Let $z = \cos\theta + i\sin\theta$. Then the value of $\sum_{m=1}^{15} \text{Im}(z^{2m-1})$ at $\theta = 2^\circ$ is

$z = \cos\theta + i\sin\theta$ $\text{To find: } \sum_{m=1}^{15} \text{Im}\left(z^{2m-1}\right) \text{ at } \theta = 2^\circ$ $z^n = \cos(n\theta) + i\sin(n\theta)$ $z^{2m-1} = \cos\big((2m-1)\theta\big) + i\sin\big((2m-1)\theta\big)$ $\text{Im}\left(z^{2m-1}\right) = \sin\big((2m-1)\theta\big)$ $\sum_{m=1}^{15} \text{Im}\left(z^{2m-1}\right) = \sum_{m=1}^{15} \sin\big((2m-1)\theta\big)$ $\theta = 2^\circ: \ \sin 2^\circ + \sin 6^\circ + \sin 10^\circ + \cdots + \sin 58^\circ$ $\text{common difference } d = 4^\circ, \quad n = 15$ $1.\ \text{write the sum as sines}$ $2.\ \text{use the AP sine-sum formula}$ $3.\ \text{substitute } \theta = 2^\circ$ $\sum_{k=0}^{n-1} e^{i(a+kd)} = e^{ia}\,\dfrac{e^{ind} - 1}{e^{id} - 1}$ $\sum_{k=0}^{n-1} \sin(a+kd) = \dfrac{\sin\!\left(\frac{nd}{2}\right)}{\sin\!\left(\frac{d}{2}\right)}\,\sin\!\left(a+\tfrac{(n-1)d}{2}\right)$ $a = \theta, \quad d = 2\theta, \quad n = 15$ $\sum = \dfrac{\sin(15\theta)}{\sin\theta}\cdot\sin(15\theta) = \dfrac{\sin^{2}(15\theta)}{\sin\theta}$ $\theta = 2^\circ: \quad \dfrac{\sin^{2}(30^\circ)}{\sin 2^\circ} = \dfrac{\left(\frac{1}{2}\right)^{2}}{\sin 2^\circ} = \dfrac{1}{4\sin 2^\circ}$ $\boxed{\sum_{m=1}^{15} \text{Im}\left(z^{2m-1}\right) = \dfrac{1}{4\sin 2^\circ}}$ $\text{Answer: option (4)}$