Hyperbolic Functions: Definitions

The hyperbolic functions $\sinh$ , $\cosh$ and $\tanh$ are close cousins of the trigonometric functions, but built from the exponential function rather than the unit circle. They share almost all the algebraic structure of $\sin$ , $\cos$ and $\tan$ — with a handful of strategic sign changes — and they unlock a whole family of integrals and differential equations later in the course. This lesson defines them, derives the fundamental identity $\cosh^2 x-\sinh^2 x=1$ , sketches the three graphs, and shows the exponential techniques that solve hyperbolic equations exactly.

1 Spec mapping

Hyperbolic functions are compulsory pure content for Papers 1 and 2 of AQA 7367. This opening lesson is the foundation for everything that follows in the strand: hyperbolic identities, inverse hyperbolic functions, and their differentiation and integration. The work is largely AO1 (apply the definitions and the identity), with the proofs of the identity and the logarithmic-form derivations reaching firmly into AO2 (construct a rigorous argument). It builds directly on the exponential and logarithm work and the laws of indices from A-Level Maths. Because the hyperbolic functions reappear in almost every later calculus topic — they are the key to a large class of standard integrals — time spent making these definitions second nature pays off repeatedly. Examiners particularly favour "prove from the definitions" questions here, precisely because the proofs are short, self-contained, and a clean test of whether a candidate truly understands the exponential basis rather than merely memorising results.

2 Core theory — definitions

The two basic hyperbolic functions are defined as the even and odd parts of $e^x$ :

$\boxed{\ \sinh x = \frac{e^x - e^{-x}}{2}, \qquad \cosh x = \frac{e^x + e^{-x}}{2}\ }$

( $\sinh$ is pronounced "shine" or "sinch"; $\cosh$ rhymes with "gosh".) Their quotient defines the hyperbolic tangent:

$\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}.$

The three reciprocal functions complete the family:

$\operatorname{cosech} x = \frac{1}{\sinh x}\ (x\ne0), \qquad \operatorname{sech} x = \frac{1}{\cosh x}, \qquad \coth x = \frac{\cosh x}{\sinh x}\ (x\ne0).$

A useful way to remember which is which: $\cosh$ takes the + sign (it is the even part of $e^x$ , so adding $e^x$ and $e^{-x}$ keeps it symmetric), while $\sinh$ takes the − sign (the odd part). Everything else follows from these two.

It is worth dwelling on why these particular combinations deserve names. Any function $f$ can be split uniquely into an even part $\tfrac12(f(x)+f(-x))$ and an odd part $\tfrac12(f(x)-f(-x))$ . Apply this to $f(x)=e^x$ , noting $f(-x)=e^{-x}$ : the even part is exactly $\cosh x$ and the odd part is exactly $\sinh x$ . So $e^x=\cosh x+\sinh x$ is nothing more than the even–odd decomposition of the exponential, which is why the hyperbolic functions inherit so much of the exponential's beautiful algebra. This single observation explains every recovery formula and every addition identity you will meet — they are all just facts about $e^x$ in disguise. The trigonometric functions are the same decomposition applied to $e^{\mathrm ix}$ , which is precisely why the two families are so closely related.

3 Key values and the exponential recovery formulae

A handful of special values anchor the graphs and act as quick sanity checks. Evaluating at $x=0$ : $\sinh 0=\tfrac{1-1}{2}=0$ , $\cosh 0=\tfrac{1+1}{2}=1$ , $\tanh 0=0$ . The contrast with the circular functions is immediate — $\cos 0=1$ matches $\cosh 0=1$ , but whereas $\sin$ and $\cos$ oscillate forever, the hyperbolic values grow without bound as $x$ increases. The table also shows the parity at a glance: $\sinh(-1)=-\sinh(1)$ (odd) while $\cosh(-1)=\cosh(1)$ (even).

$x$	$\sinh x$	$\cosh x$	$\tanh x$
$0$	$0$	$1$	$0$
$1$	$\tfrac{e-e^{-1}}{2}\approx1.175$	$\tfrac{e+e^{-1}}{2}\approx1.543$	$\approx0.762$
$-1$	$\approx-1.175$	$\approx1.543$	$\approx-0.762$

Adding and subtracting the two definitions gives the two recovery formulae, which are constantly useful:

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

Multiplying these two together is, incidentally, a one-line proof of the fundamental identity: $(\cosh x+\sinh x)(\cosh x-\sinh x)=e^x e^{-x}=1$ , i.e. $\cosh^2 x-\sinh^2 x=1$ . This is a much slicker route than the brute-force squaring in §4, and it is worth having both in your toolkit — the squaring proof is what examiners usually ask for under "prove from the definitions", but the factorised version makes the structure transparent. The recovery formulae are also the key to raising hyperbolic expressions to powers, as Worked Example 3 will show, because they convert $\cosh x\pm\sinh x$ into a single exponential that obeys the ordinary index laws.

4 The fundamental identity

$\boxed{\ \cosh^2 x - \sinh^2 x = 1\ }$

Proof from the definitions.

$\begin{aligned} \cosh^2 x - \sinh^2 x &= \left(\frac{e^x+e^{-x}}{2}\right)^2 - \left(\frac{e^x-e^{-x}}{2}\right)^2 \\ &= \frac{e^{2x}+2+e^{-2x}}{4} - \frac{e^{2x}-2+e^{-2x}}{4} = \frac{4}{4} = 1. \end{aligned}$

This is the hyperbolic analogue of $\sin^2 x+\cos^2 x=1$ , but with a minus sign — the single most important structural difference between the two families, and the source of every sign change you will meet via Osborn's rule in the next lesson. The name "hyperbolic" comes from exactly this identity: the point $(\cosh t,\sinh t)$ satisfies $X^2-Y^2=1$ and so traces a hyperbola, just as $(\cos t,\sin t)$ traces the unit circle $X^2+Y^2=1$ .

