1966 IMO Problems/Problem 4: Difference between revisions
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
Assume that <math>\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^{n}x}}=\cot{x}-\cot{2^{n}x}</math> is true, then we use <math>n=1</math> and get \cot x - \cot 2x = \frac {1}{\sin 2x}. | Assume that <math>\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^{n}x}}=\cot{x}-\cot{2^{n}x}</math> is true, then we use <math>n=1</math> and get <math>\cot x - \cot 2x = \frac {1}{\sin 2x}</math>. | ||
First, we prove <math>\cot x - \cot 2x = \frac {1}{\sin 2x}</math> | First, we prove <math>\cot x - \cot 2x = \frac {1}{\sin 2x}</math> | ||
Revision as of 15:08, 27 September 2010
I can't add latex code on this page. Somebody add it for me.
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 
