Inverse Hyperbolic Cosine Calculator

Inverse hyperbolic cosine is the function that returns the non-negative number y whose hyperbolic cosine equals x. It is written cosh⁻¹(x), arcosh(x), arccosh(x) or acosh(x), and it reverses cosh for y ≥ 0: cosh(1.317) = 2, so cosh⁻¹(2) = 1.317.

The hyperbolic cosine is cosh(y) = (eʸ + e⁻ʸ)/2. Its smallest value is cosh(0) = 1, so cosh⁻¹ accepts only x ≥ 1:

• Input: a number of 1 or more.
• Output: a number of 0 or more.
• Special values: cosh⁻¹(1) = 0, cosh⁻¹(5/4) = ln 2 = 0.6931, cosh⁻¹(2) = ln(2 + √3) = 1.317.

cosh is an even function: cosh(1.317) = cosh(−1.317) = 2. cosh⁻¹ returns the non-negative solution, called the principal value.

On the unit hyperbola x² − y² = 1, the point (cosh t, sinh t) has horizontal coordinate cosh t. cosh⁻¹ reads that coordinate and returns t, twice the shaded area. The −1 in cosh⁻¹(x) marks the inverse, not an exponent; 1/cosh(x) is sech(x).

To calculate inverse hyperbolic cosine, square x, subtract 1, take the square root, add x, then take the natural log. Follow these 5 steps for x = 3:

1. Check that x ≥ 1. A value below 1 has no real cosh⁻¹.
2. Square x and subtract 1: 3² − 1 = 8.
3. Take the square root: √8 = 2.828427.
4. Add x: 3 + 2.828427 = 5.828427.
5. Take the natural log: ln(5.828427) = 1.762747.

Check the answer with cosh: cosh(1.762747) = (5.828427 + 0.171573)/2 = 3, which matches x.

Fractions p/q with p² − q² a perfect square give exact logs. For x = 5/4, √(25/16 − 1) = 3/4, so cosh⁻¹(5/4) = ln(5/4 + 3/4) = ln 2. For x = 5/3, the root is 4/3 and cosh⁻¹(5/3) = ln 3 = 1.098612.

Type a value of 1 or more and the calculator returns cosh⁻¹(x) as you type, with the log formula, the exact value and the steps. The steps:

1. Type x in the field. It accepts decimals (1.5), fractions (5/4), roots (√2), π, e and expressions.
2. Choose the rounding, from whole numbers to 15 decimal places.
3. Read the result, the exact log form, the second solution −y, the 6 hyperbolic functions of the result and eʸ.
4. Switch the diagram between the unit hyperbola, the curve cosh(y) = x with both solutions, and the graph of cosh⁻¹.
5. Open the steps to see the log formula worked out for your value.

The calculator rejects values below 1 and states the reason. Results can be copied, shared as a link, printed or kept in the history list.

Cable span mode uses the catenary. Enter the catenary parameter a and the sag h below 2 supports at the same height. The span is 2a·arcosh(1 + h/a): a = 50 m and a sag of 8 m give 2 × 50 × arcosh(1.16) = 55.84 m.

The inverse hyperbolic cosine formula is cosh⁻¹(x) = ln(x + √(x² − 1)) for x ≥ 1.

Where the formula comes from: set x = cosh y = (eʸ + e⁻ʸ)/2 and let u = eʸ. Then u² − 2xu + 1 = 0, so u = x ± √(x² − 1). The 2 roots multiply to 1, so they give ±y. The + sign gives y ≥ 0, the principal value.

Related forms:

• Identity: cosh(cosh⁻¹ x) = x for x ≥ 1, and cosh⁻¹(cosh y) = |y| for every y.
• Reciprocal link: cosh⁻¹(x) = sech⁻¹(1/x).
• Through sinh⁻¹: cosh⁻¹(x) = sinh⁻¹(√(x² − 1)).
• Half-argument form: cosh⁻¹(x) = 2 ln(√((x + 1)/2) + √((x − 1)/2)).
• Both solutions: cosh y = x gives y = ±ln(x + √(x² − 1)).

The domain of inverse hyperbolic cosine is x ≥ 1, and the range is y ≥ 0.

cosh y never drops below 1, so cosh⁻¹(0.5) and cosh⁻¹(−2) have no real value. cosh takes each value above 1 twice, at y and −y; restricting cosh to y ≥ 0 makes it one-to-one and gives the principal value.

