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 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 n=1 and get \cot x - \cot 2x = \frac {1}{\sin 2x} | ||
First, we prove \cot x - \cot 2x = \frac {1}{\sin 2x} | First, we prove \cot x - \cot 2x = \frac {1}{\sin 2x} | ||
Revision as of 14:11, 26 September 2010
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Solution
Assume that 
is true, then we use n=1 and get \cot x - \cot 2x = \frac {1}{\sin 2x}
First, we prove \cot x - \cot 2x = \frac {1}{\sin 2x}
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
