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 \( \cos^2(\phi/2) \) over one complete cycle (from \( 0 \) to \( 2\pi \)), we can follow these steps:

### Step 1: Set up the average value formula
The average value \( \langle f \rangle \) of a function \( f(\phi) \) over the interval from \( a \) to \( b \) is given by:
\[ \langle f \rangle = \frac{1}{b-a} \int_a^b f(\phi) \, d\phi \]
In this case, \( a = 0 \), \( b = 2\pi \), and \( f(\phi) = \cos^2(\phi/2) \).

### Step 2: Write the integral for the average value
Substituting the values into the formula, we get:
\[ \langle \cos^2(\phi/2) \rangle = \frac{1}{2\pi} \int_0^{2\pi} \cos^2(\phi/2) \, d\phi \]

### Step 3: Use the trigonometric identity
To simplify \( \cos^2(\phi/2) \), we can use the identity:
\[ \cos^2(x) = \frac{1 + \cos(2x)}{2} \]
Applying this to our function:
\[ \cos^2(\phi/2) = \frac{1 + \cos(\phi)}{2} \]

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

### Step 5: Evaluate the integral
Now we can evaluate the integral:
\[ \int_0^{2\pi} (1 + \cos(\phi)) \, d\phi = \int_0^{2\pi} 1 \, d\phi + \int_0^{2\pi} \cos(\phi) \, d\phi \]
The first integral is:
\[ \int_0^{2\pi} 1 \, d\phi = 2\pi \]
The second integral is:
\[ \int_0^{2\pi} \cos(\phi) \, d\phi = 0 \]
Thus:
\[ \int_0^{2\pi} (1 + \cos(\phi)) \, d\phi = 2\pi + 0 = 2\pi \]

### Step 6: Substitute back into the average value formula
Now substituting back into the average value formula:
\[ \langle \cos^2(\phi/2) \rangle = \frac{1}{4\pi} \cdot 2\pi = \frac{1}{2} \]

### Final Result
The average value of \( \cos^2(\phi/2) \) in one cycle is:
\[ \boxed{\frac{1}{2}} \]