• Domain: [1, ∞)
• Range: [0, ∞)
• Monotonic: increasing over the whole domain
• Start point: cosh⁻¹(1) = 0, where the graph has a vertical tangent
• Growth: cosh⁻¹(x) ≈ ln(2x) for large x
• Symmetry: none; the function is neither odd nor even

With complex numbers, cosh⁻¹ extends below 1: cosh⁻¹(0.5) = iπ/3, since cosh(iθ) = cos θ.

The graph of y = cosh⁻¹(x) starts at (1, 0) with a vertical tangent and rises like ln(2x) to the right.

• Passes through (1, 0), (2, 1.317) and (3, 1.7627).
• Exists only for x ≥ 1; the dashed line marks the domain edge x = 1.
• Concave down over the whole domain: the slope 1/√(x² − 1) falls from ∞ at x = 1 toward 0.
• Mirror image of the right half of y = cosh x across the line y = x.
• No horizontal asymptote: cosh⁻¹(100) = 5.3 and cosh⁻¹(10⁶) = 14.51.

Near x = 1, cosh⁻¹(x) ≈ √(2(x − 1)): the curve leaves (1, 0) like a sideways parabola.

The derivative of cosh⁻¹(x) is 1/√(x² − 1) for x > 1. The slope is infinite at x = 1 and falls toward 0 as x grows.

Proof: let y = cosh⁻¹ x, so x = cosh y with y ≥ 0. Differentiate: 1 = sinh y · dy/dx. For y ≥ 0, sinh y = √(cosh² y − 1) = √(x² − 1), so dy/dx = 1/√(x² − 1).

• Chain rule: d/dx cosh⁻¹(u) = u′/√(u² − 1). For cosh⁻¹(2x): 2/√(4x² − 1).
• Integral it solves: ∫ dx/√(x² − a²) = cosh⁻¹(x/a) + C for x > a > 0.
• Integral of cosh⁻¹: ∫ cosh⁻¹ x dx = x·cosh⁻¹ x − √(x² − 1) + C.

On a scientific calculator, press hyp, then SHIFT (or 2nd), then cos, type the value and press =. On a Casio fx-82 or fx-991MS, the keys for cosh⁻¹(2) are hyp, SHIFT, cos, 2, ), = and the screen shows 1.316957897.

1. Press hyp. The cos key now means cosh.
2. Press SHIFT (Casio, Sharp) or 2nd (Texas Instruments) to reach cosh⁻¹.
3. Type the value, close the bracket and press =.

Math ERROR (Casio) or ERR:DOMAIN (TI) means the value is below 1. On Casio fx-991EX models, hyperbolic functions sit in the OPTN menu; on a TI-84, choose acosh( from the CATALOG.

The Android (Google) Calculator has no cosh⁻¹ key, so type the formula ln(x + √(x² − 1)). For x = 2, x² − 1 = 3, so type ln(2 + √3) and tap =: the result is 1.316957897.

1. Tap the expand arrow, or turn the phone to landscape, to show the scientific keys.
2. Tap ln. The display shows ln(.
3. Type 2, tap +, tap √ and type 3.
4. Close the bracket and tap =.

For another x, type the full form, such as ln(3 + √(3^2 − 1)) for x = 3.

On the iPhone Calculator, switch to Scientific, type the value, tap 2nd, then tap cosh⁻¹. Type 2, tap 2nd and tap cosh⁻¹: the display shows 1.316957897.

1. Turn the iPhone sideways, or tap the calculator icon and choose Scientific (iOS 18 and later).
2. Type 2.
3. Tap 2nd. The cosh key changes to cosh⁻¹.
4. Tap cosh⁻¹. The result appears at once.

A value below 1 shows Error. The Rad label has no effect on hyperbolic functions.

In Windows Calculator, choose Scientific, type the value, open Trigonometry, select 2nd and hyp, then select cosh⁻¹. For 2, the result is 1.316957897.

1. Open the menu (☰) and choose Scientific.
2. Type 2.
3. Open the Trigonometry list and select 2nd, then hyp.
4. Select cosh⁻¹. The result is 1.316957897.

A value below 1 shows “Invalid input”.

