(sin x+i cos x)n =cos π2-x+i sinπ2-xn =cosn π2-x+i sin nπ2-x =cis nπ2-xoption (2) is correct

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${(\sin x + i cosx)}^{n} =$

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$cis n (\frac{π}{2} - x)$

$cis n (\frac{π}{3} + x)$

$cis n (\frac{π}{2} + x)$

$cis n (\frac{π}{3} - x)$

answer is B.

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Detailed Solution

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

option (2) is correct

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