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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.

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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 7 '11 at 1:51
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$e^{2y}-2e^y+1=0$ I think. –  user138246 Oct 7 '11 at 1:52
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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 7 '11 at 1:59
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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 7 '11 at 5:53
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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 8 '11 at 6:46

5 Answers 5

up vote 10 down vote accepted

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

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I don't follow at all what happened. I am assuming you are using hyperbolic identities which I have not memorized. –  user138246 Oct 7 '11 at 2:56
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And you are looking up the definition and identities for hyperbolic functions right now .. correct? –  ja72 Oct 7 '11 at 11:32
    
I have them written down on an index card. –  user138246 Oct 7 '11 at 21:18

Let, $cosh^{-1}(y)=x\implies cosh(x)=y\implies e^x+e^{-x}=2y$, Let $t=e^x$, therefore $t^2-2yt+1=0$ therefore solution of $t=y+\sqrt{y^2-1}$ or $y-\sqrt{y^2-1}$

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

since $y=1\implies x=0$

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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 ...

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What does exp in this stand for? –  user138246 Oct 7 '11 at 21:19
    

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

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(+1) In my view: very, very nice! –  Gottfried Helms Oct 7 '11 at 15:46

$$ e^y+e^{-y}=2 $$ 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.

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