Difference between revisions of "1966 IMO Problems/Problem 4"
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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> | ||
| − | LHS=\frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x} | + | LHS=<math>\frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x}</math> |
| − | = \frac{2\cos^2 x}{2\cos x \sin x}-\frac{2\cos^2 x -1}{\sin 2x} | + | <math>= \frac{2\cos^2 x}{2\cos x \sin x}-\frac{2\cos^2 x -1}{\sin 2x}</math> |
| − | =\frac{2\cos^2 x}{\sin 2x}-\frac{2\cos^2 x -1}{\sin 2x} | + | <math>=\frac{2\cos^2 x}{\sin 2x}-\frac{2\cos^2 x -1}{\sin 2x}</math> |
| − | =\frac {1}{\sin 2x} | + | <math>=\frac {1}{\sin 2x}</math> |
Using the above formula, we can rewrite the original series as | 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 | + | <math>\cot x - \cot 2x + \cot 2x - \cot 4x + \cot 4x \cdot \cdot \cdot + \cot 2^{n-1} x - \cot 2^n x </math> |
| − | Which gives us the desired answer of \cot x - \cot 2^n x | + | Which gives us the desired answer of <math>\cot x - \cot 2^n x</math> |
Revision as of 16:08, 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=
Using the above formula, we can rewrite the original series as
Which gives us the desired answer of 
