The Inverse Hyperbolic Functions, and Why They Are Logarithms
Chapter Thirty-Four
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 76 to 77 of 303
In one line
Solve the defining equation for the exponential, which is a quadratic, and the inverse function comes out as a logarithm.
The three results
These are MU's label "Inverse hyperbolic functions", and they are what she wants.
asinh(x) = log(x + sqrt(x^2 + 1))
acosh(x) = log(x + sqrt(x^2 - 1))
atanh(x) = log((1 + x)/(1 - x))/2
The second holds for x of one or more and the third strictly between minus one and one, which are the ranges of cosh and tanh. Outside those ranges there is nothing to invert, and this book's checker is told the domain of each so that the claims are proved where they are made and not where they are not.
Each is derived below rather than quoted, because a question asking you to "express sinh inverse x in logarithmic form" is asking for the derivation.
Deriving the inverse sine
Let y be the number whose hyperbolic sine is x. So x = sinh y, and we want y in terms of x.
x = sinh(y)
Write sinh y out in exponentials and multiply through by 2.
sinh(y) = (e^y - e^(-y))/2
So 2x equals e to the y minus e to the minus y. Multiply everything by e to the y to clear the negative exponent, and set u = e to the y.
The equation becomes u squared minus 2xu minus 1 = 0, which is an ordinary quadratic in u. Solve it by the formula.
u = (2x + sqrt(4x^2 + 4))/2
u = x + sqrt(x^2 + 1)
The quadratic has two roots, x plus the root and x minus the root. The second must be rejected, and this is the step marks are given for: u is e to the y, which is positive for every real y, while x minus the square root of x squared plus one is always negative, because the square root exceeds the size of x. So only the plus sign survives.
Finally, y is the logarithm of u.
asinh(x) = log(x + sqrt(x^2 + 1))
The domain is every real x, and the function is defined for all of them, because x squared plus one is always positive.
Deriving the inverse cosine
The same method, starting from x = cosh y, gives the quadratic u squared minus 2xu plus 1 = 0, whose roots are x plus and x minus the square root of x squared minus one.
acosh(x) = log(x + sqrt(x^2 - 1))
Two differences from the sine, and both are asked about.
The domain is x of one or more. The hyperbolic cosine never takes a value below one, so there is nothing to invert below one, and the square root of x squared minus one would not be real there either.
The Inverse Hyperbolic Functions, and Why They Are Logarithms
Both roots are legitimate, because both are positive when x is at least one, and their product is one. So cosh inverse is genuinely two-valued: for any x above one there are two values of y, one positive and one negative, differing only in sign. The principal value is the one with the plus sign, which gives the non-negative answer, and the other is its negative.
log(x + sqrt(x^2 - 1)) = -log(x - sqrt(x^2 - 1))
That identity is the two-valuedness written down: the two answers are negatives of each other. It is what you would expect from the graph of cosh, which is symmetric about the vertical axis, so a horizontal line crosses it twice.
Deriving the inverse tangent
From x = tanh y, write the tangent in the form the hyperbolic chapter gave.
tanh(y) = (e^(2y) - 1)/(e^(2y) + 1)
So x(e to the 2y plus 1) equals e to the 2y minus 1. Collecting the exponential on one side gives e to the 2y times (1 minus x) equal to 1 plus x.
e^(2y) = (1 + x)/(1 - x)
atanh(x) = log((1 + x)/(1 - x))/2
The domain is strictly between minus one and one, because outside that range the fraction is negative or undefined and its real logarithm does not exist. That matches the range of tanh, which the hyperbolic chapter gave as strictly between minus one and one.
The other three
For completeness, and because a question occasionally asks.
acoth(x) = log((x + 1)/(x - 1))/2
asech(x) = log((1 + sqrt(1 - x^2))/x)
acsch(x) = log((1 + sqrt(1 + x^2))/x)
Worked values
asinh(0) = 0
asinh(1) = log(1 + sqrt(2))
acosh(1) = 0
atanh(0) = 0
atanh(1/2) = log(3)/2
asinh(3/4) = log(2)
The last one is a favourite: three quarters plus the square root of nine sixteenths plus one is three quarters plus five quarters, which is two, so the answer is exactly log 2. Questions are usually built backwards from a neat answer like that, so if your logarithm's argument is not tidy, check the arithmetic under the root.
The derivatives, which Module 2 needs
diff(asinh(x), x) = 1/sqrt(x^2 + 1)
diff(atanh(x), x) = 1/(1 - x^2)
These are the reason the inverse hyperbolic functions appear in integration tables, and the second is the reason a partial fraction with two linear factors can produce an inverse hyperbolic tangent instead of two logarithms.
Check yourself
asinh(x) = log(x + sqrt(x^2 + 1))
asinh(-x) = -asinh(x)
asinh(sinh(2)) = 2
atanh(tanh(1)) = 1
acosh(cosh(3)) = 3
atanh(x) = log((1 + x)/(1 - x))/2
The third line says sinh inverse is odd, which it must be because sinh is. Confirming it from the logarithmic form takes one line of algebra with the conjugate surd, and it is a common short question.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.