For any $θ \in (\frac{π}{4}, \frac{π}{2})$ ,the expression $3 (\sin θ - \cos θ)^{4}$ $+ 6 (\sin θ + \cos θ)^{2} + 4 \sin^{6} θ$ equals

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$13 - 4 \cos^{2} θ + 6 \sin^{2} θ \cos^{2} θ$

$13 - 4 \cos^{6} θ$

$13 - 4 \cos^{4} θ + 2 \sin^{2} θ \cos^{2} θ$

$13 - 4 \cos^{2} θ + 6 \cos^{4} θ$

answer is B.

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Detailed Solution

Given, $3 (\sin θ - \cos θ)^{4} + 6 (\sin θ + \cos θ)^{2} + 4 \sin^{6} θ$

$= 3 ({(\sin^{2} θ + \cos^{2} θ - 2 \sin θ \cos θ)}^{2})$ $+ 6 (\sin^{2} θ + \cos^{2} θ + 2 \sin θ \cos θ) + 4 \sin^{6} θ$

$\begin{array}{l} = 3 ((1 - \sin 2 θ)^{2}) + 6 (1 + \sin 2 θ) + 4 \sin^{6} θ \\ = 3 (1 + \sin^{2} 2 θ - 2 \sin 2 θ) + 6 + 6 \sin 2 θ + 4 \sin^{6} θ \end{array}$

$\begin{array}{l} = 9 + 12 \sin^{2} θ \cdot \cos^{2} θ + 4 {(1 - \cos^{2} θ)}^{3} \\ = 9 + 12 (1 - \cos^{2} θ) \cos^{2} θ + 4 (1 - \cos^{6} θ - 3 \cos^{2} θ + 3 \cos^{4} θ) \\ = 13 - 4 \cos^{6} θ \end{array}$