# Find the equation of tangent to the curve $$x-\sin\: 3t, y=\cos\: 2t \: at \: t=\large\frac{\pi}{4}.$$

Toolbox:
• Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
Step 1:
$x=\sin 3t$
Differentiating w.r.t $t$ we get,
$\large\frac{dx}{dt}$$=3\cos 3t y=\cos 2t Differentiating w.r.t t we get, \large\frac{dy}{dt}$$=-2\sin 2t$
$\large\frac{dy}{dx}=\frac{-2\sin 2t}{3\cos 3t}$
$\big(\large\frac{dy}{dx}\big)_{t=\Large\frac{\pi}{4}}=\frac{-2\sin 2(\pi/4)}{3\cos 3(\pi/4)}$
$\qquad\qquad=\large\frac{-2\sin 2(\pi/2)}{3\cos 3(\pi/4)}$
But $\sin\large\frac{\pi}{2}$$=1 \cos\large\frac{3\pi}{4}$$=\large\frac{-1}{\sqrt 2}$
$\therefore \large\frac{dy}{dx}$$=m \Rightarrow \large\frac{-2}{3\times -1/\sqrt 2} \Rightarrow \large\frac{2\sqrt 2}{3} Step 2: Equation of the tangent is y-y_1=m(x-x_1) Here x_1=\sin 3t \sin 3(\large\frac{\pi}{4})=\large\frac{-1}{\sqrt 2} y_1=\cos 2\big(\large\frac{\pi}{4}\big)$$=0$
Equation of the tangent is $y-0=\large\frac{2\sqrt 2}{3}$$(x+\large\frac{1}{\sqrt 2}\big)$
$\Rightarrow 3y=2\sqrt 2(\sqrt x+1)$
$\therefore 4x-3y+2\sqrt 2=0$