In one line
Because hyperbolic functions are built from \(e^x\), their inverses are logarithms: \(\sinh^{-1}x=\log(x+\sqrt{x^2+1})\), \(\cosh^{-1}x=\log(x+\sqrt{x^2-1})\), \(\tanh^{-1}x=\dfrac12\log\dfrac{1+x}{1-x}\).
Explained simply
The inverse of "multiply by 3" is "divide by 3". The inverse of \(e^x\) is \(\log x\). Since \(\sinh x\) is made of exponentials, undoing it must involve a logarithm. Solving for the exponent turns into a quadratic equation in \(e^y\).
If \(y=\sinh^{-1}x\), then \(x=\sinh y\). Here "log" always means the natural logarithm (base \(e\)).
Why we need these forms: a calculator may not have an \(\tanh^{-1}\) key, but it always has \(\log\). More importantly, in Module 2 problems the answer often comes out as \(\tanh^{-1}(\text{something})\), and examiners expect you to convert it into a neat logarithm. The same log forms are used later in integration, for example \(\displaystyle\int\frac{dx}{\sqrt{x^2+1}}=\sinh^{-1}x+c=\log(x+\sqrt{x^2+1})+c\).
Notice the domain conditions. \(\sinh x\) takes every real value, so \(\sinh^{-1}x\) works for all \(x\). But \(\cosh x\ge1\) and \(-1<\tanh x<1\), so \(\cosh^{-1}x\) needs \(x\ge1\) and \(\tanh^{-1}x\) needs \(|x|<1\).
How it works
- Let \(y=\sinh^{-1}x\), so \(x=\dfrac{e^y-e^{-y}}{2}\).
- Multiply by \(2e^y\): \(e^{2y}-2xe^y-1=0\), a quadratic in \(e^y\).
- Solve: \(e^y=x\pm\sqrt{x^2+1}\). Since \(e^y>0\) and \(x-\sqrt{x^2+1}<0\), take the plus sign.
- Take log: \(y=\log(x+\sqrt{x^2+1})\).
- For \(\tanh^{-1}\): \(x=\dfrac{e^{2y}-1}{e^{2y}+1}\Rightarrow e^{2y}=\dfrac{1+x}{1-x}\Rightarrow y=\dfrac12\log\dfrac{1+x}{1-x}\).
Key formula
- \(\sinh^{-1}x=\log\left(x+\sqrt{x^2+1}\right)\), all real \(x\)
- \(\cosh^{-1}x=\log\left(x+\sqrt{x^2-1}\right)\), \(x\ge1\) (principal, non-negative value)
- \(\tanh^{-1}x=\dfrac12\log\dfrac{1+x}{1-x}\), \(|x|<1\)
- \(\coth^{-1}x=\dfrac12\log\dfrac{x+1}{x-1}\), \(|x|>1\)
Quick example
(a) \(\sinh^{-1}\dfrac34=\log\left(\dfrac34+\sqrt{\dfrac9{16}+1}\right)=\log\left(\dfrac34+\dfrac54\right)=\log2\). Check: \(\sinh(\log2)=\dfrac{2-\frac12}{2}=\dfrac34\).
(b) \(\tanh^{-1}\dfrac12=\dfrac12\log\dfrac{3/2}{1/2}=\dfrac12\log3\approx0.549\).
Common mistakes
- Keeping both signs \(\pm\) in \(e^y=x\pm\sqrt{x^2+1}\); the minus root is negative and must be rejected with a reason.
- Using \(\tanh^{-1}x\) for \(|x|\ge1\); the log argument becomes zero or negative there.
- Confusing \(\sinh^{-1}x\) (inverse function) with \(\dfrac1{\sinh x}=\mathrm{cosech}\,x\).
How MU asks it
3-mark Q.1 part: "Prove that \(\tanh^{-1}x=\dfrac12\log\dfrac{1+x}{1-x}\)" or "Show that \(\sinh^{-1}x=\log(x+\sqrt{x^2+1})\)". Also used inside 6-mark problems to simplify an answer like \(\phi=\dfrac12\tanh^{-1}(\sin\alpha)\). Write the defining equation, form the quadratic, solve, reject the negative root with a reason, take log.