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 4 of 5: Confirm no term is undefined

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In plain words

The domain restriction stated in the problem is not an extra hypothesis to prove — it is precisely the condition needed for every division used in the proof to be legal.

x\ne \frac{k\pi}{2^t}\ (t=0,1,\dots,n,\ k\in\mathbb{Z})

Detailed analysis

Every step used $\sin(2^tx)\ne 0$ for $t=1,\dots,n$ (to divide by $\sin 2^kx$ ) and for $t=0,\dots,n-1$ (so the cotangents are defined). The restriction on $x$ given in the problem excludes exactly the values where some $2^tx$ is a multiple of $\pi$ , which is precisely what makes every denominator above nonzero.