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

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

LHS$\ =\ \frac{\cos \theta}{\sin \theta}-\frac{\cos 2\theta}{\sin 2\theta}$

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

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

$=\frac {1}{\sin 2\theta}$

Using the above formula, we can rewrite the original series as

$\cot x - \cot 2x + \cot 2x - \cot 4x + \cot 4x + \dots + \cot 2^{n-1} x - \cot 2^n x$.

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

See Also

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