1966 IMO Problems/Problem 4: Difference between revisions
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== Solution == | == 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< | 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 08:11, 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
\cot x - \cot 2x = \frac {1}{\sin 2x}
