`underset(r=1)overset(n-1)(sum)cos^(2)""(rpi)/(n)` is equal to

Options

A. `n/2`
B. `(n-1)/2`
C. `n/2-1`
D. `(n+1)/2`

Answer

To solve the problem, we need to evaluate the summation:

\[ \frac{1}{n} \sum_{r=1}^{n-1} \cos^2\left(\frac{r\pi}{n}\right) \]

### Step 1: Rewrite \(\cos^2\) using the double angle formula
We know the identity:

\[ \cos^2 a = \frac{1 + \cos(2a)}{2} \]

Applying this to our summation:

\[ \cos^2\left(\frac{r\pi}{n}\right) = \frac{1 + \cos\left(\frac{2r\pi}{n}\right)}{2} \]

### Step 2: Substitute into the summation
Now we can substitute this back into the summation:

\[ \sum_{r=1}^{n-1} \cos^2\left(\frac{r\pi}{n}\right) = \sum_{r=1}^{n-1} \frac{1 + \cos\left(\frac{2r\pi}{n}\right)}{2} \]

This simplifies to:

\[ \sum_{r=1}^{n-1} \cos^2\left(\frac{r\pi}{n}\right) = \frac{1}{2} \sum_{r=1}^{n-1} 1 + \frac{1}{2} \sum_{r=1}^{n-1} \cos\left(\frac{2r\pi}{n}\right) \]

### Step 3: Evaluate the first summation
The first summation is straightforward:

\[ \sum_{r=1}^{n-1} 1 = n - 1 \]

### Step 4: Evaluate the second summation
The second summation involves the cosine function. We use the formula for the sum of cosines:

\[ \sum_{r=0}^{n-1} \cos\left(\frac{2r\pi}{n}\right) = 0 \]

Thus,

\[ \sum_{r=1}^{n-1} \cos\left(\frac{2r\pi}{n}\right) = -1 \quad \text{(since we are missing the term for } r=0\text{)} \]

### Step 5: Combine the results
Now substituting back, we have:

\[ \sum_{r=1}^{n-1} \cos^2\left(\frac{r\pi}{n}\right) = \frac{1}{2} (n - 1) + \frac{1}{2} (-1) = \frac{n - 2}{2} \]

### Step 6: Final result
Now we divide by \(n\):

\[ \frac{1}{n} \sum_{r=1}^{n-1} \cos^2\left(\frac{r\pi}{n}\right) = \frac{1}{n} \cdot \frac{n - 2}{2} = \frac{n - 2}{2n} \]

Thus, the final answer is:

\[ \frac{n - 2}{2n} \]