Hyperbolic functions.

HookGalileo called the hanging chain a parabola — the Gateway Arch says otherwise

ModelBuilt from exponentials, on the blueprint of trig

The definitions are the section: \(\sinh x=\frac{e^x-e^{-x}}{2}\), \(\cosh x=\frac{e^x+e^{-x}}{2}\), and \(\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^{2x}-1}{e^{2x}+1}\). The graphs follow directly. \(\sinh\) is odd, passes through the origin, and grows like \(\frac{e^x}{2}\) for large \(x\): domain and range both all of \(\mathbb{R}\). \(\cosh\) is even, is the average of \(e^x\) and \(e^{-x}\), and has minimum value \(\cosh 0=1\) — its range is \(y\ge 1\), and the curve is exactly the catenary. \(\tanh\) is odd and squashed: since \(\cosh x>|\sinh x|\) everywhere, \(\tanh\) is trapped in \(-1<\tanh x<1\), rising through the origin with gradient \(1\) and flattening onto horizontal asymptotes \(y=\pm 1\). That saturating S-shape is why \(\tanh\) reappears everywhere from special relativity (velocities add via \(\tanh\)) to neural networks (an activation that caps its output).

The three reciprocals complete the family: \(\operatorname{sech}x=\frac{1}{\cosh x}\) with range \(0<y\le 1\), \(\operatorname{cosech}x=\frac{1}{\sinh x}\) undefined at \(x=0\), and \(\coth x=\frac{1}{\tanh x}\), also undefined at \(0\), with \(|\coth x|>1\). Domains and ranges are not decoration here — AQA asks for them directly, and they control which inverse functions exist and where the log forms are valid.

ModelIdentities and the hyperbola: where the name comes from

Square the definitions and subtract, and the cross terms do the work: \(\cosh^2 x-\sinh^2 x=\frac{e^{2x}+2+e^{-2x}}{4}-\frac{e^{2x}-2+e^{-2x}}{4}=1\). This is the section's Pythagoras, and it explains the name: the point \((\cos t,\sin t)\) traces the unit circle \(x^2+y^2=1\), while \((\cosh t,\sinh t)\) traces a branch of the unit hyperbola \(x^2-y^2=1\) — the same conic that appears in the further-algebra section, now parametrised. The double-angle results follow the trig blueprint: \(\sinh 2x=2\sinh x\cosh x\) exactly as in trig, and \(\cosh 2x=\cosh^2 x+\sinh^2 x=2\cosh^2 x-1=1+2\sinh^2 x\), where the middle form has a plus sign trig does not.

The sign pattern is captured by Osborn's rule: take any trig identity and flip the sign of every term containing a product of two sinhs — \(\sinh^2\), or \(\sinh A\sinh B\), including the hidden product inside \(\tanh^2=\frac{\sinh^2}{\cosh^2}\). So \(\cos^2+\sin^2=1\) becomes \(\cosh^2-\sinh^2=1\), \(1+\tan^2=\sec^2\) becomes \(1-\tanh^2=\operatorname{sech}^2\), and \(\sin 2x=2\sin x\cos x\) is untouched because it contains only one sinh factor. Osborn's rule is a memory aid, not a proof — in a 'show that' question the exponential definitions are the argument, and the rule is merely how you predict what you are about to prove.

MechanismCalculus without the sign flip

Differentiate the definitions and the exponentials shuffle: \(\frac{d}{dx}\sinh x=\cosh x\) and \(\frac{d}{dx}\cosh x=\sinh x\) — no minus sign. Where trig's derivative cycle takes four steps to return (\(\sin\to\cos\to-\sin\to-\cos\to\sin\)), the hyperbolic cycle takes two: the pair just swap. Quotient rule then gives \(\frac{d}{dx}\tanh x=\operatorname{sech}^2 x\), the exact analogue of \(\sec^2\). Reversing: \(\int\sinh x\,dx=\cosh x+c\), \(\int\cosh x\,dx=\sinh x+c\), and \(\int\tanh x\,dx=\int\frac{\sinh x}{\cosh x}\,dx=\ln(\cosh x)+c\) — a function-over-its-derivative pattern, with no modulus needed since \(\cosh x\ge 1>0\).

These feed every larger technique unchanged: chain rule (\(\frac{d}{dx}\cosh 3x=3\sinh 3x\)), product rule, parts (\(\int x\sinh x\,dx=x\cosh x-\sinh x+c\)), and the identity-then-integrate move for even powers — \(\int\cosh^2 x\,dx\) surrenders to \(\cosh^2 x=\frac{\cosh 2x+1}{2}\), mirroring how \(\cos^2\) surrenders to its double angle. The one place trig instinct actively betrays you is the sign on \(\frac{d}{dx}\cosh\): writing \(-\sinh x\) is the single most common error in the topic, and it silently wrecks every later line of a curve-sketching or arc-length part. When any doubt creeps in, differentiate the exponential form — \(\frac{e^x+e^{-x}}{2}\) visibly differentiates to \(\frac{e^x-e^{-x}}{2}\), and the question is settled in one line.

