Problem 4

Prove the following equality for any natural number

n

and any real number

x

for which no denominator vanishes (that is,

x\ne \frac{k\pi}{2^t}

for

t=0,1,\dots,n

and integer

k

\frac{1}{\sin 2x}+\frac{1}{\sin 4x}+\frac{1}{\sin 8x}+\cdots+\frac{1}{\sin 2^nx} = \cot x - \cot 2^nx.

Step 1 of 5: Prove the key lemma

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In plain words

This single identity, applied with $y=2^{k-1}x$ , produces the telescoping term for every summand in the problem.

\cot y - \cot 2y = \frac{1}{\sin 2y}

Detailed analysis

Write $\cot y-\cot 2y=\frac{\cos y}{\sin y}-\frac{\cos 2y}{\sin 2y}=\frac{\cos y}{\sin y}-\frac{2\cos^2y-1}{2\sin y\cos y}=\frac{2\cos^2y-(2\cos^2y-1)}{2\sin y\cos y}=\frac{1}{2\sin y\cos y}=\frac{1}{\sin 2y}$ , using $\cos 2y=2\cos^2y-1$ and $\sin 2y=2\sin y\cos y$ .