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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.

J. M. ain't a mathematician
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    Multiply both sides by $e^y$, then subtract the right-hand side from the left-hand side. What does that give you? – anon Oct 07 '11 at 01:51
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    $e^{2y}-2e^y+1=0$ I think. –  Oct 07 '11 at 01:52
  • Why do you set $x$ to $1$? Are you looking for $\cosh^{-1}(1)$ in particular? – hmakholm left over Monica Oct 07 '11 at 01:52
  • Because it is the inverse of cosh1. –  Oct 07 '11 at 01:53
  • $\cosh(1)$ is just a number, not a function. It doesn't have an inverse (in the sense relevant here). – hmakholm left over Monica Oct 07 '11 at 01:54
  • How am I supposed to know that? –  Oct 07 '11 at 01:56
  • Sorry, it was yunone's edit that made the argument disappear and confused me. – hmakholm left over Monica Oct 07 '11 at 01:59
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    Look up what I did in detail for the inverse $\sinh$. Almost everything will be the same except for one change of sign, and the fact that $x=1$. – André Nicolas Oct 07 '11 at 01:59
  • The proper way of saying it is that you want to evaluate $\cosh^{-1}(x)$ at $x=1$; in any event, you now have a quadratic you can solve after your last move... – J. M. ain't a mathematician Oct 07 '11 at 02:01
  • So I am suppose to find the inverse of this? Because I am getting $e^{2y}-2e^y+1=0$ –  Oct 07 '11 at 02:02
  • You're correct, you get the same thing as $(e^y)^2-2e^y+1=0$. This is a quadratic equation in $e^y$ (that is, we have $e^y$ squared plus a multiple of it plus a constant equals zero, as all quadratic equations go). You can factor this as $(e^y-1)^2=0$, or use the quadratic formula so find $e^y=1$. If you take the natural logarithm of both sides, you get $y=\ln1=0$. – anon Oct 07 '11 at 02:03
  • Wow, I had no idea that factored down into a square. Was there any easy way of knowing that? Also how does that help me get an answer? I am trying to make this fit into the quadractic formula but I am not sure what sort of numbers it will accept, I don't know how exact the form has to be to make it work. –  Oct 07 '11 at 02:06
  • @Jordan: Well, you do have $(e^y)^2=e^{2y}$, through the laws of exponents... – J. M. ain't a mathematician Oct 07 '11 at 02:10
  • I forgot those rules. I can't even make the quadratic formula work, I keep getting math error. –  Oct 07 '11 at 02:13
  • For the quadratic formula to work you should plug in $a=1,b=-2,c=1$. Is that what you're doing? The easy way of seeing that it can be factored is that I have things like $u^2-2u+1=(u-1)^2$ memorized, and I recognized it was the same form only with $e^y$ instead of a $u$. – anon Oct 07 '11 at 02:21
  • I was using e in the quadratic formula. –  Oct 07 '11 at 02:22
  • The equation is $1(e^y)^2-2(e^y)+1=0$, so $e$ is not any of the coefficients. – anon Oct 07 '11 at 02:28
  • Oh I didn't know that is how the quadratic formula worked. I don't know how I will remember that. –  Oct 07 '11 at 02:32
  • I am getting 2+/-0/2a Is that right? –  Oct 07 '11 at 02:40
  • Remember $a=1$ so that's $(2\pm0)/2$, which is just $1$. But yes, that's right. – anon Oct 07 '11 at 02:48
  • Oh, I didn't see that for some reason. Quadratic formula goes over my head. –  Oct 07 '11 at 02:50
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    @Jordan: If even such a basic thing as the quadratic formula goes over your head, why on earth are you tormenting yourself by trying to learn calculus? You're just wasting your time. You would be better off spending your time on something that you're good at... (If you really really need to learn calculus, then – as many people have already pointed out in comments to your previous questions – you should first learn the pre-calculus stuff reasonably well.) – Hans Lundmark Oct 07 '11 at 05:53
  • @Hans Thanks, but I know I am stupid. –  Oct 07 '11 at 21:18
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    @Jordan: Please don't think of yourself as stupid just because you're struggling with math. At the university where I work, I meet lots of students that are obviously intelligent and ambitious, but still have a hard time with the math courses. In many cases, I think it can be blamed on their being exposed to years of bad teaching in elementary school. When I see the types of errors and misunderstandings that you display here, I don't think "That's one stupid fellow!", but rather "How is it possible that no math teacher has explained this and set it straight years ago?". – Hans Lundmark Oct 08 '11 at 06:46
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    (Cont.) That said, of course we're different. To some, math comes easy, while others have to work ten times harder. No matter what's the case with you, I think it's clear that it will be very hard for you to pass calculus unless you first get a grip on the basic stuff. – Hans Lundmark Oct 08 '11 at 06:50
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    @Hans, Nice try. – Did Oct 09 '11 at 09:48

5 Answers5

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

John Alexiou
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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 $y\ge1$.

APW
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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 $\cosh(0)={1\over2}(e^0+e^{-0})=1$.

Christian Blatter
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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 ...

Gottfried Helms
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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.

Michael Hardy
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