Sum of Imaginary Parts of Odd Powers of Complex Number

To find the value of the given expression, we start with the complex number:
$z = \cos \theta + i \sin \theta = e^{i\theta}$

By De Moivre's Theorem (or Euler's formula), the power $z^{2m-1}$ is given by:
$z^{2m-1} = \cos((2m-1)\theta) + i \sin((2m-1)\theta)$

The imaginary part of $z^{2m-1}$ is therefore:
$\text{Im}(z^{2m-1}) = \sin((2m-1)\theta)$

Now, we need to evaluate the sum:
$S = \sum_{m=1}^{15} \text{Im}(z^{2m-1}) = \sum_{m=1}^{15} \sin((2m-1)\theta) = \sin \theta + \sin 3\theta + \sin 5\theta + \dots + \sin 29\theta$

This is a standard trigonometric sum of sines whose angles are in an arithmetic progression with first term $\alpha = \theta$, common difference $\beta = 2\theta$, and number of terms $n = 15$.

Using the identity:
$\sum_{k=1}^{n} \sin(\alpha + (k-1)\beta) = \frac{\sin\left(\frac{n\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} \sin\left(\alpha + \frac{(n-1)\beta}{2}\right)$

Substituting $\alpha = \theta$, $\beta = 2\theta$, and $n = 15$:
$S = \frac{\sin\left(\frac{15 \cdot 2\theta}{2}\right)}{\sin\left(\frac{2\theta}{2}\right)} \sin\left(\theta + \frac{14 \cdot 2\theta}{2}\right) = \frac{\sin(15\theta)}{\sin\theta} \sin(15\theta) = \frac{\sin^2(15\theta)}{\sin\theta}$

We are given that $\theta = 2^\circ$, which implies:
$15\theta = 15 \times 2^\circ = 30^\circ$

Since $\sin(30^\circ) = \frac{1}{2}$, we have:
$\sin^2(15\theta) = \sin^2(30^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

Substituting this into the expression for $S$:
$S = \frac{\frac{1}{4}}{\sin 2^\circ} = \frac{1}{4\sin 2^\circ}$

Therefore, the correct option is (D).