Deriving the formula for arcosh(x)

Let \(y= \cosh(x)\):

\[ y = \frac{e^x + e^{-x}}{2}\]

In order to find the formula for the inverse of \(\cosh\) we need to make \(x\) the subject of the above formula.

\[ 2y = e^x + e^{-x}\]

If we multiply both sides by \(e^x\):

\[\begin{gathered} 2ye^x = (e^x)^2 + 1 \\ (e^x)^2 - 2ye^x + 1 = 0 \end{gathered}\]

Let \(v = e^x\):

\[v^2 -2yv + 1 = 0 \]

We can use the quadratic formula here:

\[v=\frac{-(-2y) ± \sqrt{(-2y)^2 - 4(1)(1)}}{2(1)} = \frac{2y ± \sqrt{4y^2 - 4}}{2} \]

Simplifying:

\[v = \frac{2y ± 2\sqrt{y^2 - 1}}{2} = y ∓ \sqrt{y^2 - 1} \]

Since \(v=e^x\):

\[x = \ln(y ∓ \sqrt{y^2 - 1}) \]

The function \(y=\cosh(x)\) has a domain of \((-\infty,\infty)\) and is an even function, so it's inverse will fail the vertical line test. To solve this, \(\operatorname{arcosh}(y)\) is defined as the inverse of \(\cosh(x)\) where \(x \in [0, \infty)\). This means the range of \(\operatorname{arcosh}(y)\) is \([0,\infty)\):

\[\begin{gathered} x = \ln(y ∓ \sqrt{y^2 - 1}) \ge 0 \\ y ∓ \sqrt{y^2 - 1} \ge 1 \end{gathered}\]

Let \( \sqrt{y^2 - 1}\) always be negative.

\[\begin{gathered} y - \sqrt{y^2 - 1} \ge 1 \\ -\sqrt{y^2 - 1} \ge 1-y \\ \sqrt{y^2 - 1} \le y-1 \end{gathered}\]

If we square both sides:

\[\begin{gathered} y^2 - 1 \le (y-1)^2 \\ y^2 - 1 \le y^2 -2y +1 \\ -2 \le -2y \end{gathered}\]

The range of \(y=\cosh(x)\) is \([1,\infty)\). This means the above inequality is false. Therefore, \( \sqrt{y^2 - 1}\) cannot be negative. Let's say it's positive instead:

\[\begin{gathered} y + \sqrt{y^2 - 1} \ge 1 \\ \sqrt{y^2 - 1} \ge 1-y \\ y^2 - 1 \ge 1-2y+y^2 \end{gathered}\]

This means:

\[\begin{gathered} y^2 - 1 \ge 1-2y+y^2 \\ -2 \ge -2y \end{gathered}\]

The above inequality is true for \(y \in [1,\infty)\). Therefore, \( \sqrt{y^2 - 1}\) is always positive:

\[\operatorname{arcosh}(y) = \ln(y + |\sqrt{y^2 - 1})| \]

Deriving The Formula For arcosh(x)