Question

If $\omega$ is a complex cube root of unity, then sin $\omega^{10} + + \omega^{23} \pi - \frac{\pi}{4}$ is equal to

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. 1
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (A)

Since, $\omega$ is a cube root of unity,
$\therefore \sin \omega^{10}+\omega^{23} \pi-\frac{\pi}{4}$
$=\sin \omega+\omega^{2} \pi-\frac{\pi}{4}$
$=\sin -\pi-\frac{\pi}{4}$
$=\sin -\pi-\frac{\pi}{4}=\sin \frac{\pi}{4}$
$=\frac{1}{\sqrt{2}}$