${[(\cos α - cosβ) + i (sinα - \sin β)]}^{n} + {[(\cos α - cosβ) - i (\sin α - \sin β)]}^{n} =$

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$2^{n + 1} . \sin^{n} [\frac{α - β}{2}] \cos [n [\frac{π}{2} + \frac{α + β}{2}]]$

$2^{n - 1} . \sin^{n} [\frac{α - β}{2}] \cos [n [\frac{π}{2} + \frac{α + β}{2}]]$

$2^{n + 1} . \sin^{n} [\frac{α + β}{2}] \cos [n [\frac{π}{2} + \frac{α - β}{2}]]$

$2^{n - 1} . \sin^{n} [\frac{α + β}{2}] \cos [n [\frac{π}{2} + \frac{α - β}{2}]]$

answer is A.

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Detailed Solution

$\begin{array}{rcl} {((\cos α - cosβ) + i (sinα - \sin β))}^{n} + {((\cos α - cosβ) - i (\sin α - \sin β))}^{n} \\ {(- 2 \sin \frac{α + β}{2} \sin \frac{α - β}{2} + i (2 \cos \frac{α + β}{2} \sin \frac{α - β}{2}))}^{n} + {(- 2 \sin \frac{α + β}{2} \sin \frac{α - β}{2} - i (2 \cos \frac{α + β}{2} \sin \frac{α - β}{2}))}^{n} \\ 2^{n} \sin^{n} \frac{α - β}{2} [{(- \sin (\frac{α + β}{2}) + i \cos (\frac{α + β}{2}))}^{n} + {(- \sin (\frac{α + β}{2}) - i \cos (\frac{α + β}{2}))}^{n}] \\ 2^{n} \sin^{n} \frac{α - β}{2} [{(\cos (\frac{π}{2} + \frac{α + β}{2}) + i \sin (\frac{π}{2} + \frac{α + β}{2}))}^{n} + {(\cos (\frac{π}{2} + \frac{α + β}{2}) - i \sin (\frac{π}{2} + \frac{α + β}{2}))}^{n}] \\ 2^{n} \sin^{2} \frac{α - β}{2} [\cos n (\frac{π}{2} + \frac{α + β}{2}) + i \sin (\frac{π}{2} + \frac{α + β}{2}) + \cos n (\frac{π}{2} + \frac{α + β}{2}) - i \sin (\frac{π}{2} + \frac{α + β}{2})] \\ 2^{n} \sin^{n} \frac{α - β}{2} 2 \cos n (\frac{π}{2} + \frac{α + β}{2}) \\ = & 2^{n + 1} \sin^{n} \frac{α - β}{2} \cos n (\frac{π}{2} + \frac{α + β}{2}) \end{array}$