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Hyperbolic functions and identities

The hyperbolic functions are defined by exponentials. These definitions give their parity, identities, graphs and inverse functions. Derive identities algebraically or apply Osborn's rule with the correct sign change for each pair of sine factors, and state the domain and range of each function and each inverse.

Year FMEDEXCEL 9FM0 CP2 8.1, 8.3, 8.4

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.
State the domain and range of all six hyperbolic functions and of the three inverses.
Prove and use cosh²x − sinh²x = 1 and its relatives.
Apply Osborn's rule to convert a trigonometric identity into the corresponding hyperbolic identity.
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.

None of them is periodic. They are the even and odd halves of the exponential: cosh x = (e^x + e^−x)/2 has minimum value 1 at x = 0, sinh x = (e^x − e^−x)/2 increases through the origin and takes every real value, and tanh x approaches 1 and −1 without attaining either. The name comes from the curve: (cosh t, sinh t) lies on x² − y² = 1.

Splitting the exponential

The hyperbolic functions split e^x into its even and odd halves:

\text{cosh } x = \frac{e^{x} + e^{-x}}{2}, \text{sinh } x = \frac{e^{x} - e^{-x}}{2}

NOT IN THE BOOKLET: LEARN IT

Check the current formula booklet for your awarding organisation; whether or not the definitions, identities or logarithmic forms are supplied, you must know when each applies and how to derive it. tanh x is the ratio sinh x / cosh x. None of these functions is periodic: cosh has a minimum value 1 at x = 0, sinh is increasing through the origin and takes every real value, and tanh tends to 1 and −1 without attaining either. For large positive x both cosh x and sinh x are close to e^x/2.

The cosh valley bottoming out at nought comma one and the sinh curve sweeping through the origin, with a dashed half-exponential that both curves squeeze against on the right. Nothing repeats anywhere: the resemblance to sine and cosine is in name and algebra, not shape. — FIG. 1cosh and sinh alongside exp(x)/2: the even and odd parts of the exponential, neither of which is periodic.

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.

Written as ranges: cosh x ≥ 1; sinh x takes every real value; and −1 < tanh x < 1, with both bounds approached and neither attained. Each reciprocal's range follows by inverting the parent's. Dividing 1 by a value not less than 1 gives 0 < sech x ≤ 1, the greatest value occurring where cosh is least, at x = 0, so the graph of sech has a single maximum of height 1 and tends to zero in both directions.

The other two reciprocals are undefined at x = 0, so their graphs have two branches. On the domain x ≠ 0, sinh takes every real value except 0, so cosech takes every real value but 0 as well, drawn as two branches, one in the first quadrant and one in the third, with the y-axis as an asymptote between them and the x-axis as an asymptote far out. coth inverts a number lying strictly between −1 and 1 and away from 0, so coth x > 1 for positive x and coth x < −1 for negative x, and its two branches fall from the vertical asymptote at x = 0 towards the horizontal lines y = 1 and y = −1, approaching them from outside.

The identities, and Osborn's rule

Square the definitions and subtract. The cross terms cancel, leaving cosh²x − sinh²x = 1, so the point (cosh t, sinh t) lies on the hyperbola x² − y² = 1, which is the origin of the name. At t = ln 3, cosh t = 5/3 and sinh t = 4/3, and 25/9 − 16/9 = 1.

The right branch of the hyperbola x squared minus y squared equals one, opening from its vertex at one comma nought, with an amber point at five thirds comma four thirds. cosh t and sinh t are that point's coordinates, exactly as cos and sin coordinate the circle. — FIG. 2The hyperbola x² − y² = 1 with the point (5/3, 4/3) at t = ln 3: cosh and sinh are its coordinates, as cos and sin are the coordinates on the unit circle.

Dividing that identity by cosh²x gives 1 − tanh²x = sech²x, and dividing by sinh²x gives coth²x − 1 = cosech²x. Osborn's rule gives the remaining identities. To convert a trigonometric identity using Osborn's rule, replace cos by cosh and sin by sinh, then reverse the sign for each pair of sine factors. Thus a term with two sine factors changes sign; a term with four does not. 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. A tan² term contains two sine factors, so 1 + tan²θ = sec²θ becomes 1 − tanh²x = sech²x.

The inverse functions in logarithmic form

Because the functions are defined by e^x, their inverses can be written as logarithms. Setting y = sinh x and clearing the fraction gives a quadratic in e^x whose positive root gives arsinh x = ln(x + √(x² + 1)). The same method 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 used to solve hyperbolic equations.

An inverse exchanges domain and range, so each condition above is the parent function's range read as a domain, and each inverse's range is the parent's domain. arsinh has domain and range the whole real line. artanh has domain −1 < x < 1 and range all real numbers.

arcosh requires a restriction, because cosh is not one-to-one. The branch used here is x ≥ 0, so the range of arcosh is y ≥ 0. That choice corresponds to taking the positive root of the quadratic in the derivation, and the formula reflects it: for x ≥ 1 the expression x + √(x² − 1) is at least 1, so its logarithm comes out at 0 or above, with the value 0 at x = 1.

cosh on x ≥ 0 reflected in y = x to give arcosh: the inverse is the logarithm ln(x + √(x² − 1)) — FIG. 3cosh restricted to x ≥ 0, reflected in y = x to give arcosh: the restriction makes the reflection a function, and the marked pair of points shows cosh(ln(2 + √3)) = 2.

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) = 2.00 to three significant figures, which checks the logarithmic form.

Solving hyperbolic equations

An equation involving two hyperbolic functions is solved by the same method. Use an identity to express everything in terms of one function, solve the resulting quadratic, then convert each root using a logarithmic form. Count the solutions: cosh x = k has two solutions for k > 1, while sinh x = k has exactly one for every real k.

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 rather than quoting them.
cosh x = k has two solutions for k > 1, and sinh x = k has exactly one for every real k. State which case applies.
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 range or domain, such as cosh x = −2 or arcosh of anything below 1.
A question asking for domain and range requires both sets to be stated. arcosh has domain x ≥ 1 and range y ≥ 0, the branch chosen when cosh was restricted.

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

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, which is cosh²x − sinh²x = 1.

cosh and sinh are the even and odd parts of e^x, and cosh²x − sinh²x = 1 follows from the definitions.

Ranges: cosh x ≥ 1, 0 < sech x ≤ 1, |tanh x| < 1 and |coth x| > 1; sinh takes every real value and cosech takes every real value except 0.

Osborn's rule replaces cos by cosh and sin by sinh, reversing the sign once for each pair of sine factors.

arsinh x = ln(x + √(x² + 1)); arcosh x = ln(x + √(x² − 1)) for x ≥ 1, returning y ≥ 0; 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.
State the domain and range of all six hyperbolic functions and of the three inverses.
Prove and use cosh²x − sinh²x = 1 and its relatives.
Apply Osborn's rule to convert a trigonometric identity into the corresponding hyperbolic identity.
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.