Find `(d^(2)y)/(dx^(2))` when :
`x=2costheta-cos2thetaandy=2sintheta-sin2theta`

Answer

To find the second derivative \(\frac{d^2y}{dx^2}\) given the parametric equations \(x = 2\cos\theta - \cos 2\theta\) and \(y = 2\sin\theta - \sin 2\theta\), we will follow these steps:

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

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

Differentiating \(x\) with respect to \(\theta\):
\[ \frac{dx}{d\theta} = -2\sin\theta + 2\sin 2\theta \]

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

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

Differentiating \(y\) with respect to \(\theta\):
\[ \frac{dy}{d\theta} = 2\cos\theta - 2\cos 2\theta \]

### 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 \(\frac{dy}{dx}\)

Factoring out the common terms:
\[ \frac{dy}{dx} = \frac{2(\cos\theta - \cos 2\theta)}{2(-\sin\theta + \sin 2\theta)} = \frac{\cos\theta - \cos 2\theta}{-\sin\theta + \sin 2\theta} \]

### Step 5: Differentiate \(\frac{dy}{dx}\) with respect to \(\theta\)

Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta}} \]

Let \(u = \cos\theta - \cos 2\theta\) and \(v = -\sin\theta + \sin 2\theta\):
\[ \frac{d^2y}{dx^2} = \frac{v\frac{du}{d\theta} - u\frac{dv}{d\theta}}{v^2} \cdot \frac{1}{\frac{dx}{d\theta}} \]

### Step 6: Find \(\frac{du}{d\theta}\) and \(\frac{dv}{d\theta}\)

Calculating:
\[ \frac{du}{d\theta} = -\sin\theta + 2\sin 2\theta \]
\[ \frac{dv}{d\theta} = -\cos\theta + 2\cos 2\theta \]

### Step 7: Substitute back into the formula

Substituting \(u\), \(v\), \(\frac{du}{d\theta}\), and \(\frac{dv}{d\theta}\) into the expression for \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{(-\sin\theta + 2\sin 2\theta)(-\sin\theta + \sin 2\theta) - (\cos\theta - \cos 2\theta)(-\cos\theta + 2\cos 2\theta)}{(-\sin\theta + \sin 2\theta)^2} \cdot \frac{1}{\frac{dx}{d\theta}} \]

### Step 8: Final expression

After simplification, we can express \(\frac{d^2y}{dx^2}\) in terms of \(\theta\).