If \(x = 3\cos\theta - 2\cos^{3}\theta\) and \(y = 3\sin\theta - 2\sin^{3}\theta\), where \(\sin\theta \ne 0\) and \(\cos 2\theta \ne 0\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\cot\theta\)
Differentiating each with respect to \(\theta\) and taking out the common factor:
\( \displaystyle \frac{dx}{d\theta} = -3\sin\theta + 6\cos^{2}\theta\sin\theta = 3\sin\theta(2\cos^{2}\theta - 1) \)
\( \displaystyle \frac{dy}{d\theta} = 3\cos\theta - 6\sin^{2}\theta\cos\theta = 3\cos\theta(1 - 2\sin^{2}\theta) \)
Using the double-angle formula \(\cos 2\theta = 2\cos^{2}\theta - 1 = 1 - 2\sin^{2}\theta\):
\( \displaystyle \frac{dx}{d\theta} = 3\sin\theta\cos 2\theta, \quad \frac{dy}{d\theta} = 3\cos\theta\cos 2\theta \)
In parametric form:
\( \displaystyle \frac{dy}{dx} = \frac{3\cos\theta\cos 2\theta}{3\sin\theta\cos 2\theta} = \cot\theta \)