Why inverse hyperbolic cosine is just $\ln(x + \sqrt{x^2 - 1})$ and not $\ln(x - \sqrt{x^2 - 1})$ (source: Wikipedia)?

Question

I am an amateur who loves elementary mathematics. I am not too good by any means, and never considered myself a genius or a math pro etc.

Can I please ask a question which I am really curious about? Googling or asking on some other sites did not help me too much.

My task is to find the inverse of hyperbolic cosine: $$ y = \cosh x = \frac{e^x + e^{-x}}{2}. $$

We must interchange $y$ with $x$ and then express the variable $y$ in terms of $x$. So we have: \begin{align} x &= \frac{e^y + e^{-y}}{2} \\ x &= \frac{e^y + \frac{1}{e^y}}{2} \\ x &= \frac{e^{2y} + 1}{2e^y} \\ 2x &= \frac{e^{2y} + 1}{e^y}. \end{align}

Here, we must replace the variable. Let $e^y = z$. Then $e^{2y} = z^2$. Note: $z > 0$ (strictly greater than zero) as the value of any exponential function is always positive. \begin{align} 2x &= \frac{z^2 + 1}{z} \\ z^2 + 1 &= 2xz \\ z^2 - 2xz + 1 &= 0. \end{align}

Here, we have a parametric quadratic equation where the unknown variable is $z$ (and $x$ is a parameter). Coefficients are $$ a = 1; \quad b = -2x; \quad c = 1 $$ and discriminant is $$ D = b^2 - 4ac = (-2x)^2 - 4 \cdot 1 \cdot 1 = 4x^2 - 4 = 4(x^2 - 1). $$ So the solutions are \begin{align} x_1 &= \frac{-b + \sqrt{D}}{2a} = \frac{2x + 2\sqrt{x^2 - 1}}{2} = x + \sqrt{x^2 - 1}; \\ x_1 &= \frac{-b - \sqrt{D}}{2a} = \frac{2x - 2\sqrt{x^2 - 1}}{2} = x - \sqrt{x^2 - 1} \end{align}

So here is my problem! I obtained two roots and I do not know which one to choose. Must I take the first, the second, or both?

Wikipedia is very consistent and only took the positive root. So, according to Wikipedia, $$ \operatorname{arcosh} x = \ln \bigl(x + \sqrt{x^2 - 1}\bigr). $$

No alternatives.

Maybe it's something I miss, but I want to try to understand it. I tried to ask a random person, and he responded: "Well, in the case of finding the inverse of hyperbolic sine $y = \sinh x$, we obviously took the plus sign root. So we must be consistent, thus we must do exactly the same thing with cosine." However, I doubt a bit about such kind of answer.

Sorry for grammar mistakes if any.

Wikipedia: $\operatorname{arcosh} x = \ln \bigl( x + \sqrt{x^2 - 1} \bigr)$ and not $\operatorname{arcosh} x = \ln \bigl( x - \sqrt{x^2 - 1} \bigr)$

Answer 1

Short answer. It's an arbitrary choice, and we prefer positive numbers if given the choice.

But why do we have to make a choice at all? Well, the hyperbolic cosine function is not one-to-one since $$ \cosh(-x) = \cosh(x) $$ for any $x \in \mathbb{R}$. In other words, if $b = \cosh(a)$ and we know $b$, how do we recover $a$? Well, the problem is that there are two possible answers: if $a$ works, then its opposite $-a$ also works! In order for this inverse to be a function there must be one definitive output for any given input. Hence, we must ignore half of the real numbers that could possibly be inputs for the hyperbolic cosine, and by convention we ignore the negative ones and keep the non-negative ones.

So, the inverse hyperbolic cosine is defined to give those same non-negative real numbers as outputs. $$ \operatorname{arcosh} x = y \; \iff \; x = \cosh y \text{ and } y \geq 0. $$

In other words, this $\operatorname{arcosh} x$ function is the inverse to the function defined only for $x \geq 0$ by the rule $y = \cosh x$. Here you can see the graphs of these two mutually inverse functions—which, as always, are reflections $(x, y) \leftrightarrow (y, x)$ of one another across the line $y=x$.

