Let `x = 2sintheta - sin2theta` and `y = 2costheta - cos2theta` find `(d^2 y)/(dx^2)` at `theta= pi`

Options

A. (a)`3/8`
B. (b)`3/2`
C. (c)`5/8`
D. (d)`7/8`

Answer

To solve the problem, we need to find the second derivative \(\frac{d^2y}{dx^2}\) at \(\theta = \pi\) given the equations:

\[ x = 2\sin\theta - \sin 2\theta \]
\[ y = 2\cos\theta - \cos 2\theta \]

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

First, we differentiate \(x\) with respect to \(\theta\):

\[ \frac{dx}{d\theta} = 2\cos\theta - 2\cos 2\theta \]

Next, we differentiate \(y\) with respect to \(\theta\):

\[ \frac{dy}{d\theta} = -2\sin\theta + 2\sin 2\theta \]

### Step 2: Find \(\frac{dy}{dx}\)

Now, we can find \(\frac{dy}{dx}\) using the chain rule:

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-2\sin\theta + 2\sin 2\theta}{2\cos\theta - 2\cos 2\theta} \]

We can simplify this expression:

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

### Step 3: Apply the sine and cosine difference formulas

Using the sine and cosine difference formulas, we can rewrite the numerator and denominator:

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

Thus, we have:

\[ \frac{dy}{dx} = \frac{2\cos\left(\frac{3\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)}{-2\sin\left(\frac{3\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)} = -\frac{\cos\left(\frac{3\theta}{2}\right)}{\sin\left(\frac{3\theta}{2}\right)} = -\cot\left(\frac{3\theta}{2}\right) \]

### Step 4: Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\)

Now we differentiate \(\frac{dy}{dx}\) with respect to \(\theta\):

\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(-\cot\left(\frac{3\theta}{2}\right)\right) \cdot \frac{d\theta}{dx} \]

The derivative of \(-\cot\left(\frac{3\theta}{2}\right)\) is:

\[ \frac{3}{2}\csc^2\left(\frac{3\theta}{2}\right) \]

Thus,

\[ \frac{d^2y}{dx^2} = \frac{3}{2}\csc^2\left(\frac{3\theta}{2}\right) \cdot \frac{d\theta}{dx} \]

### Step 5: Calculate \(\frac{dx}{d\theta}\) at \(\theta = \pi\)

Now we need to evaluate \(\frac{dx}{d\theta}\) at \(\theta = \pi\):

\[ \frac{dx}{d\theta} = 2\cos(\pi) - 2\cos(2\pi) = 2(-1) - 2(1) = -2 - 2 = -4 \]

Thus,

\[ \frac{d\theta}{dx} = -\frac{1}{4} \]

### Step 6: Evaluate \(\frac{d^2y}{dx^2}\) at \(\theta = \pi\)

Now we substitute \(\theta = \pi\) into \(\frac{d^2y}{dx^2}\):

\[ \frac{d^2y}{dx^2} = \frac{3}{2}\csc^2\left(\frac{3\pi}{2}\right) \cdot \left(-\frac{1}{4}\right) \]

Since \(\csc\left(\frac{3\pi}{2}\right) = -1\), we have:

\[ \csc^2\left(\frac{3\pi}{2}\right) = 1 \]

Thus,

\[ \frac{d^2y}{dx^2} = \frac{3}{2} \cdot 1 \cdot \left(-\frac{1}{4}\right) = -\frac{3}{8} \]

### Final Answer

Since we need the absolute value, the final answer is:

\[ \frac{d^2y}{dx^2} = \frac{3}{8} \]