Inverse Hyperbolic Cotangent Calculator

Inverse hyperbolic cotangent is the function that returns the number y whose hyperbolic cotangent equals x, for |x| > 1. It is written coth⁻¹(x), arcoth(x), arccoth(x) or acoth(x): coth(0.5493) = 2, so coth⁻¹(2) = 0.5493.

The hyperbolic cotangent is coth(y) = (eʸ + e⁻ʸ)/(eʸ − e⁻ʸ) = 1/tanh(y). Its values lie outside [−1, 1], and the inverse is half a natural log:

• Input: a number below −1 or above 1.
• Output: any real number except 0, with the same sign as x.
• Special values: coth⁻¹(2) = ½ ln 3 = 0.5493, coth⁻¹(5/3) = ln 2 = 0.6931, coth⁻¹(3) = ½ ln 2 = 0.3466.
• Shortcut: coth⁻¹(x) = tanh⁻¹(1/x).

Large inputs give small outputs: coth⁻¹(10) = 0.1003 and coth⁻¹(x) ≈ 1/x for large x. Inputs close to 1 give large outputs: coth⁻¹(1.01) = 2.6517.

The −1 in coth⁻¹(x) marks the inverse, not an exponent. 1/coth(x) is tanh(x).

To calculate inverse hyperbolic cotangent, divide x + 1 by x − 1, take the natural log, then halve it. Follow these 4 steps for x = 2:

1. Check that |x| > 1. Values from −1 to 1 have no real coth⁻¹.
2. Divide: (2 + 1)/(2 − 1) = 3.
3. Take the natural log: ln 3 = 1.098612.
4. Halve it: 1.098612/2 = 0.549306.

Check the answer with coth: coth(0.549306) = (3 + 1)/(3 − 1) = 2, since e^(2·0.549306) = 3.

A second route uses the reciprocal: coth⁻¹(2) = tanh⁻¹(1/2) = 0.549306. Exact cases: coth⁻¹(5/3) = ½ ln 4 = ln 2, coth⁻¹(5/4) = ½ ln 9 = ln 3 and coth⁻¹(13/12) = ½ ln 25 = ln 5.

Type a value with |x| > 1 and the calculator returns coth⁻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 (5/3), roots (√2) and expressions.
2. Choose the rounding, from whole numbers to 15 decimal places.
3. Read the result, the exact log form, the 6 hyperbolic functions of the result and eʸ.
4. Switch the diagram between the unit hyperbola, the curve coth(y) = x and the graph of coth⁻¹.
5. Open the steps to see the log formula worked out for your value.

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

Rapidity mode is the relativity use of arcoth. Enter a speed in km/s; the calculator forms c/v and returns the rapidity φ = arcoth(c/v) and the Lorentz factor γ = cosh φ. Half the speed of light gives φ = arcoth(2) = 0.5493 and γ = 1.1547.

The inverse hyperbolic cotangent formula is coth⁻¹(x) = ½ ln((x + 1)/(x − 1)) for |x| > 1.

Where the formula comes from: set x = coth y and let u = e²ʸ. Then x = (u + 1)/(u − 1), so u = (x + 1)/(x − 1). Taking the log gives 2y = ln((x + 1)/(x − 1)).

Related forms:

• Reciprocal link: coth⁻¹(x) = tanh⁻¹(1/x).
• Identity: coth(coth⁻¹ x) = x for |x| > 1, and coth⁻¹(coth y) = y for y ≠ 0.
• Odd function: coth⁻¹(−x) = −coth⁻¹(x).
• Split log: coth⁻¹(x) = ½[ln(x + 1) − ln(x − 1)] for x > 1.
• Convention note: some texts extend coth⁻¹ to |x| < 1 with complex values; this calculator returns real results only.

The domain of inverse hyperbolic cotangent is x < −1 or x > 1, and the range is all real numbers except 0.

coth y = 1/tanh y, and |tanh y| < 1, so |coth y| > 1 for every y ≠ 0. That leaves a gap from −1 to 1 where coth⁻¹ has no real value.

• Domain: (−∞, −1) ∪ (1, ∞)
• Range: (−∞, 0) ∪ (0, ∞)
• Vertical asymptotes: x = −1 and x = 1
• Horizontal asymptote: y = 0, as x goes to ±∞
• Monotonic: decreasing on each of the 2 pieces
• Symmetry: odd, coth⁻¹(−x) = −coth⁻¹(x)

The graph of y = coth⁻¹(x) has 2 separate branches, one for x > 1 and one for x < −1, each falling toward the x-axis.

• Right branch: starts near +∞ at x = 1, passes (2, 0.5493) and (3, 0.3466), and flattens toward 0.
• Left branch: the right branch rotated 180° about the origin, passing (−2, −0.5493).
• Vertical asymptotes at x = ±1; horizontal asymptote y = 0.
• No graph between −1 and 1.
• Mirror image of y = coth x across the line y = x.

For large |x|, the curve follows y ≈ 1/x: coth⁻¹(10) = 0.1003, close to 0.1.

The derivative of coth⁻¹(x) is 1/(1 − x²) for |x| > 1, which is always negative there. The formula matches the derivative of tanh⁻¹, but the 2 live on different domains.

Proof: let y = coth⁻¹ x, so x = coth y. Differentiate: 1 = −csch² y · dy/dx = (1 − coth² y)·dy/dx = (1 − x²)·dy/dx.

