Value of expression `sin(pi/9)+sin((2pi)/9)+sin((3pi)/9)+...+sin((17pi)/9)=`

Options

A. 0
B. `-1`
C. 1
D. `-3//2`

Explanation

To solve the expression \( \sin\left(\frac{\pi}{9}\right) + \sin\left(\frac{2\pi}{9}\right) + \sin\left(\frac{3\pi}{9}\right) + \ldots + \sin\left(\frac{17\pi}{9}\right) \), we can use the properties of sine and some trigonometric identities.

### Step-by-step Solution:

1. **Identify the Terms**:
The expression consists of sine terms from \( \sin\left(\frac{\pi}{9}\right) \) to \( \sin\left(\frac{17\pi}{9}\right) \).

2. **Pairing the Terms**:
Notice that \( \sin\left(\frac{17\pi}{9}\right) \) can be paired with \( \sin\left(\frac{\pi}{9}\right) \), \( \sin\left(\frac{16\pi}{9}\right) \) with \( \sin\left(\frac{2\pi}{9}\right) \), and so on. The pairs can be written as:
\[ \sin\left(\frac{17\pi}{9}\right) + \sin\left(\frac{\pi}{9}\right), \quad \sin\left(\frac{16\pi}{9}\right) + \sin\left(\frac{2\pi}{9}\right), \quad \ldots, \quad \sin\left(\frac{9\pi}{9}\right) \]

3. **Using the Sine Identity**:
We can use the identity:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
For the first pair:
\[ A = \frac{17\pi}{9}, \quad B = \frac{\pi}{9} \]
Thus,
\[ A + B = \frac{18\pi}{9} = 2\pi, \quad A - B = \frac{16\pi}{9} \]
Therefore,
\[ \sin\left(\frac{17\pi}{9}\right) + \sin\left(\frac{\pi}{9}\right) = 2 \sin\left(\pi\right) \cos\left(\frac{8\pi}{9}\right) \]
Since \( \sin(\pi) = 0 \), this sum equals \( 0 \).

4. **Continuing the Pairing**:
By applying the same identity to all pairs, we find that each pair sums to zero:
\[ \sin\left(\frac{16\pi}{9}\right) + \sin\left(\frac{2\pi}{9}\right) = 0, \quad \sin\left(\frac{15\pi}{9}\right) + \sin\left(\frac{3\pi}{9}\right) = 0, \quad \ldots \]
The middle term \( \sin\left(\frac{9\pi}{9}\right) = \sin(\pi) = 0 \).

5. **Final Result**:
Adding all these results together gives:
\[ 0 + 0 + 0 + \ldots + 0 = 0 \]

Thus, the value of the expression is \( \boxed{0} \).