Finding period of $\sin\left(2\left(\tan^{-1}\left(\frac{\tan(x)}k\right)\right)\right)$

Could anyone kindly help me evaluate the period of the following function:

$$f(x)=\sin\left(2\left(\tan^{-1}\left(\frac{\tan(x)}k\right)\right)\right)$$

where $k$ is some positive real constant.

Had it been only $\sin2(\tan^{-1}(\tan(x)))$, I could have found its period but this constant $k$ is proving to be a challenge for me.

I guess, for certain values of $k$, f may not be periodic at all, can we find under what condition this function will be periodic?

Thanks for your time.

PS: I am in electrical engineering and I used to find the period of function ten years back, but now I have really forgotten the concepts and formulas and I don't have any book at my disposal either.

• It is of the form $f(x)=g(\tan(x))$. Since $\tan$ is periodic $\pi$, so is $f$. – Simply Beautiful Art Aug 22 '17 at 23:50
• @SimplyBeautifulArt $0$ is of the form $g(\tan x)$. It has no (fundamental) period at all, though. – Cauchy Aug 23 '17 at 0:29
• @Cauchy Well, its no answer, but its certainly close – Simply Beautiful Art Aug 23 '17 at 0:33
• @SimplyBeautifulArt: Thanks for editing the question and for your comment. But, assuming a fundamental period of $\pi$ somehow doesn't support analysis result of my circuit that I am getting from my simulator. And why the period will not be dependent on $k$. What if $k=\pi$, for example? – MoM Aug 23 '17 at 0:39

The period of sin is 2π. Therefore $$\tan^{-1}\left(\frac{\tan(x)}k\right)$$ will need to change in increments of π - {0,π,0,π}; $$\left(\frac{\tan(x)}k\right)$$ will be {tan(0), tan(π)} (both 0). tan(x)/k is only 0 when x is a multiple of π. Thus the period of $$f(x)=\sin\left(2\left(\tan^{-1}\left(\frac{\tan(x)}k\right)\right)\right)$$ is π.
As written the function $$f(x):=\sin\left(2\arctan{\tan x\over k}\right)$$ is undefined at odd multiples of ${\pi\over2}$. But it turns out that these singularities are removable, and that $f$ is real analytic on all of ${\mathbb R}$ when $k>0$. Put $$\arctan{\tan x\over k}=:\alpha\in\left]{-{\pi\over2}},{\pi\over2}\right[\ .$$ Then $${\tan x\over k}=\tan\alpha$$ whenever $\tan x$ is defined, and we 0btain $$f(x)=\sin(2\alpha)={2\tan\alpha\over1+\tan^2\alpha}={2k\tan x\over k^2+\tan^2 x}={k\sin(2x)\over k^2\cos^2 x+\sin^2 x}\ ,$$ whereby now the RHS makes sense for all $x\in{\mathbb R}$. This RHS is of period $\pi$ by inspection, and it is easy to see that no number of the form ${\pi\over n}$, $n\geq2$, can be a period, whatever the value of $k>0$.