Hyperbolic Identities

The hyperbolic functions obey a complete algebra of identities that runs almost exactly parallel to ordinary trigonometry — addition formulae, double-angle formulae, factor formulae, the lot. The single recurring difference is a sign change attached to every product of two $\sinh$sinh terms, a pattern captured by the mnemonic Osborn's rule. This lesson derives the family of identities from the exponential definitions $\sinh x=\tfrac{e^x-e^{-x}}{2}$sinhx=2ex−e−x​ and $\cosh x=\tfrac{e^x+e^{-x}}{2}$coshx=2ex+e−x​, explains why Osborn's rule works, and shows how examiners separate a "quote the formula" answer from a genuine "prove from definitions" answer.

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1 Spec mapping

Hyperbolic identities are compulsory pure content for Papers 1 and 2 of AQA Further Mathematics (7367) — each paper 2 hours, 100 marks, $33\tfrac13\%$3331​% of the qualification. The work divides cleanly by assessment objective: applying a stated identity to simplify or evaluate is AO1, while proving an identity from the exponential definitions is squarely AO2 (construct a rigorous, well-communicated argument). It builds directly on the previous lesson's definitions and on the exponential and index laws of A-Level Maths, and it is the toolkit you will reach for constantly in the differentiation, integration and equation-solving lessons that follow. "Show that" and "prove that" commands dominate this topic precisely because the proofs are short and self-contained — a clean test of whether you understand the exponential basis or have merely memorised a results sheet.

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2 Core theory — the three fundamental identities

Everything in this lesson hangs off the master identity proved in the previous lesson,

$\boxed{\ \cosh^2 x - \sinh^2 x = 1\ }$ cosh2x−sinh2x=1 ​

the hyperbolic analogue of $\cos^2 x+\sin^2 x=1$cos2x+sin2x=1 but with a minus sign. Dividing it through by $\cosh^2 x$cosh2x and by $\sinh^2 x$sinh2x in turn produces the other two members of the family:

$\begin{aligned} \frac{\cosh^2 x}{\cosh^2 x}-\frac{\sinh^2 x}{\cosh^2 x}=\frac{1}{\cosh^2 x} &\ \implies\ \boxed{\,1-\tanh^2 x=\operatorname{sech}^2 x\,}, \\[4pt] \frac{\cosh^2 x}{\sinh^2 x}-\frac{\sinh^2 x}{\sinh^2 x}=\frac{1}{\sinh^2 x} &\ \implies\ \boxed{\,\coth^2 x-1=\operatorname{cosech}^2 x\,}. \end{aligned}$cosh2xcosh2x​−cosh2xsinh2x​=cosh2x1​sinh2xcosh2x​−sinh2xsinh2x​=sinh2x1​​ ⟹ 1−tanh2x=sech2x​, ⟹ coth2x−1=cosech2x​.​

These are the mirror images of $1+\tan^2 x=\sec^2 x$1+tan2x=sec2x and $\cot^2 x+1=\operatorname{cosec}^2 x$cot2x+1=cosec2x. Notice how the minus sign migrates: it sits in front of $\tanh^2$tanh2 where trig has a plus, and in front of $1$1 where trig has a plus again. That migrating sign is exactly what Osborn's rule will let you predict without re-deriving anything.

It pays to keep a consistency check in mind for each of these three. The reciprocal hyperbolic functions satisfy $|\tanh x|<1$∣tanhx∣<1, $\operatorname{sech} x\in(0,1]$sechx∈(0,1], and $|\coth x|>1$∣cothx∣>1, so the signs in the identities are forced: $1-\tanh^2 x$1−tanh2x must be a positive quantity less than $1$1 (it equals $\operatorname{sech}^2 x\le1$sech2x≤1), and $\coth^2 x-1$coth2x−1 must be positive (it equals $\operatorname{cosech}^2 x>0$cosech2x>0). If you ever write down an identity whose two sides cannot agree in sign or size, you have made an Osborn error — a thirty-second check that saves marks. The same three identities, incidentally, are the engine behind almost every "given one hyperbolic value, find another" question: $\cosh^2 x-\sinh^2 x=1$cosh2x−sinh2x=1 converts between $\sinh$sinh and $\cosh$cosh, and $1-\tanh^2 x=\operatorname{sech}^2 x$1−tanh2x=sech2x converts between $\tanh$tanh and $\cosh$cosh, so between them they connect any one hyperbolic ratio to all the others without ever solving for $x$x.

