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Hyperbolic Functions

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WFM03 · Further Pure Mathematics 3 · Chapter 1

Hyperbolic Functions

Big idea: hyperbolic functions are the "exponential cousins" of sin, cos and tan — built from eˣ and e⁻ˣ instead of circles, and almost every trig identity you know has a hyperbolic twin.

Summary

  • sinh x, cosh x, tanh x are defined directly from ex and e−x — no circles involved.
  • The three reciprocal functions cosech x, sech x, coth x behave like cosec, sec, cot.
  • Hyperbolic functions obey addition/double-angle formulae that look almost identical to trig ones.
  • Osborne's Rule is the shortcut: convert any trig identity to a hyperbolic one, flipping the sign in front of any term that is (or hides) a product of two sines.
  • The inverse functions arsinh, arcosh, artanh undo sinh, cosh, tanh, and each has a closed logarithmic form.
  • Equations mixing sinh/cosh can be solved either by converting to exponentials, or by using hyperbolic identities + the log form of the inverse.

1. Definitions and Graphs

Forget circles for a second. Hyperbolic functions are just combinations of ex and e−x, built to mimic the shape and algebra of sin, cos and tan without actually being trigonometric at all. That's why they're called "hyperbolic" — they relate to the hyperbola x² − y² = 1 the way sin/cos relate to the circle x² + y² = 1.

Core definitions
sinh x = ½(eˣ − e⁻ˣ)   cosh x = ½(eˣ + e⁻ˣ)   tanh x = sinh x / cosh x = (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ)

Because they're built from exponentials, you can read off their behaviour instantly:

  • sinh x is an odd function (sinh(−x) = −sinh x), passes through the origin, and grows like ½eˣ for large x.
  • cosh x is an even function (cosh(−x) = cosh x), has a minimum of 1 at x = 0 — it never dips below 1 — and looks like a smile (this is the actual shape of a hanging chain, a "catenary").
  • tanh x squeezes everything between −1 and 1: it flattens out to horizontal asymptotes y = 1 and y = −1 as x → ±∞.
y = sinh x  → odd, S-shaped, passes through (0,0), no asymptotes
y = cosh x  → even, U-shaped, minimum point (0,1), never negative
y = tanh x  → odd, S-shaped, bounded between y = −1 and y = 1

The reciprocal functions

Exactly as sec = 1/cos, cosec = 1/sin, cot = 1/tan, we define:

Reciprocal hyperbolic functions
cosech x = 1/sinh x   sech x = 1/cosh x   coth x = 1/tanh x

Their graphs follow directly from "flipping" sinh, cosh, tanh:

  • cosech x has a vertical asymptote at x = 0 (since sinh 0 = 0) and two separate branches, one above and one below the x-axis.
  • sech x has a maximum value of 1 at x = 0 (since cosh x ≥ 1 always) and tends to 0 as x → ±∞.
  • coth x has a vertical asymptote at x = 0, and horizontal asymptotes at y = 1 and y = −1.
Quick sanity check You should be able to sketch all six graphs from memory in under a minute — examiners love asking for a quick sketch as a "free" first mark. The shape of cosech/sech/coth always follows from the shape of sinh/cosh/tanh: asymptotes appear wherever the original function was zero, and vice versa.
Practice: Show that cosh²x − sinh²x = 1 directly from the exponential definitions.

2. Addition Formulae, Double Angle Formulae, and Osborne's Rule

Here's the genuinely useful shortcut of this chapter: you almost never need to derive hyperbolic identities from scratch. Every trig identity has a hyperbolic twin, and there's a simple rule for converting between them.

Osborne's Rule

Osborne's Rule Take a trig identity. Replace every trig function with its hyperbolic namesake (sin → sinh, cos → cosh, tan → tanh, etc.). Then, wherever the original identity contains a product of two sines (explicitly, or hidden inside tan² = sin²/cos² or sec² = 1/cos²), flip the sign of that term.

Why does this work? Because sin x = (unit circle) involves i somewhere when you connect it to exponentials (sin x = (eix − e−ix)/2i), and every time you multiply two sines together you pick up an i² = −1 that a pure hyperbolic identity never has. You don't need to know the proof to use the rule — just spot "product of two sines" and flip that one sign.

