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

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(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

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(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

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