We can evaluate $\cos(\arcsin x)$ with simple geometric intuition. Write $y = \cos(\arcsin x)$. Letting $\theta = \arcsin x$, $y = \cos \theta$ corresponds to a right triangle with angle $\theta$, adjacent side length $y$, and hypotenuse $1$. Pythagoras gives that the opposite side has length $\sqrt{1-y^2}$. Then $\sin \theta = \sqrt{1-y^2}/1 = \sqrt{1-y^2}$. Now $\sin \theta = \sin(\arcsin x) = x$ with appropriate bounds, so $x = \sqrt{1 - y^2}$ with appropriate bounds, so $y = \cos (\arcsin x) = \sqrt{1 - x^2}$ with appropriate bounds.
It can be shown that $\cosh (\mathrm{arcsinh} \ x) = \sqrt{1+x^2}$. Can we derive this using a similar intuitive geometric construction, perhaps on a hyperbola? If so, how?
Solved it myself. We can mirror the argument above, but using the hyperbolic identity $\cosh^2 - \sinh^2 = 1$ in place of Pythagoras.
Write $y = \cosh(\mathrm{arcsinh}\ x) = \cosh(\theta)$. Then $y = \cosh \theta, x = \sinh \theta$, so $y^2 - x^2 = 1 \implies y = \sqrt{1 + x^2}$.