Trig functions of inverse trig functions identities. I found these identities online.
$$\cos(\arcsin(\omega)) = \sqrt{1-\omega^2}$$
$$\sin(\arccos(\omega)) = \sqrt{1-\omega^2}$$
$$\cos(\arctan(\omega)) = \frac 1 {\sqrt{1+\omega^2}}$$
$$\sin(\arctan(\omega)) = \frac{\omega}{\sqrt{1+\omega^2}}$$
$$\tan(\arccos(\omega)) = \frac {\sqrt{1-\omega^2}}{\omega}$$
$$\tan(\arcsin(\omega)) = \frac{\omega} {\sqrt{1 - \omega^2}}$$
I'm wondering how can these be found or proved
 A: Remember: when you take the arcsine, arccosine, and arctangent of a number, the output is an angle.
For the first one, for example, draw yourself a right triangle with an angle $\theta = \arcsin(\omega)$; i.e. a triangle with an angle $\theta$ whose sine is $\omega$—simply choose lengths for the opposite side and the hypotenuse so that the ratio of these two comes out to be $\omega$ (label these side lengths on your picture). No need to worry that there are many such triangles: one can prove that all right triangles with that ratio of lengths between a leg and the hypotenuse are similar (therefore the ratios between any two sides—in particular, the sine, cosine, and tangent of $\theta$—of any such triangle will be the same).
Once you've drawn such a triangle, use the Pythagorean theorem$^\dagger$ if necessary to deduce the length of the missing side, and then, looking at your picture, simply take the cosine of that angle (divide the length of the side adjacent to $\theta$ by the length of the hypotenuse).

$^\dagger$Indeed, the Pythagorean theorem is the source of the $\sqrt{1 \pm \omega^2}$ in all of these identities, since using it to find the missing side involves solving for one of the letters in $a^2 + b^2 = c^2$ (where $a$, $b$, and $c$ are side lengths of a right triangle).
A: For example we know that if an inverse function exists it is unique.
$$\cos(\omega)=\pm\sqrt{1-\sin^2(\omega)}$$
Putting $\omega \to \arcsin(\omega)$ we have the plus sign (definition of inverse function that you should remember)
$$\cos(\arcsin(\omega))=\sqrt{1-\sin^2(\arcsin(\omega))}=\sqrt{1-\omega^2}$$
Hence being $\sin(\omega)= \sqrt{1-\cos^2(\omega)}$
$$\sin(\arccos(\omega))= \sqrt{1-\cos^2(\arccos(\omega))}=\sqrt{1-\omega^2}$$
and after I suggest also this video on YouTube https://www.youtube.com/watch?v=A3eIhBlLx0Y&ab_channel=IntegralsForYou for the other identities.
