S= & \frac{1}{\sin 8^{\circ}}+\frac{1}{\sin 16^{\circ}}+\frac{1}{\sin 32^{\circ}}+\cdots+\frac{1}{\sin 4096^{\circ}}+\frac{1}{\sin 8192^{\circ}} \\
= & \left(\cot 4^{\circ}-\cot 8^{\circ}\right)+\left(\cot 8^{\circ}-\cot 16^{\circ}\right)+\left(\cot 16^{\circ}-\cot 32^{\circ}\right)+ \\
& \cdots+\left(\cot 2048^{\circ}-\cot 4096^{\circ}\right)+\left(\cot 4096^{\circ}-\cot 8192^{\circ}\right) \\
= & \cot 4^{\circ}-\cot 8192^{\circ}
\end{aligned}
$$

since the sum "telescopes".

Since the cotangent function has a period of $180^{\circ}$, and $8100^{\circ}$ is a multiple of $180^{\circ}$, then $\cot 8192^{\circ}=\cot 92^{\circ}$.

Therefore,

$$
\begin{aligned}
S & =\cot 4^{\circ}-\cot 92^{\circ} \\
& =\frac{\cos 4^{\circ}}{\sin 4^{\circ}}-\frac{\cos 92^{\circ}}{\sin 92^{\circ}} \\
& =\frac{\cos 4^{\circ}}{\sin 4^{\circ}}-\frac{-\sin 2^{\circ}}{\cos 2^{\circ}} \\
& =\frac{\cos 4^{\circ}}{2 \sin 2^{\circ} \cos 2^{\circ}}+\frac{\sin 2^{\circ}}{\cos 2^{\circ}} \\
& =\frac{\cos 4^{\circ}+2 \sin ^{2} 2^{\circ}}{2 \sin 2^{\circ} \cos 2^{\circ}} \\
& =\frac{\left(1-2 \sin ^{2} 2^{\circ}\right)+2 \sin ^{2} 2^{\circ}}{\sin 4^{\circ}} \\
& =\frac{1}{\sin 4^{\circ}}
\end{aligned}
$$

Therefore, $\alpha=4^{\circ}$.

Final answer: $4^{\circ}$