Finding inverse cosh

Question

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.

Answer 1

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

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

Answer 3

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 ...

Answer 4

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

Answer 5

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 $y=1\implies x=0$