Q60 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
If $\omega$ is a complex cube root of unity, then the value of $\sin\left[\pi(\omega^{10} + \omega^{23}) - \dfrac{\pi}{4}\right] =$
- A$-\dfrac{\sqrt{3}}{2}$
- B$-\dfrac{1}{\sqrt{2}}$
- Correct answerC$\dfrac{1}{\sqrt{2}}$
- D$\dfrac{\sqrt{3}}{2}$
Solution
Since $\omega$ is a complex cube root of unity, $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
$\omega^{10} = (\omega^3)^3 \cdot \omega = \omega$
$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = \omega^2$
$\omega^{10} + \omega^{23} = \omega + \omega^2 = -1$
Substituting this into the given expression:
$\sin\left[\pi(\omega^{10} + \omega^{23}) - \dfrac{\pi}{4}\right] = \sin\left(-\pi - \dfrac{\pi}{4}\right)$
$= \sin\left(-\dfrac{5\pi}{4}\right)$
$= -\sin\left(\pi + \dfrac{\pi}{4}\right)$
$= -\left(-\sin\dfrac{\pi}{4}\right)$
$= \dfrac{1}{\sqrt{2}}$
Answer: $\dfrac{1}{\sqrt{2}}$
$\omega^{10} = (\omega^3)^3 \cdot \omega = \omega$
$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = \omega^2$
$\omega^{10} + \omega^{23} = \omega + \omega^2 = -1$
Substituting this into the given expression:
$\sin\left[\pi(\omega^{10} + \omega^{23}) - \dfrac{\pi}{4}\right] = \sin\left(-\pi - \dfrac{\pi}{4}\right)$
$= \sin\left(-\dfrac{5\pi}{4}\right)$
$= -\sin\left(\pi + \dfrac{\pi}{4}\right)$
$= -\left(-\sin\dfrac{\pi}{4}\right)$
$= \dfrac{1}{\sqrt{2}}$
Answer: $\dfrac{1}{\sqrt{2}}$
Hence, the correct answer is option: 'C'.