Question

If ${\cosh ^{ - 1}}x = \log (2 + \sqrt 3 )$, then $x = $?

Answer 1

Hint: Here we have given a hyperbolic trigonometric function. They are similar to the trigonometric function but defined using the hyperbola rather than the circle. Here we use the formula of cosine hyperbolic inverse function which is ${\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right)$ to find the value of $x$. We will compare the given function with the above formula and find the value of $x$.

Complete step by step solution:
Here we have given a hyperbolic cosine function. hyperbolic function are similar to trigonometric function but are defined using the hyperbola rather than the circle as the points $(\sin t,\,\,\cos t)$ in trigonometry form a unit circle with radius the points $(\sinh \,t,\,\,\cosh \,t)$ form the right half of the unit parabola.
Hyperbolic functions are shown up in the calculation of angles and distance in hyperbolic geometry. They are also used in the solutions of many linear equations, cubic equations and Laplace’s equation in cartesian coordinates.
The properties of hyperbolic function are similar to the properties of trigonometric functions such as $\sinh ( - y) = - \sinh (y),\,\,\cosh ( - y) = \cosh (y)$ or \[\sinh 2y = 2\sinh y\cosh y,\,\,\,\cosh 2y = {\cosh ^2}y - 1\].
Here, we have given a cosine hyperbolic function.
we have ${\cosh ^{ - 1}}x = \log (2 + \sqrt 3 )$
we know that ${\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right)$
comparing the given function with the above formula. we get,
$ \Rightarrow 2 + \sqrt 3 = \left( {x + \sqrt {{x^2} - 1} } \right)$
The rational part is equal on both sides. We get,
$ \Rightarrow x = 2$
Hence, $x = 2$ for the function ${\cosh ^{ - 1}}x = \log (2 + \sqrt 3 )$.

Note: Hyperbolic functions can also be derived with the help of the trigonometric function with complex arguments such as $\sinh y = i\sin (iy),\,\,\cosh y = \cos (iy),\,\,\tanh y = - i\tan \left( {iy} \right)$. Hyperbolic functions are derived in terms of ${e^x}$ and ${e^{ - x}}$, so we can easily derives rules for their integration such as $\int {\cosh y\,dy = \sinh y + C,\,\,\int {\sinh y\,dy = \cosh y + C} } $.