Inverse Hyperbolic Functions

An inverse hyperbolic function simply undoes its hyperbolic function: $y = \text{arsinh}\,x \iff \sinh y = x$ and similarly for $\text{arcosh}$ and $\text{artanh}$.

The domain of each inverse equals the range of the original function:

• $\sinh$ has range all reals → $\text{arsinh}$ accepts any $x$.
• $\cosh$ has range $y \geq 1$ → $\text{arcosh}$ accepts only $x \geq 1$.
• $\tanh$ has range $-1 < y < 1$ → $\text{artanh}$ accepts only $|x| < 1$.

The arcosh complication. $\cosh$ is even, so it is not one-to-one — both $x = c$ and $x = -c$ give the same $\cosh$ value. To define a single-valued inverse, we restrict to the principal branch $y \geq 0$. That is why $\text{arcosh}$ always returns a non-negative answer and uses the positive square root in its log form.

Cambridge tip. Whenever you invert $\cosh$, expect a $\pm$ pair of answers in the original variable, even though $\text{arcosh}$ itself returns only the non-negative value.

Each inverse hyperbolic function can be written as a single natural logarithm. These three are printed in MF19:

$\text{arsinh}\,x = \ln\left(x + \sqrt{x^2 + 1}\right) \qquad (\text{all } x)$

$\text{arcosh}\,x = \ln\left(x + \sqrt{x^2 - 1}\right) \qquad (x \geq 1)$

$\text{artanh}\,x = \frac{1}{2}\ln\frac{1 + x}{1 - x} \qquad (|x| < 1)$

Three details that earn or lose marks:

• arsinh has $+1$, arcosh has $-1$ under the root. (Memory hook: arsinh pairs with $\cosh^2 = 1 + \sinh^2$.)
• arcosh takes the positive root (principal branch).
• artanh has the factor $\frac{1}{2}$ out the front — easy to drop.

Using them to solve equations. Solving $\sinh x = k$ is just $x = \text{arsinh}\,k$, evaluated via the log form. For $\cosh x = k$ ($k > 1$), $x = \pm\text{arcosh}\,k$ (the $\pm$ because cosh is even).

Worked check. $\text{arsinh}\,0 = \ln(0 + \sqrt{0 + 1}) = \ln 1 = 0$ — correct, since $\sinh 0 = 0$.

Cambridge tip. Even though the forms are in MF19, copy them carefully: the most common slip is writing $x^2 - 1$ for arsinh (it should be $+1$).

'Show that $\text{arsinh}\,x = \ln(x + \sqrt{x^2 + 1})$' is a classic Paper 2 'show that'. The method is always the same four steps:

1. Set up. Let $y = \text{arsinh}\,x$, so $\sinh y = x$. Substitute the exponential definition: $\frac{e^y - e^{-y}}{2} = x \;\Rightarrow\; e^y - e^{-y} = 2x$

2. Form a quadratic in $e^y$. Multiply through by $e^y$ and let $u = e^y$: $e^{2y} - 2x e^y - 1 = 0 \;\Rightarrow\; u^2 - 2xu - 1 = 0$

3. Solve and choose the right root. By the quadratic formula: $u = x \pm \sqrt{x^2 + 1}$ Since $u = e^y > 0$ and $\sqrt{x^2 + 1} > |x|$, the minus root is negative — reject it. So $e^y = x + \sqrt{x^2 + 1}$.

4. Take logarithms: $y = \ln\left(x + \sqrt{x^2 + 1}\right) \qquad \blacksquare$

The arcosh and artanh derivations are the same idea:

• For arcosh, the quadratic is $u^2 - 2xu + 1 = 0$ giving $u = x \pm \sqrt{x^2 - 1}$. Both roots are positive (they are reciprocals), so you choose the one $\geq 1$ (the principal branch $y \geq 0$).
• For artanh, use the form $\tanh y = \frac{e^{2y} - 1}{e^{2y} + 1} = x$, cross-multiply to get $e^{2y} = \frac{1+x}{1-x}$, then take logs and halve to get the $\frac{1}{2}$.

Cambridge tip. On a 'show that', the root-rejection step must be justified in words ('$e^y > 0$, so reject the negative root' / 'principal branch, so take the root $\geq 1$'). That sentence is worth a mark.