Inverse Trigonometric Functions

Find the value of: $\;$ $\cos \left[2 \cos^{-1} \left(x\right) + \sin^{-1} \left(x\right)\right]$ $\;$ when $\;$ $x = \dfrac{1}{3}$


$\left[\text{Note: } \sin^{-1} \left(x\right) = \cos^{-1} \left(\sqrt{1 - x^2}\right) \right.$

$\left. 2 \cos^{-1} \left(x\right) = \cos^{-1} \left(2 x^2 - 1\right) \right.$

$\left. \cos^{-1} \left(x\right) + \cos^{-1} \left(y\right) = \cos^{-1} \left(x y - \sqrt{1 - x^2} \sqrt{1 - y^2}\right), \;\; -1 \leq x, \; y \leq 1, \; x + y \geq 0 \right.$

$\left. \cos^{-1} \left(- x\right) = \pi - \cos^{-1} \left(x\right) \right]$

When $\;$ $x = \dfrac{1}{3}$,

$\begin{aligned} \cos \left[2 \cos^{-1} \left(x\right) + \sin^{-1} \left(x\right)\right] & = \cos \left[2 \cos^{-1} \left(\dfrac{1}{3}\right) + \sin^{-1} \left(\dfrac{1}{3}\right)\right] \\\\ & = \cos \left[2 \cos^{-1} \left(\dfrac{1}{3}\right) + \cos^{-1} \left(\sqrt{1 - \dfrac{1}{9}}\right)\right] \\\\ & = \cos \left[2 \cos^{-1} \left(\dfrac{1}{3}\right) + \cos^{-1} \left(\dfrac{\sqrt{8}}{3}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{2}{9} - 1\right) + \cos^{-1} \left(\dfrac{\sqrt{8}}{3}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(- \dfrac{7}{9}\right) + \cos^{-1} \left(\dfrac{\sqrt{8}}{3}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{-7 \sqrt{8}}{27} - \sqrt{1 - \dfrac{49}{81}} \times \sqrt{1 - \dfrac{8}{9}}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{- 7 \sqrt{8}}{27} - \dfrac{\sqrt{32}}{9} \times \dfrac{1}{3}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{- 7 \sqrt{8} - 2 \sqrt{8}}{27}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{-9 \sqrt{8}}{27}\right)\right] \\\\ & = \cos \left[\cos^{-1} \left(\dfrac{- \sqrt{8}}{3}\right)\right] \\\\ & = \cos \left[\pi - \cos^{-1} \left(\dfrac{\sqrt{8}}{3}\right)\right] \\\\ & = - \cos \left[\cos^{-1} \left(\dfrac{\sqrt{8}}{3}\right)\right] \\\\ & = - \dfrac{\sqrt{8}}{3} \end{aligned}$