Derivatives of inverse functions, $\cos\bigl(\arcsin(x)\bigr) = 1/\sqrt{1-x^2}$?

Question

I studied inverse derivatives, I had(probably classic question) asked to prove using the law of inverse derivatives that:

$\arcsin'(x) = 1/\sqrt{1-x^2}$

I got into:

$\arcsin'(x) = 1/\cos(\arcsin(x)) $

Because I did everything as I should I set $x = 0.5$ and checked and it looks like:

$\ 1/\sqrt{1-x^2} = 1/\cos(\arcsin(x)) \Rightarrow \sqrt{1-x^2} = \cos(\arcsin(x)) $

What I'm trying to understand is how could I know without checking the raw expressions with some x value that they are equals?, is there a method for transforming from $\cos(\arcsin(x))$ to $\sqrt{1-x^2}$?

Answer 1

See the following picture and use Pythagoras theorem.

Answer 2

Since the restriction of $\sin$ to the interval $\left[-\frac\pi2,\frac\pi2\right]$ is increasing, $\arcsin$ must be increasing too. Therefore, $\bigl(\forall x\in(-1,1)\bigr):\arcsin'(x)\geqslant0$.

Answer 3

By definition, $\arcsin x$ is the one and only angle in $y\in\left[-\frac\pi2,\frac\pi2\right]$ such that $\sin y=x$.

By Pythagoras' theorem, $\cos^2y+\sin^2y=1$, and thus we obtain $$\lvert\cos\arcsin x\rvert=\sqrt{1-x^2}$$

Since $-\frac\pi2\le\arcsin x\le \frac\pi2$, it holds $\cos\arcsin x\ge 0$, thus $$\cos\arcsin x=\sqrt{1-x^2}.$$

One can similarly prove that $\sin\arccos x=\sqrt{1-x^2}$.