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 $\omega$: $\o

Vedclass · Accepted Answer

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

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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 $\omega$: $\omega^{10} = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$ $\omega^{23} = (\omega^3)^7 \cdot \omega^2 = 1^7 \cdot \omega^2 = \omega^2$ Substituting these into the expression: $\sin \left\{(\omega + \omega^2) \pi - \frac{\pi}{4}ight\}$ Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$. Thus,the expression becomes: $\sin \left\{(-1) \pi - \frac{\pi}{4}ight\} = \sin \left(-\pi - \frac{\pi}{4}ight)$ $= \sin \left(-\left(\pi + \frac{\pi}{4}ight)ight) = -\sin \left(\pi + \frac{\pi}{4}ight)$ Using the identity $\sin(\pi + 	heta) = -\sin 	heta$: $-(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

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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