Difference between revisions of "1966 IMO Problems/Problem 4"
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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 == | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <math>=\frac{2\cos^2 \theta}{\sin 2\theta}-\frac{2\cos^2 \theta -1}{\sin 2\theta}</math> | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | 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)
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 |