If omega is a complex cube root of unity, the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that $\omega$ is a complex cube root of unity, we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.We simplify the powers of $\omega$:$\o

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(C) Given that $\omega$ is a complex cube root of unity, we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.We simplify the powers of $\omega$:$\omega^{10} = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$.$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = 1^7 \cdot \omega^2 = \omega^2$.Substitute these into the expression:$\sin [\pi(\omega + \omega^2) - \frac{\pi}{4}]$.Since $1 + \omega + \omega^2 = 0$, we have $\omega + \omega^2 = -1$.Now, substitute this value:$\sin [\pi(-1) - \frac{\pi}{4}] = \sin [-\pi - \frac{\pi}{4}] = \sin [-\frac{5\pi}{4}]$.Using the property $\sin(-	heta) = -\sin(	heta)$:$-\sin(\frac{5\pi}{4}) = -\sin(\pi + \frac{\pi}{4}) = -(-\sin(\frac{\pi}{4})) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

If $\omega$ is a complex cube root of unity, then the value of $\sin [\pi(\omega^{10} + \omega^{23}) - \frac{\pi}{4}] = $

Similar Questions

The product of the four values of $(1+i \sqrt{3})^{3/4}$ is

If $\alpha, \beta$ are the roots of the equation $1+x+x^2=0$,then $(2-\alpha)(2-\beta)(2-\alpha^{10})(2-\alpha^{20})=$

If $i = \sqrt{-1}$,then $4 + 5\left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)^{334} + 3\left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)^{365}$ is equal to

If $\alpha$ and $\beta$ are imaginary cube roots of unity,then $\alpha^4 + \beta^4 + \frac{1}{\alpha\beta} = $

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^{4}+x^{3}+x^{2}+x+1=0$,then $\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module