If omega is a complex cube root of unity,then | vedclass

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(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.First,simplify the powers of $\omega$:$

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$\frac{1}{\sqrt{2}}$

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(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.First,simplify the powers of $\omega$:$\omega^{10} = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = 1^7 \cdot \omega^2 = \omega^2$Now,substitute these into the expression:$\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\}$Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.Substituting this value:$= \sin \left\{-\pi - \frac{\pi}{4}ight\} = \sin \left\{-\left(\pi + \frac{\pi}{4}ight)ight\}$Using the property $\sin(-	heta) = -\sin(	heta)$:$= -\sin \left(\pi + \frac{\pi}{4}ight)$Using the property $\sin(\pi + 	heta) = -\sin(	heta)$:$= -(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

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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