sinh, cosh and tanh and their graphs
The three main hyperbolic functions are made from eˣ and e⁻ˣ:
- cosh x = (eˣ + e⁻ˣ)/2 (say “cosh”)
- sinh x = (eˣ − e⁻ˣ)/2 (say “shine”)
- tanh x = sinh x / cosh x = (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ) (say “than”)
Graphs
- cosh x: U-shaped, minimum point (0, 1), even (cosh(−x) = cosh x), range y ≥ 1.
- sinh x: S-shaped, through (0, 0), odd (sinh(−x) = −sinh x), range all real numbers.
- tanh x: through (0, 0), odd, horizontal asymptotes y = 1 and y = −1, range −1 < y < 1.
Also sech x = 1/cosh x, cosech x = 1/sinh x, coth x = 1/tanh x.
The key identity: cosh² x − sinh² x = 1
Proof from the definitions: cosh²x = (e²ˣ + 2 + e⁻²ˣ)/4 and sinh²x = (e²ˣ − 2 + e⁻²ˣ)/4. Subtract: 4/4 = 1.
So the point (cosh t, sinh t) lies on the hyperbola x² − y² = 1, just as (cos t, sin t) lies on the circle x² + y² = 1. That is where the name “hyperbolic” comes from (3D step 3).
Dividing by cosh²x gives 1 − tanh²x = sech²x. Dividing by sinh²x gives coth²x − 1 = cosech²x.
Inverse hyperbolic functions and their log forms
The inverse of sinh is arsinh (also written sinh⁻¹). Its graph is sinh reflected in y = x.
- arsinh x = ln(x + √(x² + 1)), all real x
- arcosh x = ln(x + √(x² − 1)), x ≥ 1 (we take the positive value, because cosh is not one-to-one)
- artanh x = ½ ln((1 + x)/(1 − x)), −1 < x < 1
Where the log form comes from
Let y = arsinh x, so x = sinh y = (eʸ − e⁻ʸ)/2. Multiply by 2eʸ: e²ʸ − 2x eʸ − 1 = 0. This is a quadratic in eʸ: eʸ = x ± √(x² + 1). eʸ must be positive, so take +. Then y = ln(x + √(x² + 1)).
Further hyperbolic identities (Osborn's rule)
Hyperbolic identities look like trig identities with some signs changed:
- sinh 2x = 2 sinh x cosh x
- cosh 2x = cosh²x + sinh²x = 2cosh²x − 1 = 1 + 2sinh²x
- sinh(A ± B) = sinh A cosh B ± cosh A sinh B
- cosh(A ± B) = cosh A cosh B ± sinh A sinh B
Osborn's rule: take a trig identity, change cos to cosh and sin to sinh, and flip the sign of every term that contains a product of two sines (sin² counts, and so does tan² because it hides sin²). Always check with the exponential definitions if unsure.
Solving equations: write in terms of eˣ, or use identities. Example: cosh x = 2 gives x = ±arcosh 2 = ±ln(2 + √3).
Differentiating and integrating hyperbolic functions
- d/dx sinh x = cosh x
- d/dx cosh x = sinh x (no minus sign, unlike cos!)
- d/dx tanh x = sech² x
So ∫ sinh x dx = cosh x + c, ∫ cosh x dx = sinh x + c, ∫ sech²x dx = tanh x + c. Chain rule works as usual: d/dx cosh 3x = 3 sinh 3x.
Derivatives of the inverses
- d/dx arsinh x = 1/√(x² + 1)
- d/dx arcosh x = 1/√(x² − 1), x > 1
- d/dx artanh x = 1/(1 − x²), |x| < 1
Integrals that give inverse hyperbolic functions
- ∫ 1/√(x² + a²) dx = arsinh(x/a) + c = ln(x + √(x² + a²)) + c
- ∫ 1/√(x² − a²) dx = arcosh(x/a) + c = ln(x + √(x² − a²)) + c, x > a
- Compare: ∫ 1/√(a² − x²) dx = arcsin(x/a) + c
Substitutions to use: x = a sinh u for √(x² + a²), and x = a cosh u for √(x² − a²). For a quadratic like x² + 4x + 13, first complete the square: (x + 2)² + 9.
Try it: hang a chain
Hang a thin chain or necklace between two fingers at the same height. The curve you see is a cosh curve. Now move your fingers closer: the curve gets deeper but keeps the same family shape. Then in the 3D free play, move t and check that cosh² − sinh² stays 1 every time.
Key formulas and definitions
- cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ − e⁻ˣ)/2, tanh x = sinh x/cosh x
- cosh²x − sinh²x = 1; 1 − tanh²x = sech²x
- sinh 2x = 2 sinh x cosh x; cosh 2x = 2cosh²x − 1
- arsinh x = ln(x + √(x² + 1)); arcosh x = ln(x + √(x² − 1)), x ≥ 1; artanh x = ½ ln((1+x)/(1−x))
- d/dx sinh x = cosh x; d/dx cosh x = sinh x; d/dx tanh x = sech²x
- ∫ dx/√(x² + a²) = arsinh(x/a) + c; ∫ dx/√(x² − a²) = arcosh(x/a) + c
Worked examples
1. Find cosh(ln 3) as a fraction.
e^(ln 3) = 3, e^(−ln 3) = 1/3. cosh = (3 + 1/3)/2 = (10/3)/2 = 5/3.
2. Solve sinh x = 2, giving x in log form.
x = arsinh 2 = ln(2 + √5) ≈ 1.444.
3. Prove that cosh 2x = 1 + 2 sinh²x.
1 + 2sinh²x = 1 + 2(e²ˣ − 2 + e⁻²ˣ)/4 = 1 + (e²ˣ + e⁻²ˣ)/2 − 1 = (e²ˣ + e⁻²ˣ)/2 = cosh 2x.
4. Solve 2cosh x − sinh x = 2.
Use exponentials: (eˣ + e⁻ˣ) − (eˣ − e⁻ˣ)/2 = 2. Multiply by 2: eˣ + 3e⁻ˣ = 4. Multiply by eˣ: e²ˣ − 4eˣ + 3 = 0, so (eˣ − 1)(eˣ − 3) = 0. eˣ = 1 or 3, so x = 0 or x = ln 3.
5. Differentiate y = x cosh 2x.
Product rule: dy/dx = cosh 2x + x · 2 sinh 2x = cosh 2x + 2x sinh 2x.
6. Find ∫ 1/√(x² + 9) dx.
a = 3: arsinh(x/3) + c = ln(x + √(x² + 9)) + c (the constant ln 3 is absorbed into c).
7. Evaluate ∫ from 2 to 3 of 1/√(x² − 4x + 5) dx.
Complete the square: x² − 4x + 5 = (x − 2)² + 1. Integral = [arsinh(x − 2)] from 2 to 3 = arsinh 1 − arsinh 0 = ln(1 + √2) ≈ 0.881.
Common mistakes
- Writing d/dx cosh x = −sinh x. Unlike cos, there is no minus sign.
- Mixing the signs: cosh has + (eˣ + e⁻ˣ), sinh has − (eˣ − e⁻ˣ).
- Copying a trig identity without changing signs: cosh²x + sinh²x is cosh 2x, not 1.
- Taking both signs in arcosh or arsinh log forms: eʸ must be positive, and arcosh is defined as the positive value.