When I sketch sin(x), arcsin(x) and arcsin(arcsin(x)) on Geogebra, there is a slight difference between the second and third graphs. I thought that perhaps because you're doing an 'inverse of an inverse' that you might get back to the original function but there is something different going on. Can anybody please explain why the gradients are slightly different (endpoints are the same) and how I can show this algebraically?
$\endgroup$ 42 Answers
$\begingroup$You could have them compute the Taylor series as far as they want. Alpha gives$$x + x^3/3 + (7 x^5)/30 + (64 x^7)/315 + (4477 x^9)/22680 + (28447 x^{11})/138600 + O(x^{12})$$ I think the intuitive terms are the first two. We know $\arcsin x \approx x$ and the correction is $+\frac {x^3}6$, so taking two arcsins should give twice the correction.
$\endgroup$ 2 $\begingroup$Of course $\sin^{-1} x$ and $\sin^{-1} (\sin^{-1} x)$ have different slopes: they are different functions!
You know the latter goes through $(-\pi/4, -\pi/2)$, $(0,0)$ and $(\pi/4, \pi/2)$ and is monotonically increasing and antisymmetric. That gets you close.
But why not just plot it?
