How to find cosh(arcsinh(f(x)))?

Question

With the regular trig functions, if I ever end up with something like $\operatorname{trig}_1(\operatorname{arctrig}_2(f(x))$, where $\text{trig}_1$ and $\text{trig}_2$ are two arbitrary trigonometric functions, I can draw a right triangle to find a formula for this that doesn't involve any trigonmetric functions.

How do I find a similar result for hyperbolic functions? For instance, when working a problem recently, I ended up with $\cosh(\operatorname{arcsinh}(3x))$. WolframAlpha told me that it was $\sqrt{1+9x^2}$, but how do I figure that out?

What picture can I draw? I'm not sure of the geometry here. I'm pretty sure that hyperbolic functions are related to hyperbolas the way that trig functions are related to circles, but I don't figure out the trig(arctrig) expressions by looking at circles -- I draw a triangle. Is there something similar I can do with hyperbolic functions?

Answer 1

Hint: Use $\sinh^{-1}(x)=\log(x+\sqrt{1+x^2})$ and use $\cosh(x)=\dfrac{e^{x}+e^{-x}}{2}$

When plugging in $3x$, you should get $\cosh(\sinh^{-1}(3x))=\dfrac{3x+\sqrt{1+9x^2}+\frac{1}{3x+\sqrt{1+9x^2}}}{2}$.

That simplifies to $\dfrac{2\sqrt{9x^2+1}}{2}=\sqrt{9x^2+1}$

Answer 2

The Gudermannian function $\theta = gd(x) = arcsin(tanh\ x)$ transforms between the hyperbolic trigonometric functions and the circular ones. For all $x \in \mathbb{R}$, it gives us a unique $\theta \in (-\pi/2, \pi/2)$ such that:

• $sinh\ x = tan\ \theta$
• $cosh\ x = sec\ \theta$
• $tanh\ x = sin\ \theta$
• $cosech\ x = cot\ \theta$*
• $sech\ x = cos\ \theta$
• $coth\ x = cosech\ \theta$*

*When $x = 0$, $\theta = 0$ meaning $cosech\ x$, $coth\ x$, $cosec\ \theta$ and $cot\ \theta$ are all undefined. But if we allow "undefined = undefined" these still hold.

As you probably know, there are three Pythagorean triples relating the circular trigonometric functions:

• $sin^2\ \theta + cos^2\ \theta = 1$
• $tan^2\ \theta + 1 = sec^2\ \theta$
• $1 + cot^2\ \theta = cosec^2\ \theta$

Translating these via the Gudermann relationships, we get three Pythagorean triples which can be used to label the sides of a right-angled triangle:

• $tanh^2\ x + sech^2\ x = 1$
• $sinh^2\ x + 1 = cosh^2\ x$
• $1 + cosech^2\ x = coth^2\ x$

NB With a bit of handwaving about undefined values / infinity, both the function $gd(x)$ and its inverse $x = gd^{-1}(\theta) = artanh(sin\ \theta)$ can take a domain of the entire complex plane. So for any $x$ [or $\theta$] in $\mathbb{C}$, there is a corresponding $\theta$ [or $x$] such that the above relationships hold.

Answer 3

This answer may be a little late, but I was wondering the same thing, and I think I may have come up with an answer.

Let's say we want to find $\sinh(\operatorname{artanh}(x))$. Draw your triangle as per usual, putting x on the opposite, and 1 on the adjacent.

However, from here on out, consider the adjacent side is the hypotenuse, and carry out the pythagorean theorem that way. This should give $\sqrt{1 - x^2}$ on the "regular" hypotenuse.

Then, as you would per usual, use the $\sinh$ part to figure out the value. You end up with $\sinh(\operatorname{artanh}(x)) = \frac{x}{\sqrt{1 - x^2}}$ which can be confirmed by graphing.

In short, dealing with something like $\tanh(\operatorname{arcosh}(x))$ is the exact same as dealing with $\tan(\arccos(x))$, except you consider the adjacent side of the triangle as the hypotenuse when dealing with the pythagorean theorem.

This is not something I have proved, just observed in multiple cases. So, there may be an unforeseen counter example, but I doubt it. I think this works because of the parallel between $\sin^2(x) + \cos^2(x) = 1$ and $\tanh^2(x) + \operatorname{sech}^2(x) = 1$.

Answer 4

Hyperbolic functions satisfy the fundamental identity $$\cosh(x)^2 - \sinh(x)^2 = 1.$$

This identity comes from the interpretation of hyperbolic functions in terms of hyperbolic triangles, cf. Wikipedia for example.

So now $$\cosh(\operatorname{argsinh}(y))^2 - \sinh(\operatorname{argsinh}(y))^2 = 1 \implies \cosh(\operatorname{argsinh}(y)) = \sqrt{1 + y^2}$$ because $\cosh$ is always nonnegative. Replace $y$ by $f(x)$ to get the answer: $\sqrt{1+(3x)^2} = \sqrt{1+9x^2}$.

Answer 5

You can draw a triangle with an imaginary side !

Indeed, $\cos(x)=\cosh(y)$ and $\sin(x)=i\sinh(y)$, where $y=ix$. Then all the rules known for the trigonometric functions follow for the hyperbolic ones.

For example

$$\cos^2(x)+\sin^2(x)=1\leftrightarrow\cosh^2(y)-\sinh^2(y)=1,\\ \arccos(\sin(t))=\sqrt{1-t^2}\leftrightarrow \text{arcosh}(\sinh(t))=\sqrt{1+t^2}.$$