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, tanh and their reciprocals from exponentials, and sketch the graphs.
- Prove and use cosh²x − sinh²x = 1 and its relatives.
- Apply Osborn's rule to convert a trig identity into its hyperbolic twin.
- Solve hyperbolic equations exactly via the logarithmic forms of the inverses.
COMMON MISCONCEPTION
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:
The booklet gives you the hyperbolic identities and the logarithmic forms of the inverses, but not these two definitions. Memorise them, and tanh x as their ratio, which is not printed either. Nothing repeats. cosh is a valley with minimum 1 at x = 0, sinh climbs through the origin taking every real value, and tanh flattens towards ±1 without reaching either. For large positive x both cosh and sinh hug ex/2. The names advertise a family resemblance to trig, not periodicity.
Three reciprocals complete the set. sech x = 1/cosh x, cosech x = 1/sinh x and coth x = 1/tanh x. Only sech is defined everywhere, since the other two need sinh x ≠ 0 and so exclude x = 0.
The identities, and Osborn's rule
Square the definitions and subtract. The cross terms collapse and you are left with cosh²x − sinh²x = 1, so the point (cosh t, sinh t) rides the hyperbola x² − y² = 1. That 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.
Divide that identity by cosh²x to get 1 − tanh²x = sech²x, and by sinh²x to get coth²x − 1 = cosech²x. Osborn's rule generates the rest. Take any trig identity, swap cos for cosh and sin for sinh, then flip the sign of every term containing a product of two sinh factors. So cos 2θ = 1 − 2sin²θ becomes cosh 2x = 1 + 2sinh²x, while sin 2θ = 2 sin θ cos θ becomes sinh 2x = 2 sinh x cosh x with no change at all. A tanh² term counts as a hidden product of two sinhs, so 1 + tan²θ = sec²θ arrives as 1 − tanh²x = sech²x.
Inverses that are logarithms
Because the functions are built from ex, their inverses come out as logarithms. Set y = sinh x, clear the fraction, and you have a quadratic in ex whose positive root gives arsinh x = ln(x + √(x² + 1)). The same route gives arcosh x = ln(x + √(x² − 1)) for x ≥ 1, and artanh x = ½ln((1 + x)/(1 − x)) 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.
Solving hyperbolic equations
Most exam equations mix two hyperbolic functions, and the route is always the same. Use an identity to get everything in terms of one of them, solve the resulting quadratic, then convert each root with a logarithmic form. Count the solutions as you go, because cosh gives two and sinh gives one.
WORKED EXAMPLE
A quadratic in disguise
Solve cosh 2x − 3 sinh x = 3, giving exact answers.
Osborn on cos 2θ = 1 − 2sin²θ gives cosh 2x = 1 + 2sinh²x.
The equation becomes 2sinh²x − 3 sinh x − 2 = 0, so (2 sinh x + 1)(sinh x − 2) = 0.
sinh x = −1/2 or sinh x = 2, and sinh is one-to-one, so each gives exactly one root.
x = ln((√5 − 1)/2) or ln(2 + √5), roughly −0.481 and 1.444.
GUIDED PRACTICE
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 and no ± is needed.
ASSESSMENT FOCUS
- Prove identities from the exponential definitions. That derivation is the expected working, not a quotation.
- cosh x = k has two solutions for k > 1 and sinh x = k always has exactly one. Say which case you are in.
- Reduce a mixed equation to a quadratic in one hyperbolic function before solving anything.
- Quote arsinh, arcosh and artanh in log form whenever the question says exact.
- Reject roots that fall outside a domain, such as cosh x = −2 or arcosh of anything below 1.
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
eln 3 = 3 and e−ln 3 = 1/3.
Show the answer
cosh(ln 3) = (3 + 1/3)/2 = 5/3 and sinh(ln 3) = (3 − 1/3)/2 = 4/3. Then 25/9 − 16/9 = 1, as cosh² − sinh² promises.
cosh and sinh are the even and odd halves of ex, and cosh² − sinh² = 1 follows in two lines.
Osborn's rule converts trig identities by flipping the sign on products of two sinhs.
arsinh x = ln(x + √(x² + 1)); arcosh x = ln(x + √(x² − 1)) for x ≥ 1; artanh x = ½ln((1 + x)/(1 − x)) for −1 < 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.
Or read them with their worked answers on the hyperbolic functions and identities questions page.
CHECK YOUR PROGRESS
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- Define sinh, cosh, tanh and their reciprocals from exponentials, and sketch the graphs.
- Prove and use cosh²x − sinh²x = 1 and its relatives.
- Apply Osborn's rule to convert a trig identity into its hyperbolic twin.
- 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.