frac{(sin frac{pi}{8} + i cos frac{pi}{8})^8} | vedclass

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

Question

(C) We have the expression $E = \frac{(\sin \frac{\pi}{8} + i \cos \frac{\pi}{8})^8}{(\sin \frac{\pi}{8} - i \cos \frac{\pi}{8})^8}$.Factor out $i$ fr

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) We have the expression $E = \frac{(\sin \frac{\pi}{8} + i \cos \frac{\pi}{8})^8}{(\sin \frac{\pi}{8} - i \cos \frac{\pi}{8})^8}$.Factor out $i$ from the numerator and $-i$ from the denominator:$E = \frac{[i(\cos \frac{\pi}{8} - i \sin \frac{\pi}{8})]^8}{[(-i)(\cos \frac{\pi}{8} + i \sin \frac{\pi}{8})]^8}$.Since $i^8 = 1$ and $(-i)^8 = 1$,the expression simplifies to:$E = \frac{(\cos \frac{\pi}{8} - i \sin \frac{\pi}{8})^8}{(\cos \frac{\pi}{8} + i \sin \frac{\pi}{8})^8}$.Using Euler's formula $e^{i	heta} = \cos 	heta + i \sin 	heta$,we have:$E = \frac{(e^{-i\pi/8})^8}{(e^{i\pi/8})^8} = \frac{e^{-i\pi}}{e^{i\pi}}$.Since $e^{i\pi} = \cos \pi + i \sin \pi = -1$ and $e^{-i\pi} = \cos(-\pi) + i \sin(-\pi) = -1$,$E = \frac{-1}{-1} = 1$.

$\frac{(\sin \frac{\pi}{8} + i \cos \frac{\pi}{8})^8}{(\sin \frac{\pi}{8} - i \cos \frac{\pi}{8})^8}$ is equal to

Similar Questions

If $\omega$ is a complex cube root of unity,then find the value of $\sum_{x=1}^{10} ((\omega x+2)(\omega^2 x+2)-3)$.

The imaginary part of $(\sqrt{3}-i)^{2016}+(-\sqrt{3}-i)^{2019}$ is

If $\omega ( \neq 1)$ is a cube root of unity and $(1 + \omega )^7 = A + B\omega$,then $A$ and $B$ are respectively,the numbers

The product of all the values of $(\sqrt{3}-i)^{2/5}$ is

If $1, \omega, \omega^2$ are the three cube roots of unity,then the value of $(a + b\omega + c\omega^2)^3 + (a + b\omega^2 + c\omega)^3$ is equal to,given that $a + b + c = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module