`sum_(r=1)^(n-1)sin^(2)"" (r pi )/( n ) ` equals

Options

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

Answer

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

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

### Step 1: Rewrite the Sine Square Function
We can use the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \). Thus, we can rewrite our sum as:

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

### Step 2: Factor Out the Constant
Now, we can factor out the constant \( \frac{1}{2} \):

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

### Step 3: Separate the Sum
We can separate the sum into two parts:

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

### Step 4: Evaluate the First Sum
The first sum \( \sum_{r=1}^{n-1} 1 \) simply counts the number of terms from 1 to \( n-1 \):

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

### Step 5: Evaluate the Second Sum
The second sum \( \sum_{r=1}^{n-1} \cos\left(\frac{2r \pi}{n}\right) \) can be evaluated using the formula for the sum of cosines:

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

because \( \sin\left(\frac{(n-1) \pi}{n}\right) = \sin\left(\pi - \frac{\pi}{n}\right) = \sin\left(\frac{\pi}{n}\right) \).

### Step 6: Combine the Results
Putting it all together, we have:

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

### Final Result
Thus, the value of the sum is:

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