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?

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    $\begingroup$ Hint: $\sqrt{x^2+1}$ is the result. Why? $\endgroup$ – Amzoti Aug 19 '13 at 5:38

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.


  • $\begingroup$ Thanks, I should have thought of that. This solves it nicely, but relies on the identity relating $\sinh$ and $\cosh$; is there a more general method that would work for, say, $\mathrm{csch}\ \mathrm{arctanh}\ x$? $\endgroup$ – David Zhang Aug 19 '13 at 5:55
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    $\begingroup$ I don't know of a more general method but I usually try to reduce the expression so that it is in terms of $\sinh(x)$, $\cosh(x)$, $\tanh(x)$, $\mathrm{arcsinh(x)}$, $\mathrm{arccosh(x)}$, and $\mathrm{arctanh(x)}$. Then I would apply identities to solve. So for your example I would first note that $csch(\mathrm{arctanh(x)})=\frac{1}{\sinh(\mathrm{arctanh(x)})}$. Then I would note that $x=\tanh(\mathrm{arctanh(x)})=\frac{\sinh(\mathrm{arctanh(x)})}{\cosh(\mathrm{arctanh(x)})}=\frac{\sinh(\mathrm{arctanh(x)})}{\sqrt{1+\sinh^{2}(\mathrm{arctanh(x)})}}$. Solve for $\sinh(\mathrm{arctanh(x)})$. $\endgroup$ – user71352 Aug 19 '13 at 6:44
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    $\begingroup$ It occurred to me that to solve your example I could have just noted that $csch^{2}(\mathrm{arctanh(x)})=\cot^{2}(\mathrm{arctanh(x)})-1=\frac{1}{\tanh^{2}(\mathrm{arctanh(x)})}-1=\frac{1}{x^{2}}-1$. You will have to specify cases for the formula since you need to take a square root and $csch(x)$ changes sign. $\endgroup$ – user71352 Aug 19 '13 at 22:11

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