Paper 2 · Further Pure 2 ✦

Hyperbolic Functions

Built from $e^{x}$ and $e^{- x}$ , they behave almost exactly like sine and cosine — with a handful of sign flips that decide several marks a paper.

Syllabus 2.1Log forms given by MF19…but you must be able to derive them

What the examiner expects you to do

Know the definitions of $\sinh x, \cosh x, \tanh x, sech x, cosech x, \coth x$ in terms of the exponential function.
Sketch the graphs of the hyperbolic functions.
Prove and use identities involving hyperbolic functions.
Understand and use the inverse hyperbolic functions, and derive and use the logarithmic forms.

§1 The definitions — everything comes from these

◆ The six functions

\sinh x = \frac{e^{x} - e^{- x}}{2}, \cosh x = \frac{e^{x} + e^{- x}}{2}, \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}},

sech x = \frac{1}{\cosh x}, cosech x = \frac{1}{\sinh x}, \coth x = \frac{1}{\tanh x} = \frac{\cosh x}{\sinh x} .

These definitions are not on MF19. Everything else in this topic can be rebuilt from them, so learn them first and learn them cold.

◇ Two consequences worth memorising separately

\cosh x + \sinh x = e^{x}, \cosh x - \sinh x = e^{- x} .

Adding and subtracting the definitions gives them instantly, and they turn a great many "solve this hyperbolic equation" questions into a quadratic in $e^{x}$ .

Parity: $\cosh$ is even ( $\cosh (- x) = \cosh x$ ); $\sinh$ and $\tanh$ are odd. That is why $\cosh x = k$ has two solutions $\pm$ , while $\sinh x = k$ has exactly one.

★ Where the name comes from

$\cos^{2} t + \sin^{2} t = 1$ parametrises the circle $x^{2} + y^{2} = 1$ ; $\cosh^{2} t - \sinh^{2} t = 1$ parametrises the hyperbola $x^{2} - y^{2} = 1$ . One sign, one whole family of functions.

§2 The graphs

◇ What each sketch must show

Function	Shape and key features	Range
$\cosh x$	even, U-shaped, minimum at $(0,1)$ ; grows like $\frac{1}{2} e^{x}$ as $x \to \infty$	$y \geq 1$
$\sinh x$	odd, through the origin, always increasing; also grows like $\frac{1}{2} e^{x}$	all real $y$
$\tanh x$	odd, through the origin, increasing, S-shaped, with horizontal asymptotes $y = \pm 1$	$- 1 < y < 1$
$sech x$	even, bell-shaped, maximum at $(0,1)$ , $\to 0$ both ways	$0 < y \leq 1$
$cosech x$	odd, undefined at $x = 0$ ; asymptotes $x = 0$ and $y = 0$	$y \neq 0$
$\coth x$	odd, undefined at $x = 0$ ; asymptotes $x = 0$ and $y = \pm 1$	$\| y \| > 1$

The most-marked features are the minimum at $(0,1)$ for $\cosh$ , and the $y = \pm 1$ asymptotes for $\tanh$ . Sketch them and label them.

★ A* tip — the shape of cosh, explained in one line

$\cosh x = \frac{1}{2} e^{x} + \frac{1}{2} e^{- x}$ is the sum of two exponential curves, so far to the right the $\frac{1}{2} e^{x}$ dominates and far to the left the $\frac{1}{2} e^{- x}$ does. $\sinh x$ is the same two curves subtracted, which is why it dives below the axis on the left. Watch this happen in the visualiser.

§3 Identities — and Osborn's rule

◆ The core identities MF19 gives the starred three

\cosh^{2} x - \sinh^{2} x \equiv 1 ⋆, \sinh 2 x \equiv 2 \sinh x \cosh x ⋆, \cosh 2 x \equiv \cosh^{2} x + \sinh^{2} x ⋆,

1 - \tanh^{2} x \equiv {sech}^{2} x, \coth^{2} x - 1 \equiv {cosech}^{2} x,

\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 .

The two rearrangements of $\cosh 2 x$ do most of the work in equations and integrals:

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

Proof of

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

— the template for every other proof

Go straight back to the exponential definitions and expand:

\cosh^{2} x - \sinh^{2} x = {(\frac{e^{x} + e^{- x}}{2})}^{2} - {(\frac{e^{x} - e^{- x}}{2})}^{2} = \frac{(e^{2 x} + 2 + e^{- 2 x}) - (e^{2 x} - 2 + e^{- 2 x})}{4} = \frac{4}{4} = 1 . ■

Notice that $e^{x} \cdot e^{- x} = 1$ is what produces the middle terms $\pm 2$ . Every hyperbolic proof in this course works the same way: write both sides in terms of $e^{x}$ and $e^{- x}$ , and simplify. There is no cleverness required.

