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

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Answer 1 · 2016-05-04T21:41:01

see below

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

$x=cosh y =(e^y+e^-y)/2$

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

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

$e^y-2x+1/e^y=0$

$e^(2y)-2xe^y+1=0$

Let $u=e^y$ then we have

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

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

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

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

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

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

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

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

$cosh ^-1 x = ln (x+sqrt(x^2-1))$

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