Let `z=costheta+isintheta`. Then the value of `sum_(m->1-15)Img(z^(2m-1))` at `theta=2^@` is:

Options

A. `1/(sin2^(@))`
B. `1/(3sin2^(@))`
C. `1/(2sin2^(@))`
D. `1/(4sin2^(@))`

Answer

To solve the given problem, we need to find the value of the sum:

\[ \sum_{m=1}^{15} \text{Img}(z^{2m-1}) \]

where \( z = \cos(\theta) + i \sin(\theta) \) and \( \theta = 2^\circ \).

### Step-by-Step Solution:

1. **Express \( z^{2m-1} \)**:
We know that \( z = \cos(\theta) + i \sin(\theta) \). By De Moivre's theorem, we can express \( z^n \) as:
\[ z^n = \cos(n\theta) + i \sin(n\theta) \]
Therefore, for \( z^{2m-1} \):
\[ z^{2m-1} = \cos((2m-1)\theta) + i \sin((2m-1)\theta) \]

2. **Find the Imaginary Part**:
The imaginary part of \( z^{2m-1} \) is:
\[ \text{Img}(z^{2m-1}) = \sin((2m-1)\theta) \]

3. **Set Up the Sum**:
Now, we can rewrite the sum:
\[ \sum_{m=1}^{15} \text{Img}(z^{2m-1}) = \sum_{m=1}^{15} \sin((2m-1)\theta) \]
Substituting \( \theta = 2^\circ \):
\[ = \sum_{m=1}^{15} \sin((2m-1) \cdot 2^\circ) = \sum_{m=1}^{15} \sin((2m-1) \cdot 2^\circ) = \sum_{m=1}^{15} \sin((2m-1) \cdot 2^\circ) \]

4. **Calculate the Angles**:
The angles will be:
\[ \sin(1 \cdot 2^\circ), \sin(3 \cdot 2^\circ), \sin(5 \cdot 2^\circ), \ldots, \sin(29 \cdot 2^\circ) \]

5. **Use the Sine Sum Formula**:
The sum of sines of angles in arithmetic progression can be calculated using the formula:
\[ \sum_{k=0}^{n-1} \sin(a + kd) = \frac{\sin\left(\frac{nd}{2}\right) \sin\left(a + \frac{(n-1)d}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]
Here, \( a = 2^\circ \), \( d = 4^\circ \), and \( n = 15 \).

6. **Apply the Formula**:
Plugging in the values:
\[ \sum_{m=1}^{15} \sin((2m-1) \cdot 2^\circ) = \frac{\sin\left(\frac{15 \cdot 4^\circ}{2}\right) \sin\left(2^\circ + \frac{(15-1) \cdot 4^\circ}{2}\right)}{\sin\left(\frac{4^\circ}{2}\right)} \]
Simplifying:
\[ = \frac{\sin(30^\circ) \sin(30^\circ)}{\sin(2^\circ)} \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ = \frac{\left(\frac{1}{2}\right)^2}{\sin(2^\circ)} = \frac{\frac{1}{4}}{\sin(2^\circ)} \]

7. **Final Result**:
Thus, the value of the sum is:
\[ \frac{1/4}{\sin(2^\circ)} \]