Simplifying $\cosh \mathrm{arcsinh} \ x$

Question

How can I simplify the following: $$\cosh \mathrm{arcsinh} \ x$$ I know that an expression of the form $f(g^{-1}(x))$ where $f$ and $g$ are trigonometric functions can be simplified by constructing a right triangle. Is there an analgous construction for the hyperbolic functions?

Answer 1

Recall that $\cosh^{2}(x)-\sinh^{2}(x)=1$ and take $x=\mathrm{arcsinh(y)}$.

On the hyperbola $x^{2}-y^{2}=1$ you can interpret the $x$ and $y$ components as $\cosh(A)$ and $\sinh(A)$ respectively (where $A$ is the area described in the diagram). Putting $A=\mathrm{arcsinh(t)}$ and recalling that the $x$ and $y$ components satisfy $x^{2}-y^{2}=1$ gives you the formula. This is analogous to drawing triangles to solve for an identity.