Proof of

\cosh 2 x \equiv 1 + 2 \sinh^{2} x

1 + 2 \sinh^{2} x = 1 + 2 {(\frac{e^{x} - e^{- x}}{2})}^{2} = 1 + \frac{e^{2 x} - 2 + e^{- 2 x}}{2} = \frac{2 + e^{2 x} - 2 + e^{- 2 x}}{2} = \frac{e^{2 x} + e^{- 2 x}}{2} = \cosh 2 x . ■

★ Osborn's rule — for remembering, never for proving

Take any standard trigonometric identity, replace $\cos \to \cosh$ and $\sin \to \sinh$ , and then change the sign of every term containing a product of two sines.

Trigonometric	Hyperbolic	Sign flipped?
$\cos^{2} x + \sin^{2} x \equiv 1$	$\cosh^{2} x - \sinh^{2} x \equiv 1$	yes — $\sin^{2}$ is $\sin \times \sin$
$\sin 2 x \equiv 2 \sin x \cos x$	$\sinh 2 x \equiv 2 \sinh x \cosh x$	no — only one $\sinh$
$\cos 2 x \equiv \cos^{2} x - \sin^{2} x$	$\cosh 2 x \equiv \cosh^{2} x + \sinh^{2} x$	yes
$\sec^{2} x \equiv 1 + \tan^{2} x$	${sech}^{2} x \equiv 1 - \tanh^{2} x$	yes — $\tanh^{2}$ hides $\sinh \times \sinh$

Watch $\tanh^{2}$ , $\coth^{2}$ and ${cosech}^{2}$ : each secretly contains $\sinh \times \sinh$ , so each flips.

⚠ If the question says "prove", Osborn scores nothing. Go back to $e^{x}$ and $e^{- x}$ . Osborn is for checking your memory in the margin.

§4 Inverse hyperbolic functions and their logarithmic forms

◇ Domains and ranges

Inverse	Domain	Range	Why
$arsinh x$	all real $x$	all real $y$	$\sinh$ is one-to-one already
$arcosh x$	$x \geq 1$	$y \geq 0$	$\cosh$ is even, so we take the positive branch only
$artanh x$	$\| x \| < 1$	all real $y$	$\tanh$ never reaches $\pm 1$

◆ The logarithmic forms all three on MF19

arsinh x = \ln (x + \sqrt{x^{2} + 1}), arcosh x = \ln (x + \sqrt{x^{2} - 1}) (x \geq 1),

artanh x = \frac{1}{2} \ln (\frac{1 + x}{1 - x}) (| x | < 1) .

They are printed — but the syllabus says derive and use, so the derivation is examinable in its own right. It is the same three-step move every time.

Deriving

arsinh x = \ln (x + \sqrt{x^{2} + 1})

Let $y = arsinh x$ , so $x = \sinh y = \frac{e^{y} - e^{- y}}{2}$ , giving $2 x = e^{y} - e^{- y}$ .
Multiply through by $e^{y}$ to clear the negative power — this is the key step: $2 x e^{y} = e^{2 y} - 1 ⟹ (e^{y})^{2} - 2 x e^{y} - 1 = 0,$ a quadratic in $e^{y}$ .
By the quadratic formula, $e^{y} = \frac{2 x \pm \sqrt{4 x^{2} + 4}}{2} = x \pm \sqrt{x^{2} + 1}$ .
Discard the negative root. Since $\sqrt{x^{2} + 1} > \sqrt{x^{2}} = | x |$ , the minus sign would make $e^{y} < 0$ , which is impossible. So $e^{y} = x + \sqrt{x^{2} + 1}$ and

y = \ln (x + \sqrt{x^{2} + 1}) . ■

Step 4 is worth a mark on its own — you must say why the other root is rejected, not simply drop it.

Deriving

arcosh x

and

artanh x

arcosh. With $y = arcosh x$ and $x \geq 1$ : $2 x = e^{y} + e^{- y}$ , so $(e^{y})^{2} - 2 x e^{y} + 1 = 0$ and $e^{y} = x \pm \sqrt{x^{2} - 1}$ . Here both roots are positive, so the sign is decided by the range instead: the principal branch has $y \geq 0$ , i.e. $e^{y} \geq 1$ , which selects the $+$ . Hence $arcosh x = \ln (x + \sqrt{x^{2} - 1})$ .

(The two roots multiply to $1$ , so $\ln (x - \sqrt{x^{2} - 1}) = - arcosh x$ — the negative branch, and a favourite wrong answer.)

artanh. With $y = artanh x$ and $| x | < 1$ , write $\tanh y$ with $e^{2 y}$ :

x = \frac{e^{y} - e^{- y}}{e^{y} + e^{- y}} = \frac{e^{2 y} - 1}{e^{2 y} + 1} ⟹ x e^{2 y} + x = e^{2 y} - 1 ⟹ e^{2 y} (1 - x) = 1 + x,

⟹ e^{2 y} = \frac{1 + x}{1 - x} ⟹ y = \frac{1}{2} \ln (\frac{1 + x}{1 - x}) . ■

The condition $| x | < 1$ is exactly what keeps $\frac{1 + x}{1 - x}$ positive, so the logarithm is defined.

