Difference between revisions of "1966 IMO Problems/Problem 4"
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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 \cot x - \cot 2x = \frac {1}{\sin 2x}. | ||
− | First, we prove \cot x - \cot 2x = \frac {1}{\sin 2x} | + | First, we prove <math>\cot x - \cot 2x = \frac {1}{\sin 2x}</math> |
LHS=\frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x} | LHS=\frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x} |
Revision as of 15:07, 27 September 2010
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Solution
Assume that is true, then we use
and get \cot x - \cot 2x = \frac {1}{\sin 2x}.
First, we prove
LHS=\frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x}
= \frac{2\cos^2 x}{2\cos x \sin x}-\frac{2\cos^2 x -1}{\sin 2x}
=\frac{2\cos^2 x}{\sin 2x}-\frac{2\cos^2 x -1}{\sin 2x}
=\frac {1}{\sin 2x}
Using the above formula, we can rewrite the original series as
\cot x - \cot 2x + \cot 2x - \cot 4x + \cot 4x \cdot \cdot \cdot + \cot 2^{n-1} x - \cot 2^n x
Which gives us the desired answer of \cot x - \cot 2^n x