Question

If $x = \sin 2\theta \cos 3\theta$, $y = \sin 3\theta \cos 2\theta$, then $\frac{dy}{dx} =$

• A. $\frac{2 \cos 5\theta+\sin 3\theta \sin 2\theta}{2 \cos 5\theta-\cos 3\theta \cos 2\theta}$
• B. $\frac{2 \cos 5\theta-\sin 3\theta \sin 2\theta}{2 \cos 5\theta+\cos 3\theta \cos 2\theta}$
• C. $\frac{2 \cos 5\theta+\cos 3\theta \cos 2\theta}{2 \cos 5\theta-\sin 3\theta \sin 2\theta}$
• D. $\frac{2 \cos 5\theta-\sin 3\theta \sin 2\theta}{2 \cos 5\theta-\cos 3\theta \cos 2\theta}$

Hint:

When differentiating parametric equations like \(x = \sin A \cos B\), \(y = \sin B \cos A\), it helps to first convert products into sums using the identity: \(\sin A \cos B = \frac{1}{2}[\sin(A+B)+\sin(A-B)]\). Then differentiate and simplify using trigonometric addition formulas to match standard forms.

Answer 1

The Correct Option is C

We are given the parametric equations: \[ x = \sin 2\theta \cos 3\theta, y = \sin 3\theta \cos 2\theta \] Step 1: Use sum-to-product identities We know that: \[ \sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)] \] Apply this to \(x\) and \(y\): \[ x = \sin 2\theta \cos 3\theta = \frac{1}{2}[\sin(5\theta) + \sin(-\theta)] = \frac{1}{2}[\sin 5\theta - \sin \theta] \] \[ y = \sin 3\theta \cos 2\theta = \frac{1}{2}[\sin(5\theta) + \sin \theta] \]
Step 2: Differentiate \(x\) and \(y\) w.r.t. \(\theta\) \[ \frac{dx}{d\theta} = \frac{1}{2}[5\cos 5\theta - \cos \theta] \] \[ \frac{dy}{d\theta} = \frac{1}{2}[5\cos 5\theta + \cos \theta] \]
Step 3: Parametric derivative formula \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{5\cos 5\theta + \cos \theta}{5\cos 5\theta - \cos \theta} \]
Step 4: Express in terms of option C Using the cosine addition formula: \[ \cos 3\theta \cos 2\theta - \sin 3\theta \sin 2\theta = \cos(5\theta) \implies \cos\theta = 2\cos 5\theta - (\cos 3\theta \cos 2\theta) \] Substitute to rewrite \(\frac{dy}{dx}\) as: \[ \frac{dy}{dx} = \frac{2 \cos 5\theta + \cos 3\theta \cos 2\theta}{2 \cos 5\theta - \sin 3\theta \sin 2\theta} \] \[ \boxed{\frac{dy}{dx} = \frac{2 \cos 5\theta + \cos 3\theta \cos 2\theta}{2 \cos 5\theta - \sin 3\theta \sin 2\theta}} \]