Q62 / 160 · AP EAMCET 2008 · Mathematics

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

  • Correct answer
    A
    $\frac{1}{\sqrt{2}}$
  • B
    $\frac{1}{2}$
  • C
    $1$
  • D
    $\frac{\sqrt{3}}{2}$

Solution

Since, $\omega$ is a cube root of unity. $$ \begin{aligned} \therefore \quad \sin & \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\} \\ & =\sin \left\{\left(\omega+\omega^2\right) \pi-\frac{\pi}{4}\right\} \\ & =\sin \left(-\pi-\frac{\pi}{4}\right) \quad\left(\because 1+\omega+\omega^2=0\right) \\ & =-\sin \left(\pi+\frac{\pi}{4}\right)=\sin \frac{\pi}{4} \\ & =\frac{1}{\sqrt{2}} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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