3
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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 $y^2 - x^2 = 1 \implies y = \sqrt{1 + x^2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.