Difference between revisions of "1966 IMO Problems/Problem 4"
| Line 1: | Line 1: | ||
| − | + | == Solution == | |
| + | |||
| + | Assume that \frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^{n}x}}=\cot{x}-\cot{2^{n}x is true, then we use n=1</math> and get <math>\cot x - \cot 2x = \frac {1}{\sin 2x | ||
Revision as of 09:10, 24 September 2010
Solution
Assume that \frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^{n}x}}=\cot{x}-\cot{2^{n}x is true, then we use n=1</math> and get <math>\cot x - \cot 2x = \frac {1}{\sin 2x
