If omega is an imaginary cube root of unity,t | vedclass

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

Question

(C) Given that $\omega$ is an imaginary cube root of unity,we know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.First,simplify the powers of $\

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given that $\omega$ is an imaginary cube root of unity,we know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.First,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 \left[ (\omega + \omega^2)\pi - \frac{\pi}{4} ight]$Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.So,the expression becomes $\sin \left[ -1 \cdot \pi - \frac{\pi}{4} ight] = \sin \left( -\pi - \frac{\pi}{4} ight)$.Using the property $\sin(-	heta) = -\sin(	heta)$:$-\sin \left( \pi + \frac{\pi}{4} ight) = -(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

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

Similar Questions

Let $\alpha$ and $\beta$ be the roots of $x^2 - \sqrt{2}x + 1 = 0$. Then the value of $\alpha^{50} + \beta^{50}$ is:

$n$ is a positive integer and not a multiple of $3$. If $\omega$ is a non-real cube root of unity,then $\omega^n + \omega^{2n}$ is equal to

If $\alpha$ is a complex number satisfying the equation $\alpha^{2}+\alpha+1=0$,then $\alpha^{31}$ is equal to

${\left( \frac{-1 + i\sqrt{3}}{2} \right)^{20}} + {\left( \frac{-1 - i\sqrt{3}}{2} \right)^{20}} = $

If $(\sqrt{3}+i)^{10}=a+bi$,where $a, b \in R$,then the values of $a$ and $b$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module