# How do you prove sinh^-1 (t) = ln(t+ sqrt(t^2 + 1))?

May 4, 2016

see below

#### Explanation:

Let $y = {\sinh}^{-} 1 t$ then by definition

$t = \sinh y = \frac{{e}^{y} - {e}^{-} y}{2}$

$2 t = {e}^{y} - {e}^{-} y$

${e}^{y} - 2 t - {e}^{-} y = 0$

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

${e}^{2 y} - 2 t {e}^{y} - 1 = 0$

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

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

$x = \frac{2 t \pm \sqrt{4 {t}^{2} + 4}}{2}$

${e}^{y} = \frac{2 t \pm \sqrt{4 {t}^{2} + 4}}{2}$

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

$2 {e}^{y} = 2 t \pm 2 \sqrt{{t}^{2} + 1}$

${e}^{y} = t \pm \sqrt{{t}^{2} + 1}$

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

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

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

$=$Right side