Inverse Hyperbolic Secant Calculator

Inverse hyperbolic secant is the function that returns the non-negative number y whose hyperbolic secant equals x, for 0 < x ≤ 1. It is written sech⁻¹(x), arsech(x), arcsech(x) or asech(x): sech(1.317) = 0.5, so sech⁻¹(0.5) = 1.317.

The hyperbolic secant is sech(y) = 1/cosh(y) = 2/(eʸ + e⁻ʸ). It peaks at sech(0) = 1 and falls toward 0, so sech⁻¹ accepts only 0 < x ≤ 1:

• Input: a number greater than 0 and at most 1.
• Output: a number of 0 or more.
• Special values: sech⁻¹(1) = 0, sech⁻¹(4/5) = ln 2 = 0.6931, sech⁻¹(0.5) = ln(2 + √3) = 1.317.
• Shortcut: sech⁻¹(x) = cosh⁻¹(1/x).

sech is even, so sech(y) = x has 2 solutions, ±y; sech⁻¹ returns the non-negative one. Small inputs give large outputs: sech⁻¹(0.01) = 5.2983.

On the unit hyperbola, the tangent line at (cosh t, sinh t) crosses the x-axis at sech t. sech⁻¹ turns that crossing point into t, twice the shaded area.

To calculate inverse hyperbolic secant, square x, subtract it from 1, take the square root, add 1, divide by x, then take the natural log. Follow these 5 steps for x = 0.5:

1. Check that 0 < x ≤ 1.
2. Compute 1 − x²: 1 − 0.25 = 0.75.
3. Take the square root and add 1: 1 + √0.75 = 1.866025.
4. Divide by x: 1.866025/0.5 = 3.732051.
5. Take the natural log: ln(3.732051) = 1.316958.

Shortcut: sech⁻¹(0.5) = cosh⁻¹(1/0.5) = cosh⁻¹(2) = ln(2 + √3), the same 1.316958.

Fractions p/q with q² − p² a perfect square give exact logs: sech⁻¹(4/5) = ln((5 + 3)/4) = ln 2, sech⁻¹(3/5) = ln((5 + 4)/3) = ln 3 and sech⁻¹(12/13) = ln((13 + 5)/12) = ln(3/2).

Type a value above 0 and up to 1, and the calculator returns sech⁻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 (0.5), fractions (4/5), roots (1/√2) 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 sech(y) = x with both solutions, and the graph of sech⁻¹.
5. Open the steps to see the log formula worked out for your value.

The calculator rejects 0, negative values and values above 1, and states the reason.

Tractrix mode uses the curve traced by an object dragged on a leash. Enter the leash length a and the distance y of the object from your path; the calculator returns how far you have walked, x = a·arsech(y/a) − √(a² − y²). A 10 m leash with the object 4 m away gives 6.50 m.

The inverse hyperbolic secant formula is sech⁻¹(x) = ln((1 + √(1 − x²))/x) for 0 < x ≤ 1.

Where the formula comes from: sech y = x means cosh y = 1/x. Apply the cosh⁻¹ formula to 1/x: ln(1/x + √(1/x² − 1)) = ln((1 + √(1 − x²))/x).

Related forms:

• Reciprocal link: sech⁻¹(x) = cosh⁻¹(1/x).
• Identity: sech(sech⁻¹ x) = x for 0 < x ≤ 1, and sech⁻¹(sech y) = |y|.
• Both solutions: sech y = x gives y = ±sech⁻¹(x).
• Through tanh⁻¹: sech⁻¹(x) = tanh⁻¹(√(1 − x²)) for 0 < x ≤ 1.
• Small x: sech⁻¹(x) ≈ ln(2/x) as x approaches 0.

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

sech y = 1/cosh y, and cosh y ≥ 1, so sech y lies in (0, 1]. sech⁻¹(0), sech⁻¹(−0.5) and sech⁻¹(2) have no real value.

• Domain: (0, 1]
• Range: [0, ∞)
• Monotonic: decreasing over the whole domain
• Vertical asymptote: x = 0, where sech⁻¹(x) goes to +∞
• End point: sech⁻¹(1) = 0
• Symmetry: none; the function is neither odd nor even

The domain of sech⁻¹ is the reciprocal of the domain of cosh⁻¹: x ≥ 1 there becomes 0 < x ≤ 1 here.

The graph of y = sech⁻¹(x) falls from +∞ near x = 0 to 0 at x = 1, where it ends with a vertical tangent.

• Passes through (0.25, 2.0634), (0.5, 1.317), (0.8, 0.6931) and (1, 0).
• Vertical asymptote at x = 0; the dashed line x = 1 marks the end of the domain.
• The slope −1/(x√(1 − x²)) is steep at both ends and flattest at x = 1/√2, where it equals −2.
• Mirror image of the right half of y = sech x across the line y = x.

Near x = 0 the curve follows ln(2/x): sech⁻¹(0.01) = 5.2983 and ln 200 = 5.2983.

The derivative of sech⁻¹(x) is −1/(x√(1 − x²)) for 0 < x < 1. The slope is negative everywhere, so sech⁻¹ falls over its whole domain.

Proof: let y = sech⁻¹ x, so x = sech y with y ≥ 0. Differentiate: 1 = −sech y · tanh y · dy/dx. For y ≥ 0, tanh y = √(1 − sech² y) = √(1 − x²), so dy/dx = −1/(x√(1 − x²)).

