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Inverse hyperbolic functions

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

arsinh, arcosh and artanh undo sinh, cosh and tanh. Because the hyperbolic functions are built from e^x, their inverses can be written with natural logarithms. You must know, and be able to derive, these log forms.

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Domains and ranges

y = arsinh x: defined for all x, an odd function through the origin.
y = arcosh x: cosh is not one-one, so we restrict to x ≥ 0. arcosh x is defined for x ≥ 1 and gives y ≥ 0 (the principal value).
y = artanh x: defined for -1 < x < 1, with vertical asymptotes x = ±1.

Each graph is the reflection of the original (restricted) graph in y = x.

Function	Domain	Log form
arsinh x	all x	ln(x + sqrt(x² + 1))
arcosh x	x ≥ 1	ln(x + sqrt(x² - 1))
artanh x	-1 < x < 1	(1/2) ln((1 + x)/(1 - x))

Deriving the log form of arsinh

Let y = arsinh x, so x = sinh y = (e^y - e^-y)/2.

Multiply by 2e^y: e^2y - 2x e^y - 1 = 0, a quadratic in e^y.

e^y = x ± sqrt(x² + 1). Since sqrt(x² + 1) > x, the minus sign gives a negative value, which e^y cannot be. So y = ln(x + sqrt(x² + 1)).

arcosh and artanh

For arcosh the same method gives e^y = x ± sqrt(x² - 1). Both are positive, but they multiply to 1, so their logs are negatives of each other; the principal value y ≥ 0 uses the + sign.

For artanh: x = (e^2y - 1)/(e^2y + 1) rearranges to e^2y = (1 + x)/(1 - x), so y = (1/2) ln((1 + x)/(1 - x)).

Worked example

Find the exact value of artanh(1/3).

artanh x = (1/2) ln((1 + x)/(1 - x))
(1 + 1/3)/(1 - 1/3) = (4/3)/(2/3) = 2

Answer: (1/2) ln 2

Key idea: arsinh x = ln(x + sqrt(x² + 1)) for all x; arcosh x = ln(x + sqrt(x² - 1)) for x ≥ 1; artanh x = (1/2) ln((1 + x)/(1 - x)) for |x| < 1.

Check yourself

Find arsinh 2, to 4 decimal places.
Show answer
1.4436. arsinh 2 = ln(2 + sqrt(5)) = 1.4436.
Find artanh(1/3), to 4 decimal places.
Show answer
0.3466. (1/2) ln 2 = 0.3466.
Find arcosh(5/3), to 4 decimal places.
Show answer
1.0986. ln(5/3 + sqrt(25/9 - 1)) = ln(5/3 + 4/3) = ln 3 = 1.0986.
What is the domain of arcosh x?
- x ≥ 0
- -1 < x < 1
- x ≥ 1
- all real x
Show answer
x ≥ 1. arcosh undoes cosh, whose range is cosh x ≥ 1.
What is the domain of artanh x?
- -1 < x < 1
- x ≥ 1
- x > 0
- all real x
Show answer
-1 < x < 1. The range of tanh x is -1 < tanh x < 1. Also (1 + x)/(1 - x) must be positive.
Deriving arsinh x, you reach e^y = x ± sqrt(x² + 1). Why is the minus sign rejected?
- It gives a complex value of y
- It only works when x is negative
- x - sqrt(x² + 1) is always negative, but e^y > 0
- It makes y equal to zero
Show answer
x - sqrt(x² + 1) is always negative, but e^y > 0. sqrt(x² + 1) > |x|, so x - sqrt(x² + 1) < 0 for every x. An exponential is always positive.
Which is the logarithmic form of arcosh x?
- ln(x + sqrt(x² - 1))
- ln(x + sqrt(x² + 1))
- (1/2) ln((1 + x)/(1 - x))
- ln(x - sqrt(x² - 1))
Show answer
ln(x + sqrt(x² - 1)). From x = cosh y: e^2y - 2x e^y + 1 = 0 gives e^y = x ± sqrt(x² - 1); the principal value takes +.

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Sources

Pearson Edexcel International Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (issue 3), Unit FP3

Last updated 8 November 2026. Spotted a mistake? Tell us.