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If you flip a hanging telephone cable upside down, you get the graph of cosh⁡(x)\cosh(x). But what happens if you want to work backwards — finding the exact xx-coordinate that produced a specific height?
Just like standard trigonometry has inverse functions, hyperbolic trigonometry has inverse hyperbolic functions.
We write them as arsinh⁡(x)\operatorname{arsinh}(x), arcosh⁡(x)\operatorname{arcosh}(x), and artanh⁡(x)\operatorname{artanh}(x), where the prefix "ar-" stands for area because these functions compute the area of hyperbolic sectors.
Can we simply flip the axes for any hyperbolic function, or do some require domain restrictions?

The Inverse of sinh⁡(x)\sinh(x)

The hyperbolic sine function, sinh⁡(x)=ex−e−x2\sinh(x) = \frac{e^x - e^{-x}}{2}, is strictly increasing across all real numbers. Because every output corresponds to exactly one input, it passes the horizontal line test across its entire domain.
We define arsinh⁡(x)\operatorname{arsinh}(x) as the inverse of sinh⁡(x)\sinh(x): if y=arsinh⁡(x)y = \operatorname{arsinh}(x), then x=sinh⁡(y)x = \sinh(y).
Both the domain and the range of arsinh⁡(x)\operatorname{arsinh}(x) are all real numbers, R\mathbb{R}.
What happens when we try to invert cosh⁡(x)\cosh(x), which is U-shaped and fails the horizontal line test?

The Inverse of cosh⁡(x)\cosh(x)

The function cosh⁡(x)=ex+e−x2\cosh(x) = \frac{e^x + e^{-x}}{2} is an even function, meaning cosh⁡(−x)=cosh⁡(x)\cosh(-x) = \cosh(x). For example, cosh⁡(1)=cosh⁡(−1)≈1.543\cosh(1) = \cosh(-1) \approx 1.543, so a direct inverse would produce two outputs for a single input.
To ensure the inverse is a proper function, we restrict the domain of cosh⁡(x)\cosh(x) to the non-negative half, x≥0x \ge 0.