Question:
If \(\omega\) is imaginary cube root of unity, then \(\sin\left\{(\omega^{13} + \omega^{20})\pi + \dfrac{\pi}{4}\right\}\) is equal to
यदि \(\omega\) इकाई का एक काल्पनिक घनमूल है, तो \(\sin\left\{(\omega^{13} + \omega^{20})\pi + \dfrac{\pi}{4}\right\}\) किसके बराबर है?
Options:
- \(-\dfrac{1}{\sqrt{2}}\)
- \(\dfrac{\sqrt{3}}{2}\)
- \(\dfrac{1}{\sqrt{2}}\)
- \(-\dfrac{\sqrt{3}}{2}\)
- \(-\dfrac{1}{\sqrt{2}}\)
- \(\dfrac{\sqrt{3}}{2}\)
- \(\dfrac{1}{\sqrt{2}}\)
- \(-\dfrac{\sqrt{3}}{2}\)
Correct Answer: a)
Explanation:
\(\omega^{13} = \omega^{12} \cdot \omega = \omega\) and \(\omega^{20} = \omega^{18} \cdot \omega^2 = \omega^2\).
\(\omega^{13} + \omega^{20} = \omega + \omega^2 = -1\).
\(\sin\left\{(-1)\pi + \dfrac{\pi}{4}\right\} = \sin\left(-\pi + \dfrac{\pi}{4}\right) = -\sin\left(\pi – \dfrac{\pi}{4}\right) = -\sin\left(\dfrac{\pi}{4}\right) = -\dfrac{1}{\sqrt{2}}\).
\(\omega^{13} = \omega\) तथा \(\omega^{20} = \omega^2\)।
\(\omega^{13} + \omega^{20} = \omega + \omega^2 = -1\)।
\(\sin\left\{-\pi + \dfrac{\pi}{4}\right\} = -\sin\left(\pi – \dfrac{\pi}{4}\right) = -\sin\left(\dfrac{\pi}{4}\right) = -\dfrac{1}{\sqrt{2}}\)।