If $x=\sin t\cos 2t$ and $y=\cos t\sin 2t$, then ${\left(\frac{dy}{dx}\right)}_{t=\frac{\pi }{4}}=$

$x=\sin t.\cos 2t$

$\frac{dx}{dt}=\sin t.(-\sin 2t).2+\cos 2t.\cos t$

$=-2\sin t.\sin 2t+\cos t.\cos 2t$

$y=\cos t.\sin 2t$

$\frac{dy}{dx}=\cos t.\cos 2t.2+\sin 2t.(-\sin t)$

$=2.\cos t.\cos 2t-\sin t.\sin 2t$

$\frac{dy}{dx}=\frac{2\cos t.\cos 2t-\sin t.\sin 2t}{-2\sin t.\sin 2t+\cos t.\cos 2t}$

${\left(\frac{dy}{dx}\right)}_{t=\frac{\pi }{4}}=\frac{2.\cos \frac{\pi }{4}\cos \frac{\pi }{2}-\sin \frac{\pi }{4}.\sin \frac{\pi }{2}}{-2\sin \frac{\pi }{4}.\sin \frac{\pi }{2}+\cos \frac{\pi }{4}.\cos \frac{\pi }{2}}$

$=\frac{1}{2}$