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If $\omega$ is a complex cube root of unity, then
$\sin \left[\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right]=$
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$\sin \left[\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right]=$
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Verified Answer
The correct answer is:
$1 / \sqrt{2}$
$\sin \left[\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right]$
$=\sin \left[\left(w+w^2\right) \pi-\frac{\pi}{4}\right] \quad\left(\because \omega^3=1\right)$
$=\sin \left[-\pi-\frac{\pi}{4}\right]=-\sin \left[\frac{5 \pi}{4}\right] \quad\left(\because \omega^2+\omega+1=0\right)$
$=-\left(-\frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}$
$=\sin \left[\left(w+w^2\right) \pi-\frac{\pi}{4}\right] \quad\left(\because \omega^3=1\right)$
$=\sin \left[-\pi-\frac{\pi}{4}\right]=-\sin \left[\frac{5 \pi}{4}\right] \quad\left(\because \omega^2+\omega+1=0\right)$
$=-\left(-\frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}$
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