Difference between revisions of "1966 IMO Problems/Problem 4"
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== 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
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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