`sum_(r=0)^(n) sin^(2)""(rpi)/(n)` is equal to

Options

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

Explanation

To solve the problem \( \sum_{r=0}^{n} \sin^2\left(\frac{r\pi}{n}\right) \), we can use a trigonometric identity and properties of summation. Here’s a step-by-step solution:

### Step 1: Use the identity for \(\sin^2\)
We know that:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]
So, we can rewrite the summation:
\[ \sum_{r=0}^{n} \sin^2\left(\frac{r\pi}{n}\right) = \sum_{r=0}^{n} \frac{1 - \cos\left(\frac{2r\pi}{n}\right)}{2} \]

### Step 2: Split the summation
This can be split into two separate summations:
\[ = \frac{1}{2} \sum_{r=0}^{n} 1 - \frac{1}{2} \sum_{r=0}^{n} \cos\left(\frac{2r\pi}{n}\right) \]
The first summation simplifies to:
\[ \sum_{r=0}^{n} 1 = n + 1 \]
Thus, we have:
\[ = \frac{1}{2}(n + 1) - \frac{1}{2} \sum_{r=0}^{n} \cos\left(\frac{2r\pi}{n}\right) \]

### Step 3: Evaluate the cosine summation
The summation \( \sum_{r=0}^{n} \cos\left(\frac{2r\pi}{n}\right) \) can be evaluated using the formula for the sum of cosines:
\[ \sum_{r=0}^{n} \cos\left(\frac{2r\pi}{n}\right) = \frac{\sin\left(\frac{(n+1)\pi}{n}\right)}{\sin\left(\frac{\pi}{n}\right)} \]
Since \( \sin\left(\frac{(n+1)\pi}{n}\right) = \sin(\pi) = 0 \), we find:
\[ \sum_{r=0}^{n} \cos\left(\frac{2r\pi}{n}\right) = 0 \]

### Step 4: Substitute back into the equation
Substituting back, we have:
\[ \sum_{r=0}^{n} \sin^2\left(\frac{r\pi}{n}\right) = \frac{1}{2}(n + 1) - \frac{1}{2}(0) = \frac{n + 1}{2} \]

### Final Result
Thus, the final result is:
\[ \sum_{r=0}^{n} \sin^2\left(\frac{r\pi}{n}\right) = \frac{n + 1}{2} \]

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