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 :
Correct Option: , 2
The value of $\left(\frac{1+\sin 2 \pi / 9+i \cos 2 \pi / 9}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)$
$=\left(\frac{1+\sin \left(\frac{\pi}{2}-\frac{5 \pi}{18}\right)+i \cos \left(\frac{\pi}{2}-\frac{5 \pi}{18}\right)}{1+\sin \left(\frac{\pi}{2}-\frac{5 \pi}{18}\right)-i \cos \left(\frac{\pi}{2}-\frac{5 \pi}{18}\right)}\right)^{3}$
$=\left(\frac{1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18}}{1+\cos \frac{5 \pi}{18}-i \sin \frac{5 \pi}{18}}\right)^{3}$
$=\left(\frac{2 \cos ^{2} \frac{5 \pi}{36}+2 i \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}{2 \cos ^{2} \frac{5 \pi}{36}-2 i \sin \frac{5 \pi}{36} \cdot \cos \frac{5 \pi}{36}}\right)^{3}$
$=\left(\frac{\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}}{\cos \frac{5 \pi}{36}-i \sin \frac{5 \pi}{36}}\right)^{3}$
$=\left(\frac{\mathrm{e}^{i 5 \pi / 36}}{\mathrm{e}^{-i 5 \pi / 36}}\right)^{3}=\left(\mathrm{e}^{i 5 \pi / 18}\right)^{3}$
$=\cos \frac{5 \pi}{6}+i \sin 5 \pi / 6$
$=-\frac{\sqrt{3}}{2}+i / 2$
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