5 The three graphs

Knowing the shapes is examinable, so internalise each one and — more importantly — how it follows from the definition, since that reasoning is what earns the marks if you are asked to justify a feature. A reliable trick for all three graphs is to think of them as combinations of the two exponential curves $\tfrac12 e^x$ (rising) and $\tfrac12 e^{-x}$ (falling): adding them gives $\cosh$ , subtracting gives $\sinh$ , and their dominance for large $|x|$ dictates the tails.

$y=\sinh x$ . Domain $\mathbb R$ , range $\mathbb R$ ; an odd function ( $\sinh(-x)=-\sinh x$ ), so its graph has rotational symmetry about the origin. It passes through the origin like a gently stretched cubic, with gradient $\cosh 0=1$ there. For large positive $x$ the falling term $-\tfrac12 e^{-x}$ dies away and $\sinh x\approx\tfrac12 e^{x}$ , so the right-hand tail climbs exponentially; by oddness the left tail plunges as $-\tfrac12 e^{-x}$ . There is no maximum or minimum — the function is strictly increasing everywhere, which is exactly why $\sinh x=k$ always has a unique solution.

$y=\cosh x$ . Domain $\mathbb R$ , range $[1,\infty)$ ; an even function ( $\cosh(-x)=\cosh x$ ), so its graph is symmetric about the $y$ -axis. It is a smooth U-shape with a single minimum value of $1$ at $x=0$ (not $0$ — a frequent slip), and for large $|x|$ it grows like $\tfrac12 e^{|x|}$ on both sides. Although it superficially resembles a parabola near the bottom, it grows far faster than any polynomial. This is the catenary, the shape a uniform flexible chain takes when hung between two points under its own weight.

$y=\tanh x$ . Domain $\mathbb R$ , range the open interval $(-1,1)$ ; an odd function. It is the classic S-shape (sigmoid) through the origin, with gradient $\operatorname{sech}^2 0=1$ there, flattening towards horizontal asymptotes $y=\pm1$ . The asymptotes follow from the exponential form: writing $\tanh x=\tfrac{e^{2x}-1}{e^{2x}+1}$ (divide numerator and denominator of the definition by $e^{-x}$ ), as $x\to+\infty$ the $e^{2x}$ terms dominate and the ratio $\to1$ ; as $x\to-\infty$ the constant terms dominate and it $\to-1$ . It never actually attains $\pm1$ , which is why the range is open — the reason $\tanh$ is so popular as a "squashing" function in machine learning, where a bounded smooth output is wanted.

6 Worked examples with mark scheme

The two recurring tasks in this strand are (i) given one hyperbolic value, find the others using the fundamental identity, and (ii) solve a hyperbolic equation by going back to exponentials. The worked examples drill both.

Worked Example 1 — using the identity

Given $\sinh x=\tfrac34$ , find $\cosh x$ and $\tanh x$ .

The fundamental identity lets us find $\cosh$ from $\sinh$ without ever solving for $x$ itself — much faster than computing $x$ and substituting back.

$\cosh^2 x = 1+\sinh^2 x = 1+\tfrac{9}{16}=\tfrac{25}{16} \ \implies\ \cosh x = \tfrac54 \ (\text{positive, since } \cosh x\ge1).$

$\tanh x = \frac{\sinh x}{\cosh x} = \frac{3/4}{5/4} = \tfrac35.$

Mark scheme: M1 use $\cosh^2 x=1+\sinh^2 x$ ; A1 $\cosh x=\tfrac54$ (taking the positive root with reason); A1 $\tanh x=\tfrac35$ .

Worked Example 2 — solving via the definition

Show that $\sinh x=2$ gives $x=\ln(2+\sqrt5)$ .

This is the archetypal "solve a hyperbolic equation" question, and the method generalises to every such problem: replace the hyperbolic function by its exponential definition, multiply through by $e^x$ to clear the negative power, and recognise the result as a quadratic in $e^x$ . Substitute the definition and clear the fraction:

$\frac{e^x-e^{-x}}{2}=2 \ \implies\ e^x - e^{-x} = 4 \ \implies\ e^{2x}-4e^{x}-1=0 \quad(\times\, e^x).$

Let $u=e^x>0$ :

$u^2-4u-1=0 \ \implies\ u=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt5.$

Since $u=e^x>0$ we reject $2-\sqrt5<0$ , leaving $e^x=2+\sqrt5$ , so $x=\ln(2+\sqrt5)$ .

Mark scheme: M1 substitute the exponential definition; M1 form the quadratic in $e^x$ ; A1 $e^x=2+\sqrt5$ (rejecting the negative root with reason); A1 $x=\ln(2+\sqrt5)$ .

Hyperbolic Functions: Definitions

Hyperbolic Functions: Definitions

1 Spec mapping

2 Core theory — definitions

3 Key values and the exponential recovery formulae

4 The fundamental identity

5 The three graphs

6 Worked examples with mark scheme

Worked Example 1 — using the identity

Worked Example 2 — solving via the definition

Worked Example 3 — the recovery formula in action

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