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 == | ||
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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 == | |||
Revision as of 10:43, 12 April 2011
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
Assume that 
is true, then we use 
and get 
.
First, we prove
LHS=
Using the above formula, we can rewrite the original series as
Which gives us the desired answer of 
