1966 IMO Problems/Problem 4
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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