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 will calculate the areas \( A_1 \) and \( A_2 \) as defined in the question.

### Step 1: Calculate \( A_1 \)

The area \( A_1 \) is bounded by the curve \( y = \cos x \) from \( x = 0 \) to \( x = \frac{\pi}{3} \). We can find this area using integration:

\[ A_1 = \int_{0}^{\frac{\pi}{3}} \cos x \, dx \]

### Step 2: Integrate \( \cos x \)

The integral of \( \cos x \) is \( \sin x \). Therefore, we have:

\[ A_1 = \left[ \sin x \right]_{0}^{\frac{\pi}{3}} \]

### Step 3: Evaluate the limits

Now, we will evaluate the limits:

\[ A_1 = \sin\left(\frac{\pi}{3}\right) - \sin(0) \]

We know that \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \) and \( \sin(0) = 0 \). Thus:

\[ A_1 = \frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{2} \]

### Step 4: Calculate \( A_2 \)

Next, we calculate the area \( A_2 \) which is bounded by the curve \( y = \cos 2x \) from \( x = 0 \) to \( x = \frac{\pi}{3} \):

\[ A_2 = \int_{0}^{\frac{\pi}{3}} \cos 2x \, dx \]

### Step 5: Integrate \( \cos 2x \)

The integral of \( \cos 2x \) is \( \frac{1}{2} \sin 2x \). Therefore, we have:

\[ A_2 = \left[ \frac{1}{2} \sin 2x \right]_{0}^{\frac{\pi}{3}} \]

### Step 6: Evaluate the limits

Now, we will evaluate the limits:

\[ A_2 = \frac{1}{2} \sin\left(2 \cdot \frac{\pi}{3}\right) - \frac{1}{2} \sin(0) \]

We know that \( \sin(0) = 0 \) and \( \sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \). Thus:

\[ A_2 = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{4} \]

### Step 7: Find the ratio \( \frac{A_1}{A_2} \)

Now we can find the ratio of the areas \( A_1 \) and \( A_2 \):

\[ \frac{A_1}{A_2} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{4}} = \frac{\sqrt{3}}{2} \cdot \frac{4}{\sqrt{3}} = \frac{4}{2} = 2 \]

### Final Answer

Thus, the areas are:

- \( A_1 = \frac{\sqrt{3}}{2} \)
- \( A_2 = \frac{\sqrt{3}}{4} \)
- The ratio \( \frac{A_1}{A_2} = 2 \)