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If $\omega$ is a complex cube root of unity, then $\sin \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\}$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{2}}$
Since, $\omega$ is a cube root of unity.
$$
\begin{aligned}
\therefore \quad \sin & \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\} \\
& =\sin \left\{\left(\omega+\omega^2\right) \pi-\frac{\pi}{4}\right\} \\
& =\sin \left(-\pi-\frac{\pi}{4}\right) \quad\left(\because 1+\omega+\omega^2=0\right) \\
& =-\sin \left(\pi+\frac{\pi}{4}\right)=\sin \frac{\pi}{4} \\
& =\frac{1}{\sqrt{2}}
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad \sin & \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\} \\
& =\sin \left\{\left(\omega+\omega^2\right) \pi-\frac{\pi}{4}\right\} \\
& =\sin \left(-\pi-\frac{\pi}{4}\right) \quad\left(\because 1+\omega+\omega^2=0\right) \\
& =-\sin \left(\pi+\frac{\pi}{4}\right)=\sin \frac{\pi}{4} \\
& =\frac{1}{\sqrt{2}}
\end{aligned}
$$
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