Calculate: $\;$ $\sin \left[2 \sin^{-1} \left(\dfrac{3}{5}\right)\right]$
Given Expression (GE): $\;$ $\sin \left[2 \sin^{-1} \left(\dfrac{3}{5}\right)\right]$
Let $\;$ $\sin^{-1} \left(\dfrac{3}{5}\right) = \theta$
Then, GE $= \sin 2 \theta = 2 \sin \theta \cos \theta$
i.e. $\;$ GE $= 2 \times \sin \left[\sin^{-1} \left(\dfrac{3}{5}\right)\right] \cos \left[\sin^{-1} \left(\dfrac{3}{5}\right)\right]$ $\;\;\;$ [substituting for $\theta$]
$= 2 \times \dfrac{3}{5} \times \cos \left[\cos^{-1} \left(\sqrt{1 - \dfrac{9}{25}}\right)\right]$
$= \dfrac{6}{5} \times \cos \left[\cos^{-1} \left(\dfrac{4}{5}\right)\right]$
$= \dfrac{6}{5} \times \dfrac{4}{5} = \dfrac{24}{25}$
Formulas used:
$\sin \left(\sin^{-1} x\right) = x, \; x \in \left[-1, 1\right]$
$\sin^{-1} x = \cos^{-1} \left(\sqrt{1 - x^2}\right), \; 0 \leq x \leq 1$
$\cos \left(\cos^{-1} x\right) = x. \; x \in \left[-1, 1\right]$