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3 Osborn's rule — and why it works

Osborn's rule is a fast way to write down a hyperbolic identity from the corresponding trigonometric one:

Replace each trig function by its hyperbolic namesake ($\sin\to\sinh$sin→sinh, $\cos\to\cosh$cos→cosh, $\tan\to\tanh$tan→tanh, …), but change the sign of any term that contains a product of two sines — including "hidden" products such as the $\tan^2=\sin^2/\cos^2$tan2=sin2/cos2 in $1+\tan^2$1+tan2.

Worked through the fundamental identity: $\cos^2 x+\sin^2 x=1$cos2x+sin2x=1 becomes $\cosh^2 x+(\text{flip the }\sin^2)\to\cosh^2 x-\sinh^2 x=1$cosh2x+(flip the sin2)→cosh2x−sinh2x=1. For $1+\tan^2 x=\sec^2 x$1+tan2x=sec2x, the $\tan^2$tan2 hides a $\sin^2$sin2, so it flips: $1-\tanh^2 x=\operatorname{sech}^2 x$1−tanh2x=sech2x.

Trigonometric identity	Hyperbolic identity (via Osborn)
$\cos^2 x+\sin^2 x=1$cos2x+sin2x=1	$\cosh^2 x-\sinh^2 x=1$cosh2x−sinh2x=1
$1+\tan^2 x=\sec^2 x$1+tan2x=sec2x	$1-\tanh^2 x=\operatorname{sech}^2 x$1−tanh2x=sech2x
$\cos(A+B)=\cos A\cos B-\sin A\sin B$cos(A+B)=cosAcosB−sinAsinB	$\cosh(A+B)=\cosh A\cosh B+\sinh A\sinh B$cosh(A+B)=coshAcoshB+sinhAsinhB
$\sin(A+B)=\sin A\cos B+\cos A\sin B$sin(A+B)=sinAcosB+cosAsinB	$\sinh(A+B)=\sinh A\cosh B+\cosh A\sinh B$sinh(A+B)=sinhAcoshB+coshAsinhB
$2\sin A\sin B=\cos(A-B)-\cos(A+B)$2sinAsinB=cos(A−B)−cos(A+B)	$2\sinh A\sinh B=\cosh(A+B)-\cosh(A-B)$2sinhAsinhB=cosh(A+B)−cosh(A−B)

Why does it work? Because of the link to complex numbers: $\cosh(\mathrm ix)=\cos x$cosh(ix)=cosx and $\sinh(\mathrm ix)=\mathrm i\sin x$sinh(ix)=isinx. Each genuine $\sin$sin carries a factor $\mathrm i$i once rewritten hyperbolically, so a product of two sines carries $\mathrm i^2=-1$i2=−1 — that hidden $-1$−1 is precisely the sign Osborn's rule flips back. A single $\sinh$sinh, or a product of two $\cosh$cosh's, picks up no such factor, so its sign is unchanged.