Software calls inverse hyperbolic cosine ACOSH or acosh, and the result is a plain number.

• Excel and Google Sheets: =ACOSH(2) returns 1.316957897.
• Python: math.acosh(2) returns 1.3169578969248166. NumPy uses np.arccosh.
• JavaScript: Math.acosh(2) returns 1.3169578969248166.
• MATLAB and Octave: acosh(2).
• C and C++: acosh(2) in C99 and C++11. Java has no Math.acosh, so use Math.log(x + Math.sqrt(x*x − 1)).
• Wolfram Language: ArcCosh[2].

Values below 1 produce errors: Excel returns #NUM!, Python raises ValueError: math domain error, and JavaScript returns NaN. MATLAB and NumPy with complex input return a complex result, such as acosh(0.5) = 1.0472i.

Without a calculator, use cosh⁻¹ x ≈ ln(2x) − 1/(4x²) − 3/(32x⁴) for x well above 1, or √(2(x − 1)) for x just above 1.

1. Large x: cosh⁻¹ x = ln(2x) − Σ (2n)! / (4ⁿ (n!)² · 2n) · x^(−2n), summed from n = 1.
2. x near 1: cosh⁻¹(1 + h) ≈ √(2h) · (1 − h/12) for small h.
3. Exact cases: fractions p/q with p² − q² a perfect square, such as 5/4, 5/3 and 13/12, give an exact log.

Example: cosh⁻¹(3) ≈ ln 6 − 1/36 − 3/2592 = 1.791759 − 0.027778 − 0.001157 = 1.762824. The true value is 1.762747.

The large-x series converges slowly near x = 1, where every term counts. The widget shows this: at x = 1 it needs many terms to approach 0.

The inverse hyperbolic cosine table lists cosh⁻¹(x) for common values of x from 1 upward, with exact log forms where they exist. Click a row to load it into the calculator.

Values to know: cosh⁻¹(1) = 0, cosh⁻¹(5/4) = ln 2 = 0.6931, cosh⁻¹(5/3) = ln 3 = 1.0986 and cosh⁻¹(2) = ln(2 + √3) = 1.317.

cosh⁻¹(x) is never negative: cosh is an even function, cosh(−y) = cosh(y), so the equation cosh y = x has 2 solutions, ±y, and cosh⁻¹ returns the one with y ≥ 0.

Example: cosh(1.317) = cosh(−1.317) = 2. The calculator returns cosh⁻¹(2) = 1.317, and the full solution of cosh y = 2 is y = ±1.317.

• A negative input has no real cosh⁻¹: cosh⁻¹(−2) does not exist, since cosh y ≥ 1 for every real y.
• cosh⁻¹(cosh y) = |y|: cosh⁻¹(cosh(−3)) = 3, not −3.
• To solve cosh y = x in an equation, write y = ±cosh⁻¹(x).

arcosh inverts the hyperbolic cosine and accepts x ≥ 1; arccos inverts the circular cosine and accepts −1 ≤ x ≤ 1. The 2 domains touch only at x = 1, where both functions return 0.

The link comes from cosh(iθ) = cos θ: for −1 ≤ x ≤ 1, arcosh(x) = i·arccos(x). So arcosh(0.5) = iπ/3 and arccos(0.5) = π/3.

Inverse hyperbolic cosine finds the span of hanging cables and arches, distances in hyperbolic geometry and speeds in relativity.

• Hanging cables: a chain or cable hangs in the catenary y = a·cosh(x/a). A point h above the lowest point lies at x = a·cosh⁻¹((h + a)/a). With a = 8 m and h = 4 m, x = 8·cosh⁻¹(1.5) = 7.7 m, so the poles stand 15.4 m apart.
• Arches: the Gateway Arch in St. Louis, 192 m tall and 192 m wide, follows a flattened catenary, and its design equation uses cosh.
• Hyperbolic geometry: in the Poincaré half-plane, the distance between 2 points is cosh⁻¹(1 + ((x₂ − x₁)² + (y₂ − y₁)²)/(2y₁y₂)).
• Relativity: an object with Lorentz factor γ has rapidity φ = cosh⁻¹(γ). At γ = 2 (v = 0.866c), φ = 1.317.
• Integrals: ∫ dx/√(x² − a²) = cosh⁻¹(x/a) + C appears in orbit and cable problems.