Difference between revisions of "1966 IMO Problems/Problem 4"
(One intermediate revision by one other user not shown) | |||
Line 1: | Line 1: | ||
+ | == 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 == | ||
Line 18: | Line 22: | ||
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 == | ||
+ | {{IMO box|year=1966|num-b=3|num-a=5}} |
Latest revision as of 12:53, 29 January 2021
Problem
Prove that for every natural number , and for every real number (; any integer)
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
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 |