# How do you prove Arc cosh(x) = ln(x + (x^2 - 1)^(1/2))?

May 4, 2016

see below

#### Explanation:

Let $y = {\cosh}^{-} 1 x$ then by definition

$x = \cosh y = \frac{{e}^{y} + {e}^{-} y}{2}$

$2 x = {e}^{y} + {e}^{-} y$

${e}^{y} - 2 x + {e}^{-} y = 0$

${e}^{y} - 2 x + \frac{1}{e} ^ y = 0$

${e}^{2 y} - 2 x {e}^{y} + 1 = 0$

Let $u = {e}^{y}$ then we have

${u}^{2} - 2 x u + 1 = 0$--> Now use quadratic formula to solve

$u = \frac{2 x \pm \sqrt{4 {x}^{2} - 4}}{2}$

${e}^{y} = \frac{2 x \pm \sqrt{4 {x}^{2} - 4}}{2}$

${e}^{y} = \frac{2 x \pm \sqrt{4 \left({x}^{2} - 1\right)}}{2}$

$2 {e}^{y} = 2 x \pm 2 \sqrt{{x}^{2} - 1}$

${e}^{y} = x \pm \sqrt{{x}^{2} - 1}$

$\ln {e}^{y} = \ln \left(x \pm \sqrt{{x}^{2} - 1}\right)$

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

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