🌉 Hyperbolic Functions
Built from eˣ instead of angles, hyperbolic functions describe the shape of hanging cables, the voltage profile along transmission lines, and the stress in curved shells.
2Topics
8Worked Examples
4Eng. Cases
12Practice Qs
🌉Definitions & Identitiessinh, cosh, tanh
🔍Solving EquationsUsing inverse hyperbolic forms
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Define sinh, cosh and tanh from exponentials; apply the hyperbolic identity cosh²x − sinh²x = 1; and solve hyperbolic equations using the logarithmic (inverse hyperbolic) forms.
💡 Hyperbolic functions look like trig functions and share similar-looking identities, but they come from eˣ, not angles — always double-check the sign in any identity you transfer across.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
🌉
Definitions & Identities
Built from eˣ instead of angles — the shape of hanging cables and open-ended transmission lines.
🔧Hyperbolic Function CalculatorEnter x, then press Calculate
Enter a value of x, then press Calculate to see sinh x, cosh x, tanh x and the identity check…
📖 Key Vocabulary
Hyperbolic sine (sinh) — an odd function built from eˣ − e⁻ˣ.
Hyperbolic cosine (cosh) — an even function, minimum value 1 at x=0.
Catenary — the y = cosh(x) shaped curve a hanging cable or chain naturally forms.
tanh x — sinh x ÷ cosh x; bounded between −1 and 1, useful as a smooth "switch" function.
👁️ Visual aid: a freely hanging chain or power cable settles into exactly this cosh(x) curve — not a parabola, though the two look similar near the bottom.
✏️Worked Example 1 — Evaluating from the exponential definitionType: Direct evaluation
Question Evaluate sinh(1) and cosh(1) to 4 d.p.
sinh
(e¹ − e⁻¹)/2 = (2.7183 − 0.3679)/2cosh
(e¹ + e⁻¹)/2 = (2.7183 + 0.3679)/2Answer sinh(1) ≈ 1.1752, cosh(1) ≈ 1.5431
🔗Worked Example 2 — Verifying the hyperbolic identityType: Identity check
Question Verify cosh²(1) − sinh²(1) = 1 using the previous example's values.
Square each
1.5431² = 2.3812, 1.1752² = 1.3811Subtract
2.3812 − 1.3811 = 1.0001Answer ≈ 1 ✓ (rounding accounts for the small error) — the identity holds for every x.
➗Worked Example 3 — Working with tanhType: Derived function
Question Evaluate tanh(2) to 4 d.p. and state which asymptote it approaches as x→∞.
sinh(2)
(e²−e⁻²)/2 = (7.389−0.1353)/2 = 3.6269cosh(2)
(e²+e⁻²)/2 = (7.389+0.1353)/2 = 3.7622Answer tanh(2) = 3.6269/3.7622 = 0.9640 — approaching the asymptote y = 1 as x→∞.
⚙️Worked Example 4 — Engineering context (catenary cable sag)Type: Engineering
Question A power cable hangs as y = 50cosh(x/50), x in metres from the lowest point. Find the sag at x = 20 m.
Substitute
y(20) = 50cosh(20/50) = 50cosh(0.4)Compute cosh(0.4)
(e^0.4+e^−0.4)/2 = (1.4918+0.6703)/2 = 1.0811Height
y(20) = 50 × 1.0811 = 54.05 mAnswer Sag = 54.05 − 50 = 4.05 m over a 20 m horizontal span from the lowest point.
🧠 Common slip: mixing up the identity sign — it's cosh²x minus sinh²x = 1, the opposite sign convention from sin²+cos²=1.
🔍
Solving Hyperbolic Equations
Every hyperbolic function has a logarithmic inverse — the key to solving equations directly.
🔧Hyperbolic Equation SolverSolve sinhx = k or coshx = k for x
Choose sinh or cosh, enter k, then press Solve to see the inverse hyperbolic (log) working…
📖 Key Vocabulary
Inverse hyperbolic function — the function that undoes sinh, cosh or tanh, always expressible using ln.
