Maths › Hyperbolic functions › Hyperbolic functions and identities
Hyperbolic functions and identities
Build trig's cousins from exponentials and everything about them becomes checkable: graphs, identities, and inverses that turn out to be plain logarithms once the definition is unpacked.
Builds on Exponential functions and e and Trigonometric graphs and equations.
IN THIS TOPIC
- Define sinh, cosh and tanh from exponentials and sketch their graphs.
- Prove and use cosh²x − sinh²x = 1 and its relatives.
- Solve hyperbolic equations exactly via the logarithmic forms of the inverses.
WHAT YOU PROBABLY THINK
Functions with names like sinh and cosh must be periodic like sine and cosine.
Splitting the exponential
The hyperbolic functions split ex into its even and odd halves:
with tanh x their ratio. Nothing repeats: cosh is a valley with minimum 1 at x = 0, sinh climbs through the origin, and both hug ex/2 for large x. The names advertise a family resemblance to trig, not periodicity, so the opening claim reads too much into them.
Squaring the definitions and subtracting collapses the cross terms to give cosh²x − sinh²x = 1: the point (cosh t, sinh t) rides the hyperbola x² − y² = 1, which is where the family name comes from. At x = ln 3, for instance, cosh = 5/3 and sinh = 4/3, and 25/9 − 16/9 = 1 on the nose.
Inverses that are logarithms
Because the functions are built from ex, their inverses come out as logarithms. Setting y = sinh x and solving the resulting quadratic in ex gives arsinh x = ln(x + √(x² + 1)), and similarly arcosh x = ln(x + √(x² − 1)) for x ≥ 1. These exact forms are how hyperbolic equations are solved on paper.
WORKED EXAMPLE
Solving cosh x = 2 exactly
Solve cosh x = 2, giving both answers in exact form.
arcosh 2 = ln(2 + √3) is one solution.
cosh is even, so x = ±ln(2 + √3).
Numerically that is ±1.317, and cosh(1.317) returns 2.00: the log form is checkable in seconds on a calculator.
YOUR TURN
A sinh equation with a tidy answer
Solve sinh x = 3/4 exactly.
Show the working
x = arsinh(3/4) = ln(3/4 + √(9/16 + 1)) = ln(3/4 + 5/4).
= ln 2.
Check: sinh(ln 2) = (2 − 1/2)/2 = 3/4. sinh is one-to-one, so this is the only solution, no ± needed.
THE EXAM BIT
- Prove identities from the exponential definitions; that derivation is the expected working.
- cosh x = k has two solutions for k > 1, sinh x = k always exactly one. Say which and why.
- Osborn's rule: trig identities transfer with a sign flip on any product of two sinh terms.
- Quote arsinh and arcosh in log form when exact answers are demanded.
CHECK YOURSELF
Using the definitions, find the exact values of cosh(ln 3) and sinh(ln 3), and verify the identity connecting them.
Show a hint
e^(ln 3) = 3 and e^(−ln 3) = 1/3.
Show the answer
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cosh and sinh are the even and odd halves of ex; cosh² − sinh² = 1 follows in two lines.
arsinh x = ln(x + √(x² + 1)); arcosh x = ln(x + √(x² − 1)) for x ≥ 1.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Define sinh, cosh and tanh from exponentials and sketch their graphs.
- Prove and use cosh²x − sinh²x = 1 and its relatives.
- Solve hyperbolic equations exactly via the logarithmic forms of the inverses.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.