We have, $x=\sin \left(2 \tan ^{-1} 2\right)$ and $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$ $\Rightarrow x=\sin \left(2 \tan ^{-1} 2\right)$
Let $\tan ^{-1} 2-\alpha \Rightarrow \tan \alpha=2$
$\therefore \quad x=\sin 2 \alpha=\frac{2 \tan \alpha}{1+\tan ^2 \alpha}=\frac{2(2)}{1+(2)^2}=\frac{4}{5}$
Now, $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$
Let $\tan ^{-1} \frac{4}{3}=\beta$
$\tan \beta=\frac{4}{3}, \cos \beta=\frac{1}{\sqrt{1+\tan ^2 \beta}}=\frac{3}{5}$
Then, $y=\sin \left(\frac{\beta}{2}\right)$
$\begin{aligned}
& =\sqrt{\frac{1-\cos \beta}{2}}=\sqrt{\frac{1-\frac{3}{5}}{2}}=\sqrt{\frac{1}{5}} \\
& \therefore \quad x>y
\end{aligned}$