(1+sin θ+i cos θ)n 1+cos π2-θ+i sin π2-θn 2cos2π4-θ2+i2sin π4-θ2cosπ4-θ2n =2cosπ4-θ2n cosπ4-θ2+i sinπ4-θ2n =2n cosn π4-θ2 cos nπ4-θ2+i sin nπ4-θ2 =2n cosn π4-θ2 cis nπ4-θ2

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${(1 + \sin θ + i \cos θ)}^{n} =$

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$2^{n} . \cos^{n} [\frac{π}{6} - \frac{θ}{2}] . cis [n (\frac{π}{6} - \frac{θ}{2})]$

$2^{n} . \cos^{n} [\frac{π}{3} - \frac{θ}{2}] . cis [n (\frac{π}{3} - \frac{θ}{2})]$

$2^{n} . \cos^{n} [\frac{π}{4} - \frac{θ}{2}] . cis [n (\frac{π}{4} - \frac{θ}{2})]$

$2^{n} . \cos^{n} [\frac{π}{5} - \frac{θ}{2}] . cis [n (\frac{π}{5} - \frac{θ}{2})]$

answer is B.

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Detailed Solution

$\begin{array}{rcl} {(1 + \sin θ + i \cos θ)}^{n} \\ {(1 + \cos (\frac{π}{2} - θ) + i \sin (\frac{π}{2} - θ))}^{n} \\ {(2 \cos^{2} (\frac{π}{4} - \frac{θ}{2}) + i (2 \sin (\frac{π}{4} - \frac{θ}{2}) \cos (\frac{π}{4} - \frac{θ}{2})))}^{n} \\ = & {(2 \cos (\frac{π}{4} - \frac{θ}{2}))}^{n} {(\cos (\frac{π}{4} - \frac{θ}{2}) + i \sin (\frac{π}{4} - \frac{θ}{2}))}^{n} \\ = & 2^{n} \cos^{n} (\frac{π}{4} - \frac{θ}{2}) (\cos n (\frac{π}{4} - \frac{θ}{2}) + i \sin n (\frac{π}{4} - \frac{θ}{2})) \\ = & 2^{n} \cos^{n} (\frac{π}{4} - \frac{θ}{2}) cis n (\frac{π}{4} - \frac{θ}{2}) \end{array}$