TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Logarithm formulas for inverse sinh, inverse cosh, and inverse tanh on their natural domains

Statement

The strictly increasing bijections $\sinh : R \to R$ , $\cosh : [0, \infty) \to [1, \infty)$ , and $\tanh : R \to (- 1, 1)$ have inverse functions satisfying $arsinh u = \log (u + \sqrt{u^{2} + 1}),$ $arcosh u = \log (u + \sqrt{u^{2} - 1}) (u \geq 1),$ $artanh u = \frac{1}{2} \log \frac{1 + u}{1 - u} (∣ u ∣ < 1) .$

Facts & Assumptions

Given: A real $u$ in the stated domain.

[L1]

The hyperbolic identities hold, and $\sinh : R \to R$ , $\cosh : [0, \infty) \to [1, \infty)$ , and $\tanh : R \to (- 1, 1)$ are strictly increasing bijections (Addition formulas, identities, parity, and derivatives of the hyperbolic functions, The six hyperbolic functions and their natural domains).

[L2]

$\log$ is the inverse of $\exp$ and satisfies its product and reciprocal laws (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[L3]

Positive-base real powers are continuous, and every nonnegative real has its nonnegative square root (Continuity and derivatives of positive-base real powers, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Proof

technique · direct

1.1

Solving $\sinh z = u$ after putting $v = \exp z > 0$ gives $v^{2} - 2 u v - 1 = 0$ , hence $v = u + \sqrt{u^{2} + 1} > 0$ and $z = \log (u + \sqrt{u^{2} + 1})$ .

L1L2L3

1.2

Solving $\cosh z = u$ with $z \geq 0$ gives $v^{2} - 2 u v + 1 = 0$ and the allowed root $v = u + \sqrt{u^{2} - 1}$ , hence the displayed arcosh formula.

L1L2L3

1.3

Solving $\tanh z = u$ gives $v^{2} = (1 + u) / (1 - u) > 0$ , so $z = \frac{1}{2} \log ((1 + u) / (1 - u))$ .

L1L2

2.1

The strict monotonicity and stated ranges in [L1] make each algebraic solution the unique inverse value on its declared domain.

step 1.1step 1.2step 1.3L1∎

Depends on

Used by

The surface generated by rotating y=sin x on [0,π] has area 2π(√2+arsinh1) Example

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis, Logarithm and Exponential (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)