Question

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

Answer 1

$\mathrm{We}\mathrm{have},x=2\cos \theta -\cos 2\theta$

$$\begin{gathered} \Rightarrow \frac{dx}{d\theta }=2\left(-\sin \theta \right)-\left(-\sin 2\theta \right)\frac{d}{d\theta }\left(2\theta \right) \\ \Rightarrow \frac{dx}{d\theta }=-2\sin \theta +2\sin 2\theta \\ \Rightarrow \frac{dx}{d\theta }=2\left(\sin 2\theta -\sin \theta \right)...\left(i\right) \\ \mathrm{and}, \\ y=2\sin \theta -\sin 2\theta \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dy}{d\theta }=2\cos \theta -\cos 2\theta \frac{d}{d\theta }\left(2\theta \right) \\ \Rightarrow \frac{dy}{d\theta }=2\cos \theta -\cos 2\theta \left(2\right) \\ \Rightarrow \frac{dy}{d\theta }=2\cos \theta -2\cos 2\theta \\ \Rightarrow \frac{dy}{d\theta }=2\left(\cos \theta -\cos 2\theta \right)...\left(\mathrm{ii}\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{by}\mathrm{equation}\left(\mathrm{i}\right), \\ \frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}=\frac{2\left(\cos \theta -\cos 2\theta \right)}{2\left(\sin 2\theta -\sin \theta \right)} \\ \Rightarrow \frac{dy}{dx}=\frac{\cos \theta -\cos 2\theta }{\sin 2\theta -\sin \theta } \\ \Rightarrow \frac{dy}{dx}=\frac{-2\sin \left({\displaystyle \frac{\theta +2\theta }{2}}\right)\sin \left({\displaystyle \frac{\theta -2\theta }{2}}\right)}{2\cos \left({\displaystyle \frac{2\theta +\theta }{2}}\right)\sin \left({\displaystyle \frac{2\theta -\theta }{2}}\right)}\left[\begin{array}{c}\because \sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)\\ \mathrm{and}\cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)\end{array}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{-\sin \left({\displaystyle \frac{3\theta }{2}}\right)\sin \left({\displaystyle \frac{-\theta }{2}}\right)}{\cos \left({\displaystyle \frac{3\theta }{2}}\right)\sin \left({\displaystyle \frac{\theta }{2}}\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{-\sin \left({\displaystyle \frac{3\theta }{2}}\right)\left(-\sin {\displaystyle \frac{\theta }{2}}\right)}{\cos \left({\displaystyle \frac{3\theta }{2}}\right)\sin \left({\displaystyle \frac{\theta }{2}}\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{\sin \left({\displaystyle \frac{3\theta }{2}}\right)}{\cos \left({\displaystyle \frac{3\theta }{2}}\right)} \\ \Rightarrow \frac{dy}{dx}=\tan \left(\frac{3\theta }{2}\right) \end{gathered}$$