Calculate: $\;$ $\sin \left[2 \tan^{-1} \left(\dfrac{1}{3}\right) \right] + \cos \left[\tan^{-1} \left(2 \sqrt{2}\right)\right]$
Given Expression (GE): $\;$ $\sin \left[2 \tan^{-1} \left(\dfrac{1}{3}\right) \right] + \cos \left[\tan^{-1} \left(2 \sqrt{2}\right)\right]$
$= 2 \sin \left[\tan^{-1} \left(\dfrac{1}{3}\right)\right] \cos \left[\tan^{-1} \left(\dfrac{1}{3}\right)\right] + \cos \left[\tan^{-1} \left(2 \sqrt{2}\right)\right]$
$= 2 \times \dfrac{\dfrac{1}{3}}{\sqrt{1 + \dfrac{1}{9}}} \times \dfrac{1}{\sqrt{1 + \dfrac{1}{9}}} + \dfrac{1}{\sqrt{1 + 8}}$
$= \dfrac{\dfrac{2}{3}}{\dfrac{\sqrt{10}}{3}} \times \dfrac{1}{\dfrac{\sqrt{10}}{3}} + \dfrac{1}{3}$
$= \dfrac{2}{\sqrt{10}} \times \dfrac{3}{\sqrt{10}} + \dfrac{1}{3}$
$= \dfrac{3}{5} + \dfrac{1}{3} = \dfrac{14}{15}$
Formulas used:
$\sin 2 \theta = 2 \sin \theta \cos \theta$
$\cos \left(\tan^{-1} x\right) = \dfrac{1}{\sqrt{1 + x^2}}, \;\; 0 \leq x \leq 1$
$\sin \left(\tan^{-1} x\right) = \dfrac{x}{\sqrt{1 + x^2}}, \;\; x > 0$