Could somebody please prove these identities and explain a little bit ? I just can't figure them out.
sinh^-1x = ln(x + sq.rt of(x^2 + 1))
cosh^-1x = ln(x + sq.rt of (x^2 + 1))
tanh^-1x = (1/2)ln ( (1+x)/(1 - x) )
Could somebody please prove these identities and explain a little bit ? I just can't figure them out.
sinh^-1x = ln(x + sq.rt of(x^2 + 1))
cosh^-1x = ln(x + sq.rt of (x^2 + 1))
tanh^-1x = (1/2)ln ( (1+x)/(1 - x) )
I prove just one, you prove the rest.Originally Posted by yakkow
Since,
$\displaystyle y=\sinh x$ is a bijective function it has an inverse.
Thus you interchange the x and y in,
$\displaystyle y=\frac{e^x-e^{-x}}{2}$
To get,
$\displaystyle x=\frac{e^y-e^{-y}}{2}$
Thus,
$\displaystyle 2x=e^y-e^{-y}$
Thus,
$\displaystyle e^{2y}-1=2xe^{y}$
Thus,
$\displaystyle e^{2y}-2xe^y-1=0$
Allow, $\displaystyle t=e^y$ thus,
$\displaystyle t^2-2xt-1=0$
Thus,
$\displaystyle t=\frac{2x\pm\sqrt{4x^2+4}}{2}$
Thus,
$\displaystyle t=\frac{2x\pm 2\sqrt{x^2+1}}{2}$
Which is,
$\displaystyle t=x\pm \sqrt{x^2+1}$
Since, $\displaystyle t>0$ we have,
$\displaystyle t=x+\sqrt{x^2+1}$
Thus,
$\displaystyle e^y=x+\sqrt{x^2+1}$
Thus,
$\displaystyle \sinh^{-1}x=\ln (x+\sqrt{x^2+1})$