--- title: "Hyperbolic Functions and Their Inverses" description: "Build sinh and cosh from the exponential function, prove the identity that replaces the Pythagorean one, and write the inverses as logarithms." canonical: https://lightmysky.com/learn/mathematics/hyperbolic-functions-and-their-inverses-mt_gCUl1sWi9M source: https://lightmysky.com/learn/mathematics/hyperbolic-functions-and-their-inverses-mt_gCUl1sWi9M.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # 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. Subject: Mathematics · Area: Algebra · Ages 17 to 18 Page: https://lightmysky.com/learn/mathematics/hyperbolic-functions-and-their-inverses-mt_gCUl1sWi9M ## Ready when they can - 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 ## Lesson: Exponentials wearing trig costumes 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. *(drawing: Use the same two exponential values in both definitions.)* 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. *(drawing: The cross terms remain after the pure squares cancel.)* 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. *(drawing: The other quadratic root is negative, so it cannot equal e^t.)* 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. *(drawing: A real logarithm needs a positive input.)* 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. **Recap.** 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.* ## Practice 18 questions on this page, each with its working shown. ## Needs first - [The exponential function and the number e](https://lightmysky.com/learn/mathematics/the-exponential-function-and-the-number-e-mt_2hyjN4JfJJ) - [Logarithms as the Inverse of Exponentials](https://lightmysky.com/learn/mathematics/logarithms-as-the-inverse-of-exponentials-mt_skLQLUan0T) ## Optional connections - [The Pythagorean and Quotient Identities](https://lightmysky.com/learn/mathematics/the-pythagorean-and-quotient-identities-mt_NXBq37vxQh)