Hyperbolic functions – further maths revision notes to download as PDF

Starter activity

Hold a necklace or a length of chain by its two ends and let it hang. The curve it makes looks like a parabola, but it is not. Galileo thought it was a parabola, but a more careful analysis in 1691 by Leibniz, Huygens and Johann Bernoulli showed that it is the graph of \(y=a\cosh\dfrac xa\). Sketch the graphs of \(e^x\) and \(e^{-x}\) on the same axes and add them together. Divide by two. What shape do you get? What is its lowest point, and is the graph symmetrical?

Note: Osborn's rule

To convert a trigonometric identity to a hyperbolic one, replace each \(\cos\) by \(\cosh\) and each \(\sin\) by \(\sinh\), but change the sign wherever there is a product (or implied product) of two sines. For example \(1+\tan^2x=\sec^2x\) becomes \(1-\tanh^2x=\operatorname{sech}^2x\).

Note: proof for arsinh

Let \(y=\operatorname{arsinh}x\), so \(x=\sinh y=\frac{e^y-e^{-y}}2\). Then \(e^{2y}-2xe^y-1=0\), so \(e^y=x+\sqrt{x^2+1}\) (the positive root).

Note: even and odd functions

The function \(\cosh x\) is even, which means that \(\cosh(-x)=\cosh x\), so its graph is symmetric about the \(y\)-axis. The function \(\sinh x\) is odd, which means that \(\sinh(-x)=-\sinh x\), so its graph is symmetric under a half turn about the origin. Together they split the exponential function into its even part and its odd part, since \(e^x=\cosh x+\sinh x\) and \(e^{-x}=\cosh x-\sinh x\). This is the reason for the identity \(\cosh^2x-\sinh^2x=1\): multiplying the two equations gives \(e^xe^{-x}=1\). Recognising symmetry also helps when sketching graphs and when evaluating integrals over a symmetric interval, since the integral of an odd function from \(-a\) to \(a\) is zero.

Applications of hyperbolic functions

Hyperbolic functions occur naturally whenever a quantity grows or decays exponentially in two directions. The shape of a hanging cable, the Gateway Arch in St Louis, and many suspension bridge cables follow the catenary curve, because a flexible chain under its own weight takes the shape of the graph of the hyperbolic cosine. The graph of \(\tanh\) describes saturating processes, such as the speed of a falling object subject to air resistance, the magnetisation of a material, and the activation function used in artificial neural networks. In physics, the rapidity of special relativity is the inverse hyperbolic tangent of the speed divided by the speed of light.

Why the hyperbola and not the circle

The names of the hyperbolic functions reflect a close analogy with the trigonometric functions. For the circular functions, the point with coordinates cosine and sine of an angle lies on the unit circle, and the angle measures twice the area of the sector cut off. For the hyperbolic functions, the point with coordinates hyperbolic cosine and hyperbolic sine of a number lies on the right-hand branch of the unit hyperbola, and the number measures twice the area of the corresponding hyperbolic sector. Because the trigonometric functions are built from imaginary exponentials and the hyperbolic functions from real ones, the two families are linked by the relations \(\cosh ix=\cos x\) and \(\sinh ix=i\sin x\). This is the deeper reason why Osborn's rule works: converting a trigonometric identity into a hyperbolic one is equivalent to replacing \(x\) by \(ix\), and the sign changes arise from the powers of \(i\).

Reading graphs and asymptotes

When sketching graphs involving hyperbolic functions, mark the key features carefully. For \(y=\cosh x\), label the minimum point \((0,1)\) and note that the curve is symmetric. For \(y=\sinh x\), mark the origin and show that the gradient there is 1, and that there is no maximum or minimum. For \(y=\tanh x\), draw the horizontal asymptotes \(y=1\) and \(y=-1\) as dashed lines, and show that the curve is steepest at the origin. For the inverse functions, remember that their graphs are reflections of the originals in the line \(y=x\): the graph of \(\operatorname{arsinh}x\) grows slowly like a logarithm, the graph of \(\operatorname{arcosh}x\) starts at the point \((1,0)\) with a vertical tangent, and the graph of \(\operatorname{artanh}x\) has vertical asymptotes at \(x=1\) and \(x=-1\).

Linking hyperbolic functions to series and complex numbers

The exponential function has a power series that holds for every real number, and the hyperbolic functions inherit series from it. Taking the even terms of the series for \(e^x\) gives \(\cosh x=1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\cdots\), and taking the odd terms gives \(\sinh x=x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\cdots\). These resemble the series for cosine and sine, but with all the signs positive. You will meet these series in the chapter on Maclaurin expansions. For small values of \(x\) the first term is a very good approximation, so \(\sinh x\approx x\) and \(\cosh x\approx1+\tfrac12x^2\), which explains why the graph of the catenary looks like a parabola near its lowest point.

Exam technique for hyperbolic questions

Questions on this topic often begin with a "show that" step, for example proving an identity or deriving a logarithmic form. Always set out the proof from the definitions in terms of exponentials, because this is the method that the mark scheme expects, and write the final line exactly as the result stated in the question. When solving an equation, state the range of the function in order to justify rejecting a root, since this earns a mark. When integrating, check the form of the integral against the formula booklet: if the square root has a plus sign inside, the answer involves arsinh, and if it has a minus sign, the answer involves arcosh. Finally, always check an answer by estimating it numerically with a calculator, using the exponential definitions if necessary.

Final advice

The hyperbolic functions reward careful, tidy work. Because the identities are so close to the familiar trigonometric ones, it is easy to write a wrong sign without noticing, so make a habit of checking each identity with a simple value such as \(x=0\) or \(x=\ln2\). Practise the derivations of the logarithmic forms, since a question may ask for them as a proof. Use the definitions in terms of exponentials whenever you are unsure, because they always work. With a secure understanding of this chapter you will find that integrals involving square roots of quadratics, which looked difficult before, become routine applications of a single substitution.

Did you know?

The word hyperbolic comes from the hyperbola, because the point \((\cosh t,\sinh t)\) lies on the hyperbola \(x^2-y^2=1\) in the same way as \((\cos t,\sin t)\) lies on the circle \(x^2+y^2=1\). The hyperbolic functions were introduced by the Italian Jesuit Vincenzo Riccati in the 1760s and developed by Johann Heinrich Lambert. The first accurate analysis of the hanging chain was one of the great problems of the late seventeenth century, solved independently by Leibniz, Huygens and Johann Bernoulli in 1691, only a few years after the invention of calculus.