The average value of `cos^2(phi/2)` in one cycle is

Options

A. Zero
B. `1/3`
C. `1/2`
D. `1/4`

Answer

To find the average value of \( \cos^2\left(\frac{\phi}{2}\right) \) over one complete cycle (from \( 0 \) to \( 2\pi \)), we can follow these steps:

### Step 1: Set up the formula for average value
The average value \( A \) of a function \( f(\phi) \) over the interval \( [a, b] \) is given by:
\[ A = \frac{1}{b-a} \int_a^b f(\phi) \, d\phi \]
In our case, \( f(\phi) = \cos^2\left(\frac{\phi}{2}\right) \), and we will integrate from \( 0 \) to \( 2\pi \).

### Step 2: Write the integral
Thus, the average value becomes:
\[ A = \frac{1}{2\pi - 0} \int_0^{2\pi} \cos^2\left(\frac{\phi}{2}\right) \, d\phi \]
This simplifies to:
\[ A = \frac{1}{2\pi} \int_0^{2\pi} \cos^2\left(\frac{\phi}{2}\right) \, d\phi \]

### Step 3: Use the trigonometric identity
We can use the trigonometric identity:
\[ \cos^2 x = \frac{1 + \cos(2x)}{2} \]
Applying this identity, we have:
\[ \cos^2\left(\frac{\phi}{2}\right) = \frac{1 + \cos(\phi)}{2} \]

### Step 4: Substitute the identity into the integral
Now substituting this into the integral:
\[ A = \frac{1}{2\pi} \int_0^{2\pi} \frac{1 + \cos(\phi)}{2} \, d\phi \]
This simplifies to:
\[ A = \frac{1}{4\pi} \int_0^{2\pi} (1 + \cos(\phi)) \, d\phi \]

### Step 5: Split the integral
We can split the integral into two parts:
\[ A = \frac{1}{4\pi} \left( \int_0^{2\pi} 1 \, d\phi + \int_0^{2\pi} \cos(\phi) \, d\phi \right) \]

### Step 6: Evaluate the integrals
1. The first integral:
\[ \int_0^{2\pi} 1 \, d\phi = 2\pi \]
2. The second integral:
\[ \int_0^{2\pi} \cos(\phi) \, d\phi = 0 \]
(since the integral of cosine over one complete cycle is zero).

### Step 7: Substitute back into the average formula
Now substituting these values back:
\[ A = \frac{1}{4\pi} \left( 2\pi + 0 \right) = \frac{2\pi}{4\pi} = \frac{1}{2} \]

### Final Answer
Thus, the average value of \( \cos^2\left(\frac{\phi}{2}\right) \) over one cycle is:
\[ \boxed{\frac{1}{2}} \]