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Hyperbolic functions

The exponential definitions explain the signs in hyperbolic identities.

sinh x = (eˣ − e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2. Subtracting their squares gives cosh²x − sinh²x = 1. Unlike the trigonometric identity, this is a difference of squares.

cosh is even, sinh is odd, and cosh x ≥ 1 for real x. Use these facts to find the domains and ranges of inverse hyperbolic functions. The derivative of cosh is sinh, with no minus sign.

Worked example

Given sinh x = 3 and x is real, find cosh x.

  1. Use cosh²x − sinh²x = 1.
  2. cosh²x = 10.
  3. For real x, cosh x is positive.

Answer: √10

Revise first: Fractional and negative indices.

Use the exponential definitions to settle signs

sinh x = (eˣ − e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2. Subtracting their squares gives cosh²x − sinh²x = 1. The minus sign differs from the circular identity.

cosh is even and is at least 1 for real x. Recovering x from cosh x therefore needs a branch condition. sinh is strictly increasing and has one real inverse value.

If cosh x = 5/4 and x > 0, find eˣ

  1. Put u = eˣ > 0. Then u + 1/u = 5/2.
  2. Multiplication by 2u gives 2u² − 5u + 2 = 0.
  3. The roots are u = 2 and u = 1/2.
  4. The condition x > 0 means u > 1, so eˣ = 2. Without that condition both roots would be possible.
Why is cosh x = 1/2 impossible for real x?

u + 1/u ≥ 2 for u > 0, so cosh x ≥ 1.

Take the idea further

Given sinh x = 3/4 and cosh x = 5/4, sinh(2x) = 15/8 and cosh(2x) = 17/8.

More advanced methods

Practise hyperbolic functions

Course mapping

These specification references show where the topic occurs. The questions cover only some parts of each topic.

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