The value of `sum_(r=1)^(8)(sin((2rpi)/9)+icos((2rpi)/9))`, is

Options

A. `-1`
B. 1
C. i
D. `-i`

Answer

To solve the problem \( \sum_{r=1}^{8} \left( \sin\left(\frac{2r\pi}{9}\right) + i \cos\left(\frac{2r\pi}{9}\right) \right) \), we can follow these steps:

### Step 1: Rewrite the Sum
We can rewrite the sum to include \( r = 0 \) for simplification:
\[ \sum_{r=1}^{8} \left( \sin\left(\frac{2r\pi}{9}\right) + i \cos\left(\frac{2r\pi}{9}\right) \right) = \sum_{r=0}^{8} \left( \sin\left(\frac{2r\pi}{9}\right) + i \cos\left(\frac{2r\pi}{9}\right) \right) - \left( \sin(0) + i \cos(0) \right) \]
This gives us:
\[ = \sum_{r=0}^{8} \left( \sin\left(\frac{2r\pi}{9}\right) + i \cos\left(\frac{2r\pi}{9}\right) \right) - i \]

### Step 2: Use Euler's Formula
Using Euler's formula, we have:
\[ \sin\theta + i \cos\theta = i \left( \cos\theta - i \sin\theta \right) = i e^{-i\theta} \]
Thus:
\[ \sin\left(\frac{2r\pi}{9}\right) + i \cos\left(\frac{2r\pi}{9}\right) = i e^{-i\frac{2r\pi}{9}} \]

### Step 3: Substitute into the Sum
Now substituting this back into our sum:
\[ \sum_{r=0}^{8} \left( \sin\left(\frac{2r\pi}{9}\right) + i \cos\left(\frac{2r\pi}{9}\right) \right) = i \sum_{r=0}^{8} e^{-i\frac{2r\pi}{9}} \]

### Step 4: Recognize the Geometric Series
The sum \( \sum_{r=0}^{8} e^{-i\frac{2r\pi}{9}} \) is a geometric series with:
- First term \( a = 1 \)
- Common ratio \( r = e^{-i\frac{2\pi}{9}} \)
- Number of terms \( n = 9 \)

The formula for the sum of a geometric series is:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
Substituting the values:
\[ S = \frac{1 \left( 1 - \left( e^{-i\frac{2\pi}{9}} \right)^9 \right)}{1 - e^{-i\frac{2\pi}{9}}} \]
Since \( \left( e^{-i\frac{2\pi}{9}} \right)^9 = e^{-i2\pi} = 1 \):
\[ S = \frac{1 - 1}{1 - e^{-i\frac{2\pi}{9}}} = 0 \]

### Step 5: Final Calculation
Thus, we have:
\[ \sum_{r=0}^{8} \left( \sin\left(\frac{2r\pi}{9}\right) + i \cos\left(\frac{2r\pi}{9}\right) \right) = i \cdot 0 = 0 \]
Finally, we subtract \( i \):
\[ \sum_{r=1}^{8} \left( \sin\left(\frac{2r\pi}{9}\right) + i \cos\left(\frac{2r\pi}{9}\right) \right) = 0 - i = -i \]

### Conclusion
The value of the sum is:
\[ \boxed{-i} \]