If `S=cos^2(pi/n)+cos^2((2pi)/(n))+....+cos^2(((n-1)pi)/(n)),` then S equals

Options

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

Answer

To solve the problem \( S = \cos^2\left(\frac{\pi}{n}\right) + \cos^2\left(\frac{2\pi}{n}\right) + \cdots + \cos^2\left(\frac{(n-1)\pi}{n}\right) \), we can follow these steps:

### Step 1: Rewrite the expression for S
We can express \( S \) as:
\[ S = \sum_{k=1}^{n-1} \cos^2\left(\frac{k\pi}{n}\right) \]

### Step 2: Use the identity for cosine squared
Using the identity \( \cos^2 x = \frac{1 + \cos(2x)}{2} \), we can rewrite each term in the sum:
\[ S = \sum_{k=1}^{n-1} \frac{1 + \cos\left(\frac{2k\pi}{n}\right)}{2} \]
This simplifies to:
\[ S = \frac{1}{2} \sum_{k=1}^{n-1} \left(1 + \cos\left(\frac{2k\pi}{n}\right)\right) \]

### Step 3: Separate the summation
We can separate the sum into two parts:
\[ S = \frac{1}{2} \left( \sum_{k=1}^{n-1} 1 + \sum_{k=1}^{n-1} \cos\left(\frac{2k\pi}{n}\right) \right) \]
The first sum, \( \sum_{k=1}^{n-1} 1 \), counts the number of terms, which is \( n-1 \):
\[ \sum_{k=1}^{n-1} 1 = n - 1 \]

### Step 4: Evaluate the cosine sum
The second sum, \( \sum_{k=1}^{n-1} \cos\left(\frac{2k\pi}{n}\right) \), can be evaluated using the formula for the sum of cosines:
\[ \sum_{k=0}^{n-1} \cos\left(\frac{2k\pi}{n}\right) = 0 \]
Thus, we have:
\[ \sum_{k=1}^{n-1} \cos\left(\frac{2k\pi}{n}\right) = -1 \quad \text{(since we are excluding the term for } k=0\text{)} \]

### Step 5: Substitute back into S
Now substituting back into our expression for \( S \):
\[ S = \frac{1}{2} \left( (n - 1) - 1 \right) = \frac{1}{2} (n - 2) \]

### Step 6: Final simplification
Thus, we can simplify \( S \) to:
\[ S = \frac{n}{2} \]

### Conclusion
Therefore, the final result is:
\[ S = \frac{n}{2} \]