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By the way, this is the same arbitrary choice that defines the more familiar square-root function: $$ \sqrt{x} = y \; \iff \; x = y^2 \text{ and } y \geq 0. $$

Answer 2

The graph of the cosh function $y=\cosh(x)$ does not pass the horizontal line test. To put this another way, the cosh function is not one-to-one.

When that happens, an inverse function does not exist per se.

However, what one can do in this situation is to throw away a minimal portion of the graph until what is left does pass the horizontal line test, and once that is done then the restricted function whose graph is the leftover portion now does have an inverse function.

For a function whose graph fails the horizontal line test, there is always more than one way to choose which portion to throw away. For functions that are particularly special or important to the mathematical community, the community has usually settled on one choice as the canonical one.

For a more familiar example, consider the graph of the squaring function, i.e. the graph of $y=x^2$. This example has two choices of what to keep/throw away: either throw away $x<0$ and keep $x \ge 0$; or throw away $x>0$ and keep $x \le 0$. The community has chosen to throw away $x<0$ and keep $x \ge 0$ (because we like positive numbers? Maybe...). The resulting inverse function is the square root function $y=\sqrt{x}$, whose value is never negative.

For the inverse of the cosh function, one encounters the same two choices as the squaring function, and the mathematics community has made this choice similarly: throw away $x<0$ and keep $x \ge 0$. When that is done, in order to follow that choice when deriving a formula for the inverse function, you have to pick the formula which yields a positive output. As indicated in the comment of @user10354138, that restriction leads to the formula with the plus sign: $$\cosh^{-1}(x) = \ln\bigl(x + \sqrt{x^2-1}\bigr) $$

Answer 3

That we must choose $\cosh^{-1} x = \ln ( x + \sqrt{x^2 - 1} )$ and not $\cosh^{-1} x = \ln ( x - \sqrt{x^2 - 1} )$ becomes clear if we consider the limiting case $x \to +\infty$. Since, a function inverse is defined generally by $f^{-1}(f(x))=x$, and we know that $$\lim_{x\to+\infty}\cosh x$$ diverges to $+\infty$, we expect that

$$ \lim_{x\to+\infty}\cosh^{-1}(x)$$

also diverges to $+\infty$. However, $$\lim_{x\to+\infty}\ln(x-\sqrt{x^2 - 1})$$ diverges to $-\infty$. So, it cannot be $\cosh^{-1}x$. A quick look at the other formula should convince you that it diverges in the way we expect. So, by process of elimination, $$\cosh^{-1} x = \ln(x+\sqrt{x^2 - 1}), \qquad x \ge 1.$$

Of course, as others have pointed out, all this assumes that you are seeking the inverse for $\cosh x$ when it is restricted to the domain $0 \le x < +\infty,$ which is a very natural assumption to make.

You may apply the same line of reasoning to discover that the inverse of $\cosh x$, when restricted to the domain $-\infty < x \le 0$, is $\ln(x-\sqrt{x^2-1}).$ To summarize everyone and perhaps more directly answer your question, albeit at the risk of abusing notation in the process, if $y=\cosh x,$ then

$$\cosh^{-1}y = \begin{cases}\ln(y+\sqrt{y^2 - 1}), & 0 \le x < +\infty \\ \ln(y-\sqrt{y^2 - 1}), & -\infty < x < 0\end{cases} $$

Those who know analysis can confirm the above by checking the relevant sign of the derivative, recalling that $\text{sgn}\,(\mathrm dy/\mathrm dx) = \text{sgn}\,(\mathrm dx/\mathrm dy)$ for $0 < \left\lvert\frac{\mathrm dy}{\mathrm dx}\right\rvert < \infty.$