Question

If $x=2\cos \theta -\cos 2\theta$ and $y=2\sin \theta -\sin 2\theta$, then prove that $\frac{dy}{dx}=\tan \left(\frac{3\theta }{2}\right)$

Answer 1

Differentiate x and y with respect to $\theta$.

$\frac{dx}{d\theta }=-2sin\theta +2sin2\theta$

$\frac{dy}{d\theta }=2cos\theta -2cos2\theta$

$\frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}$

$=\frac{2cos\theta -2cos2\theta }{-2sin\theta +2sin2\theta }$

=$\frac{cos\theta -cos2\theta }{-sin\theta +sin2\theta }$

We use trignometric identities $cosA-CosB=-sin\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right)$, and $sinA-sinB=2cos\left(\frac{A+B}{2}\right)sin\frac{(A-B)}{2}$

$\frac{cos\theta -cos2\theta }{-sin\theta +sin2\theta }=\frac{2sin\frac{3\theta }{2}sin\frac{\theta }{2}}{2cos\frac{3\theta }{2}sin\frac{\theta }{2}}$

Hence we obtain$\frac{dy}{dx}=tan\frac{3\theta }{2}$