How might you normalize the "skewed sine wave" function: $y=\arctan\frac{t \sin{x}}{1-t\cos{x}}$?

Question

While trying to find a formula that generalizes sine waves with sawtooth waves by producing skewed or asymmetrical waves, I found the question Equation of a "tilted" sine, and gareth-mccaughan's answer https://math.stackexchange.com/a/2430837/1613058:

$$ \frac{1}{t}\tan^{-1}\frac{t\sin{x}}{1-t\cos{x}} $$

I've graphed this expression for various values of $t$: Graph of the function family. It produces left-facing sawtooth and right-facing sawtooth waves at $t=-1$ and $t=1$ and a pure sine wave in the limit $t\rightarrow0$. For this function to be useful to me, however, it would need to have a constant amplitude for all values of $t$. It is already partially normalized: the factor $1/t$ out front prevents the amplitude from going to zero as $t$ goes to zero, but as it stands the sine waves ($t\rightarrow0$) have amplitude $A=1$ but the sawtooth waves ($t=\pm1$) have amplitude $A=\frac{\pi}{2}$.

Numerically, it appears that while the amplitude of the function approaches $\pi/2$ as $t\rightarrow1$, the derivative of the amplitude w.r.t. $t$, $dA/dt$, diverges as $t\rightarrow1$. Graph of amplitude as a function of t. This led me to try using a power law with exponent between 0 and 1 as a normalization, . However when I actually try that, it typically improves the normalization but still diverges at $t\rightarrow1$.

Is there a way to normalize the above function so that its amplitude $A=1$ for $-1<t<1$? Or, how might I go about finding such a normalization?

Answer 1

A normalization can be found by dividing by the maximum. Setting the derivative to $0$:

$$\frac d{dx} \tan^{-1}\left(\frac{t\sin{x}}{1-t\cos{x}}\right)=\frac{t(\cos(x)-t)}{t^2-2t\cos(x)+1}=0\implies \cos(x)=t,\sqrt{1-t^2}=\sin(x)$$

of which the locations match graphically. Substituting:

$$\max\left(\tan^{-1}\left(\frac{t\sin{x}}{1-t\cos{x}}\right)\right)=\tan^{-1}\left(\frac t{\sqrt{1-t^2}}\right)=\sin^{-1}(t)$$

where one can use a right angle triangle with leg lengths of $1$ and $t$ for the last equality. Finally, the normalization is:

$$y=\frac1{\sin^{-1}(t)}\tan^{-1}\left(\frac{t\sin{x}}{1-t\cos{x}}\right)$$

which has amplitude $1$ as graphed here: