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) )

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- Jun 21st 2006, 05:37 AMyakkowCan somebody prove - Hyperbolic functions
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) ) - Jun 21st 2006, 05:53 AMThePerfectHackerQuote:

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,

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