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

• A. $\frac{1}{2}(\sqrt{3}-i)$
• B. $-\frac{1}{2}(\sqrt{3}-i)$
• C. $-\frac{1}{2}(1-i\sqrt{3})$
• D. $\frac{1}{2}(1-i\sqrt{3})$

Answer 1

Answer: (B)

The value of $\left(\frac{1+\sin 2\pi /9+i\cos 2\pi /9}{1+\sin \frac{2\pi }{9}-i\cos \frac{2\pi }{9}}\right)$
$={\left(\frac{1+\sin \left(\frac{\pi }{2}-\frac{5\pi }{18}\right)+i\cos \left(\frac{\pi }{2}-\frac{5\pi }{18}\right)}{1+\sin \left(\frac{\pi }{2}-\frac{5\pi }{18}\right)-i\cos \left(\frac{\pi }{2}-\frac{5\pi }{18}\right)}\right)}^3$
$={\left(\frac{1+\cos \frac{5\pi }{18}+i\sin \frac{5\pi }{18}}{1+\cos \frac{5\pi }{18}-i\sin \frac{5\pi }{18}}\right)}^3$
$={\left(\frac{2{\cos }^2\frac{5\pi }{36}+2i\sin \frac{5\pi }{36}\cos \frac{5\pi }{36}}{2{\cos }^2\frac{5\pi }{36}-2i\sin \frac{5\pi }{36}\cdot \cos \frac{5\pi }{36}}\right)}^3$
$={\left(\frac{\cos \frac{5\pi }{36}+i\sin \frac{5\pi }{36}}{\cos \frac{5\pi }{36}-i\sin \frac{5\pi }{36}}\right)}^3$
$={\left(\frac{e^{i5\pi /36}}{e^{-i5\pi /36}}\right)}^3={\left(e^{i5\pi /18}\right)}^3$
$=\cos \frac{5\pi }{6}+i\sin 5\pi /6$
$=-\frac{\sqrt{3}}{2}+i/2$