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?
 A: 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: $\sqrt{1+(3x)^2} = \sqrt{1+9x^2}$.
A: 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
$$\cos^2(x)+\sin^2(x)=1\leftrightarrow\cosh^2(y)-\sinh^2(y)=1,\\
\arccos(\sin(t))=\sqrt{1-t^2}\leftrightarrow \text{arcosh}(\sinh(t))=\sqrt{1+t^2}.$$
A: 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 $\dfrac{2\sqrt{9x^2+1}}{2}=\sqrt{9x^2+1}$
A: 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 $\tanh^2(x) + \operatorname{sech}^2(x) = 1$.
