Question

To determine \({\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right)\) where \(x \ge 1\).

Short answer

The proof is below.

Step-by-step solution

Given data

The equation to be proved is \({\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right)\).

Identity of hyperbolic function

Hyperbolic function:

\({\cosh ^2}y - {\sinh ^2}y = 1\)

\(\cosh y + \sinh y = {e^y}\)

Prove of the equation \({\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right)\)

To prove the above equation let\(y = {\cosh ^{ - 1}}x\).

\( \Rightarrow x = \cosh y\)When\(x \ge 0\).

Now, substitute \(x = \cosh y\) in \({\cosh ^2}y - {\sinh ^2}y = 1\).

\(\begin{array}{c}{x^2} - {\sinh ^2}y = 1\\{\sinh ^2}y = {x^2} - 1\\\sinh y = \sqrt {{x^2} - 1} \end{array}\)

Now, substitute \(\sinh y = \sqrt {{x^2} - 1} \) and \(\cosh y = x\) in \(\cosh y + \sinh y = {e^y}\).\(\begin{array}{c}\sqrt {{x^2} - 1} + x = {e^y}\\{e^y} = x + \sqrt {{x^2} - 1} \end{array}\)

Take natural logarithm on both sides.

\(\begin{array}{c}y = \ln \left( {x + \sqrt {{x^2} - 1} } \right)\\{\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right)\end{array}\)

If the values\(0 < x < 1\), then negative sign occur in root and the function\({\cosh ^{ - 1}}x\)is not defined.

\({\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right)\)Is defined when\(x \ge 1\).

Therefore,\({\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right)\).

Where,\(x \ge 1\).

Hence the result is proved.