Question

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

Answer 1

\[\text{ We have, x } = 2 \cos\theta - \cos2\theta\]

\[\Rightarrow \frac{dx}{d\theta} = 2\left( - \sin\theta \right) - \left( - \sin2\theta \right)\frac{d}{d\theta}\left( 2\theta \right)\]

\[ \Rightarrow \frac{dx}{d\theta} = - 2\sin\theta + 2 \sin2\theta\]

\[ \Rightarrow \frac{dx}{d\theta} = 2\left( \sin2\theta - \sin\theta \right) . . . \left( i \right)\]

\[\text{ and }, \]

\[y = 2 \sin\theta - \sin2\theta\]

\[\Rightarrow \frac{dy}{d\theta} = 2 \cos\theta - \cos2\theta\frac{d}{d\theta}\left( 2\theta \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = 2 \cos\theta - \cos2\theta\left( 2 \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = 2 \cos\theta - 2 \cos2\theta\]
\[ \Rightarrow \frac{dy}{d\theta} = 2\left( \cos\theta - \cos2\theta \right) . . . \left( ii \right)\]
\[\text{ Dividing equation } \left( ii \right) \text{ by equation } \left( i \right), \]
\[\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2\left( \cos\theta - \cos2\theta \right)}{2\left( \sin2\theta - \sin\theta \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\cos\theta - \cos2\theta}{\sin2\theta - \sin\theta}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 2\sin\left( \frac{\theta + 2\theta}{2} \right)\sin\left( \frac{\theta - 2\theta}{2} \right)}{2\cos\left( \frac{2\theta + \theta}{2} \right)\sin\left( \frac{2\theta - \theta}{2} \right)} ............[\because \sin A - \sin B = 2 \cos\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right)\text{ and } \cos A - \cos B = - 2\sin\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right)]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- \sin\left( \frac{3\theta}{2} \right)\sin\left( \frac{- \theta}{2} \right)}{\cos\left( \frac{3\theta}{2} \right)\sin\left( \frac{\theta}{2} \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- \sin\left( \frac{3\theta}{2} \right)\left( - \sin\frac{\theta}{2} \right)}{\cos\left( \frac{3\theta}{2} \right)\sin\left( \frac{\theta}{2} \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin\left( \frac{3\theta}{2} \right)}{\cos\left( \frac{3\theta}{2} \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \tan\left( \frac{3\theta}{2} \right)\]