Difference between revisions of "1966 IMO Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | First, we prove <math>\cot | + | First, we prove <math>\cot \theta - \cot 2\theta = \frac {1}{\sin 2\theta}</math>. |
| − | LHS | + | LHS<math>\ =\ \frac{\cos \theta}{\sin \theta}-\frac{\cos 2\theta}{\sin 2\theta}</math> |
| − | <math>= \frac{2\cos^2 | + | <math>= \frac{2\cos^2 \theta}{2\cos \theta \sin \theta}-\frac{2\cos^2 \theta -1}{\sin 2\theta}</math> |
| − | <math>=\frac{2\cos^2 | + | <math>=\frac{2\cos^2 \theta}{\sin 2\theta}-\frac{2\cos^2 \theta -1}{\sin 2\theta}</math> |
| − | <math>=\frac {1}{\sin | + | <math>=\frac {1}{\sin 2\theta}</math> |
Using the above formula, we can rewrite the original series as | Using the above formula, we can rewrite the original series as | ||
| − | <math>\cot x - \cot 2x + \cot 2x - \cot 4x + \cot 4x \ | + | <math>\cot x - \cot 2x + \cot 2x - \cot 4x + \cot 4x + \dots + \cot 2^{n-1} x - \cot 2^n x </math>. |
| − | Which gives us the desired answer of <math>\cot x - \cot 2^n x</math> | + | Which gives us the desired answer of <math>\cot x - \cot 2^n x</math>. |
== See Also == | == See Also == | ||
{{IMO box|year=1966|num-b=3|num-a=5}} | {{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 
, 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 | ||