• Example: at x = 0.5, the slope is −1/(0.5 · 0.866) = −2.309.
• Integral it solves: ∫ dx/(x√(1 − x²)) = −sech⁻¹ x + C.
• Integral of sech⁻¹: ∫ sech⁻¹ x dx = x·sech⁻¹ x + sin⁻¹ x + C.

Most scientific calculators have no sech⁻¹ key, so use sech⁻¹(x) = cosh⁻¹(1/x): press hyp, SHIFT, cos, type 1 ÷ x and press =. On a Casio fx-82 or fx-991MS, sech⁻¹(0.5) is hyp, SHIFT, cos, 1, ÷, 0.5, ), = and the screen shows 1.316957897.

1. Press hyp, then SHIFT (or 2nd), then cos for cosh⁻¹.
2. Type 1 ÷ x, the reciprocal of the value.
3. Close the bracket and press =.

On a TI-84, use acosh(1/x) from the CATALOG. Math ERROR means x is 0, negative or above 1.

The Android (Google) Calculator has no sech⁻¹ key, so type the log formula. For x = 0.5, (1 + √(1 − 0.25))/0.5 = 2 + √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 ln((1 + √(1 − x^2)) ÷ x) with your value, such as ln((1 + √(1 − 0.6^2)) ÷ 0.6) = ln 3 for x = 0.6.

On the iPhone Calculator, type the value, tap 1/x, tap 2nd, then tap cosh⁻¹. sech⁻¹(x) = cosh⁻¹(1/x), so for 0.5 the display shows 1.316957897.

1. Turn the iPhone sideways, or tap the calculator icon and choose Scientific (iOS 18 and later).
2. Type 0.5.
3. Tap 1/x. The display shows 2.
4. Tap 2nd, then tap cosh⁻¹.

In Windows Calculator, choose Scientific, type the value, open Trigonometry, select 2nd and hyp, then select sech⁻¹. For 0.5, the result is 1.316957897.

1. Open the menu (☰) and choose Scientific.
2. Type 0.5.
3. Open the Trigonometry list and select 2nd, then hyp.
4. Select sech⁻¹.

Windows Calculator has a direct sech⁻¹ key, so no reciprocal step is needed.

Excel, Python and JavaScript have no asech, so use acosh(1/x); MATLAB and Wolfram have it built in.

• Excel and Google Sheets: =ACOSH(1/0.5) returns 1.316957897.
• Python: math.acosh(1/0.5) returns 1.3169578969248166. NumPy: np.arccosh(1/x).
• JavaScript: Math.acosh(1/0.5).
• MATLAB and Octave: asech(0.5).
• C and C++: acosh(1.0/x).
• Wolfram Language: ArcSech[1/2].

For x above 1, 1/x falls below 1 and acosh fails: Excel returns #NUM!, Python raises ValueError and JavaScript returns NaN.

Without a calculator, use sech⁻¹ x ≈ ln(2/x) − x²/4 − 3x⁴/32 for small x.

1. Series: sech⁻¹ x = ln(2/x) − Σ (2n)! / (4ⁿ (n!)² · 2n) · x^(2n), summed from n = 1, for 0 < x ≤ 1.
2. Near x = 1: sech⁻¹(1 − h) ≈ √(2h) for small h.
3. Exact cases: fractions such as 4/5, 3/5, 12/13 and 8/17 give plain logs.

Example: sech⁻¹(0.5) ≈ ln 4 − 0.0625 − 0.005859 = 1.317935. The true value is 1.316958.

The series is the cosh⁻¹ large-x series with 1/x in place of x, so it converges fast for small x and slowly near x = 1.

The inverse hyperbolic secant table lists sech⁻¹(x) for common values of x from 0 to 1, with exact log forms. Click a row to load it into the calculator.

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

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

Example: sech(1.317) = sech(−1.317) = 0.5. The calculator returns sech⁻¹(0.5) = 1.317, and the full solution of sech y = 0.5 is y = ±1.317.

• Negative inputs have no real sech⁻¹, since sech y > 0 for every real y.
• sech⁻¹(sech y) = |y|: sech⁻¹(sech(−2)) = 2.
• To solve sech y = x in an equation, write y = ±sech⁻¹(x).

arsech accepts 0 < x ≤ 1 and returns y ≥ 0; arcsec accepts |x| ≥ 1 and returns an angle from 0° to 180°. The 2 domains touch only at x = 1, where both return 0.

Both functions come from the reciprocal of a cosine: arsech undoes 1/cosh and arcsec undoes 1/cos.

Inverse hyperbolic secant traces the tractrix, the path of an object dragged on a leash, and measures the width of sech-shaped laser pulses and solitons.

• Tractrix: a toy pulled by a 5 m string whose holder walks along a straight path follows x = a·sech⁻¹(y/a) − √(a² − y²). When the toy is 2 m from the path, it has moved x = 3.251 m forward.
• Pseudosphere: spinning the tractrix about its path gives Beltrami's pseudosphere, a surface of constant negative curvature used as a model of hyperbolic geometry.
• Laser pulses: mode-locked lasers produce pulses with intensity sech²(t/T). The half-maximum points sit at t = T·sech⁻¹(1/√2), so the pulse width is 2 × 0.8814 × T = 1.7627T.
• Solitons: water and optical solitons have sech or sech² profiles, so their widths come from sech⁻¹ in the same way.