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.
 A: 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 $k>0$.
A: 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.