Domain restriction — cosh⁻¹ needs y ≥ 1; tanh⁻¹ needs −1 < y < 1, since these match the original functions' ranges.
Two solutions for cosh — cosh x = k (k>1) has both a positive and negative x solution, since cosh is even.
Quadratic substitution — some hyperbolic equations reduce to a quadratic in eˣ.
👁️ Visual aid: the inverse hyperbolic formula converts the equation straight into something a calculator can evaluate in one line.
✏️Worked Example 1 — Solving coshx = kType: Basic solve
Question Solve cosh x = 2 for x > 0, using x = ln(k + √(k²−1)).
Apply the inverse formula
x = ln(2 + √(4−1)) = ln(2 + √3)Compute
= ln(2 + 1.732) = ln(3.732)Answer x ≈ 1.317
🔄Worked Example 2 — Solving sinhx = kType: Sinh solve
Question Solve sinh x = 1.5.
Apply the inverse formula
x = ln(1.5 + √(1.5²+1)) = ln(1.5 + √3.25)Compute
= ln(1.5 + 1.803) = ln(3.303)Answer x ≈ 1.195
🧮Worked Example 3 — Quadratic-in-eˣ approachType: From first principles
Question Solve 2cosh x = 5 by substituting u = eˣ.
Expand
2 × (eˣ+e⁻ˣ)/2 = 5 → eˣ + e⁻ˣ = 5Substitute u=eˣ
u + 1/u = 5 → u² − 5u + 1 = 0Solve quadratic
u = (5±√21)/2 = 4.791 or 0.2087Answer x = ln(4.791) = 1.567 or x = ln(0.2087) = −1.567 (the two solutions are negatives of each other, since cosh is even).
⚙️Worked Example 4 — Engineering context (cable length calculation)Type: Engineering
Question A cable follows y = 40cosh(x/40). At what x does the cable reach a height of 50 m?
Set up
50 = 40cosh(x/40) → cosh(x/40) = 1.25Apply inverse formula
x/40 = ln(1.25+√(1.5625−1)) = ln(1.25+0.75)Compute
x/40 = ln(2) = 0.6931 → x = 27.7 mAnswer The cable reaches 50 m height at x ≈ 27.7 m from the lowest point (on either side, by symmetry).
🧠 Common slip: forgetting cosh⁻¹ gives two x values (±) since cosh is an even function — sinh⁻¹ gives only one, since sinh is one-to-one.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🌉 Suspension Cable — Catenary Sag
A cable hanging under its own weight follows y = a·cosh(x/a), where a depends on the cable's weight and horizontal tension.
y(x) = a cosh(x/a); Sag = y(x) − a
📟 Try it — Cable Sag Calculator
Enter a and x, then press Calculate…
📡 Transmission Line — Voltage Profile
On a lossy transmission line of length l, the voltage distribution can involve cosh and sinh of the propagation constant γ times distance.
V(x) = V_s cosh(γx) − I_s Z₀ sinh(γx)
📟 Try it — Line Voltage at Distance x
Enter V_s, I_s, Z₀ and γx, then press Calculate…
🏗️ Arch Design — Catenary Arch Height
An arch built to the inverted catenary shape carries load in pure compression — the same cosh curve as a hanging cable, flipped upside down.
Rise from springing = a[cosh(L/2a) − 1]
📟 Try it — Arch Rise Calculator
Enter parameter a and span L, then press Calculate…
🔥 Fin Heat Transfer — Temperature Profile
Temperature along a cooling fin of length L follows a cosh profile: θ(x)/θ_b = cosh(m(L−x)) / cosh(mL).
θ(x) = θ_b · cosh(m(L−x)) ÷ cosh(mL)
📟 Try it — Fin Temperature Calculator
Enter θ_b, m, L and x, then press Calculate…
🎮
Quick-Fire Hyperbolic Game
12 questions on definitions, identities and solving. How many can you get?
