# Equivalence of Definitions of Real Inverse Hyperbolic Cosine

## Theorem

The following definitions of the concept of Real Inverse Hyperbolic Cosine are equivalent:

### Definition 1

The inverse hyperbolic cosine $\cosh^{-1}: S \to \R$ is a real function defined on $S$ as:

$\forall x \in S: \cosh^{-1} \left({x}\right) := y \in \R_{\ge 0}: x = \cosh \left({y}\right)$

where $\cosh \left({y}\right)$ denotes the hyperbolic cosine function.

### Definition 2

The inverse hyperbolic cosine $\cosh^{-1}: S \to \R$ is a real function defined on $S$ as:

$\forall x \in S: \map {\cosh^{-1} } x := \map \ln {x + \sqrt {x^2 - 1} }$

where:

$\sqrt {x^2 - 1}$ denotes the positive square root of $x^2 - 1$
$\ln$ denotes the natural logarithm of a (strictly positive) real number.

## Proof

### Definition 1 implies Definition 2

Let $x = \cosh y$, where $y > 0$.

Let $z = e^y$.

Then:

 $\displaystyle x$ $=$ $\displaystyle \frac {e^y + e^{-y} } 2$ Definition of Hyperbolic Cosine $\displaystyle \leadsto \ \$ $\displaystyle 2 x$ $=$ $\displaystyle e^y + e^{-y}$ $\displaystyle \leadsto \ \$ $\displaystyle 2 x e^y$ $=$ $\displaystyle e^{2 y} + 1$ $\displaystyle \leadsto \ \$ $\displaystyle z^2 - 2 x z + 1$ $=$ $\displaystyle 0$ Power of Power $\displaystyle \leadsto \ \$ $\displaystyle z$ $=$ $\displaystyle \frac {2 x \pm \sqrt {\paren {-2 x}^2 - 4} } 2$ Quadratic Formula $\displaystyle$ $=$ $\displaystyle x \pm \sqrt{x^2 - 1}$ $\displaystyle \leadsto \ \$ $\displaystyle e^y$ $=$ $\displaystyle x \pm \sqrt{x^2 - 1}$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \map \ln {x \pm \sqrt {x^2 - 1} }$

Also, from Minimum of Real Hyperbolic Cosine Function:

$x = \cosh y \ge 1$

Also:

 $\displaystyle x$ $=$ $\displaystyle \sqrt {x^2}$ $x$ is positive $\displaystyle$ $>$ $\displaystyle \sqrt {x^2 - 1}$ Square Root is Strictly Increasing $\displaystyle \leadsto \ \$ $\displaystyle x - \sqrt {x^2 - 1}$ $>$ $\displaystyle 0$

Thus $x - \sqrt {x^2 - 1}$ is (strictly) positive.

Aiming for a contradiction, suppose $x - \sqrt {x^2 - 1} > 1$.

Then:

 $\displaystyle x - \sqrt {x^2 - 1}$ $>$ $\displaystyle 1$ $\displaystyle \leadsto \ \$ $\displaystyle x - 1$ $>$ $\displaystyle \sqrt {x^2 - 1}$ Both sides are (strictly) positive $\displaystyle \leadsto \ \$ $\displaystyle \paren {x - 1}^2$ $>$ $\displaystyle x^2 - 1$ right hand side is (strictly) positive because $x \ge 1$ $\displaystyle \leadsto \ \$ $\displaystyle x^2 - 2 x + 1$ $>$ $\displaystyle x^2 - 1$ $\displaystyle \leadsto \ \$ $\displaystyle 2$ $>$ $\displaystyle 2 x$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $<$ $\displaystyle 1$

Therefore:

$x - \sqrt {x^2 - 1} < 1$
$y = \map \ln {x - \sqrt {x^2 - 1} } < \ln 1 = 0$

Since $y$ is (strictly) positive from the first definition of real inverse hyperbolic cosine:

$y = \map \ln {x + \sqrt {x^2 - 1} }$

$\Box$

### Definition 2 implies Definition 1

Let $z = x + \sqrt {x^2 - 1}$.

Then:

$y = \ln z$
 $\displaystyle \map \cosh {\map \ln {x + \sqrt {x^2 - 1} } }$ $=$ $\displaystyle \map \cosh {\ln z}$ $\displaystyle$ $=$ $\displaystyle \frac {e^{\ln z} + e^{-\ln z} } 2$ Definition of Hyperbolic Cosine $\displaystyle$ $=$ $\displaystyle \frac {z + \frac 1 z} 2$ $\displaystyle$ $=$ $\displaystyle \frac {z^2 + 1} {2 z}$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {x + \sqrt {x^2 - 1} }^2 + 1} {2 x + 2 \sqrt {x^2 - 1} }$ $\displaystyle$ $=$ $\displaystyle \frac {x^2 + 2 x \sqrt {x^2 - 1} + \paren {x^2 - 1} + 1} {2 x + 2 \sqrt {x^2 - 1} }$ $\displaystyle$ $=$ $\displaystyle \frac {2 x^2 + 2 x \sqrt {x^2 - 1} } {2 x + 2 \sqrt {x^2 - 1} }$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {2 x + 2 \sqrt {x^2 - 1} } x} {2 x + 2 \sqrt {x^2 - 1} }$ $\displaystyle$ $=$ $\displaystyle x$

If $-1 < x < 1$, $z$ is not defined.

If $x \le -1$:

 $\displaystyle \sqrt {x^2 - 1}$ $<$ $\displaystyle \sqrt {x^2}$ Square Root is Strictly Increasing $\displaystyle$ $=$ $\displaystyle -x$ $x$ is negative $\displaystyle \leadsto \ \$ $\displaystyle x + \sqrt{x^2 - 1}$ $<$ $\displaystyle 0$

If $x \ge 1$, $z \ge 1$.

Therefore, $y = \ln z \ge \ln 1 = 0$.

$\Box$

Therefore:

 $\text {(1)}: \quad$ $\displaystyle y > 0 \land x = \cosh y$ $\implies$ $\displaystyle y = \map \ln {x + \sqrt {x^2 + 1} }$ Definition 1 implies Definition 2 $\text {(2)}: \quad$ $\displaystyle y = \map \ln {x + \sqrt {x^2 + 1} }$ $\implies$ $\displaystyle x = \cosh y \land y > 0$ Definition 2 implies Definition 1 $\displaystyle \leadsto \ \$ $\displaystyle y > 0 \land x = \cosh y$ $\iff$ $\displaystyle y = \map \ln {x + \sqrt {x^2 + 1} }$

$\blacksquare$