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

Question

2155 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-04T21:32:49

see below

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

$t=sinh y =(e^y-e^-y)/2$

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

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

$e^y-2t-1/e^y=0$

$e^(2y)-2te^y-1=0$

Let $x=e^y$ then we have

$x^2-2tx-1=0$ --> Now use quadratic formula to solve

$x=(2t+-sqrt(4t^2+4))/2$

$e^y = (2t+-sqrt(4t^2+4))/2$

$e^y = (2t+-sqrt(4(t^2+1)))/2$

$2e^y = 2t+-2sqrt(t^2+1)$

$e^y=t+-sqrt(t^2+1)$

$ln e^y = ln (t+-sqrt(t^2+1))$

$y=ln (t+sqrt(t^2+1))$

$sinh ^-1 t = ln (t+sqrt(t^2+1))$

$=$ Right side

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