1+cosπ8-i sinπ81+cosπ8+i sinπ88=

1+cosπ8-i sinπ81+cosπ8+i sinπ88 =2cos2π16-i2sinπ16 cosπ162cos2π16+i 2sinπ16 cosπ168 =2cosπ16cosπ16-i sinπ162cosπ16cosπ16+i sinπ168 =cosπ16-i sinπ1616 =cos π-i sin π =-1

1+cosπ8-i sinπ81+cosπ8+i sinπ88=

1+cosπ8-i sinπ81+cosπ8+i sinπ88 =2cos2π16-i2sinπ16 cosπ162cos2π16+i 2sinπ16 cosπ168 =2cosπ16cosπ16-i sinπ162cosπ16cosπ16+i sinπ168 =cosπ16-i sinπ1616 =cos π-i sin π =-1

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${(\frac{1 + \cos \frac{π}{8} - i \sin \frac{π}{8}}{1 + \cos \frac{π}{8} + i \sin \frac{π}{8}})}^{8} =$

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Detailed Solution

$\begin{array}{rcl} {(\frac{1 + \cos \frac{π}{8} - i \sin \frac{π}{8}}{1 + \cos \frac{π}{8} + i \sin \frac{π}{8}})}^{8} \end{array}$

$\begin{array}{rcl} = & {(\frac{2 \cos^{2} \frac{π}{16} - i (2 \sin \frac{π}{16} \cos \frac{π}{16})}{2 \cos^{2} \frac{π}{16} + i (2 \sin \frac{π}{16} \cos \frac{π}{16})})}^{8} \end{array}$

$\begin{array}{rcl} = & {(\frac{2 \cos \frac{π}{16} (\cos \frac{π}{16} - i \sin \frac{π}{16})}{2 \cos \frac{π}{16} (\cos \frac{π}{16} + i \sin \frac{π}{16})})}^{8} \end{array}$