[JEE Advanced 1994] if omega is an imaginary cube root of unity then sin omega 10 omega
If $\omega$ is an imaginary cube root of unity, then $\sin\!\left((\omega^{10}+\omega^{23})\pi-\frac\pi4\right)$ equals
(a) $\frac{\sqrt3}2$
(b) $\frac1{\sqrt2}$
(c) $\frac1{\sqrt2}$
(d) $\frac{\sqrt3}2$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
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Correct answer: (c) $\frac1{\sqrt2}$
$\omega^{10}=\omega,\ \omega^{23}=\omega^2$, so $\omega^{10}+\omega^{23}=\omega+\omega^2=-1$. Then $\sin(-\pi-\frac\pi4)=\sin\frac\pi4=\frac1{\sqrt2}$.
JEE Advanced 1994 · Trigonometry — verified solution by the Vidaara Team.
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