• Example: at x = 2, the slope is 1/(1 − 4) = −1/3.
• Integral it solves: ∫ dx/(1 − x²) = coth⁻¹ x + C for |x| > 1, the partner of tanh⁻¹ x for |x| < 1. Both equal ½ ln|(1 + x)/(1 − x)| + C.
• Integral of coth⁻¹: ∫ coth⁻¹ x dx = x·coth⁻¹ x + ½ ln(x² − 1) + C.

Most scientific calculators have no coth⁻¹ key, so use coth⁻¹(x) = tanh⁻¹(1/x): press hyp, SHIFT, tan, type 1 ÷ x and press =. On a Casio fx-82 or fx-991MS, coth⁻¹(2) is hyp, SHIFT, tan, 1, ÷, 2, ), = and the screen shows 0.5493061443.

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

A TI-84 works the same way with atanh(1/x) from the CATALOG. Math ERROR means |x| ≤ 1.

The Android (Google) Calculator has no coth⁻¹ key, so type ½ ln((x + 1)/(x − 1)). For x = 2, (2 + 1)/(2 − 1) = 3, so type ln(3) ÷ 2 and tap =: the result is 0.5493061443.

1. Tap the expand arrow, or turn the phone to landscape, to show the scientific keys.
2. Tap ln and type 3, the value of (x + 1)/(x − 1).
3. Close the bracket, tap ÷ and type 2.
4. Tap =.

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

On the iPhone Calculator, type the value, tap 1/x, tap 2nd, then tap tanh⁻¹. coth⁻¹(x) = tanh⁻¹(1/x), so for 2 the display shows 0.5493061443.

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

The iPhone has no coth key, so the 1/x step does the work.

In Windows Calculator, choose Scientific, type the value, open Trigonometry, select 2nd and hyp, then select coth⁻¹. For 2, the result is 0.5493061443.

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

Windows Calculator is one of the few calculators with a direct coth⁻¹ key.

Excel and MATLAB have a direct ACOTH or acoth; Python, JavaScript and C use atanh(1/x).

• Excel and Google Sheets: =ACOTH(2) returns 0.549306144 (Excel 2013 and later).
• Python: math.atanh(1/2) returns 0.5493061443340548. NumPy: np.arctanh(1/x).
• JavaScript: Math.atanh(1/2).
• MATLAB and Octave: acoth(2).
• C and C++: atanh(1.0/x).
• Wolfram Language: ArcCoth[2].

For |x| ≤ 1, Excel's ACOTH returns #NUM! and Python's math.atanh(1/x) raises ValueError. At x = 0, 1/x itself fails first.

Without a calculator, use the series coth⁻¹ x = 1/x + 1/(3x³) + 1/(5x⁵) + … for |x| > 1. It is the tanh⁻¹ series with 1/x in place of x.

1. Series: coth⁻¹ x = Σ 1/((2n + 1)·x^(2n+1)), summed from n = 0.
2. Large x: 1/x alone is a good estimate: coth⁻¹(10) ≈ 0.1, true value 0.100335.
3. Exact cases: when (x + 1)/(x − 1) is a perfect square, coth⁻¹ x is a plain log, such as coth⁻¹(5/3) = ln 2.

Example: coth⁻¹(2) ≈ 0.5 + 0.041667 + 0.00625 + 0.001116 = 0.549033. The true value is 0.549306.

Machin-style formulas for logarithms use the same series: ln 2 = 2·coth⁻¹(3) = 2(1/3 + 1/81 + 1/1215 + …).

The inverse hyperbolic cotangent table lists coth⁻¹(x) for common values of x with |x| > 1, with exact log forms. Click a row to load it into the calculator.

Values to know: coth⁻¹(2) = ½ ln 3 = 0.5493, coth⁻¹(3) = ½ ln 2 = 0.3466, coth⁻¹(5/3) = ln 2 = 0.6931 and coth⁻¹(5/4) = ln 3 = 1.0986.

The inverse hyperbolic cotangent of a negative number is negative: coth⁻¹(−x) = −coth⁻¹(x). coth⁻¹(−2) = −0.5493 and coth⁻¹(−5/3) = −ln 2.

coth⁻¹ is an odd function, so the left branch of the graph is the right branch rotated 180° about the origin. Negative inputs from −1 to 0, such as −0.5, are outside the domain.

What is coth⁻¹(−2)?

coth⁻¹(−2) = ½ ln(1/3) = −0.549306. (−2 + 1)/(−2 − 1) = 1/3, and ½ ln(1/3) = −½ ln 3.

arcoth accepts |x| > 1 and returns any nonzero real number; arccot accepts every real number and returns an angle between 0° and 180°.

For large x, both functions shrink like 1/x: arcoth(10) = 0.1003 and arccot(10) = 0.0997 rad.

Inverse hyperbolic cotangent turns speeds into rapidities in relativity, sizes transmission-line lengths and computes logarithms quickly.

• Relativity: the rapidity of speed v is φ = coth⁻¹(c/v). Rapidities add. A ship at 0.6c launching a probe at 0.7c gives φ = coth⁻¹(1/0.6) + coth⁻¹(1/0.7) = 0.6931 + 0.8673, so the probe moves at tanh(1.5604)c = 0.9155c, not 1.3c.
• Transmission lines: an open-ended line has input impedance Z = Z₀·coth(γℓ), so its electrical length is γℓ = coth⁻¹(Z/Z₀).
• Logarithms: ln 2 = 2·coth⁻¹(3) and ln 3 = 2·coth⁻¹(2). Fast log routines in computer algebra use these series.
• Integrals: ∫ dx/(x² − a²) = −(1/a)·coth⁻¹(x/a) + C for |x| > a.