Basic Hyperbolic Simplification b

• Sep 30th 2010, 11:31 PM
ugkwan
Basic Hyperbolic Simplification b
second question:
$cosh^{-1}2=?$
I tried
$\frac{1}{cosh2}=sech2=\frac{2}{e^2+e^{-2}}$

The answer given is $ln(1+\sqrt{3})$
How did they get to a answer w/ natural log?
• Sep 30th 2010, 11:35 PM
Prove It
Because $\cosh^{-1}{x} = \ln{(x + \sqrt{x^2 - 1})}$.
• Oct 1st 2010, 02:21 AM
HallsofIvy
Unfortunate notation- $f^{-1}(x)$ means the inverse function, not the reciprocal!

Specifically, $cosh^{-1}(2)$ is the value of x such that $cosh(x)= \frac{e^x+ e^{-x}}{2}= 2$. Solve that equation for x: $e^x+ e^{-x}= 4$. Let $y= e^x$ so the equation becomes $y+ y^{-1}= 4$. Multiply both sides by y: $y^2+ 1= 4y$ or $y^2- 4y+ 1= 0$. Solve that equation for y then take the logarithm to find x.
• Oct 1st 2010, 02:23 AM
Prove It
And you should note that

$\cosh^{-1}{2} = \ln{(2 + \sqrt{3})}$, not $\ln{(1 + \sqrt{3})}$.
• Oct 1st 2010, 12:05 PM
ugkwan
Thanks for the clarification on this post and the last post. You guys sure know how to explain things in details better than some professors.