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Difference between revisions of "1966 IMO Problems/Problem 4"

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== Problem ==
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Prove that for every natural number <math>n</math>, and for every real number <math>x \neq \frac{k\pi}{2^t}</math> (<math>t=0,1, \dots, n</math>; <math>k</math> any integer)
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<cmath> \frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^nx}}=\cot{x}-\cot{2^nx}  </cmath>
  
 
== Solution ==
 
== Solution ==
  
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Assume that <math>\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^{n}x}}=\cot{x}-\cot{2^{n}x}</math> is true, then we use <math>n=1</math> and get <math>\cot x - \cot 2x = \frac {1}{\sin 2x}</math>.
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First, we prove <math>\cot \theta - \cot 2\theta = \frac {1}{\sin 2\theta}</math>.
  
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First, we prove <math>\cot x - \cot 2x = \frac {1}{\sin 2x}</math>
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LHS<math>\ =\ \frac{\cos \theta}{\sin \theta}-\frac{\cos 2\theta}{\sin 2\theta}</math>
  
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LHS=<math>\frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x}</math>
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<math>= \frac{2\cos^2 \theta}{2\cos \theta \sin \theta}-\frac{2\cos^2 \theta -1}{\sin 2\theta}</math>
  
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<math>= \frac{2\cos^2 x}{2\cos x \sin x}-\frac{2\cos^2 x -1}{\sin 2x}</math>
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<math>=\frac{2\cos^2 \theta}{\sin 2\theta}-\frac{2\cos^2 \theta -1}{\sin 2\theta}</math>
  
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<math>=\frac{2\cos^2 x}{\sin 2x}-\frac{2\cos^2 x -1}{\sin 2x}</math>
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<math>=\frac {1}{\sin 2\theta}</math>
  
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<math>=\frac {1}{\sin 2x}</math>
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Using the above formula, we can rewrite the original series as
  
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Using the above formula, we can rewrite the original series as
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<math>\cot x - \cot 2x + \cot 2x - \cot 4x + \cot 4x + \dots + \cot 2^{n-1} x - \cot 2^n x </math>.
  
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<math>\cot x - \cot 2x + \cot 2x - \cot 4x + \cot 4x \cdot \cdot \cdot + \cot 2^{n-1} x - \cot 2^n x </math>
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Which gives us the desired answer of <math>\cot x - \cot 2^n x</math>.
  
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Which gives us the desired answer of <math>\cot x - \cot 2^n x</math>
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== See Also ==
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{{IMO box|year=1966|num-b=3|num-a=5}}

Latest revision as of 12:17, 24 September 2024

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
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