arccosh_증명

Theorem

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

Proof

$\displaystyle y=\cosh^{-1}x$
$\displaystyle \cosh y=x$
$\displaystyle \frac{e^y+e^{-y}}2=x$
$\displaystyle e^y-2x+e^{-y}=0$
$\displaystyle t=e^y$ 치환
$\displaystyle t-2x+\frac1t=0$
$\displaystyle t^2-2xt+1=0$
$\displaystyle t=e^y=x\pm\sqrt{x^2-1}$
여기서 $\displaystyle \pm$ 에서 $\displaystyle +$ 만 남는 이유는 TBW
$\displaystyle e^y>0$ 이므로
$\displaystyle e^y=x+\sqrt{x^2-1}$

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