MechanismInverses and their logarithmic forms

Because \(\sinh\) is strictly increasing over all of \(\mathbb{R}\), it inverts cleanly: \(\operatorname{arsinh}x\) has domain and range \(\mathbb{R}\). \(\tanh\) inverts to \(\operatorname{artanh}x\), defined only for \(-1<x<1\) — the range of \(\tanh\) becomes the domain of its inverse. \(\cosh\) is the awkward one: it is even, so it fails to be one-to-one, and the convention is to invert only the right-hand branch, giving \(\operatorname{arcosh}x\) with domain \(x\ge 1\) and range \(y\ge 0\). That restriction is not small print — it is the fact that decides a \(\pm\) sign in the derivation below, and AQA's spec statement H4 says 'derive', so the derivation itself is examinable, not just the result.

Each inverse has a closed logarithmic form, obtained the same way every time: set \(y\) equal to the inverse, rewrite in exponentials, and solve a hidden quadratic in \(e^y\). The results: \(\operatorname{arsinh}x=\ln\big(x+\sqrt{x^2+1}\big)\), \(\operatorname{arcosh}x=\ln\big(x+\sqrt{x^2-1}\big)\) for \(x\ge 1\), and \(\operatorname{artanh}x=\frac{1}{2}\ln\frac{1+x}{1-x}\) for \(|x|<1\). These are what make 'give an exact answer' possible: a solution like \(x=\operatorname{arsinh}\frac{3}{4}\) converts to \(\ln 2\) — see the exam-technique block — and the log forms are the bridge.

MechanismThe integrals the section was built for

The hyperbolic substitutions exist to kill square roots of quadratics. In \(\int\frac{dx}{\sqrt{x^2+a^2}}\), substitute \(x=a\sinh u\): then \(\sqrt{x^2+a^2}=\sqrt{a^2\sinh^2 u+a^2}=a\cosh u\) by the Pythagorean identity, \(dx=a\cosh u\,du\), and the integral collapses to \(\int du=u+c\). Hence \[\int\frac{dx}{\sqrt{x^2+a^2}}=\operatorname{arsinh}\frac{x}{a}+c,\qquad\int\frac{dx}{\sqrt{x^2-a^2}}=\operatorname{arcosh}\frac{x}{a}+c,\] the second via \(x=a\cosh u\). Compare the further-calculus section: \(\sqrt{a^2-x^2}\) called for a trig substitution because \(1-\sin^2=\cos^2\) kills a difference from \(a^2\). The sign pattern under the root is the dispatcher: \(a^2-x^2\) trig, \(x^2+a^2\) sinh, \(x^2-a^2\) cosh.

Exam integrals hide the pattern behind a general quadratic, and completing the square is the reveal: \(\sqrt{x^2+4x+13}=\sqrt{(x+2)^2+9}\) is a \(u^2+a^2\) with \(u=x+2\), \(a=3\). Definite versions either convert the limits into \(u\)-values or finish with the log forms for an exact answer. The same substitutions also handle \(\int\sqrt{x^2+a^2}\,dx\) and relatives — there the collapse is to \(\int a^2\cosh^2 u\,du\), dispatched by the double-angle identity, which is why the identities block is a prerequisite rather than an aside.

CaseProof and equation questions: exponentials are the master key

AQA's H7 asks you to construct proofs involving hyperbolic functions, and the proofs come in a recognisable repertoire: verify an identity from the definitions, derive a log form (as above), or solve an equation exactly. For identities, the strategy is mechanical and reliable — replace every hyperbolic function by its exponential definition, expand, and watch the algebra close. The style requirement is direction: start from one side and work to the other, or manipulate both sides into a visibly identical expression; a proof that assumes the result and simplifies to \(0=0\) earns nothing.

Equations offer a genuine choice of route. An equation mixing \(\sinh x\) and \(\cosh x\) linearly goes straight to exponentials and becomes a quadratic in \(e^x\); an equation mixing \(\cosh 2x\) with \(\sinh x\) first uses \(\cosh 2x=1+2\sinh^2 x\) to become a quadratic in \(\sinh x\), solved and then converted by \(x=\operatorname{arsinh}s=\ln\big(s+\sqrt{s^2+1}\big)\). Two pruning rules do the tidying: \(e^x>0\) always, so negative roots for \(e^x\) are discarded with a stated reason; and \(\cosh x\ge 1\), so any root demanding \(\cosh x<1\) dies the same way. The exact answers these methods produce — clean logarithms like \(\ln 4\) — are the point: this is a non-calculator-friendly topic, and 'exact value' in the question is the instruction to travel via exponentials or log forms rather than decimals.

VocabularyKey terms the mark scheme pays for

TrapsMisconceptions that cost marks

ExamWhat examiners want