Inverse hyperbolic functions

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Hyperbolic sine and cosine already give a real height for every real input. For a chosen height $y$ , does $y = sinh x$ name one real $x$ ? Unlike sine, sinh on $R$ needs no freeze: every height already has exactly one preimage.

Reflect sinh across the diagonal

Drag the real height $y$ . The green point sits on $y = sinh x$ . Reflect it across the diagonal $y = x$ : the blue point's height is that unique real solution.

Drag the real height y. The green point sits on y = sinh x. The blue point is its reflection across y = x, so its height is the unique real solution arsinh y.

For every real $y$ , the equation $y = sinh x$ with $x \in R$ has exactly one solution. Call that solution

$x = arsinh y .$

Swap the letters formally: for every real $x$ ,

$y = arsinh x$

means $sinh y = x$ and $y \in R$ . Reflecting the graph of sinh across the diagonal produces this graph.

The Latin names area sinus hyperbolicus, area cosinus hyperbolicus, area tangens hyperbolicus, and area cotangens hyperbolicus record that these inverses measure a hyperbolic area; the prefix $ar$ is that "area" reading.

Freeze where the partner is not one-to-one

Each partner is that reflection after any cut the forward map still needs:

function	principal domain	principal range
$arsinh x$	$R$	$R$
$arcosh x$	$[1, + \infty)$	$[0, + \infty)$
$artanh x$	$(- 1, 1)$	$R$
$arcoth x$	$(- \infty, - 1) \cup (1, + \infty)$	$R ∖ {0}$

Cosh is even, so two real inputs share each height $h \geq 1$ . Freeze cosh to $[0, + \infty)$ . Drag $h \geq 1$ . The green point sits on that nonnegative branch. The blue point is the reflection, so its height is the principal $arcosh h \geq 0$ .

Drag the height h ≥ 1. The green point sits on the nonnegative branch of cosh. The blue point is its reflection across y = x, so its height is the principal arcosh h ≥ 0.

The unrestricted equation $y = cosh x$ with $y \geq 1$ then recovers both signs as $x = \pm arcosh y$ . The negative partner is not a second function name.

Tanh maps $R$ onto the open interval $(- 1, 1)$ . Reflect that graph: $artanh$ takes every height strictly between $\pm 1$ , and the vertical lines $x = \pm 1$ stay out of reach. Drag $t$ inside $(- 1, 1)$ . The green point sits on tanh. The blue point is the reflection.

Drag the height t between −1 and 1. The green point sits on y = tanh x. The blue point is its reflection, so its height is artanh t; the blue point’s x stays between the dashed vertical lines x = ±1.

Coth maps away from $0$ onto $∣ y ∣ > 1$ . The same reflection gives $arcoth$ , whose graph has vertical asymptotes at $x = \pm 1$ and never meets height $0$ .

The same solution as an explicit logarithm

The unique real solution of each equation also has a closed form in $ln$ . For every real $y$ ,

$arsinh y = ln (y + y^{2} + 1) .$

For $y \geq 1$ ,

$\pm arcosh y = \pm ln (y + y^{2} - 1) .$

For $- 1 < y < 1$ ,

$artanh y = \frac{1}{2} ln \frac{1 + y}{1 - y} .$

For $∣ y ∣ > 1$ ,

$arcoth y = \frac{1}{2} ln \frac{y + 1}{y - 1} .$

Drag a real height $y$ . The green height is $asinh (y)$ . The blue height is $ln (y + y^{2} + 1)$ . They stay equal.

Drag the real height y. The green height is asinh(y). The blue height is ln(y + √(y² + 1)). They stay equal.

Arsinh, arcosh, artanh, and arcoth are the unique values on those principal branches, each also written with $ln$ . The next question is what their slopes are, and how the four branches rewrite into one another.