Question

Prove that the inverse hyperbolic functions can be written in terms of logarithms as shown in Exercises 97–99. (Hint for the first problem: Solve $\sin hy=x$for $y$by using algebra to get an expression that is quadratic in $e^y$(i.e., of the form $a{e}^{2y}+b{e}^y+c$) and then applying the quadratic formula.) (These exercises involve hyperbolic functions.)

$\sin h^{-1}x=\ln \left(x+\sqrt{x^2+1}\right)$, for any$x$

Short answer

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

Step-by-step solution

Step 1. Given Information

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

Step 2. Proving the statement

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

Solving the quadratic we get,

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