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Inverse hyperbolic functionsAQA A-Level Further Maths: Revision notes

Section 1

Definition of the inverse hyperbolic functions

The inverse of sinh⁡\sinh is written sinh⁡−1x\sinh^{-1}x or arsinh⁡x\operatorname{arsinh}x, and similarly cosh⁡−1x=arcosh⁡x\cosh^{-1}x=\operatorname{arcosh}x and tanh⁡−1x=artanh⁡x\tanh^{-1}x=\operatorname{artanh}x. They are inverse functions, not reciprocals: y=sinh⁡−1x  ⟺  x=sinh⁡y,y=\sinh^{-1}x\iff x=\sinh y, and likewise for cosh⁡\cosh and tanh⁡\tanh. Their graphs are reflections of the originals in the line y=xy=x. For cosh⁡\cosh, which is not one-to-one, the inverse cosh⁡−1x\cosh^{-1}x is defined as the non-negative value y≥0y\ge0 with cosh⁡y=x\cosh y=x, so it needs x≥1x\ge1. For tanh⁡−1x\tanh^{-1}x we need −1<x<1-1<x<1.

Key termsinverse functionarsinharcoshartanh
Common mistake

Reading sinh⁡−1x\sinh^{-1}x as 1sinh⁡x\frac{1}{\sinh x}. It is the inverse function, not the reciprocal.

Section 2

Logarithmic form of sinh⁡−1x\sinh^{-1}x

Let y=sinh⁡−1xy=\sinh^{-1}x, so x=12(ey−e−y)x=\frac12(e^y-e^{-y}). Multiply by 2ey2e^y: (ey)2−2xey−1=0(e^y)^2-2xe^y-1=0, a quadratic in eye^y. Then ey=2x±4x2+42=x±x2+1.e^y=\frac{2x\pm\sqrt{4x^2+4}}{2}=x\pm\sqrt{x^2+1}. Since ey>0e^y>0 and x−x2+1<0x-\sqrt{x^2+1}<0, reject the minus sign, giving sinh⁡−1x=ln⁡(x+x2+1),\sinh^{-1}x=\ln\left(x+\sqrt{x^2+1}\right), valid for all real xx. Example: sinh⁡−134=ln⁡(34+54)=ln⁡2\sinh^{-1}\frac34=\ln\left(\frac34+\frac54\right)=\ln2.

Key termsquadratic in $e^y$
Exam tip

Always justify rejecting a root: eye^y is positive, so the logarithm of a negative number is not allowed.

Section 3

Logarithmic form of cosh⁡−1x\cosh^{-1}x

Let y=cosh⁡−1xy=\cosh^{-1}x with y≥0y\ge0 and x≥1x\ge1. From x=12(ey+e−y)x=\frac12(e^y+e^{-y}) we get (ey)2−2xey+1=0(e^y)^2-2xe^y+1=0, so ey=x±x2−1e^y=x\pm\sqrt{x^2-1}. Because y≥0y\ge0, we need ey≥1e^y\ge1, which selects the plus sign: cosh⁡−1x=ln⁡(x+x2−1),x≥1.\cosh^{-1}x=\ln\left(x+\sqrt{x^2-1}\right),\quad x\ge1. The other root, x−x2−1=1x+x2−1x-\sqrt{x^2-1}=\frac{1}{x+\sqrt{x^2-1}}, gives −cosh⁡−1x-\cosh^{-1}x, the negative solution of cosh⁡y=x\cosh y=x. This is why cosh⁡x=53\cosh x=\frac53 has two solutions, x=±ln⁡3x=\pm\ln3. Example: cosh⁡−153=ln⁡(53+43)=ln⁡3\cosh^{-1}\frac53=\ln\left(\frac53+\frac43\right)=\ln3.

Common mistake

Using x2+1x^2+1 in the cosh⁡−1\cosh^{-1} formula or x2−1x^2-1 in the sinh⁡−1\sinh^{-1} formula. The sign inside the root matches cosh⁡2−sinh⁡2=1\cosh^2-\sinh^2=1.

Section 4

Logarithmic form of tanh⁡−1x\tanh^{-1}x

Let y=tanh⁡−1xy=\tanh^{-1}x, so x=ey−e−yey+e−y=e2y−1e2y+1x=\frac{e^y-e^{-y}}{e^y+e^{-y}}=\frac{e^{2y}-1}{e^{2y}+1}. Then x(e2y+1)=e2y−1x\left(e^{2y}+1\right)=e^{2y}-1, so e2y(1−x)=1+xe^{2y}(1-x)=1+x and tanh⁡−1x=12ln⁡(1+x1−x),−1<x<1.\tanh^{-1}x=\frac12\ln\left(\frac{1+x}{1-x}\right),\quad-1<x<1. The fraction 1+x1−x\frac{1+x}{1-x} is positive only when ∣x∣<1|x|<1, which explains the domain. Examples: tanh⁡−135=12ln⁡8/52/5=12ln⁡4=ln⁡2\tanh^{-1}\frac35=\frac12\ln\frac{8/5}{2/5}=\frac12\ln4=\ln2 and tanh⁡−1(−x)=−tanh⁡−1x\tanh^{-1}\left(-x\right)=-\tanh^{-1}x.

Common mistake

Forgetting the factor 12\frac12 in tanh⁡−1x\tanh^{-1}x.

Section 5

Using the logarithmic forms

To evaluate an inverse hyperbolic function, substitute into the correct log formula and simplify the number inside the root; exact values often involve ln⁡2\ln2, ln⁡3\ln3 or ln⁡(1+2)\ln\left(1+\sqrt2\right). To solve sinh⁡−1x=k\sinh^{-1}x=k, write x=sinh⁡k=12(ek−e−k)x=\sinh k=\frac12\left(e^k-e^{-k}\right). For example sinh⁡−1x=ln⁡3\sinh^{-1}x=\ln3 gives x=12(3−13)=43x=\frac12\left(3-\frac13\right)=\frac43. Combine logarithms using ln⁡p+ln⁡q=ln⁡pq\ln p+\ln q=\ln pq and nln⁡p=ln⁡pnn\ln p=\ln p^n.

Exam tip

Check by substituting back: sinh⁡(ln⁡2)=12(2−12)=34\sinh\left(\ln2\right)=\frac12\left(2-\frac12\right)=\frac34, consistent with sinh⁡−134=ln⁡2\sinh^{-1}\frac34=\ln2.

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Exam questions on Inverse hyperbolic functions

  1. Let a=sinh⁡−1(34)a=\sinh^{-1}\left(\frac34\right) and b=tanh⁡−1(12)b=\tanh^{-1}\left(\frac12\right).
    Express a+2ba+2b as a single natural logarithm.2 marks
  2. Let c=cosh⁡−1(53)c=\cosh^{-1}\left(\frac53\right).
    Solve cosh⁡x=53\cosh x=\frac53.2 marks
  3. A student sets y=sinh⁡−1xy=\sinh^{-1}x for a real number xx, so that x=sinh⁡yx=\sinh y.
    Show that y=ln⁡(x+x2+1)y=\ln\left(x+\sqrt{x^2+1}\right).3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).