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,
$\therefore \sin {\omega }^{10}+{\omega }^{23}\pi -\frac{\pi }{4}$
$=\sin \omega +{\omega }^2\pi -\frac{\pi }{4}$
$=\sin -\pi -\frac{\pi }{4}$
$=\sin -\pi -\frac{\pi }{4}=\sin \frac{\pi }{4}$
$=\frac{1}{\sqrt{2}}$