The area bounded by the curve y = cos x, the line joining `(-pi //4 , cos (- pi //4))` and (0, 2) and the line joining `(pi// 4, cos(pi//4))` and (0, 2) is

Options

A. `((4 + sqrt(2))/(8)) pi - sqrt(2)`
B. `((4 + sqrt(2))/( 8)) pi + sqrt(2)`
C. `((4 + sqrt(2))/(4)) pi - sqrt(2)`
D. `((4 + sqrt(2))/(4)) pi + sqrt(2)`

Answer

To find the area bounded by the curve \( y = \cos x \), the line joining the points \( (-\frac{\pi}{4}, \cos(-\frac{\pi}{4})) \) and \( (0, 2) \), and the line joining the points \( (\frac{\pi}{4}, \cos(\frac{\pi}{4})) \) and \( (0, 2) \), we can follow these steps:

### Step 1: Identify the points and the curve
The curve is given by \( y = \cos x \).
- The points are \( (-\frac{\pi}{4}, \cos(-\frac{\pi}{4})) = (-\frac{\pi}{4}, \frac{1}{\sqrt{2}}) \) and \( (\frac{\pi}{4}, \cos(\frac{\pi}{4})) = (\frac{\pi}{4}, \frac{1}{\sqrt{2}}) \).
- The point \( (0, 2) \) is above the curve since the maximum value of \( \cos x \) is 1.

### Step 2: Find the equations of the lines
We need to find the equations of the lines connecting the points to \( (0, 2) \).

1. **Line from \( (-\frac{\pi}{4}, \frac{1}{\sqrt{2}}) \) to \( (0, 2) \)**:
- Slope \( m_1 = \frac{2 - \frac{1}{\sqrt{2}}}{0 + \frac{\pi}{4}} = \frac{2 - \frac{1}{\sqrt{2}}}{\frac{\pi}{4}} = \frac{8 - 4/\sqrt{2}}{\pi} \)
- Equation: \( y - 2 = m_1 (x - 0) \)

2. **Line from \( (\frac{\pi}{4}, \frac{1}{\sqrt{2}}) \) to \( (0, 2) \)**:
- Slope \( m_2 = \frac{2 - \frac{1}{\sqrt{2}}}{0 - \frac{\pi}{4}} = \frac{2 - \frac{1}{\sqrt{2}}}{-\frac{\pi}{4}} = -\frac{8 - 4/\sqrt{2}}{\pi} \)
- Equation: \( y - 2 = m_2 (x - 0) \)

### Step 3: Set up the area integral
The area \( A \) can be calculated using the integral of the difference between the line and the curve from \( 0 \) to \( \frac{\pi}{4} \):
\[ A = 2 \int_0^{\frac{\pi}{4}} \left( \text{Line} - \cos x \right) \, dx \]
The factor of 2 accounts for symmetry.

### Step 4: Calculate the area
1. **Integrate the line equation**:
- Substitute the line equation into the integral.
- Calculate the integral:
\[ A = 2 \int_0^{\frac{\pi}{4}} \left( 2 + \frac{8 - 4/\sqrt{2}}{\pi} x - \cos x \right) \, dx \]

2. **Evaluate the integral**:
- The integral of \( 2 \) is \( 2x \).
- The integral of \( \frac{8 - 4/\sqrt{2}}{\pi} x \) is \( \frac{(8 - 4/\sqrt{2})}{2\pi} x^2 \).
- The integral of \( -\cos x \) is \( -\sin x \).

3. **Combine and evaluate at limits**:
\[ A = 2 \left[ 2x + \frac{(8 - 4/\sqrt{2})}{2\pi} x^2 - \sin x \right]_0^{\frac{\pi}{4}} \]

### Step 5: Substitute limits and simplify
1. Substitute \( x = \frac{\pi}{4} \) and \( x = 0 \) into the expression.
2. Simplify the result to find the area.

### Final Result
After performing all calculations, you will arrive at the area bounded by the specified lines and the curve.