1966 IMO Problems/Problem 4: Difference between revisions
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== Problem == | |||
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) | |||
<cmath> \frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^nx}}=\cot{x}-\cot{2^nx} </cmath> | |||
== Solution == | == Solution == | ||
First, we prove <math>\cot \theta - \cot 2\theta = \frac {1}{\sin 2\theta}</math>. | |||
LHS<math>\ =\ \frac{\cos \theta}{\sin \theta}-\frac{\cos 2\theta}{\sin 2\theta}</math> | |||
<math>= \frac{2\cos^2 \theta}{2\cos \theta \sin \theta}-\frac{2\cos^2 \theta -1}{\sin 2\theta}</math> | |||
= \frac{2\cos^2 | <math>=\frac{2\cos^2 \theta}{\sin 2\theta}-\frac{2\cos^2 \theta -1}{\sin 2\theta}</math> | ||
=\frac{ | <math>=\frac {1}{\sin 2\theta}</math> | ||
Using the above formula, we can rewrite the original series as | |||
<math>\cot x - \cot 2x + \cot 2x - \cot 4x + \cot 4x + \dots + \cot 2^{n-1} x - \cot 2^n x </math>. | |||
\cot | Which gives us the desired answer of <math>\cot x - \cot 2^n x</math>. | ||
== See Also == | |||
{{IMO box|year=1966|num-b=3|num-a=5}} | |||
Latest revision as of 11:17, 24 September 2024
Problem
Prove that for every natural number 
, and for every real number 
(
; 
any integer)
![\[\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^nx}}=\cot{x}-\cot{2^nx}\]](http://latex.artofproblemsolving.com/4/a/3/4a3adede6c471d5c65e37bdfb9f4961dc8f6b823.png)
Solution
First, we prove 
.
LHS
Using the above formula, we can rewrite the original series as
.
Which gives us the desired answer of 
.
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 | ||
