# 1966 IMO Problems/Problem 4

## Problem

Prove that for every natural number $n$, and for every real number $x \neq \frac{k\pi}{2^t}$ ( $t=0,1, \dots, n$; $k$ any integer) $$\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^nx}}=\cot{x}-\cot{2^nx}$$

## Solution

Assume that $\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^{n}x}}=\cot{x}-\cot{2^{n}x}$ is true, then we use $n=1$ and get $\cot x - \cot 2x = \frac {1}{\sin 2x}$.

First, we prove $\cot x - \cot 2x = \frac {1}{\sin 2x}$

LHS= $\frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x}$ $= \frac{2\cos^2 x}{2\cos x \sin x}-\frac{2\cos^2 x -1}{\sin 2x}$ $=\frac{2\cos^2 x}{\sin 2x}-\frac{2\cos^2 x -1}{\sin 2x}$ $=\frac {1}{\sin 2x}$

Using the above formula, we can rewrite the original series as $\cot x - \cot 2x + \cot 2x - \cot 4x + \cot 4x \cdot \cdot \cdot + \cot 2^{n-1} x - \cot 2^n x$

Which gives us the desired answer of $\cot x - \cot 2^n x$

## See Also

 1966 IMO (Problems) • Resources Preceded byProblem 3 1 • 2 • 3 • 4 • 5 • 6 Followed byProblem 5 All IMO Problems and Solutions
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