If `A_(1)` is the area bounded by the curve y = cos x and `A_(2)` is the area bounded by y = cos 2 x along with x - axis from x = 0 and `x = (pi)/(0)`, then

Options

A. `A_(1) = (sqrt(3))/(2), A_(2) = (sqrt(3))/(4)`
B. `A_(1) = (sqrt(3))/(2), A_(2) = (1)/(2)`
C. `(A_(1))/(A_(2)) = (2)/(1)`
D. `(A_(1))/(A_(2)) = (sqrt(3))/(1)`

Answer

To solve the problem, we need to find the areas \( A_1 \) and \( A_2 \) bounded by the curves \( y = \cos x \) and \( y = \cos 2x \) along with the x-axis from \( x = 0 \) to \( x = \frac{\pi}{3} \).

### Step-by-Step Solution:

**Step 1: Calculate \( A_1 \)**
We start with the area \( A_1 \) which is bounded by the curve \( y = \cos x \).

The area \( A_1 \) can be calculated using the definite integral:
\[ A_1 = \int_{0}^{\frac{\pi}{3}} \cos x \, dx \]

**Step 2: Integrate \( \cos x \)**
The integral of \( \cos x \) is \( \sin x \). So we evaluate:
\[ A_1 = \left[ \sin x \right]_{0}^{\frac{\pi}{3}} = \sin\left(\frac{\pi}{3}\right) - \sin(0) \]

**Step 3: Substitute the values**
We know that \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \) and \( \sin(0) = 0 \). Therefore:
\[ A_1 = \frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{2} \]

**Step 4: Calculate \( A_2 \)**
Next, we find the area \( A_2 \) which is bounded by the curve \( y = \cos 2x \).

The area \( A_2 \) can be calculated using the definite integral:
\[ A_2 = \int_{0}^{\frac{\pi}{3}} \cos 2x \, dx \]

**Step 5: Integrate \( \cos 2x \)**
The integral of \( \cos 2x \) is \( \frac{1}{2} \sin 2x \). So we evaluate:
\[ A_2 = \left[ \frac{1}{2} \sin 2x \right]_{0}^{\frac{\pi}{3}} = \frac{1}{2} \sin\left(2 \cdot \frac{\pi}{3}\right) - \frac{1}{2} \sin(0) \]

**Step 6: Substitute the values**
We know that \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \) and \( \sin(0) = 0 \). Therefore:
\[ A_2 = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{4} \]

### Final Results:
- \( A_1 = \frac{\sqrt{3}}{2} \)
- \( A_2 = \frac{\sqrt{3}}{4} \)