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I am trying to find $\cosh^{-1}1$ I end up with something that looks like $e^y+e^{-y}=2x$. I followed the formula correctly so I believe that is correct up to this point. I then plug in $1$ for $x$ and I get $e^y+e^{-y}=2$ which, according to my mathematical knowledge, is still correct. From here I have absolutely no idea what to do as anything I do gives me an incredibly complicated problem or the wrong answer.

  • 1
    @Hans, Nice try.2011-10-09

5 Answers 5

13

start with

$\cosh(y)=x$

since

$\cosh^2(y)-\sinh^2(y)=1$ or $x^2-\sinh^2(y)=1$

then

$\sinh(y)=\sqrt{x^2-1}$

now add $\cosh(y)=x$ to both sides to make

$\sinh(y)+\cosh(y) = \sqrt{x^2-1} + x $

which the left hand side simplifies to : $\exp(y)$

so the answer is $y=\ln\left(\sqrt{x^2-1}+x\right)$

2

You have found out that the unknown $y$ satisfies the equation $e^y+e^{-y}=2$. Multiply by $e^y$ and rearrange terms. You then get $e^{2y}-2e^y+1=0\ .$ Now use the following trick: Put $e^y=:u$ with a new unknown $u$. This $u$ has to satisfy the quadratic equation $u^2-2u+1=0\ ,\quad{\rm i.e.,}\quad (u-1)^2=0\ .$ The last equation has the unique solution $u=1$. The corresponding $y$ therefore satisfies the equation $e^y=1$, and there is only one such real $y$, namely $y=0$.

All in all we have shown that $\cosh^{-1}(1)=0$, which is corroborated by the fact that conversely $\cosh(0)={1\over2}(e^0+e^{-0})=1$.

2

To find the inverse for any $x$, we are looking for $ y = \cosh^{-1} x, $ i.e. $x = \cosh y = \frac{1}{2} (e^y + e^{-y}). $ Multiplying through by $2e^y$ gives $ (e^y)^2 -2x\,e^y + 1 = 0, $ which is a quadratic in $e^y$. You can then use the quadratic formula, or here completing the square $ (e^y-x)^2 - x^2 + 1 = 0, $ $ e^y -x = \sqrt{x^2 - 1}, $ $ y = \ln \left(x+\sqrt{x^2-1}\right). $ At the start I said "for any $x$", but observe that the result is only valid for $x\ge 1$. This is the inverse for the right-hand side of the (even) function $y=\cosh x$, with $x\ge0$ and $y\ge1$.

0

It may be more helpful to consider the significant hyperbolic identities first. We have in general:

$\small \begin{array} {rcllll} 1)& \exp(z) &=& \cosh(z) + \sinh(z) \\ 2)& 1 &=& \cosh(z)^2 - \sinh(z)^2 \\ &&& \implies \\ 3)&\sinh(z) &=& \pm \sqrt{\cosh(z)^2-1} & \text{ using 2)}\\ 4)& \exp(z)&=& \cosh(z) \pm \sqrt{\cosh(z)^2-1} & \text{ using 1) and 3)}\\ \end{array} $

Now the given problem is to find another expression for $\small y=\cosh^{-1}(x)$ which means $\small x = \cosh(y) $
We use 4) and insert our current y for the general z to get

$\small \begin{array} {rcllll} 5)& \exp(y)&=& \cosh(y) \pm \sqrt{\cosh(y)^2-1} & \text{ using 4)}\\ 6)& \exp(y)&=& x \pm \sqrt{x^2-1} & \text{ inserting x for } \cosh(y)\\ 7)& y&=& \log(x \pm \sqrt{x^2-1} ) & \\ 8)& \cosh^{-1}(x)&=& \log(x \pm \sqrt{x^2-1} ) &\text{ inserting } \cosh^{-1}(x) \text{ for } y \\ 9)& \cosh^{-1}(1)&=& ??? \\ \end{array} $

Now 8) can be used as a new, general hyperbolic identity like that in the list from 1) to 4) and 9) is your remaining little to-do ...

  • 0
    see htt$p$://en.wikipedia.org/wiki/Exponential_function2011-10-08
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$ e^y+e^{-y}=2 $ Letting $u = e^y$, this becomes $ u + \frac 1u = 2 $ Multiplying both sides by $u$: $ u^2 + 1 = 2u $ That's just a quadratic equation.