1966 IMO Problems/Problem 4

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}

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

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