§5 Solving hyperbolic equations

◆ Two routes — pick by what the equation contains

Mixed $\sinh$ and $\cosh$ of the same angle ⟶ replace both by exponentials and multiply by $e^{x}$ to get a quadratic in $e^{x}$ .
Double angles present ⟶ use $\cosh 2 x = 2 \cosh^{2} x - 1$ or $1 + 2 \sinh^{2} x$ to make it a quadratic in $\cosh x$ or $\sinh x$ , then finish with a logarithmic form.

✎ Worked example — route 1

Solve $3 \sinh x - 2 \cosh x = 2$ , giving the answer exactly.

3 \cdot \frac{e^{x} - e^{- x}}{2} - 2 \cdot \frac{e^{x} + e^{- x}}{2} = 2 ⟹ e^{x} - 5 e^{- x} = 4 .

Multiply by $e^{x}$ : $e^{2 x} - 4 e^{x} - 5 = 0$ , i.e. $(e^{x} - 5) (e^{x} + 1) = 0$ .

$e^{x} = - 1$ is impossible, so $e^{x} = 5$ and $x = \ln 5$ .

Check: $\sinh (\ln 5) = \frac{1}{2} (5 - \frac{1}{5}) = \frac{12}{5}$ and $\cosh (\ln 5) = \frac{1}{2} (5 + \frac{1}{5}) = \frac{13}{5}$ , so $3 (\frac{12}{5}) - 2 (\frac{13}{5}) = \frac{36 - 26}{5} = 2$ ✓

✎ Worked example — route 2

Solve $\cosh 2 x - 5 \cosh x + 4 = 0$ , giving the answers exactly.

Use $\cosh 2 x = 2 \cosh^{2} x - 1$ so that everything is in $\cosh x$ :

2 \cosh^{2} x - 1 - 5 \cosh x + 4 = 0 ⟹ 2 \cosh^{2} x - 5 \cosh x + 3 = 0 ⟹ (2 \cosh x - 3) (\cosh x - 1) = 0 .

So $\cosh x = 1$ or $\cosh x = \frac{3}{2}$ .

$\cosh x = 1 \Rightarrow x = 0$ (the minimum of $\cosh$ , so a single solution).
$\cosh x = \frac{3}{2} \Rightarrow x = \pm arcosh \frac{3}{2} = \pm \ln (\frac{3}{2} + \sqrt{\frac{9}{4} - 1}) = \pm \ln (\frac{3 + \sqrt{5}}{2})$ .

Because $\cosh$ is even, every non-minimum value gives a $\pm$ pair — dropping the negative one is the standard lost mark here. Numerically $\ln (\frac{3 + \sqrt{5}}{2}) \approx 0.962$ .

§6 Interactive: build them from exponentials

The first view shows $\frac{1}{2} e^{x}$ and $\frac{1}{2} e^{- x}$ dashed: $\cosh$ is their sum, $\sinh$ their difference. Slide $x$ and the panel checks the identities and the logarithmic forms numerically at that value.

Evaluate at

x

0.80

cosh = sum, sinh = difference ✎

§7 Examiner traps & A* checklist

⚠ The seven most common mark-losers

Using $\cosh^{2} x + \sinh^{2} x = 1$ . The identity is $\cosh^{2} x - \sinh^{2} x = 1$ , and it is printed on MF19 — there is no excuse.
Losing a solution of $\cosh x = k$ . $\cosh$ is even, so $x = \pm arcosh k$ for any $k > 1$ .
Inventing a second solution for $\sinh x = k$ . $\sinh$ is one-to-one, so there is exactly one.
Quoting Osborn's rule as a proof. Prove from $e^{x}$ and $e^{- x}$ .
Forgetting to say why the negative root is rejected when deriving $arsinh$ .
Writing $arcosh x = \ln (x - \sqrt{x^{2} - 1})$ — that is $- arcosh x$ .
Ignoring the domain: $arcosh$ needs $x \geq 1$ and $artanh$ needs $| x | < 1$ . A "solution" outside those must be rejected and seen to be rejected.

★ Before you turn the page

Did I check whether the function I inverted was even (⟹ two answers) or odd (⟹ one)?
Are my exact answers in logarithmic form, simplified, with surds tidied?
If asked to prove, did I start from the exponential definitions?

