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If $\sin β$ is geometric mean between $\sin α and \cos α, then \cos 2 β =$

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$2 \sin^{2} (\frac{π}{3} - α) or 2 \cos^{2} (\frac{π}{3} + α)$

$\sin^{2} (\frac{π}{4} - α) or \cos^{2} (\frac{π}{4} + α)$

$2 \sin^{2} (\frac{π}{4} - α) or 2 \cos^{2} (\frac{π}{4} + α)$

$\sin^{2} (\frac{π}{3} - α) or \cos^{2} (\frac{π}{3} + α)$

answer is A.

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Detailed Solution

$\begin{array}{l} \sin^{2} β = \sin αcos α \\ \cos 2 β = 1 - 2 \sin αcos α = (\sin α - \cos α)^{2} \\ (\sqrt{2})^{2} {(\frac{1}{\sqrt{2}} \sin α - \frac{1}{\sqrt{2}} \cos α)}^{2} = 2 \sin^{2} (\frac{π}{4} - α) \end{array}$

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