Inverse Trigonometric Functions

Evaluate: $\;$ $\sin \left[3 \sin^{-1} \left(\dfrac{1}{3}\right)\right]$


Let $\sin^{-1} \left(\dfrac{1}{3}\right) = \theta \;\;\; \text{ where } \theta \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

Then, $\sin \left(\theta\right) = \dfrac{1}{3}$

$\begin{aligned} \therefore \; \sin \left[3 \sin^{-1} \left(\dfrac{1}{3}\right)\right] & = \sin \left(3 \theta\right) \\\\ & = 3 \sin \theta - 4 \sin^3 \theta \\\\ & = 3 \times \dfrac{1}{3} - 4 \times \left(\dfrac{1}{3}\right)^3 \\\\ & = 1 - \dfrac{4}{27} \\\\ & = \dfrac{23}{27} \end{aligned}$