Question

Prove that the inverse hyperbolic functions can be written in terms of logarithms as shown in Exercises 97–99.

$\cos h^{-1}x=\ln \left(x+\sqrt{x^2-1}\right)$for$x\ge 1$.

Short answer

We proved $\cos h^{-1}x=\ln \left(x+\sqrt{x^2-1}\right)$for $x\ge 1$.

Step-by-step solution

Step 1. Given Information

We have $\cos h=x$and we need to prove $\cos h^{-1}x=\ln \left(x+\sqrt{x^2-1}\right)$.

Step 2. Proving the statement 

$$\begin{gathered} \cos hy=x \\ \Rightarrow \frac{e^y+e^{-y}}{2}=x \\ \Rightarrow e^y+e^{-y}=2x \\ \Rightarrow e^y+\frac{1}{e^y}=2x \\ \Rightarrow \frac{e^{2y}+1}{e^y}=2x \\ \Rightarrow e^{2y}+1=2x{e}^y \\ \Rightarrow e^{2y}-2x{e}^y+1=0 \end{gathered}$$

Solving the quadratic we get,

$$\begin{gathered} e^y=\frac{2x\pm \sqrt{4x^2-4}}{2} \\ \Rightarrow \ln \left(e^y\right)=\ln \left(\frac{2x\pm \sqrt{4x^2-4}}{2}\right) \\ \Rightarrow \ln \left(e^y\right)=\ln \left(\frac{2\left(x\pm \sqrt{x^2-1}\right)}{2}\right) \\ \Rightarrow y=\ln \left(x\pm \sqrt{x^2-1}\right) \\ \Rightarrow \cos h^{-1}x=\ln \left(x\pm \sqrt{x^2-1}\right) \end{gathered}$$