Question

Find $\frac{dy}{dx}$ , if $x=2\cos \theta -\cos 2\theta$and $y=2\sin \theta -\sin 2\theta$.

• A. $\tan \frac{3\theta }{2}$
• B. $-\tan \frac{3\theta }{2}$
• C. $cot\frac{3\theta }{2}$
• D. $-cot\frac{3\theta }{2}$

Answer 1

The correct option is A

$\tan \frac{3\theta }{2}$

Explanation for the correct option:

Step 1. Differentiate $x$ with respect to $\theta$

$x=2\cos \theta –\cos 2\theta$

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

$=2(\sin 2\theta –\sin \theta )$

Step 2. Differentiate $y$ with respect to $\theta$

$y=2\sin \theta –\sin 2\theta$

$\Rightarrow$$\frac{dy}{d\theta }=2\cos \theta –2\cos 2\theta$

$=2(\cos \theta –\cos 2\theta )$

Step 3. Divide $\frac{dy}{d\theta }$ by $\frac{dx}{d\theta }$

$\Rightarrow$$\frac{dy}{dx}=\left(\frac{dy}{d\theta }\right)\left(\frac{d\theta }{dx}\right)$

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

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

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

$=\frac{2\sin \left({\displaystyle \frac{3\theta }{2}}\right)\sin \left({\displaystyle \frac{\theta }{2}}\right)}{2\cos \left({\displaystyle \frac{3\theta }{2}}\right)\sin \left({\displaystyle \frac{\theta }{2}}\right)}$ $\left[\because \sin \theta -\sin \varphi =2\cos \frac{\theta +\varphi }{2}\sin \frac{\theta -\varphi }{2}\right]$ $\Rightarrow \frac{dy}{dx}=\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathbf{\left(}\frac{\mathbf{3}\mathbf{\theta }}{\mathbf{2}}\mathbf{\right)}$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\varphi }\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\frac{\mathbf{\theta }\mathbf{+}\mathbf{\varphi }}{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{\theta }\mathbf{-}\mathbf{\varphi }}{\mathbf{2}}\right]$

Hence, Option ‘A’ is Correct.