Question

Prove Equation 4.

Short answer

The value of \({\cosh ^{ - 1}}x\) is \(\ln \left( {x + \sqrt {{x^2} - 1} } \right)\).

Step-by-step solution

Step 1:Simplify the problem using the trigonometric identity

Let \(y = {\cosh ^{ - 1}}x\), then\(x = \cosh y\).

Using the hyperbolic relation,

\(\begin{aligned}{\cosh ^2}y - {\sinh ^2}y &= 1\\{\sinh ^2}y &= {\cosh ^2}y - 1\\\sinh y &= \sqrt {{{\cosh }^2}y - 1} \\ &= \sqrt {{x^2} - 1} \end{aligned}\)

Use the identity given in Example 13

The given relation in example 13 is \({e^y} = \sinh y + \cosh y\).

Substitute \(\sqrt {{x^2} - 1} \) for \(\sinh y\) and \(x\) for \(\cosh y\).

\(\begin{aligned}{e^y} &= \sqrt {{x^2} - 1} + x\\y &= \ln \left( {x + \sqrt {{x^2} - 1} } \right)\end{aligned}\)

So, the value of \({\cosh ^{ - 1}}x\) is \(\ln \left( {x + \sqrt {{x^2} - 1} } \right)\).