How Does Arcsinh Relate to Arccosh in Trigonometric Transformations?

[mathmari]

Hey!

We have that $s=\frac{\sqrt{1-y^2}}{y}$ and $x=\text{arcsinh} (s)=\text{arcsinh} \frac{\sqrt{1-y^2}}{y}$. How can we simplify this?
According to my book it is $x=\text{arccosh} \frac{1}{y}$. How do we get that? (Wondering)

 

[I like Serena]

Hey!

We have that $s=\frac{\sqrt{1-y^2}}{y}$ and $x=\text{arcsinh} (s)=\text{arcsinh} \frac{\sqrt{1-y^2}}{y}$. How can we simplify this?
According to my book it is $x=\text{arccosh} \frac{1}{y}$. How do we get that? (Wondering)

Hey mathmari! (Smile)

We have:
$$\sinh x = s = \frac{\sqrt{1-y^2}}{y}$$
which is bijective.

Furthermore, we have:
$$\cosh^2 x - \sinh^2 x = 1$$

It follows that:
$$\cosh x= \sqrt{1+\sinh^2 x} = \frac{1}{|y|}$$
(Note that $\cosh$ is always positive.)

So that:
$$x = \pm\arcosh \frac{1}{|y|}$$

Checking the cases where $y$ is either positive or negative, we find:
$$x = \operatorname{sgn}(y)\arcosh \frac{1}{|y|}$$
(Mmm)

 

[mathmari]

I see... (Nerd) We have the curve $$\left (\arsinh s-\frac{s}{\sqrt{s^2-1}}, \frac{1}{\sqrt{1+s^2}}\right ), s>0$$

If $(x,y)$ is a point of the curve then $$x=\arsinh s-\frac{s}{\sqrt{s^2-1}} \ \ \text{ and } \ \ y=\frac{1}{\sqrt{1+s^2}}.$$

We have that $$y=\frac{1}{\sqrt{1+s^2}} \Rightarrow \sqrt{1+s^2}=\frac{1}{y} \Rightarrow 1+s^2=\frac{1}{y^2} \Rightarrow s^2=\frac{1}{y^2}-1 \Rightarrow s=\pm \sqrt{\frac{1}{y^2}-1} \Rightarrow s=\pm \frac{\sqrt{1-y^2}}{|y|}$$

We set: $$w=\arsinh s \Rightarrow \sinh w = s = \pm \frac{\sqrt{1-y^2}}{|y|}$$

We have $$\cosh^2 w - \sinh^2 w = 1 \Rightarrow \cosh w= \sqrt{1+\sinh^2 w} = \frac{1}{|y|} \Rightarrow w = \pm\arcosh \frac{1}{|y|}$$

So we get $$x=\pm\arcosh \frac{1}{|y|}-\left (\pm \frac{\sqrt{1-y^2}}{|y|}\right )y$$

How do we get rid of the $\pm$ and the absolute value, since the result should be $$x=\arcosh \frac{1}{y}- \sqrt{1-y^2}$$ ? (Wondering)

 

[mathmari]

Since we have taken $$y=\frac{1}{\sqrt{1+s^2}}$$ do we maybe assume that $y>0$ ? (Wondering)

 

[I like Serena]

Since we have taken $$y=\frac{1}{\sqrt{1+s^2}}$$ do we maybe assume that $y>0$ ? (Wondering)

Yep. (Nod)
Although rather than "assuming", we're "deducing". (Nerd)
Furthermore, in the problem statement it is given that $s>0$, so we can immediately pick a $+$ sign for $s$ as well.

 

[mathmari]

Yep. (Nod)
Although rather than "assuming", we're "deducing". (Nerd)
Furthermore, in the problem statement it is given that $s>0$, so we can immediately pick a $+$ sign for $s$ as well.

Ok... Thank you very much! (Yes)

 

[Greg]

Hey!

We have that $s=\frac{\sqrt{1-y^2}}{y}$ and $x=\text{arcsinh} (s)=\text{arcsinh} \frac{\sqrt{1-y^2}}{y}$. How can we simplify this?
According to my book it is $x=\text{arccosh} \frac{1}{y}$. How do we get that? (Wondering)

Note that \(\displaystyle \cosh^2(x)-1=\sinh^2(x),\cosh^2(x)=\sinh^2(x)+1\) and \(\displaystyle \cosh(x)+\sinh(x)=e^x\).

Then \(\displaystyle \arsinh(x)=\ln\left(x+\sqrt{x^2+1}\right)\) and \(\displaystyle \arcosh(x)=\ln\left(x+\sqrt{x^2-1}\right)\).

\(\displaystyle \arsinh\left(\frac{\sqrt{1-y^2}}{y}\right)=\ln\left(\frac{\sqrt{1-y^2}}{y}+\sqrt{\frac{1-y^2}{y^2}+1}\right)=\ln\left(\sqrt{\frac{1}{y^2}-1}+\frac{1}{|y|}\right)=\arcosh\frac{1}{|y|}\)