IMO 1966 · Problem 4

Prove that for every natural number \(n\), and for every real number \(x \ne k\pi/2^t\) (\(t = 0, 1, \ldots, n;\ k\) any integer) \[ \frac{1}{\sin 2x}+\frac{1}{\sin 4x}+\cdots+\frac{1}{\sin 2^n x} =\cot x-\cot 2^n x. \]

Official text of the problem. Official problem paper (PDF)

Solution

The hypothesis ensures that \(\sin(2^j x)\ne0\) for every \(j=0,1,\ldots,n\), since otherwise \(x=k\pi/2^j\) for some integer \(k\).

Whenever \(\sin y\) and \(\sin 2y\) are nonzero, the sine subtraction formula gives \[ \begin{aligned} \cot y-\cot 2y &=\frac{\cos y\sin 2y-\sin y\cos 2y}{\sin y\sin 2y}\\ &=\frac{\sin(2y-y)}{\sin y\sin 2y} =\frac{1}{\sin 2y}. \end{aligned} \] Apply this identity with \(y=2^{j-1}x\), for \(j=1,\ldots,n\), and sum. All intermediate cotangent terms cancel: \[ \begin{aligned} \sum_{j=1}^{n}\frac{1}{\sin(2^j x)} &=\sum_{j=1}^{n} \left(\cot(2^{j-1}x)-\cot(2^j x)\right)\\ &=\cot x-\cot(2^n x). \end{aligned} \] This is the required identity. If \(0\) is included among the natural numbers, the case \(n=0\) also holds under the empty-sum convention, since both sides are zero.

The solution was checked against John Scholes' IMO solution archive (Kalva).