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 2 of 5: Apply the lemma at each scale

y=2^{k-1}x

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In plain words

Rewriting each fraction as a difference is exactly what makes a telescoping sum possible.

\frac{1}{\sin 2^kx} = \cot\left(2^{k-1}x\right)-\cot\left(2^kx\right),\quad k=1,\dots,n

Detailed analysis

Substitute $y=2^{k-1}x$ into the lemma of step 1. As $k$ ranges over $1,\dots,n$ , this rewrites every term $\frac{1}{\sin 2^kx}$ of the left-hand sum as a difference of two consecutive cotangents.