The value of
Question:

The value of $\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^{3}$ is :

1. (1) $\frac{1}{2}(1-i \sqrt{3})$

2. (2) $\frac{1}{2}(\sqrt{3}-i)$

3. (3) $-\frac{1}{2}(\sqrt{3}-i)$

4. (4) $-\frac{1}{2}(1-i \sqrt{3})$

Correct Option: , 3

Solution:

$\left(\frac{1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18}}{1+\cos \frac{5 \pi}{18}-i \sin \frac{5 \pi}{18}}\right)^{3}$

$=\left(\frac{2 \cos ^{2} \frac{5 \pi}{36}+i 2 \sin \frac{5 \pi}{36} \cdot \cos \frac{5 \pi}{36}}{2 \cos ^{2} \frac{5 \pi}{36}-i 2 \sin \frac{5 \pi}{36} \cdot \cos \frac{5 \pi}{36}}\right)^{3}$

$=\left(\frac{\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}}{\cos \frac{5 \pi}{36}-i \sin \frac{5 \pi}{36}}\right)^{3}=\left(\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}\right)^{6}$

$=\cos \left(6 \times \frac{5 \pi}{36}\right)+i \sin \left(6 \times \frac{5 \pi}{36}\right)=\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}$

$=-\frac{\sqrt{3}}{2}+i \frac{1}{2}=-\frac{1}{2}(\sqrt{3}-i)$