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Question:

Find $\frac{d y}{d x}$, when

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

Solution:

$a s x=2 \cos \theta-\cos 2 \theta$

Differentiating it with respect to $\theta$ using chain rule,

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=2(-\sin \theta)-(-\sin 2 \theta) \frac{\mathrm{d}}{\mathrm{d} \theta}(2 \theta)$

$=-2 \sin \theta+2 \sin 2 \theta$

$\frac{d x}{d \theta}=2(\sin 2 \theta-\sin \theta)$ ....(1)

And, $y=2 \sin \theta-\sin 2 \theta$

Differentiating it with respect to $\theta$ using chain rule,

$\frac{\mathrm{dy}}{\mathrm{d} \theta}=2 \cos \theta-\cos 2 \theta \frac{\mathrm{d}}{\mathrm{d} \theta}(2 \theta)$

$=2 \cos \theta-\cos 2 \theta(2)$

$=2 \cos \theta-2 \cos 2 \theta$

dividing equation (2)by equation (1),

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=\frac{2(\cos \theta-\cos 2 \theta)}{2(\sin 2 \theta-\sin \theta)}$

$=\frac{(\cos \theta-\cos 2 \theta)}{(\sin 2 \theta-\sin \theta)}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2 \sin \left(\frac{\theta+2 \theta}{2}\right) \sin \left(\frac{\theta-2 \theta}{2}\right)}{2 \cos \left(\frac{\theta+2 \theta}{2}\right) \sin \left(\frac{2 \theta-\theta}{2}\right)}[$ since sina $-$ sinb

$\left.=2 \cos \left(\frac{a+b}{2}\right) \sin \left(\frac{a-b}{2}\right)\right]$

$\left[\cos a-\cos b=-2 \sin \left(\frac{a+b}{2}\right) \sin \left(\frac{a-b}{2}\right)\right]$

$=-\frac{\sin \left(\frac{3 \theta}{2}\right)\left(\sin \left(-\frac{\theta}{2}\right)\right)}{\cos \left(\frac{3 \theta}{2}\right) \sin \left(\frac{\theta}{2}\right)}$

$=-\frac{\sin \left(\frac{3 \theta}{2}\right)\left(-\sin \frac{\theta}{2}\right)}{\cos \left(\frac{3 \theta}{2}\right) \sin \left(\frac{\theta}{2}\right)}$

$=\frac{\sin \left(\frac{3 \theta}{2}\right)}{\cos \left(\frac{3 \theta}{2}\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\tan \left(\frac{3 \theta}{2}\right)$