How to plot hairy inverse trigonometric functions?

Question

It is easy to plot functions like $y=\sin^{ -1}(x)$, $y=\cos^{-1}(x)$ and $y=\tan^{-1}(x)$. But how to proceed with the functions like $$y = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$$ And $$y = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$$ Is there any general way to approach these problems?

Answer 1

Curve I

\begin{align*} x &= \tan t \\[5pt] \sin y &= \frac{2\tan t}{1+\tan^2 t} \\[5pt] &= \sin 2t \\[5pt] y &= n\pi+(-1)^{n}2t \\[5pt] y &= \left \{ \begin{array}{ccc} -\pi-2\tan^{-1} x & , & x \le -1 \\ 2\tan^{-1} x & , & -1 < x < 1 \\ \pi-2\tan^{-1} x & , & x \ge 1 \end{array} \right. \end{align*}

Curve II \begin{align*} x &= \tan t \\[5pt] \tan y &= \frac{3\tan t-\tan^3 t}{1-3\tan^2 t} \\[5pt] &= \tan 3t \\[5pt] y &= 3t+n\pi \\[5pt] y &= \left \{ \begin{array}{ccc} 3\tan^{-1} x+\pi & , & x \le -\frac{1}{\sqrt{3}} \\ 3\tan^{-1} x & , & -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \\ 3\tan^{-1} x-\pi & , & x \ge \frac{1}{\sqrt{3}} \end{array} \right. \end{align*}