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Problem 813

AIME late

Algebra Difficulty 5.5 Prove it

4. (YUG) Prove the following equality:
$\frac{1}{\sin 2 x}+\frac{1}{\sin 4 x}+\frac{1}{\sin 8 x}+\cdots+\frac{1}{\sin 2^{n} x}=\cot x-\cot 2^{n} x$
where $n \in \mathbb{N}$ and $x \notin \pi \mathbb{Z} / 2^{k}$ for every $k \in \mathbb{N}$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

4. It suffices to prove $1 / \sin 2^{k} x=\cot 2^{k-1} x-\cot 2^{k} x$ for any integer $k$ and real $x$ , i.e., $1 / \sin 2 x=\cot x-\cot 2 x$ for all real $x$ . We indeed have $\cot x-\cot 2 x=\cot x-\frac{\cot ^{2} x-1}{2 \cot x}=\frac{\left(\frac{\cos x}{\sin x}\right)^{2}+1}{2 \frac{\cos x}{\sin x}}=\frac{1}{2 \sin x \cos x}=\frac{1}{\sin 2 x}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.