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Hyperbolic Functions

Definitions, identities, derivatives, and inverse hyperbolic functions (log forms), with physics intuition and common use-cases.

  • GCE A-Level H2 Physics 2027
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Hyperbolic functions (sinh, cosh, tanh) behave like “exponential versions” of sine and cosine. They show up in differential equations, special relativity, and many growth/decay models.

1) Definitions (start here)

The definitions in terms of exponentials are the cleanest:

cosh x = (e^x + e^(-x))/2, sinh x = (e^x-e^(-x))/2, tanh x = (sinh x)/(cosh x).
Quick intuition

cosh x is always ≥ 1 and grows like 1/2 e^(|x|) for large |x|.
sinh x is odd (like sin x) and grows like 1/2 e^(|x|) with a sign.
tanh x is bounded between -1 and 1 (like tan is not).

2) Core identities

Hyperbolic Pythagorean identity

cosh² x- sinh² x = 1.

Useful derived forms

1- tanh² x = sech² x, coth² x-1 = csch² x.

Addition formulas

sinh(x± y) = sinh x cosh y ± cosh x sinh y,
cosh(x± y) = cosh x cosh y ± sinh x sinh y.

Double-angle formulas

sinh(2x) = 2 sinh x cosh x,
cosh(2x) = cosh² x + sinh² x = 2 cosh² x-1 = 1 + 2 sinh² x.

3) Derivatives and integrals (high frequency)

d/dx sinh x = cosh x, d/dx cosh x = sinh x.
d/dx tanh x = sech² x, (d/dx)sech x = -sechx tanh x.

Integrals:

∫ sinh x dx = cosh x + C, ∫ cosh x dx = sinh x + C.

4) Inverse hyperbolic functions (log forms)

These are useful when solving integrals or ODEs.

arsinh(u) = sinh⁻¹ (u) = ln(u + square root of (u² + 1)) (all real u).
arcosh(u) = cosh⁻¹ (u) = ln(u + square root of (u²-1)) (u ≥ 1).
artanh(u) = tanh⁻¹ (u) = 1/2 ln((1 + u)/(1-u)) (|u| < 1).

5) One physics connection: rapidity (special relativity)

Instead of velocity v, special relativity often uses rapidity η defined by:

v/c = tanh η.

This is useful because rapidities add linearly, while velocities do not.

6) Worked example (why cosh and sinh solve “exponential” ODEs)

Solve

y'' = y

with y(0) = 1 and y'(0) = 0.

A standard solution basis for y'' = y is:

y(x) = A cosh x + B sinh x.

Apply y(0) = 1:

y(0) = A cosh 0 + B sinh 0 = A = 1.

Differentiate: y' = A sinh x + B cosh x. Apply y'(0) = 0:

y'(0) = A sinh 0 + B cosh 0 = B = 0.

So the solution is:

y(x) = cosh x.

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