Hyperbolic Functions

Hyperbolic functions

Hyperbolic functions are analogues of the trig functions, defined using e:

sinh x = (eˣ − e⁻ˣ) / 2
cosh x = (eˣ + e⁻ˣ) / 2
tanh x = sinh x / cosh x

(Read as "shine", "cosh", "than".) They're called hyperbolic because they parametrise a hyperbola, just as sin/cos parametrise a circle.

Key properties

• cosh x ≥ 1 for all x (minimum of 1 at x = 0); its graph is a symmetric "U" (catenary shape).
• sinh x passes through the origin and is an odd function; tanh x ranges between −1 and 1.
• cosh is even (cosh(−x) = cosh x); sinh and tanh are odd.

Identities

Similar to trig identities but with sign changes:

cosh²x − sinh²x = 1
sinh 2x = 2 sinh x cosh x
cosh 2x = cosh²x + sinh²x

(Osborn's rule: take a trig identity and change the sign of any product of two sinh terms.)

Derivatives and integrals

d/dx(sinh x) = cosh x
d/dx(cosh x) = sinh x     (note: no minus sign, unlike cos)
d/dx(tanh x) = sech²x

Integrating reverses these.

Inverse hyperbolic functions

These can be written using logarithms:

arsinh x = ln(x + √(x² + 1))
arcosh x = ln(x + √(x² − 1))   (x ≥ 1)

They're used to evaluate certain integrals (e.g. integrals of 1/√(x²+1)).

Worked example

Show that cosh²x − sinh²x = 1.

• cosh²x − sinh²x = [(eˣ + e⁻ˣ)/2]² − [(eˣ − e⁻ˣ)/2]²
• = (1/4)[(e²ˣ + 2 + e⁻²ˣ) − (e²ˣ − 2 + e⁻²ˣ)] = (1/4)(4) = 1. ✓

Common mistakes

• Forgetting the derivative of cosh x is +sinh x (no minus, unlike cos → −sin).
• Using the trig identity signs directly (hyperbolic identities differ, e.g. cosh² − sinh² = 1).
• Getting the definitions of sinh (minus) and cosh (plus) mixed up.

Exam tips

• Learn the exponential definitions and the key identity cosh²x − sinh²x = 1.
• Know the derivatives (cosh differentiates to +sinh).
• Recall the logarithmic forms of the inverse hyperbolic functions.

Key facts to remember

• sinh x = (eˣ − e⁻ˣ)/2, cosh x = (eˣ + e⁻ˣ)/2, tanh x = sinh/cosh; cosh x ≥ 1.
• Identity cosh²x − sinh²x = 1; d/dx(sinh) = cosh, d/dx(cosh) = sinh (no minus).
• Inverses have logarithmic forms (arsinh x = ln(x + √(x²+1))).