⚡ Hyperbolic Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
Evaluate cosh(0).
cosh(0) = (e⁰+e⁰)/2 = (1+1)/2 = 1
Question 2 🥉2 marks
Evaluate sinh(0).
sinh(0) = (e⁰−e⁰)/2 = (1−1)/2 = 0
Question 3 🥉2 marks
Evaluate cosh(0.5) to 4 d.p.
(e^0.5+e^−0.5)/2 = (1.6487+0.6065)/2 = 1.1276
Question 4 🥉2 marks
State whether sinh(x) is an odd or even function, and why.
sinh(−x) = (e⁻ˣ−eˣ)/2 = −sinh(x) → odd function
Question 5 🥈3 marks
Evaluate tanh(1) to 4 d.p.
sinh(1)=1.1752, cosh(1)=1.5431
tanh(1) = 1.1752/1.5431
= 0.7616
tanh(1) = 1.1752/1.5431
= 0.7616
Question 6 🥈4 marks
Solve sinh x = 2 for x, giving your answer to 3 d.p.
x = ln(2+√(4+1)) = ln(2+2.236)
= ln(4.236)
x ≈ 1.444
= ln(4.236)
x ≈ 1.444
Question 7 🥈3 marks
Verify cosh²(0.5) − sinh²(0.5) = 1 using cosh(0.5)=1.1276, sinh(0.5)=0.5211.
1.1276² − 0.5211² = 1.2715 − 0.2715
= 1.0000 ✓
= 1.0000 ✓
Question 8 🥈4 marks
Solve cosh x = 3 for x > 0, giving your answer to 3 d.p.
x = ln(3+√(9−1)) = ln(3+2.828)
= ln(5.828)
x ≈ 1.763
= ln(5.828)
x ≈ 1.763
Question 9 🥇4 marks
Solve cosh x = 2.5 for x, giving your answer to 3 d.p.
x = ln(2.5 + √(6.25−1)) = ln(2.5+2.2913)
= ln(4.7913)
x ≈ 1.567
= ln(4.7913)
x ≈ 1.567
Question 10 🥇5 marks
A cable follows y = 60cosh(x/60). Find the sag at x = 25 m.
y(25) = 60cosh(25/60) = 60cosh(0.4167)
cosh(0.4167) = (e^0.4167+e^−0.4167)/2 ≈ 1.0880
y(25) ≈ 65.28 m
Sag = 65.28 − 60 = 5.28 m
cosh(0.4167) = (e^0.4167+e^−0.4167)/2 ≈ 1.0880
y(25) ≈ 65.28 m
Sag = 65.28 − 60 = 5.28 m
Question 11 🥇5 marks
Solve 2sinh x = 7 by first isolating sinh x, giving x to 3 d.p.
sinh x = 3.5
x = ln(3.5+√(12.25+1)) = ln(3.5+3.640)
= ln(7.140)
x ≈ 1.966
x = ln(3.5+√(12.25+1)) = ln(3.5+3.640)
= ln(7.140)
x ≈ 1.966
Question 12 🥇5 marks
Solve 3cosh x = 10 using the quadratic-in-eˣ method, giving both solutions to 3 d.p.
cosh x = 10/3 = 3.333
eˣ+e⁻ˣ = 6.667 → u+1/u=6.667
u²−6.667u+1=0
u=(6.667±√(44.44−4))/2 = (6.667±6.360)/2
u=6.514 or 0.1535
x = ln(6.514)=1.874 or x=ln(0.1535)=−1.874
eˣ+e⁻ˣ = 6.667 → u+1/u=6.667
u²−6.667u+1=0
u=(6.667±√(44.44−4))/2 = (6.667±6.360)/2
u=6.514 or 0.1535
x = ln(6.514)=1.874 or x=ln(0.1535)=−1.874