Proving the identity $\tan x+2\tan2x+4\tan4x+8\cot8x=\cot x$ by considering a more general form

Question

I came across this exercise:

Prove that $$\tan x+2\tan2x+4\tan4x+8\cot8x=\cot x$$

Proving this seems tedious but doable, I think, by exploiting double angle identities several times, and presumably several terms on the left hand side would vanish or otherwise reduce to $\cot x$.

I started to wonder if the pattern holds, and several plots for the first few powers of $2$ seem to suggest so. I thought perhaps it would be easier to prove the more general statement:

For $n\in\{0,1,2,3,\ldots\}$, prove that $$2^{n+1}\cot(2^{n+1}x)+\sum_{k=0}^n2^k\tan(2^kx)=\cot x$$

Presented this way, a proof by induction seems to be the smart way to do it.

Base case: Trivial, we have

$$\tan x+2\cot2x=\frac{\sin x}{\cos x}+\frac{2\cos2x}{\sin2x}=\frac{\cos^2x}{\sin x\cos x}=\cot x$$

Induction hypothesis: Assume that

$$2^{N+1}\cot(2^{N+1}x)+\sum_{k=0}^N2^k\tan(2^kx)=\cot x$$

Inductive step: For $n=N+1$, we have

$$\begin{align*} 2^{N+2}\cot(2^{N+2}x)+\sum_{k=0}^{N+1}2^k\tan(2^kx)&=2^{N+2}\cot(2^{N+2}x)+2^{N+1}\tan(2^{N+1}x)+\sum_{k=0}^N2^k\tan(2^kx)\\[1ex] &=2^{N+2}\cot(2^{N+2}x)+2^{N+1}\tan(2^{N+1}x)-2^{N+1}\cot(2^{N+1}x)+\cot x \end{align*}$$

To complete the proof, we need to show

$$2^{N+2}\cot(2^{N+2}x)+2^{N+1}\tan(2^{N+1}x)-2^{N+1}\cot(2^{N+1}x)=0$$

I noticed that if I ignore the common factor of $2^{N+1}$ and make the substitution $y=2^{N+1}x$, this reduces to the base case,

$$2^{N+1}\left(2\cot2y+\tan y-\cot y\right)=0$$

and this appears to complete the proof, and the original statement is true.

First question: Is the substitution a valid step in proving the identity?

Second question: Is there a nifty way to prove the special case for $n=2$?

Answer 1

Defining $t:=\tan y$, $$2\cot 2y+\tan y-\cot y=\frac{1-t^2}{t}+t-\frac{1}{t}=0.$$This will work for any $y$, so I guess I've answered your questions in reverse order (but to both I say yes).

Answer 2

Hint:

$$\cot y-\tan y=\dfrac{\cos^2y-\sin^2y}{\cos y\sin y}=\dfrac{\cos2y}{\dfrac{\sin2y}2}=?$$

Answer 3

Let $\displaystyle S =\sum^{2}_{k=0}2^k\tan(2^k x)+8\cot (8x).$

Then $\displaystyle \int Sdx=-\sum^{2}_{k=0}\ln\bigg(\cos(2^k x\bigg)+\ln(\sin 8x)$

Using $\displaystyle \prod^{n-1}_{r=0}\cos(2^r x)=\frac{1}{2^n}\frac{\sin(2^nx)}{\sin x}.$

$\displaystyle \int Sdx=-\ln\bigg(\frac{1}{2^3}\frac{\sin 8x}{\sin x}\bigg)+\ln(\sin 8x)$

$$\displaystyle \frac{d}{dx}\int Sdx=\frac{d}{dx}\bigg[\ln(\sin x)+\ln(8)\bigg]$$

$$S=\sum^{2}_{k=0}2^k\tan(2^k x)+8\cot (8x)=\cot x$$