Hyperbolic functions

Further Maths · Further Pure Mathematics 2. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

The hyperbolic functions are built from e^x and e^−x. sinh x is half their difference and cosh x is half their sum. The other four (tanh, sech, cosech, coth) are made from these two in the same way as tan, sec, cosec and cot are made from sin and cos.

Their identities look like the trigonometric ones, but some signs change. The most important is cosh² x − sinh² x = 1.

Know the graphs: cosh x is U-shaped with least value 1, sinh x passes through the origin and takes every real value, and tanh x lies between −1 and 1. Each inverse function can be written as a logarithm.

Rules to remember

sinh x = (e^x − e^−x)/2, cosh x = (e^x + e^−x)/2, tanh x = sinh x/cosh x = (e^2x − 1)/(e^2x + 1)
sech x = 1/cosh x, cosech x = 1/sinh x, coth x = 1/tanh x
cosh² x − sinh² x = 1, 1 − tanh² x = sech² x, coth² x − 1 = cosech² x
sinh 2x = 2 sinh x cosh x; cosh 2x = cosh² x + sinh² x = 2cosh² x − 1 = 1 + 2sinh² x
sinh⁻¹ x = ln(x + √(x² + 1)) for all x; cosh⁻¹ x = ln(x + √(x² − 1)) for x ≥ 1
tanh⁻¹ x = ½ ln((1 + x)/(1 − x)) for −1 < x < 1
cosh x ≥ 1 and is even; sinh x is odd; y = tanh x has asymptotes y = 1 and y = −1

Common mistake

Students copy the trigonometric identity and write cosh² x + sinh² x = 1. The correct identity has a minus sign: cosh² x − sinh² x = 1. Also remember cosh x = k has two solutions, ± cosh⁻¹ k, when k > 1.

Worked example

Solve 3 cosh x − sinh x = 3, giving exact answers.

Use the definitions: 3(e^x + e^−x)/2 − (e^x − e^−x)/2 = 3, so e^x + 2e^−x = 3.
Multiply by e^x: e^2x − 3e^x + 2 = 0, so (e^x − 1)(e^x − 2) = 0.
e^x = 1 or e^x = 2, so x = 0 or x = ln 2.
Check x = ln 2: cosh x = 5/4 and sinh x = 3/4, so 3 × 5/4 − 3/4 = 12/4 = 3.

Answer: x = 0 or x = ln 2

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Hyperbolic Functions

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