Trig identityHyperbolic identityWhy / why not flipped
sin(A+B) = sinA cosB + cosA sinB sinh(A+B) = sinhA coshB + coshA sinhB No sin×sin term → no change
cos(A+B) = cosA cosB − sinA sinB cosh(A+B) = coshA coshB + sinhA sinhB sinA sinB present → sign flips (− becomes +)
1 + tan²A = sec²A 1 − tanh²A = sech²A tan²A hides sin²A/cos²A → flips
cos²A + sin²A = 1 cosh²A − sinh²A = 1 sin²A is a product of two sines → flips
cos 2A = 1 − 2sin²A cosh 2A = 1 + 2sinh²A sin²A flips
sin 2A = 2 sinA cosA sinh 2A = 2 sinhA coshA no sin×sin term → no change
Worked example — using Osborne's Rule
Task: Write down the hyperbolic identity corresponding to tan(A − B) = (tanA − tanB)/(1 + tanA tanB).
Step 1: Replace every function: tanh(A − B) = (tanhA − tanhB)/(1 + tanhA tanhB).
Step 2: Look for hidden sin×sin. The denominator has tanA tanB = (sinA/cosA)(sinB/cosB) — this contains sinA sinB, a product of two sines.
Step 3: Flip that sign: tanh(A − B) = (tanhA − tanhB)/(1 − tanhA tanhB).
Common trap Only flip the sign in front of the sin×sin term itself — don't flip the whole equation, and don't flip terms that don't involve a product of two sines (like sin2A = 2sinAcosA, which stays unchanged).
Practice: Use Osborne's Rule to write down the hyperbolic identity for cos 2A = 2cos²A − 1.

3. Inverse Hyperbolic Functions

Graphs

Just like any inverse function, the graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x. So to sketch arsinh x, arcosh x, artanh x, just reflect the graphs from Section 1 in y = x.

y = arsinh x  → reflection of sinh x; defined for all real x; odd function
y = arcosh x  → reflection of the upper half of cosh x; defined for x ≥ 1; arcosh x ≥ 0 only
y = artanh x  → reflection of tanh x; defined for |x| < 1; vertical asymptotes at x = ±1
Careful — arcosh is single-valued by definition cosh x is not one-to-one (cosh(2) = cosh(−2)), so strictly it has no inverse over all reals. We fix this by defining arcosh x to only output the non-negative branch, so arcosh x ≥ 0 always, for x ≥ 1. But the equation cosh z = 2 genuinely has two solutions: z = +arcosh 2 and z = −arcosh 2. Don't confuse "the function arcosh" (one value) with "solving cosh z = k" (two values).

Logarithmic form

Each inverse hyperbolic function can be written as a single logarithm — genuinely useful for solving equations without a calculator, and for exam questions that ask you to "show that". The derivation pattern is always the same: set y = the inverse function, rewrite using the exponential definition, and solve the resulting quadratic in eʸ.

Worked derivation — arsinh x
Start: y = arsinh x ⇒ sinh y = x, i.e. ½(eʸ − e⁻ʸ) = x.
Rearrange: multiply by 2eʸ: e²ʸ − 2x eʸ − 1 = 0. This is a quadratic in eʸ.
Solve: eʸ = [2x ± √(4x² + 4)]/2 = x ± √(x² + 1).
Discard the invalid root: eʸ must be positive, and x − √(x²+1) is always negative, so we must take the + root only.
Conclusion: y = arsinh x = ln(x + √(x²+1)), valid for all real x.
Worked derivation — arcosh x
Start: y = arcosh x ⇒ cosh y = x, so e²ʸ − 2x eʸ + 1 = 0.
Solve: eʸ = x ± √(x² − 1). Here both roots are positive (for x ≥ 1), so both are valid values of eʸ.
Two log values, but one function: it turns out ln(x − √(x²−1)) = −ln(x + √(x²−1)), so y = ±ln(x + √(x²−1)).
Apply the definition: since arcosh x is defined to be non-negative, we take the + sign only.
Conclusion: arcosh x = ln(x + √(x²−1)), for x ≥ 1.
Logarithmic forms — memorise these
arsinh x = ln(x + √(x² + 1))  (all real x)
arcosh x = ln(x + √(x² − 1))  (x ≥ 1)
artanh x = ½ ln[(1+x)/(1−x)]  (|x| < 1)

The artanh result comes from a slightly different route — set y = artanh x, write tanh y = x in exponential form, and solve for eʸ directly (no quadratic needed this time), giving eʸ = √[(1+x)/(1−x)] and hence the ½ ln form above.

Practice: Show that arsinh(−3) = ln(√10 − 3).

