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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</math> and get <math>\cot x - \cot 2x = \frac {1}{\sin 2x
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