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I am trying to simplify this:

$$y = \cot^{-1}(\tan\theta), \ \ \ \theta = \sin^{-1}(x^n)$$

What I did was find $\tan \theta$ by representing arcsin as arctan:

$$\tan\theta = \tan\left(\tan^{-1}\left(\dfrac{x^n}{\sqrt{1-x^{2n}}}\right)\right) \\ = \dfrac{x^n}{\sqrt{1-x^{2n}}}$$

Till here it seems valid, but then I thought of substituting $x^n$ with $\cos\alpha$, though I think this may be wrong, since $\cos \alpha \in [-1,1]$, but $x^n$ is not bounded.

Anyways, I did it, and finally I got $$\tan \theta = \cot \alpha$$

So the final function looked like: $$y = x^n$$

which is obviously wrong. I need to find where I went wrong, and how to do substitutions for simplifications! Thanks!

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As $-\dfrac\pi2\le\theta\le\dfrac\pi2$

and $\arcsin u+\arccos u=\dfrac\pi2$ for $|u|\le1$

and $\arctan v+\text{arccot}v=\dfrac\pi2$ for real $v,$

$y=\dfrac\pi2-\arctan(\tan\theta)=\dfrac\pi2-\theta=\cdots=\arccos(x^n)$

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  • $\begingroup$ Oh I got it! Your approach is much simple and correct! Btw, can you tell where I went wrong, and why my answer was coming so wrong? Also, can you recommend some resource, online or offline, about inverse trigonometry? Thanks a lot. $\endgroup$ – Max Payne Mar 23 '16 at 15:50
  • $\begingroup$ @Tim, By your way: $$y=\cot^{-1}\dfrac{x^n}{\sqrt{1-x^{2n}}}$$ Now $$\arccos u=\cot^{-1}\dfrac u{\sqrt{1-u^2}}$$ $\endgroup$ – lab bhattacharjee Mar 23 '16 at 15:53
  • $\begingroup$ Yes was indeed foolish to assume $x^n$ to be $\cos \alpha$. Using u seems reasonable! Thanks a lot! $\endgroup$ – Max Payne Mar 23 '16 at 15:56

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