Hyperbolic Identities & Equations (Edexcel A Level Further Maths: Core Pure): Revision Note

Hyperbolic identities & equations

Are there identities linking the hyperbolic functions to the circular trig functions?

• Yes - these can be seen using de Moivre's Theorem to write

  • $\cos x=\frac{1}{2}\left({\mathrm{e}}^{\mathrm{i}x}+{\mathrm{e}}^{-\mathrm{i}x}\right)$

  • $\sin x=\frac{1}{2\mathrm{i}}\left({\mathrm{e}}^{\mathrm{i}x}-{\mathrm{e}}^{-\mathrm{i}x}\right)$

• Compare these with the definitions for the hyperbolic functions

  • $\cosh x=\frac{1}{2}\left({\mathrm{e}}^x+{\mathrm{e}}^{-x}\right)$

  • $\sinh x=\frac{1}{2}\left({\mathrm{e}}^x-{\mathrm{e}}^{-x}\right)$

• Therefore they can be related using the identities

  • $\cos x\equiv \cosh \left(\mathrm{i}x\right)$ or $\cosh x\equiv \cos \left(\mathrm{i}x\right)$

  • $\sin x\equiv -\mathrm{isinh}\left(\mathrm{i}x\right)$ or $\sinh x\equiv -\mathrm{isin}\left(\mathrm{i}x\right)$

What are the hyperbolic identities?

• In general, the hyperbolic identities are the same as the circular trigonometric identities except where there is a product of an even number of sinh terms, in which case the term changes sign

  • e.g. ${\cos }^{2}x+{\sin }^{2}x\equiv 1$ but  ${\cosh }^{2}x-{\sinh }^{2}x\equiv 1$

  • This is referred to as Osborn’s Rule

    • This occurs because of the connection with the imaginary number i

• All the circular trigonometric identities can be used with hyperbolic functions

• The main hyperbolic identities you are likely to need are

  • ${\cosh }^{2}x-{\sinh }^{2}x\equiv 1$

  • $\sinh 2x\equiv 2\sinh x\cosh x$

  • $\cosh 2x\equiv {\cosh }^{2}x+{\sinh }^{2}x$

    • These are listed in the formulae booklet

• Other identities include

  • $\cosh 2x\equiv 2{\cosh }^{2}x-1\equiv 1+2{\sinh }^{2}x$

  • $\sinh \left(A\pm B\right)=\sinh A\cosh B\pm \sinh B\cosh A$

  • $\cosh \left(A\pm B\right)\equiv \cosh A\cosh B\pm \sinh A\sinh B$

• The harmonic identities can also be used with hyperbolic functions

  • $a\cosh x\pm b\sinh x=R\cosh \left(x\pm \alpha \right)$

  • $a\sinh x\pm b\cosh x=R\sinh \left(x\pm \alpha \right)$

• Hyperbolic identities involving tanhx exist

  • They are not normally used as it is easier to use sinhx, coshx and their definitions

  • If you do use tanhx identities, be careful with implied or ‘hidden’ products of sinhx (e.g.  tanh²x)

• You can prove these identities by using the definitions of the hyperbolic functions in terms of e

Do reciprocal hyperbolic functions and identities exist?

• Yes! However, It is usually easier to deal with identities and equations involving these in terms of sinhx, coshx and their definitions

• $\mathrm{cosech}x=\frac{1}{\sinh x}=\frac{2}{{\mathrm{e}}^x-{\mathrm{e}}^{-x}}$ (Pronounced “coshec”)

• $\mathrm{sech}x=\frac{1}{\cosh x}=\frac{2}{{\mathrm{e}}^x+{\mathrm{e}}^{-x}}$ (Pronounced “shec”)

• $\coth x=\frac{1}{\tanh x}=\frac{\cosh x}{\sinh x}=\frac{{\mathrm{e}}^{2x}+1}{{\mathrm{e}}^{2x}-1}$(Pronounced “cough”)

How do I use hyperbolic identities to prove other identities?

• Start with the LHS and use the hyperbolic identities to rearrange into the RHS

• This approach can lead to what seems like a dead-end

  • In such cases simplify the LHS as far as possible, ideally so that it is in terms of sinhx and/or coshx only

  • Then use the sinhx and coshx definitions to write the LHS in terms of e

  • Repeat this for the RHS so that the LHS and RHS ‘meet in the middle’

How do I solve equations involving hyperbolic functions?

• Use identities to create an equation in terms of sinhx or coshx only

  • This should be a familiar equation to solve – linear, quadratic, etc

  • Find exact answers in terms of natural logarithms

    • Using the inverse hyperbolic functions definitions

  • Use your calculator if exact answers are not required

• As with circular trigonometric equations, do not cancel hyperbolic terms, rearrange so the equation equals zero and factorise

• When solving equations be careful when solving coshx = k (for constant k)

  • cosh^-1x is not necessarily the same as arcoshx

    • arcoshx is, strictly speaking, referring to the inverse function of coshx such that coshx is a one-to-one function

  • Using the graph you can see that the for k>1 there are two solutions to $\cosh x=k$

    • $x=\pm \mathrm{arcosh}k$

    • This can be written in logarithmic form as $\pm \ln \left(k+\sqrt{k^2-1}\right)$

    • This can be shown to be equivalent to $\ln \left(k\pm \sqrt{k^2-1}\right)$  

  • If k=1 then the only solution to $\cosh x=k$ is x=0

  • If k<1 then there are no real solutions to $\cosh x=k$