Find `(dy)/(dx)` if x= 3 cos theta - cos 2theta and y= sin theta - sin 2theta.`

Explanation

To find \(\frac{dy}{dx}\) given \(x = 3 \cos \theta - \cos 2\theta\) and \(y = \sin \theta - \sin 2\theta\), we will use the chain rule. We will first find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\), and then use these to find \(\frac{dy}{dx}\).

### Step 1: Differentiate \(x\) with respect to \(\theta\)

Given:
\[ x = 3 \cos \theta - \cos 2\theta \]

Differentiating \(x\) with respect to \(\theta\):
\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(3 \cos \theta) - \frac{d}{d\theta}(\cos 2\theta) \]
Using the derivative of \(\cos\):
\[ \frac{d}{d\theta}(\cos \theta) = -\sin \theta \quad \text{and} \quad \frac{d}{d\theta}(\cos 2\theta) = -\sin 2\theta \cdot 2 \]
So,
\[ \frac{dx}{d\theta} = -3 \sin \theta + 2 \sin 2\theta \]

### Step 2: Differentiate \(y\) with respect to \(\theta\)

Given:
\[ y = \sin \theta - \sin 2\theta \]

Differentiating \(y\) with respect to \(\theta\):
\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(\sin \theta) - \frac{d}{d\theta}(\sin 2\theta) \]
Using the derivative of \(\sin\):
\[ \frac{d}{d\theta}(\sin \theta) = \cos \theta \quad \text{and} \quad \frac{d}{d\theta}(\sin 2\theta) = \cos 2\theta \cdot 2 \]
So,
\[ \frac{dy}{d\theta} = \cos \theta - 2 \cos 2\theta \]

### Step 3: Find \(\frac{dy}{dx}\)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{\cos \theta - 2 \cos 2\theta}{-3 \sin \theta + 2 \sin 2\theta} \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\cos \theta - 2 \cos 2\theta}{-3 \sin \theta + 2 \sin 2\theta} \]

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