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

Options

A. `n/2`
B. `(n-1)/(2)`
C. `n/2`
D. none of these

Answer

To find the value of the expression

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

we can follow these steps:

### Step 1: Use the identity for cosine squared
We know that

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

Using this identity, we can rewrite the summation:

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

### Step 2: Simplify the summation
We can factor out the \( \frac{1}{2} \):

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

### Step 3: Evaluate the first summation
The first summation \( \sum_{r=1}^{n-1} 1 \) counts the number of terms from 1 to \( n-1 \), which is \( n-1 \):

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

### Step 4: Evaluate the second summation
The second summation \( \sum_{r=1}^{n-1} \cos\left(\frac{2r\pi}{n}\right) \) can be evaluated using the formula for the sum of cosines:

\[ \sum_{r=0}^{n-1} \cos\left(\alpha + r\beta\right) = \frac{\sin\left(\frac{n\beta}{2}\right) \cos\left(\alpha + \frac{(n-1)\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}. \]

In our case, \( \alpha = 0 \) and \( \beta = \frac{2\pi}{n} \):

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

Thus,

\[ \sum_{r=1}^{n-1} \cos\left(\frac{2r\pi}{n}\right) = -1 \quad (\text{since we need to exclude } r=0). \]

### Step 5: Combine the results
Now substituting back into our expression:

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

### Final Result
Thus, we have:

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