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

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?

• well, in this case the composition is the identity. – hjhjhj57 Dec 7 '14 at 19:37
• $\cosh^2 x - \sinh^2 x \equiv 1$ is the fundamental identity for the hyperbolic functions, just like $\cos^2 x + \sin^2 x \equiv 1$ is for the trigonometric functions. – Daniel Fischer Dec 7 '14 at 19:37
• I'm aware of that formula, but I'm not sure how to get this result from that. Trying to get $\operatorname{arcsinh}$, I get $x=\operatorname{arcsinh}(\sqrt{\cosh^2(x)-1})$. – user198366 Dec 7 '14 at 19:46
• @DanielFischer Ahh. Cool. I just figured out how to I could have gotten it from this in the case I mentioned: $3x=\sqrt{\cosh^2(\theta)-1} \implies \cosh(\theta) = \sqrt{1+9x^2}$. This isn't quite as general as the triangle method in trig -- for instance, I don't know how to get cosh(arctanh(f(x)), but ones that involves only cosh and arcsinh or sinh and arccosh should be solvable this way. Thanks. – user198366 Dec 7 '14 at 20:02
• $$\tanh (\operatorname{Ar sinh} f(x)) = \frac{\sinh (\operatorname{Ar sinh} f(x))}{\cosh (\operatorname{Ar sinh} f(x))} = \frac{f(x)}{\sqrt{1+f(x)^2}}$$ – Daniel Fischer Dec 7 '14 at 20:05

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

• So I need to memorize the logarithmic formulas for $\operatorname{arcsinh}$, $\operatorname{arccosh}$, $\operatorname{arctanh}$, $\operatorname{arccsch}$, $\operatorname{arcsech}$, and $\operatorname{arccoth}$? That is a lot more to memorize than for working with the trig functions. – user198366 Dec 7 '14 at 19:51
• There isn't a "triangle" that you can draw for hyperbolic functions, but if you remember how to derive the logarithmic forms instead of memorizing them, then you can just derive it when you need it. – Cyclohexanol. Dec 7 '14 at 19:59
• There may be something you can do in a hyperbolic plane, though (not sure) – Cyclohexanol. Dec 7 '14 at 20:01
• I actually don't know how to derive the log forms of the inverse hyperbolic functions. Can it be done straight from the exponential forms of the hyperbolic functions? – user198366 Dec 7 '14 at 20:07

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

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

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 $\sinh(arctanh(x))$. 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 $\sinh(arctanh(x)) = \frac{x}{\sqrt{1 - x^2}}$ which can be confirmed by graphing.

In short, dealing with something like $\tanh(arccosh(x))$ 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 $\tanh^2(x) + sech^2(x) = 1$.

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: $\sqrt{1+(3x)^2} = \sqrt{1+9x^2}$.