JEE Main 2020 (Online) 2nd September Morning ShiftQuestion 1
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 :
Complex Numbers
Solution
Let $\theta=\frac{2\pi}9$, $\phi=\frac\pi2-\theta=\frac{5\pi}{18}$. Then $1+\sin\theta+i\cos\theta=1+\cos\phi+i\sin\phi=2\cos\frac\phi2\,e^{i\phi/2}$. The ratio is $e^{i\phi}$; cubed: $e^{i5\pi/6}=-\frac{\sqrt3}2+\frac i2=-\frac12(\sqrt3-i)$. Answer: (D).