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If ${[1 - \cos (\frac{π}{2} + α) + \sin (\frac{3 π}{2} + α)]}^{2} + {[1 - \sin (\frac{3 π}{2} - α) - \cos (\frac{3 π}{2} + α)]}^{2} = a + b \sin^{2} (\frac{π}{4} + α) .$ then $a^{2} + b^{2} =$

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answer is B.

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Detailed Solution

$= {[1 - \cos (\frac{π}{2} + α) + \sin (\frac{3 π}{2} + α)]}^{2} + {[1 - \sin (\frac{3 π}{2} - α) - \cos (\frac{3 π}{2} + α)]}^{2}$

$= {[1 + \sin α - \cos α]}^{2} + {[1 + \cos α - \sin α]}^{2}$

$= {(1 + \sin α)}^{2} + \cos^{2} α - 2 \cos α (1 + \sin α) + {(1 + \cos α)}^{2} + \sin^{2} α - 2 \sin α (1 + \cos α)$

$= 1 + \sin^{2} α + 2 \sin α + \cos^{2} α - 2 \cos α - 2 \sin α \cos α + 1 + \cos^{2} α + 2 \cos α + \sin^{2} α - 2 \sin α - 2 \sin α \cos α$

$= 6 - \frac{4 (1 + \sin 2 α)}{2}$

$= 6 + (- 4) \sin^{2} (\frac{π}{4} + α)$

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