1. ## Inverse hyperbolic question

Hi
Basically I got $\displaystyle cosh(arsinh 2)$

How do I instantly know that the answer is $\displaystyle \sqrt{5}$ ?

PS: I actually worked it out myself the long way by letting $\displaystyle arsinh2 = ln(2+\sqrt{5})$. However, I want to know a faster way of doing this.

2. Hi

Simply use $\displaystyle \cosh x =\sqrt{1+\sinh^2x}$.

3. Originally Posted by flyingsquirrel
Hi

Simply use $\displaystyle \cosh x =\sqrt{1+\sinh^2x}$.
THANK YOU! Seriously why didn't I think of that.... I must be so dumb.
Merci (if you are French)

Thanks