To see the mechanism concretely, take the trig identity $\cos(A+B)=\cos A\cos B-\sin A\sin B$cos(A+B)=cosAcosB−sinAsinB and substitute $A\to\mathrm iA$A→iA, $B\to\mathrm iB$B→iB. The left becomes $\cos(\mathrm i(A+B))=\cosh(A+B)$cos(i(A+B))=cosh(A+B); the first product on the right becomes $\cos(\mathrm iA)\cos(\mathrm iB)=\cosh A\cosh B$cos(iA)cos(iB)=coshAcoshB; and the second becomes $\sin(\mathrm iA)\sin(\mathrm iB)=(\mathrm i\sinh A)(\mathrm i\sinh B)=-\sinh A\sinh B$sin(iA)sin(iB)=(isinhA)(isinhB)=−sinhAsinhB. The original $-\sin A\sin B$−sinAsinB therefore turns into $-(-\sinh A\sinh B)=+\sinh A\sinh B$−(−sinhAsinhB)=+sinhAsinhB, and we recover $\cosh(A+B)=\cosh A\cosh B+\sinh A\sinh B$cosh(A+B)=coshAcoshB+sinhAsinhB. That single line is a proof of Osborn's rule for this identity — and it generalises term-by-term to any trigonometric identity whatsoever. Knowing the mechanism, rather than the slogan, means you will never misapply the rule to a term that only looks like a sine product.

The terms to watch for hidden products are $\tan^2$tan2, $\cot^2$cot2, $\sec^2$sec2 (which contains no sine product — it is $1/\cos^2$1/cos2, so it does not flip), and $\operatorname{cosec}^2=1/\sin^2$cosec2=1/sin2 (which contains $1/\sin^2$1/sin2, a sine in the denominator — its sign behaviour follows from rewriting the whole identity). A safe habit is: if in any doubt, do not trust the slogan — rewrite that one term using the complex substitution above, or simply prove the identity from the exponential definitions, which is what the exam rewards anyway.

Warning. Osborn's rule is a mnemonic for recall, not a proof. If a question says "prove" or "show that", you must derive the identity from the exponential definitions; quoting Osborn earns nothing. Use the rule to check your memory, then prove properly when asked.

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4 The addition formulae

$\begin{aligned} \sinh(A\pm B) &= \sinh A\cosh B \pm \cosh A\sinh B, \\[2pt] \cosh(A\pm B) &= \cosh A\cosh B \pm \sinh A\sinh B, \\[2pt] \tanh(A\pm B) &= \frac{\tanh A\pm\tanh B}{1\pm\tanh A\tanh B}. \end{aligned}$sinh(A±B)cosh(A±B)tanh(A±B)​=sinhAcoshB±coshAsinhB,=coshAcoshB±sinhAsinhB,=1±tanhAtanhBtanhA±tanhB​.​

Note the contrast with trig in the $\cosh$cosh formula: trig has $\cos A\cos B\mp\sin A\sin B$cosAcosB∓sinAsinB (signs opposite), whereas hyperbolic has $\cosh A\cosh B\pm\sinh A\sinh B$coshAcoshB±sinhAsinhB (signs the same as the bracket). This is the most frequently misremembered sign in the whole topic. A memorable way to fix it: the $\sinh$sinh formula reads exactly like $\sin$sin (a lone $\sinh$sinh is never flipped), but the $\cosh$cosh formula reads opposite to $\cos$cos (the $\sinh\sinh$sinhsinh product is flipped). So "sinh copies sin, cosh contradicts cos". The $\tanh$tanh formula then follows by dividing: $\tanh(A+B)=\dfrac{\sinh(A+B)}{\cosh(A+B)}$tanh(A+B)=cosh(A+B)sinh(A+B)​, and dividing numerator and denominator by $\cosh A\cosh B$coshAcoshB turns every term into a $\tanh$tanh, giving the quotient above with a $+\tanh A\tanh B$+tanhAtanhB in the denominator — the flip of the trig $-\tan A\tan B$−tanAtanB.