4. Equations Involving Hyperbolic Functions

There are two main strategies, and picking the right one saves a lot of time:

  • Strategy A — convert straight to exponentials. Best when the equation is a simple linear combination like a sinh x + b cosh x = c. Substitute the eˣ, e⁻ˣ definitions and you'll get a quadratic in eˣ.
  • Strategy B — use hyperbolic identities, then the log form of the inverse. Best when the equation mixes cosh 2x with sinh x, or similar — use Osborne-style identities (like cosh 2x = 1 + 2sinh²x) to get everything in terms of one function, solve as a quadratic in that function, then convert to logs at the very end using Section 3's formulae.
Worked example — Strategy A
Solve: 6 sinh x − 2 cosh x = 7
Substitute definitions: 6 × ½(eˣ − e⁻ˣ) − 2 × ½(eˣ + e⁻ˣ) = 7
Simplify: 3eˣ − 3e⁻ˣ − eˣ − e⁻ˣ = 7 ⇒ 2eˣ − 4e⁻ˣ = 7
Multiply by eˣ: 2e²ˣ − 7eˣ − 4 = 0
Factorise: (2eˣ + 1)(eˣ − 4) = 0 ⇒ eˣ = −½ (rejected, eˣ can't be negative) or eˣ = 4
Answer: x = ln 4
Worked example — Strategy B
Solve: cosh 2x + 5 sinh x − 4 = 0
Use the identity cosh 2x = 1 + 2sinh²x (from Osborne's Rule) to get everything in terms of sinh x:
1 + 2sinh²x + 5 sinh x − 4 = 0 ⇒ 2sinh²x + 5 sinh x − 3 = 0
Factorise as a quadratic in sinh x: (2 sinh x − 1)(sinh x + 3) = 0 ⇒ sinh x = ½ or sinh x = −3
Convert to logs using arsinh x = ln(x + √(x²+1)):
x = ln(½ + √(¼+1)) = ln[(1+√5)/2]  or  x = ln(−3 + √10) = ln(√10 − 3)
How to choose your strategy fast Ask: "is this a single power of sinh/cosh mixed linearly?" → Strategy A (straight to exponentials). "Is there a cosh 2x, sinh 2x, or higher power mixed in?" → Strategy B (use an identity to reduce it to one function first, then convert to logs at the end). Never mix the two halfway through — pick one and commit.
Practice: Solve 2 cosh²x − 5 sinh x − 5 = 0.

What to Memorise

Definitions

  • sinh x = ½(eˣ − e⁻ˣ)
  • cosh x = ½(eˣ + e⁻ˣ)
  • tanh x = sinh x / cosh x
  • cosech x = 1/sinh x, sech x = 1/cosh x, coth x = 1/tanh x

Graph shapes

  • sinh, tanh, cosech, coth: odd functions
  • cosh, sech: even functions
  • cosh x ≥ 1 always; tanh x bounded by ±1
  • arcosh x ≥ 0 by definition (single-valued)

Osborne's Rule

  • Swap trig → hyperbolic function names
  • Flip sign of any term hiding a product of two sines
  • cosh²x − sinh²x = 1 (note the minus)
  • 1 − tanh²x = sech²x

Logarithmic forms

  • arsinh x = ln(x + √(x²+1)) — all real x
  • arcosh x = ln(x + √(x²−1)) — x ≥ 1
  • artanh x = ½ln[(1+x)/(1−x)] — |x| < 1

Concepts Checklist

Exam Tips

!
Don't flip signs you shouldn't. Osborne's Rule only flips terms with a genuine (possibly hidden) product of two sines. Students often flip every minus sign in an identity out of habit — check each term individually.
!
Reject invalid exponential roots. When solving a quadratic in eˣ, always check whether a root is negative or zero — eˣ can never be ≤ 0, so that root must be discarded, and examiners specifically award a mark for stating this.
!
arcosh is not multi-valued — don't write "±". The function arcosh x always gives a single non-negative answer. Only write ± when you are solving an equation like cosh z = k for z, not when evaluating arcosh directly.
!
State the domain restriction with each log form. arcosh x needs x ≥ 1 and artanh x needs |x| < 1 — quoting the formula without the restriction can lose a mark in "show that" questions.
!
Pick your strategy before you start writing. Spending 10 seconds deciding "exponentials or identity-first" avoids getting halfway through a substitution and having to start over.
!
Simplify surds fully in log answers. Answers like ln[(1+√5)/2] are often expected in exact simplified surd form — don't leave a mixed decimal/surd expression if it can be tidied.
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More Further Pure Mathematics 3 topics

Definitions and graphs
The vector product and its applications
Standard techniques
The ellipse
Derivatives of hyperbolic functions
Addition and double-angle formulae (Osborne's rule)
Further Coordinate Systems
The hyperbola
Standard substitutions
Derivatives of inverse functions