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

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Aicharya · Accepted Answer

Since, $\omega$ is a cube root of unity. $ \begin{aligned} 	herefore \quad \sin & \left\{\left(\omega^{10}+\omega^{23}ight) \pi-\frac{\pi}{4}ight\} \ & =\sin \left\{\left(\omega+\omega^2ight) \pi-\frac{\pi}{4}ight\} \ & =\sin \left(-\pi-\frac{\pi}{4}ight) \quad\left(\because 1+\omega+\omega^2=0ight) \ & =-\sin \left(\pi+\frac{\pi}{4}ight)=\sin \frac{\pi}{4} \ & =\frac{1}{\sqrt{2}} \end{aligned} $

If $\omega$ is a complex cube root of unity, then $\sin \left\{\left(\omega^{10}+\omega^{23}\right)…

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