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

Options

A. `tan""(3 theta)/(2)`
B. `- tan"" (3 theta)/(2)`
C. `cot"" (3 theta )/(2)`
D. `- cot ""(3 theta )/(2)`

Explanation

To find \(\frac{dy}{dx}\) given \(x = 2 \cos \theta - \cos 2\theta\) and \(y = 2 \sin \theta - \sin 2\theta\), we will differentiate \(x\) and \(y\) with respect to \(\theta\) and then use the chain rule to find \(\frac{dy}{dx}\).

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

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

Using the differentiation rules:
- The derivative of \(\cos \theta\) is \(-\sin \theta\).
- The derivative of \(\cos 2\theta\) is \(-\sin 2\theta \cdot 2\) (using the chain rule).

Thus,
\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(2 \cos \theta) - \frac{d}{d\theta}(\cos 2\theta) \]
\[ = 2(-\sin \theta) - (-\sin 2\theta \cdot 2) \]
\[ = -2 \sin \theta + 2 \sin 2\theta \]

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

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

Using the differentiation rules:
- The derivative of \(\sin \theta\) is \(\cos \theta\).
- The derivative of \(\sin 2\theta\) is \(\cos 2\theta \cdot 2\) (using the chain rule).

Thus,
\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(2 \sin \theta) - \frac{d}{d\theta}(\sin 2\theta) \]
\[ = 2 \cos \theta - (2 \cos 2\theta) \]
\[ = 2 \cos \theta - 2 \cos 2\theta \]

### 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:
\[ \frac{dy}{dx} = \frac{2 \cos \theta - 2 \cos 2\theta}{-2 \sin \theta + 2 \sin 2\theta} \]

### Step 4: Simplify the expression

We can factor out \(2\) from both the numerator and the denominator:
\[ \frac{dy}{dx} = \frac{2(\cos \theta - \cos 2\theta)}{2(-\sin \theta + \sin 2\theta)} \]
\[ = \frac{\cos \theta - \cos 2\theta}{-\sin \theta + \sin 2\theta} \]

### Step 5: Use trigonometric identities

Using the identity for \(\cos A - \cos B\):
\[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]

For \(\cos \theta - \cos 2\theta\):
\[ = -2 \sin\left(\frac{\theta + 2\theta}{2}\right) \sin\left(\frac{\theta - 2\theta}{2}\right) \]
\[ = -2 \sin\left(\frac{3\theta}{2}\right) \sin\left(-\frac{\theta}{2}\right) \]
\[ = 2 \sin\left(\frac{3\theta}{2}\right) \sin\left(\frac{\theta}{2}\right) \]

For \(-\sin \theta + \sin 2\theta\):
\[ = \sin 2\theta - \sin \theta = 2 \cos\left(\frac{2\theta + \theta}{2}\right) \sin\left(\frac{2\theta - \theta}{2}\right) \]
\[ = 2 \cos\left(\frac{3\theta}{2}\right) \sin\left(\frac{\theta}{2}\right) \]

### Final Result

Thus, we have:
\[ \frac{dy}{dx} = \frac{2 \sin\left(\frac{3\theta}{2}\right) \sin\left(\frac{\theta}{2}\right)}{2 \cos\left(\frac{3\theta}{2}\right) \sin\left(\frac{\theta}{2}\right)} \]
\[ = \frac{\sin\left(\frac{3\theta}{2}\right)}{\cos\left(\frac{3\theta}{2}\right)} = \tan\left(\frac{3\theta}{2}\right) \]

### Conclusion

Therefore, the final answer is:
\[ \frac{dy}{dx} = \tan\left(\frac{3\theta}{2}\right) \]