Proof of $\sinh(A+B)$sinh(A+B) from the definitions. Expand the right-hand side:

$\begin{aligned} \sinh A\cosh B+\cosh A\sinh B &= \frac{(e^A-e^{-A})}{2}\cdot\frac{(e^B+e^{-B})}{2}+\frac{(e^A+e^{-A})}{2}\cdot\frac{(e^B-e^{-B})}{2} \\[4pt] &= \tfrac14\big(e^{A+B}+e^{A-B}-e^{-A+B}-e^{-A-B}\big) \\ &\quad +\tfrac14\big(e^{A+B}-e^{A-B}+e^{-A+B}-e^{-A-B}\big) \\[4pt] &= \tfrac14\big(2e^{A+B}-2e^{-A-B}\big)=\frac{e^{A+B}-e^{-(A+B)}}{2}=\sinh(A+B). \end{aligned}$sinhAcoshB+coshAsinhB​=2(eA−e−A)​⋅2(eB+e−B)​+2(eA+e−A)​⋅2(eB−e−B)​=41​(eA+B+eA−B−e−A+B−e−A−B)+41​(eA+B−eA−B+e−A+B−e−A−B)=41​(2eA+B−2e−A−B)=2eA+B−e−(A+B)​=sinh(A+B).​

The four cross-terms in $e^{A-B}$eA−B and $e^{-A+B}$e−A+B cancel in pairs; only $e^{A+B}$eA+B and $e^{-(A+B)}$e−(A+B) survive — exactly the definition of $\sinh(A+B)$sinh(A+B).

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5 Double-angle and half-angle formulae

Putting $B=A$B=A in the addition formulae gives the double-angle results:

$\boxed{\ \sinh 2x=2\sinh x\cosh x,\qquad \cosh 2x=\cosh^2 x+\sinh^2 x,\qquad \tanh 2x=\frac{2\tanh x}{1+\tanh^2 x}\ }$ sinh2x=2sinhxcoshx,cosh2x=cosh2x+sinh2x,tanh2x=1+tanh2x2tanhx​ ​

Using $\cosh^2 x-\sinh^2 x=1$cosh2x−sinh2x=1, the $\cosh 2x$cosh2x formula has the same three guises as its trig cousin:

$\cosh 2x=\cosh^2 x+\sinh^2 x=2\cosh^2 x-1=1+2\sinh^2 x.$cosh2x=cosh2x+sinh2x=2cosh2x−1=1+2sinh2x.

Rearranging the last two gives the power-reduction (half-angle) identities that are indispensable for integrating $\sinh^2$sinh2 and $\cosh^2$cosh2 in the integration lesson:

$\boxed{\ \cosh^2 x=\frac{\cosh 2x+1}{2},\qquad \sinh^2 x=\frac{\cosh 2x-1}{2}\ }$ cosh2x=2cosh2x+1​,sinh2x=2cosh2x−1​ ​

The $\sinh^2$sinh2 version has $\cosh 2x-1$cosh2x−1 on top (trig has $1-\cos 2x$1−cos2x) — another Osborn sign flip to keep straight. There is a fast structural reason the order must be this way: since $\cosh 2x\ge1$cosh2x≥1 for all real $x$x, the numerator $\cosh 2x-1\ge0$cosh2x−1≥0, so $\sinh^2 x\ge0$sinh2x≥0 as required of a square; the reversed order $1-\cosh 2x$1−cosh2x would be $\le0$≤0 and could never equal a square. The trig version genuinely uses $1-\cos 2x$1−cos2x because $\cos 2x$cos2x can exceed $1$1's opposite — it ranges over $[-1,1]$[−1,1] — so there the subtraction goes the other way. Tying each formula to a positivity argument like this is far more reliable than memorising six near-identical fractions.

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6 Worked examples with mark scheme

Worked Example 1 — prove an addition formula

Prove, from the definitions, that $\cosh(A+B)=\cosh A\cosh B+\sinh A\sinh B$cosh(A+B)=coshAcoshB+sinhAsinhB.

