how to prove this equality

Question

There are two equalities,

$\sinh({\cosh }^{-1}x)=\sqrt { {x }^2-1 }\quad (x>1)$

$\cosh({\sinh}^{-1}y)=\sqrt{1+{y}^2}$

prove this equality please.. how to prove it? i cannot try it..

also,

$x={\cosh}^{-1}y=\ln(y+\sqrt{y^2-1})$

is it true?

Answer 1

$1:$ Let $\cosh ^{-1}x=z\implies \cosh z= x$

As $\cosh^2z-\sinh^2z=1,$

$\sinh z=\sqrt{\cosh^2z-1}\implies \sinh (\cosh ^{-1}x)=\sqrt{x^2-1}$

$2:$ Similarly, if $\sinh ^{-1}x=w, \sinh w=x$

$\cosh w=\sqrt{1+\sinh^2w}\implies\cosh (\sinh ^{-1}x)=\sqrt{1+x^2}$

$3: x=\cosh^{-1}y\implies y=\cosh x=\frac{e^x+e^{-x}}2$

On simplification, $(e^x)^2-2y\cdot e^x+1=0 $ $\implies e^x=\frac{2y\pm\sqrt{(2y)^2-4}}2=y\pm\sqrt{y^2-1}$

As the principal value $\cosh^{-1}y$ lies in $\in[0,\infty)$

$\implies x\ge0, e^x\ge1$

but $y-\sqrt{y^2-1}=\frac1{y+\sqrt{y^2-1}}<1$

So, $e^x=y+\sqrt{y^2-1}\implies x=\ln(y+\sqrt{y^2-1})$

References:

1 2

Answer 2

Hint: do the change of variable $t=\cosh x$ (where $t > 0$), observing that $t> 0 \mapsto \cosh t \in \mathbb R_+^\ast$ is a bijection; for the second one, use $u = \sinh y$ ($\sinh$ being a bijection from $\mathbb R$ onto itself); then, use the hyperbolic trigonometry equalities.

Regarding your last question, yes — see this link (the expression of $\cosh^{-1}$ as $\ln(\dots)$ can only hold for $y \geq 1$, otherwise the argument of the square root is negative).