Question

Let $z = \cos \theta + i \sin \theta$. Then, the value of $\displaystyle \sum _{m=0}^{15} Im (z^{2m-1})$ at $\theta = 2^\circ$ is

• A. $\frac{1}{\sin 2^\circ}$
• B. $\frac{1}{3 \sin 2^\circ}$
• C. $\frac{1}{2 \sin 2^\circ}$
• D. $\frac{1}{4 \sin 2^\circ}$

Answer 1

Answer: (D)

Given that, $z = \cos\theta + i \sin \theta = e^{i \theta}$
$\therefore \displaystyle \sum _{m=1}^{15} Im(z^{3m-1})=\displaystyle \sum_{m=1}^{15} Im(e^{i\theta})^{2m-1}$
$ = \displaystyle \sum_{m=1}^{15} Im \, e^i{^{(2m-1)\theta}}$
$= \sin \theta + \sin 3 \theta+ \sin 5 \theta + ... + \sin 29 \theta$
$=\frac{\sin\bigg(\frac{\theta+29+\theta}{2}\sin\bigg(\frac{15\times2\theta}{2}\bigg)}{\sin\bigg(\frac{2\theta}{2}\bigg)}$
$=\frac{\sin(15\theta) \sin(15 \theta)}{\sin\theta}=\frac{1}{4 \sin 2^{\circ}}$