Logarithmic Forms of Inverse Hyperbolic Functions

$arsinh x = \ln (x + \sqrt{x^{2} + 1})$ , $x \in ℝ$
$ar \cosh x = \ln (x + \sqrt{x^{2} - 1})$ , $x \geq 1$
- Since coshx is a many-to-one function, its domain is restricted to x ≥ 0 when finding the inverse
- Therefore, its range is coshx ≥ 1
- So, the domain of the inverse function is x ≥ 1
$artanh x = \frac{1}{2} \ln (\frac{1 + x}{1 - x})$ , $| x | < 1$
These three definitions are in the formula booklet

What are the graphs of the inverse hyperbolic functions and their key features?

As they are inverse functions, they are reflections of their original functions in the line y=x
$y = arsinh x$
- Domain: $x \in ℝ$
- Range: $y \in ℝ$

edexcel-al-fm-cp-4-1-2-arsinhx-graph

edexcel-al-fm-cp-4-1-2-arcoshx-graph

edexcel-al-fm-cp-4-1-2-artanhx-graph

You need to be able to derive each inverse from the definition of the original
STEP 1
Write in terms of e
- $y = arsinh x$ and rearrange to $x = \sinh y$
- $x = \frac{1}{2} (e^{y} - e^{- y})$
STEP 2
Form a quadratic in terms of e^y
- Multiply by 2e^y and rearrange
- ${(e^{y})}^{2} - 2 x e^{y} - 1 = 0$
STEP 3
Solve the quadratic and find an expression for y
- $e^{y} = x \pm \sqrt{x^{2} + 1}$
  - Reject $x - \sqrt{x^{2} + 1}$ as this produces negative values as $\sqrt{x^{2} + 1} > x$ whereas $e^{y} > 0$
- $y = \ln (x + \sqrt{x^{2} + 1})$
The derivations of the other two formulae are similar
- For $arcosh x$ both $\ln (x + \sqrt{x^{2} - 1})$ and $\ln (x - \sqrt{x^{2} - 1})$ are well-defined for $x \geq 1$
  - We choose $arcosh x = \ln (x + \sqrt{x^{2} - 1})$ as $arcosh x \geq 0$ and it can be shown that $\ln (x - \sqrt{x^{2} - 1}) = \ln (\frac{1}{x + \sqrt{x^{2} - 1}}) = - \ln (x + \sqrt{x^{2} - 1})$

Be careful when working with the circular (“normal”) inverse trig functions and the inverse hyperbolic functions

The “c” in arcsin x, arccos x, arctan x indicates the circular functions
The hyperbolic functions have “h”, but the “h” doesn’t come immediately after the “ar”: arsinh x, arcosh x, artanh x

Be careful not to confuse these, especially when looking them up in the formulae booklet

Starting from the definition of $\tanh x$ , show that

$artanh x = \frac{1}{2} \ln (\frac{1 + x}{1 - x})$ when $| x | < 1$ .

al-fm-4-1-2-we-solution

Logarithmic Forms of Inverse Hyperbolic Functions (Edexcel A Level Further Maths: Core Pure)