Let `z = cos theta + isin theta`. Then the value of `sum_(m=1)^(15) Im(z^(2m-1) ) ` at `theta = 2^(@)` is

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the complex number
Let \( z = \cos \theta + i \sin \theta \). This can be expressed in exponential form as:
\[ z = e^{i\theta} \]

### Step 2: Substitute the value of theta
Given \( \theta = 2^\circ \), we can write:
\[ z = e^{i \cdot 2^\circ} \]

### Step 3: Express \( z^{2m-1} \)
We need to find the imaginary part of \( z^{2m-1} \):
\[ z^{2m-1} = (e^{i \cdot 2^\circ})^{2m-1} = e^{i(2m-1) \cdot 2^\circ} = e^{i(4m - 2)^\circ} \]

### Step 4: Find the imaginary part
The imaginary part of \( z^{2m-1} \) is given by:
\[ \text{Im}(z^{2m-1}) = \sin((4m - 2)^\circ) \]

### Step 5: Set up the summation
We need to evaluate the sum:
\[ \sum_{m=1}^{15} \text{Im}(z^{2m-1}) = \sum_{m=1}^{15} \sin((4m - 2)^\circ) \]

### Step 6: Rewrite the sum
This can be rewritten as:
\[ \sum_{m=1}^{15} \sin((4m - 2)^\circ) = \sum_{m=1}^{15} \sin(4m^\circ - 2^\circ) \]

### Step 7: Use the sine sum formula
Using the sine sum formula, we can express this sum as:
\[ \sum_{m=1}^{15} \sin(4m - 2) = \frac{\sin\left(\frac{n \cdot d}{2}\right) \cdot \sin\left(a + \frac{(n-1) \cdot d}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]
where \( n = 15 \), \( a = 2^\circ \), and \( d = 4^\circ \).

### Step 8: Calculate the components
Now we calculate:
- \( \frac{n \cdot d}{2} = \frac{15 \cdot 4}{2} = 30^\circ \)
- \( a + \frac{(n-1) \cdot d}{2} = 2 + \frac{14 \cdot 4}{2} = 2 + 28 = 30^\circ \)

### Step 9: Substitute into the formula
Substituting these values into the sine sum formula gives:
\[ \sum_{m=1}^{15} \sin(4m - 2) = \frac{\sin(30^\circ) \cdot \sin(30^\circ)}{\sin(2^\circ)} \]

### Step 10: Calculate the sine values
We know:
- \( \sin(30^\circ) = \frac{1}{2} \)
Thus:
\[ \sum_{m=1}^{15} \sin(4m - 2) = \frac{\frac{1}{2} \cdot \frac{1}{2}}{\sin(2^\circ)} = \frac{1/4}{\sin(2^\circ)} \]

### Final Result
The value of \( \sum_{m=1}^{15} \text{Im}(z^{2m-1}) \) at \( \theta = 2^\circ \) is:
\[ \frac{1}{4 \sin(2^\circ)} \]