The six hyperbolic functions and their exponential definitions

Hyperbolic functions are the first topic of Further Pure Mathematics 3, the A Level unit examined as paper WFM03/01.

In the June 2024, January 2025 and June 2025 papers each had one question on this unit, worth 6 to 9 marks.

FP3 assumes everything in P1 to P4 and FP1, and this section leans on the P3 work on $\mathrm{e}^x$ and $\ln x$ in almost every line: the laws of indices, $\mathrm{e}^{\ln a} = a$, and solving a quadratic in a disguised variable.

Every function defined for all $x$ can be split into an even part (unchanged when $x$ is replaced by $-x$) and an odd part (which changes sign).

For $\mathrm{e}^x$ the split is

These two parts are so useful that they have names.

The even part is the hyperbolic cosine, $\cosh x$, and the odd part is the hyperbolic sine, $\sinh x$:

So $\cosh x + \sinh x = \mathrm{e}^x$ and $\cosh x - \sinh x = \mathrm{e}^{-x}$.

Both facts drop out at once from the definitions, and both are worth remembering because they turn hyperbolic problems back into exponential ones.

The other four hyperbolic functions are built from these two in exactly the way $\tan$, $\sec$, $\operatorname{cosec}$ and $\cot$ are built from $\sin$ and $\cos$:

Multiplying the top and bottom of $\tanh x$ by $\mathrm{e}^x$ gives a second form that is often quicker: $\tanh x = \dfrac{\mathrm{e}^{2x} - 1}{\mathrm{e}^{2x} + 1}$.

In the same way $\coth x = \dfrac{\mathrm{e}^{2x} + 1}{\mathrm{e}^{2x} - 1}$.

Notice that $\operatorname{cosech} x$ and $\coth x$ are undefined at $x = 0$, because $\sinh 0 = \frac{1}{2}(1 - 1) = 0$.

The names are said "cosh", "shine" (or "sinch"), "than", "sheck", "co-sheck" and "coth".

In writing, the $h$ is not optional: $\sinh x$ and $\sin x$ are different functions, and the mark schemes give no accuracy mark to a line that writes one for the other.

The point $(\cos t, \sin t)$ moves round the circle $x^2 + y^2 = 1$, because $\cos^2 t + \sin^2 t = 1$.

The point $(\cosh t, \sinh t)$ moves along the hyperbola $x^2 - y^2 = 1$, because, as the identities section of this unit proves, $\cosh^2 t - \sinh^2 t = 1$.

The diagram shows both curves.

As $t$ runs from $-\infty$ to $\infty$, the point $(\cosh t, \sinh t)$ traces the right-hand branch only, since $\cosh t$ is never less than 1.

There is a deeper match too: on the circle, $t$ is twice the area of the sector from $(1, 0)$ to the point, and on the hyperbola $t$ is twice the area of the corresponding hyperbolic sector.

That is why the inverse functions later in this unit are written $\operatorname{arsinh}$ ("area sinh"), not "arcsinh".

FP3 uses exactly this parametrisation, $x = a\cosh t$, $y = b\sinh t$, for the hyperbola in the coordinate systems unit.

At $x = 0$: $\cosh 0 = \frac{1}{2}(1 + 1) = 1$, $\sinh 0 = 0$, $\tanh 0 = 0$ and $\operatorname{sech} 0 = 1$, while $\operatorname{cosech} 0$ and $\coth 0$ do not exist.

At $x = \ln a$ (with $a > 0$), the definitions turn into plain fractions, because $\mathrm{e}^{\ln a} = a$ and $\mathrm{e}^{-\ln a} = \frac{1}{a}$:

This is the bridge between hyperbolic functions and logarithms, and it is why FP3 answers are so often natural logarithms.

For instance $\cosh(\ln 3) = \frac{10}{6} = \frac{5}{3}$ and $\sinh(\ln 3) = \frac{8}{6} = \frac{4}{3}$.

An equation such as $\sinh x = k$ or $\coth x = k$ becomes a quadratic once you write it in exponentials and multiply through by $\mathrm{e}^x$ (or, for $\tanh$ and $\coth$, by the denominator).

With $u = \mathrm{e}^x$:

1. Replace each hyperbolic function by its definition.

2. Multiply through by $2\mathrm{e}^x$ (or by the denominator) so that $\mathrm{e}^{-x}$ disappears.

3. Solve the quadratic in $u$ by factorising, the formula or completing the square.

4. Keep only the positive roots, because $u = \mathrm{e}^x > 0$ for every real $x$, then take $x = \ln u$.

Step 4 is where the understanding sits.

A negative or zero value of $u$ is not a wrong answer to the quadratic; it is a value that $\mathrm{e}^x$ can never take, so it gives no $x$ at all.

The definitions use nothing beyond P3 exponentials, but they open a great deal.

A chain or cable hanging under its own weight takes the shape $y = c\cosh\frac{x}{c}$, called a catenary, which is why $\cosh$ is the shape of a hanging power line and, turned upside down, of the strongest arches.

In FP3 itself, the hyperbolic functions return in the hyperbola's parametric form, in differentiation and integration (they make integrals such as $\int \frac{1}{\sqrt{x^2 + 9}}\,\mathrm{d}x$ routine), and in surface areas.

In physics, $\tanh$ is how velocities combine in special relativity.

1. (a) Determine the exact values of $\sinh(\ln 6)$ and $\cosh(\ln 6)$. (2)
(b) Hence determine the exact values of $\tanh(\ln 6)$, $\operatorname{sech}(\ln 6)$ and $\coth(\ln 6)$. (3) Total [5]

2. (a) Show that, for $x \neq 0$, $\coth x - 1 = \dfrac{2}{\mathrm{e}^{2x} - 1}$. (2)
(b) Hence solve $\coth x = 2$, giving your answer as a natural logarithm. (3) Total [5]

3. A student claims that there is a value of $x$ for which $\cosh x = \sinh x$.
(a) Show that the student's claim is false. (3)
(b) Show that $(\cosh x + \sinh x)^n = \cosh nx + \sinh nx$ for every positive integer $n$. (3) Total [6]

4. Starting from the exponential definition of $\cosh x$, solve $\operatorname{sech} x = \frac{12}{13}$, giving your answers as natural logarithms. (5) Total [5]

5. It is given that $\cosh x = \frac{41}{9}$ and that $x < 0$.
(a) Starting from the exponential definition of $\cosh x$, determine the exact value of $\sinh x$. (4)
(b) Hence determine the exact values of $\tanh x$ and $\operatorname{cosech} x$. (2) Total [6]

6. (a) Show that $\sinh\left(\ln\left(2 + \sqrt{5}\right)\right) = 2$. (3)
(b) Hence determine the exact value of $\cosh\left(\ln\left(2 + \sqrt{5}\right)\right)$. (2) Total [5]