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 $\cot x - \cot 2^n x$ .

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

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