Evaluate ` sum_(r=1)^(n-1) cos^(2) ((rpi)/(n))`.

Answer

To evaluate the sum \( S = \sum_{r=1}^{n-1} \cos^2\left(\frac{r\pi}{n}\right) \), we can follow these steps:

### Step 1: Use the identity for \(\cos^2\theta\)
We know the identity:
\[ \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \]
Applying this identity to our sum:
\[ S = \sum_{r=1}^{n-1} \cos^2\left(\frac{r\pi}{n}\right) = \sum_{r=1}^{n-1} \frac{1 + \cos\left(\frac{2r\pi}{n}\right)}{2} \]

### Step 2: Split the sum
We can split the sum into two parts:
\[ S = \frac{1}{2} \sum_{r=1}^{n-1} 1 + \frac{1}{2} \sum_{r=1}^{n-1} \cos\left(\frac{2r\pi}{n}\right) \]

### Step 3: Evaluate the first sum
The first sum is straightforward:
\[ \sum_{r=1}^{n-1} 1 = n - 1 \]
Thus,
\[ \frac{1}{2} \sum_{r=1}^{n-1} 1 = \frac{n-1}{2} \]

### Step 4: Evaluate the second sum
Now we need to evaluate the second sum:
\[ \sum_{r=1}^{n-1} \cos\left(\frac{2r\pi}{n}\right) \]
This is a sum of cosines in an arithmetic progression. The sum can be evaluated using the formula:
\[ \sum_{r=0}^{m-1} \cos(a + r \cdot d) = \frac{\sin\left(\frac{md}{2}\right) \cdot \cos\left(a + \frac{(m-1)d}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]
In our case, \(a = \frac{2\pi}{n}\), \(d = \frac{2\pi}{n}\), and \(m = n-1\):
\[ \sum_{r=1}^{n-1} \cos\left(\frac{2r\pi}{n}\right) = \frac{\sin\left(\frac{(n-1)\cdot \frac{2\pi}{n}}{2}\right) \cdot \cos\left(\frac{2\pi}{n} + \frac{(n-2)\cdot \frac{2\pi}{n}}{2}\right)}{\sin\left(\frac{\frac{2\pi}{n}}{2}\right)} \]

### Step 5: Simplify the second sum
Calculating the sine and cosine:
\[ \sin\left(\frac{(n-1)\cdot \frac{2\pi}{n}}{2}\right) = \sin\left(\frac{(n-1)\pi}{n}\right) = \sin\left(\pi - \frac{\pi}{n}\right) = \sin\left(\frac{\pi}{n}\right) \]
And,
\[ \cos\left(\frac{2\pi}{n} + \frac{(n-2)\cdot \frac{2\pi}{n}}{2}\right) = \cos\left(\frac{2\pi}{n} + \frac{(n-2)\pi}{n}\right) = \cos\left(\frac{n\pi}{n}\right) = \cos(\pi) = -1 \]
Thus, we have:
\[ \sum_{r=1}^{n-1} \cos\left(\frac{2r\pi}{n}\right) = \frac{\sin\left(\frac{\pi}{n}\right) \cdot (-1)}{\sin\left(\frac{\pi}{n}\right)} = -1 \]

### Step 6: Combine the results
Putting it all together:
\[ S = \frac{n-1}{2} + \frac{1}{2}(-1) = \frac{n-1}{2} - \frac{1}{2} = \frac{n-2}{2} \]

### Final Answer
Thus, the evaluated sum is:
\[ S = \frac{n-2}{2} \]

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