§8 Video explainers (curated)

▶Definitions & graphs
Curated search — sinh, cosh, tanh from exponentials.↗ ▶Identities & Osborn's rule
Curated search — proving identities from the definitions.↗ ▶Inverse hyperbolics & log forms
Curated search — the quadratic-in-eʸ derivation.↗

§9 Check yourself

Score 0 / 5

Q1 · definitions

$\cosh x - \sinh x$ equals:

Solution

$\frac{e^{x} + e^{- x}}{2} - \frac{e^{x} - e^{- x}}{2} = \frac{2 e^{- x}}{2} = e^{- x}$ . (Adding instead gives $\cosh x + \sinh x = e^{x}$ .) Multiplying the two results gives $\cosh^{2} x - \sinh^{2} x = e^{x} e^{- x} = 1$ — a one-line proof of the main identity.

Q2 · identities

${sech}^{2} x$ equals:

Solution

Divide $\cosh^{2} x - \sinh^{2} x = 1$ through by $\cosh^{2} x$ : $1 - \tanh^{2} x = {sech}^{2} x$ . By Osborn, the trig version $\sec^{2} x = 1 + \tan^{2} x$ flips sign because $\tanh^{2}$ contains $\sinh \times \sinh$ . The last option is ${cosech}^{2} x$ .

Q3 · logarithmic form

For $x \geq 1$ , $arcosh x$ equals:

Solution

$\ln (x + \sqrt{x^{2} + 1})$ is $arsinh x$ and $\frac{1}{2} \ln (\frac{1 + x}{1 - x})$ is $artanh x$ . The third option is $- arcosh x$ , because the two roots of the quadratic multiply to $1$ . Check at $x = 1$ : $\ln (1 + 0) = 0$ ✓, and $\cosh 0 = 1$ ✓

Q4 · solving

The complete solution of $\cosh 2 x - 5 \cosh x + 4 = 0$ is:

Solution

$2 \cosh^{2} x - 5 \cosh x + 3 = 0$ factorises as $(2 \cosh x - 3) (\cosh x - 1) = 0$ , so $\cosh x = \frac{3}{2}$ or $1$ . Since $\cosh$ is even, $\cosh x = \frac{3}{2}$ gives the pair $x = \pm arcosh \frac{3}{2} = \pm \ln (\frac{3 + \sqrt{5}}{2})$ ; $\cosh x = 1$ is the minimum, so it gives only $x = 0$ . Omitting the negative branch is the trap.

Q5 · full method — try it on paper first

(i) Prove from the definitions that $\cosh 2 x \equiv 1 + 2 \sinh^{2} x$ .

(ii) Derive the logarithmic form of $arsinh x$ .

(iii) Hence solve $2 \cosh 2 x - \sinh x = 5$ , giving your answers exactly.

Solution

(i) Work on the right-hand side, in exponentials:

1 + 2 \sinh^{2} x = 1 + 2 {(\frac{e^{x} - e^{- x}}{2})}^{2} = 1 + \frac{e^{2 x} - 2 + e^{- 2 x}}{2} = \frac{e^{2 x} + e^{- 2 x}}{2} = \cosh 2 x . ■

(ii) Let $y = arsinh x$ , so $2 x = e^{y} - e^{- y}$ . Multiplying by $e^{y}$ gives the quadratic $(e^{y})^{2} - 2 x e^{y} - 1 = 0$ , so

e^{y} = \frac{2 x \pm \sqrt{4 x^{2} + 4}}{2} = x \pm \sqrt{x^{2} + 1} .

Since $\sqrt{x^{2} + 1} > | x |$ , the minus sign gives $e^{y} < 0$ , which is impossible; so $e^{y} = x + \sqrt{x^{2} + 1}$ and $arsinh x = \ln (x + \sqrt{x^{2} + 1})$ . $■$

(iii) Substituting part (i) turns the equation into a quadratic in $\sinh x$ :

2 (1 + 2 \sinh^{2} x) - \sinh x = 5 ⟹ 4 \sinh^{2} x - \sinh x - 3 = 0 ⟹ (4 \sinh x + 3) (\sinh x - 1) = 0 .

So $\sinh x = 1$ or $\sinh x = - \frac{3}{4}$ . Since $\sinh$ is one-to-one, each gives exactly one solution, and part (ii) supplies both:

x = \ln (1 + \sqrt{2}) and x = \ln (- \frac{3}{4} + \sqrt{\frac{9}{16} + 1}) = \ln (- \frac{3}{4} + \frac{5}{4}) = \ln \frac{1}{2} = - \ln 2 .

Check: $\sinh (- \ln 2) = \frac{1}{2} (\frac{1}{2} - 2) = - \frac{3}{4}$ ✓, and $\sinh (\ln (1 + \sqrt{2})) = \frac{1}{2} ((1 + \sqrt{2}) - (\sqrt{2} - 1)) = 1$ ✓ $■$

FP2 · Topic 8 · Hyperbolic Functions