Further Maths/Lessons/Hyperbolic Functions/Identities & Equations
Core Pure · Hyperbolic Functions · Year 2

Hyperbolic Identities & Equations

cosh²x − sinh²x ≡ 1, Osborn's rule, and solving equations in a single hyperbolic function.

13 min read 3 worked examples

The fundamental identity

Theorem 6.3 — cosh²x − sinh²x ≡ 1

$$\cosh^2x-\sinh^2x\equiv1$$

Note the minus sign, unlike the trigonometric identity $\cos^2x+\sin^2x\equiv1$ — this is the key structural difference between the two families.

ProofDirect from the exponential definitions
1
$\cosh^2x-\sinh^2x=\left(\dfrac{e^x+e^{-x}}2\right)^2-\left(\dfrac{e^x-e^{-x}}2\right)^2$.
2
Expand each square: $\dfrac{e^{2x}+2+e^{-2x}}4-\dfrac{e^{2x}-2+e^{-2x}}4$.
3
Subtract: the $e^{2x}$ and $e^{-2x}$ terms cancel, leaving $\dfrac{2-(-2)}4=\dfrac44=1$.
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Related identities

Dividing $\cosh^2x-\sinh^2x=1$ by $\cosh^2x$ gives $1-\tanh^2x\equiv\operatorname{sech}^2x$. Dividing instead by $\sinh^2x$ gives $\coth^2x-1\equiv\operatorname{cosech}^2x$.

Compound-angle and double-angle formulae

Osborn's rule

Every trigonometric identity has a hyperbolic counterpart obtainable by replacing $\cos\to\cosh$, $\sin\to\sinh$, and flipping the sign of any term that is (or implicitly contains) a product of two $\sin$/$\sinh$ factors. For example:

$$\sinh(A\pm B)\equiv\sinh A\cosh B\pm\cosh A\sinh B$$

$$\cosh(A\pm B)\equiv\cosh A\cosh B\pm\sinh A\sinh B$$

$$\cosh2A\equiv\cosh^2A+\sinh^2A\equiv2\cosh^2A-1\equiv1+2\sinh^2A$$

$$\sinh2A\equiv2\sinh A\cosh A$$

This is a useful shortcut for recall, but every result can also be verified directly from the exponential definitions.

ProofSinh(A + B) ≡ sinh A cosh B + cosh A sinh B
1
RHS $=\dfrac{e^A-e^{-A}}2\cdot\dfrac{e^B+e^{-B}}2+\dfrac{e^A+e^{-A}}2\cdot\dfrac{e^B-e^{-B}}2$.
2
$=\dfrac{e^{A+B}+e^{A-B}-e^{-A+B}-e^{-A-B}}4+\dfrac{e^{A+B}-e^{A-B}+e^{-A+B}-e^{-A-B}}4$.
3
The $e^{A-B}$ and $e^{-A+B}$ terms cancel between the two fractions, leaving $\dfrac{2e^{A+B}-2e^{-(A+B)}}4=\dfrac{e^{A+B}-e^{-(A+B)}}2=\sinh(A+B)$.
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Example 1 · Writing down a hyperbolic identity
Using Osborn's rule, write down the hyperbolic identity corresponding to $\sin2A\equiv2\sin A\cos A$, and the one corresponding to $\tan(A+B)\equiv\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.

$\sinh2A\equiv2\sinh A\cosh A$ (no sign change needed — only one $\sinh$ factor).

$\tanh(A+B)\equiv\dfrac{\tanh A+\tanh B}{1+\tanh A\tanh B}$ (the denominator's product term implicitly involves $\sin A\sin B$ via $\tan A\tan B=\dfrac{\sin A\sin B}{\cos A\cos B}$, so its sign flips).

Example 2 · Proving a further identity
Given that $\cosh x=\dfrac{13}5$ and $x>0$, find the exact value of $\sinh2x$.

Using $\cosh^2x-\sinh^2x=1$: $\sinh^2x=\left(\dfrac{13}5\right)^2-1=\dfrac{169}{25}-\dfrac{25}{25}=\dfrac{144}{25}$.

Since $x>0$, $\sinh x=\dfrac{12}5$ (positive, as $\sinh$ is increasing and $\sinh0=0$).

$\sinh2x=2\sinh x\cosh x=2\times\dfrac{12}5\times\dfrac{13}5=\dfrac{312}{25}$.

Solving hyperbolic equations

Method — quadratics in a single hyperbolic function

Use $\cosh^2x-\sinh^2x\equiv1$ to rewrite the equation in terms of a single hyperbolic function, factorise, then convert each resulting linear equation to a logarithm using the arsinh/arcosh/artanh formulae.

Example 3 · Solving a hyperbolic equation
Solve $3\cosh^2x-7\sinh x=9$, giving your answers as natural logarithms.

Substitute $\cosh^2x=1+\sinh^2x$: $3(1+\sinh^2x)-7\sinh x=9$, i.e. $3\sinh^2x-7\sinh x-6=0$.

Factorise: $(3\sinh x+2)(\sinh x-3)=0$, so $\sinh x=-\dfrac23$ or $\sinh x=3$.

$x=\operatorname{arsinh}\left(-\dfrac23\right)=\ln\left(-\dfrac23+\sqrt{\dfrac49+1}\right)=\ln\left(\dfrac{-2+\sqrt{13}}3\right)$.

$x=\operatorname{arsinh}3=\ln\left(3+\sqrt{10}\right)$.

Exam-style questions

Question 1 5 marks

Prove, using the definitions of $\cosh x$ and $\sinh x$ in terms of exponentials, that $\cosh2A\equiv1+2\sinh^2A$.

RHS $=1+2\left(\dfrac{e^A-e^{-A}}2\right)^2=1+\dfrac{e^{2A}-2+e^{-2A}}2$. (M1, A1)

$=\dfrac{2+e^{2A}-2+e^{-2A}}2=\dfrac{e^{2A}+e^{-2A}}2=\cosh2A$. (M1, A1, A1)

Question 2 4 marks

Given that $\sinh x=-2$, find the exact value of $\tanh x$.

$\cosh^2x=1+\sinh^2x=1+4=5$, so $\cosh x=\sqrt5$ (always positive). (M1, A1)

$\tanh x=\dfrac{\sinh x}{\cosh x}=\dfrac{-2}{\sqrt5}=-\dfrac{2\sqrt5}5$. (M1, A1)

Question 3 6 marks

Solve $16\sinh x+7\operatorname{sech}x=0$ has no real solutions — explain why, then solve $16\sinh x-6\cosh x=2$, giving your answer as a natural logarithm.

For the second equation, write in exponential form: $16\cdot\dfrac{e^x-e^{-x}}2-6\cdot\dfrac{e^x+e^{-x}}2=2$. (M1)

$8e^x-8e^{-x}-3e^x-3e^{-x}=2 \Rightarrow 5e^x-11e^{-x}=2$. (A1)

Multiply by $e^x$: $5e^{2x}-2e^x-11=0$. (M1, A1)

$e^x=\dfrac{2\pm\sqrt{4+220}}{10}=\dfrac{2\pm\sqrt{224}}{10}$; taking the positive root, $e^x=\dfrac{1+2\sqrt{14}}5$, so $x=\ln\left(\dfrac{1+2\sqrt{14}}5\right)$. (A1, A1)

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