$\begin{aligned} \cosh A\cosh B+\sinh A\sinh B &= \tfrac14(e^A+e^{-A})(e^B+e^{-B})+\tfrac14(e^A-e^{-A})(e^B-e^{-B}) \\[2pt] &= \tfrac14\big(e^{A+B}+e^{A-B}+e^{-A+B}+e^{-A-B}\big) \\ &\quad +\tfrac14\big(e^{A+B}-e^{A-B}-e^{-A+B}+e^{-A-B}\big) \\[2pt] &= \tfrac14\big(2e^{A+B}+2e^{-(A+B)}\big)=\frac{e^{A+B}+e^{-(A+B)}}{2}=\cosh(A+B). \end{aligned}$coshAcoshB+sinhAsinhB​=41​(eA+e−A)(eB+e−B)+41​(eA−e−A)(eB−e−B)=41​(eA+B+eA−B+e−A+B+e−A−B)+41​(eA+B−eA−B−e−A+B+e−A−B)=41​(2eA+B+2e−(A+B))=2eA+B+e−(A+B)​=cosh(A+B).​

Mark scheme: M1 write both products in exponential form; M1 expand and collect; A1 the $e^{A-B},e^{-A+B}$eA−B,e−A+B terms cancel; A1 conclude $\cosh(A+B)$cosh(A+B). The structure — state, expand, cancel, conclude — is what earns the AO2 communication marks.

Worked Example 2 — apply a double-angle formula

Given $\sinh x=\tfrac{5}{12}$sinhx=125​, find the exact value of $\sinh 2x$sinh2x.

First recover $\cosh x$coshx from the fundamental identity (faster than solving for $x$x):

$\cosh^2 x=1+\sinh^2 x=1+\tfrac{25}{144}=\tfrac{169}{144}\ \implies\ \cosh x=\tfrac{13}{12}\quad(\cosh x\ge1>0).$cosh2x=1+sinh2x=1+14425​=144169​ ⟹ coshx=1213​(coshx≥1>0).

$\sinh 2x=2\sinh x\cosh x=2\cdot\tfrac{5}{12}\cdot\tfrac{13}{12}=\frac{130}{144}=\boxed{\tfrac{65}{72}}.$sinh2x=2sinhxcoshx=2⋅125​⋅1213​=144130​=7265​​.

Mark scheme: M1 use $\cosh^2 x=1+\sinh^2 x$cosh2x=1+sinh2x; A1 $\cosh x=\tfrac{13}{12}$coshx=1213​ (positive root justified); M1 apply $\sinh 2x=2\sinh x\cosh x$sinh2x=2sinhxcoshx; A1 $\tfrac{65}{72}$7265​ (fully simplified).

Worked Example 3 — simplify using the identities

Express $4\cosh^2 x-1$4cosh2x−1 in terms of $\cosh 2x$cosh2x.

$4\cosh^2 x-1=4\cdot\frac{\cosh 2x+1}{2}-1=2\cosh 2x+2-1=\boxed{2\cosh 2x+1}.$4cosh2x−1=4⋅2cosh2x+1​−1=2cosh2x+2−1=2cosh2x+1​.

Mark scheme: M1 substitute $\cosh^2 x=\tfrac{\cosh 2x+1}{2}$cosh2x=2cosh2x+1​; A1 $2\cosh 2x+1$2cosh2x+1. A neat check: at $x=0$x=0, LHS $=4-1=3$=4−1=3 and RHS $=2+1=3$=2+1=3. ✓

Worked Example 4 — combine identities to evaluate

Given $\tanh x=\tfrac{1}{2}$tanhx=21​, find the exact value of $\tanh 2x$tanh2x, and of $\cosh 2x$cosh2x.

For $\tanh 2x$tanh2x use the double-angle formula directly — no need to find $x$x:

$\tanh 2x=\frac{2\tanh x}{1+\tanh^2 x}=\frac{2\cdot\tfrac12}{1+\tfrac14}=\frac{1}{\tfrac54}=\frac{4}{5}.$tanh2x=1+tanh2x2tanhx​=1+41​2⋅21​​=45​1​=54​.