Hyperbolic Functions and Their Inverses
Build sinh and cosh from the exponential function, prove the identity that replaces the Pythagorean one, and write the inverses as logarithms.
What a learner can do afterwards
- Define sinh and cosh in terms of the exponential and sketch both graphs
- Prove the identity relating cosh squared and sinh squared and compare it with the circular case
- Express an inverse hyperbolic function as a logarithm, state its domain, and evaluate the derivative of arsinh
1 · Read
Return to exponentials and logarithms: e^(-x) = 1/e^x and ln(e^x) = x for real x. Define sinh x = (e^x - e^(-x))/2 and cosh x = (e^x + e^(-x))/2. Divide each numerator by 2 once. At zero, both exponentials equal 1, so sinh 0 = 0 and cosh 0 = 1. At x = ln a, with a > 0, use e^x = a and e^(-x) = 1/a.
To sketch sinh, mark (0, 0). It increases from negative to positive values without bound. It is odd: sinh(-x) = -sinh x. Cosh is even: cosh(-x) = cosh x. It decreases to its minimum (0, 1), then increases; both ends rise without bound. A uniform flexible chain hanging at rest under only its own weight follows a shifted, scaled cosh curve called a catenary.
Set A = e^x and B = e^(-x), so AB = 1. Then cosh squared x - sinh squared x = ((A+B)^2 - (A-B)^2)/4. The pure square terms cancel; the cross terms give 2AB - (-2AB) = 4AB. Dividing by 4 gives 1. Hence cosh squared x = 1 + sinh squared x. Compare cos squared x + sin squared x = 1. Points (cosh t, sinh t) trace the right branch of X squared - Y squared = 1.
Adding the definitions gives cosh x + sinh x = e^x; subtracting gives cosh x - sinh x = e^(-x). For derivatives, d(e^x)/dx = e^x and d(e^(-x))/dx = -e^(-x). Thus d(sinh x)/dx = cosh x and d(cosh x)/dx = sinh x. Both signs are positive. Differentiate a sum term by term, keeping constant factors. At zero, use cosh 0 = 1 and sinh 0 = 0.
To invert sinh, write y = sinh t and put u = e^t > 0. Then 2y = u - 1/u, so u squared - 2yu - 1 = 0. Complete the square: (u-y)^2 = y^2+1. Thus u = y plus or minus sqrt(y squared + 1). Only the plus root is positive. Taking ln gives arsinh x = ln(x + sqrt(x squared + 1)). Every real x is allowed because sqrt(x squared + 1) > |x| makes the log input positive.
Cosh is even, so restrict its input t to t >= 0 to define its principal inverse. Solving with u = e^t gives u squared - 2xu + 1 = 0. Choose u = x + sqrt(x squared - 1), which is at least 1. Thus arcosh x = ln(x + sqrt(x squared - 1)), for x >= 1, with nonnegative outputs. Also arcosh(cosh t) = |t|. For c > 1, cosh t = c has exactly two solutions: arcosh c and -arcosh c.
Define tanh t = sinh t / cosh t = (e^(2t)-1)/(e^(2t)+1). It increases between -1 and 1 without reaching either value. Solving x = tanh t gives e^(2t) = (1+x)/(1-x). Therefore artanh x = (1/2) ln((1+x)/(1-x)), with real domain -1 < x < 1. At 1 the denominator is zero; at -1 the log input is zero.
For the derivative of arsinh, let t = arsinh x, so sinh t = x. Here t depends on x. The chain rule multiplies the derivative of sinh by t', giving cosh t times t' = 1. The identity gives cosh squared t = 1 + x squared. Since cosh t is positive, t' = 1/sqrt(1 + x squared). To evaluate this derivative, square the input, add 1, take the positive square root, then take its reciprocal.
Half difference and half sum of exponentials give sinh and cosh, whose squares always differ by 1.
This lesson and its assessments were written for this dataset. Consulted mathematical sources informed fact checks and exercise design; their passages and illustrations are not reproduced.
These readings, questions and flashcards were written for this dataset. Consulted source records informed mathematical checks and exercise design; no source passages or applet assets are reproduced.
More to read
Read the graphs from the definitions
For each real x, the numbers e^x and e^(-x) are positive and have product 1. Cosh is their average. Sinh is half their difference.
Changing x to -x swaps the two exponential values. Their sum stays the same, so cosh(-x) = cosh x. Their difference changes sign, so sinh(-x) = -sinh x. These equations describe even and odd symmetry.
At zero, both exponential values are 1. Cosh therefore passes through (0, 1), while sinh passes through (0, 0). To sketch sinh, draw an increasing curve that extends without bound in both output directions. To sketch cosh, draw matching left and right sides with their lowest point at (0, 1).
Why can cosh never fall below 1? Expand (e^(x/2) - e^(-x/2)) squared, which is nonnegative. It gives e^x + e^(-x) - 2 >= 0. Divide by 2 to obtain cosh x >= 1. Equality occurs only at x = 0.
A flexible chain with uniform mass per unit length, hanging at rest under only its own weight, is modelled by a catenary. Its equation can have the form y = a cosh((x-h)/a) + k, where a > 0. The scale and shifts depend on the chain and its supports.
Choose the input an inverse returns
To solve y = sinh t, replace e^t with u. Then u > 0 and 2y = u - 1/u. Multiplying by u gives u squared - 2yu - 1 = 0. The two quadratic roots are y plus or minus sqrt(y squared + 1). The square root exceeds |y|, so only the plus root is positive. Taking its logarithm gives t = ln(y + sqrt(y squared + 1)). This works for every real y.
Cosh needs an extra choice. Inputs t and -t give the same output. Its principal inverse selects t >= 0. With u = e^t, this means u >= 1. Solving y = cosh t gives u squared - 2yu + 1 = 0. For y >= 1, choose u = y + sqrt(y squared - 1). Thus arcosh y = ln(y + sqrt(y squared - 1)). Its outputs are nonnegative. In particular, arcosh(cosh t) returns |t|.
For tanh, write y = (e^(2t)-1)/(e^(2t)+1). Rearrangement gives e^(2t) = (1+y)/(1-y). Therefore artanh y = (1/2) ln((1+y)/(1-y)). Both parts of this ratio are positive when -1 < y < 1. The endpoints are excluded: y = 1 makes the denominator zero, and y = -1 gives a logarithm input of zero.
An inverse reverses a function on its chosen branch. It is not the reciprocal of the function.
What cancels in the hyperbolic identity
Write A = e^x and B = e^(-x). Then AB = 1. The numerator of cosh squared x is (A+B) squared, which expands to A squared + 2AB + B squared. The numerator of sinh squared x expands to A squared - 2AB + B squared. Both denominators are 4.
Subtract the second numerator from the first. The A squared terms cancel, as do the B squared terms. The cross terms do not cancel: 2AB - (-2AB) = 4AB. Dividing by 4 leaves AB = 1. This proves the identity for every real x.
The same definitions also give cosh x + sinh x = e^x and cosh x - sinh x = e^(-x). The vertical gap between the cosh and sinh graphs is therefore e^(-x). As x increases, that gap decreases toward zero.
The points (cosh t, sinh t) satisfy X squared - Y squared = 1. Their first coordinate is at least 1, so they trace the right branch of this hyperbola. They do not trace its left branch.
Checking several numerical inputs can help find an arithmetic slip. Such checks do not establish an identity for all real inputs; the algebra above does.
2 · Watch
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