cos6θ=132(cos6θ+6cos4θ+15cos2θ+10)→(1)cos6θ=132(10−15cos2θ+6cos4θ−cos6θ)→(2)(1)+(2)⇒cos6θ+sin6=132(20+12cos4θ)=18(5+3cos4θ)

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$\cos^{6} θ + \sin^{6} θ =$

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$\frac{1}{16} (10 + \cos 4 θ)$

$\frac{1}{8} (5 - 3 \cos 4 θ + \cos 6 θ)$

$\frac{1}{8} (5 + 3 \cos 4 θ)$

$\frac{1}{16} (10 - \cos 4 θ + \cos 6 θ)$

answer is B.

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Detailed Solution

$\cos^{6} θ = \frac{1}{32} (\cos 6 θ + 6 \cos 4 θ + 15 \cos 2 θ + 10) \to (1)$

$\cos^{6} θ = \frac{1}{32} (10 - 15 \cos 2 θ + 6 \cos 4 θ - \cos 6 θ) \to (2)$

$(1) + (2)$

$\Rightarrow \cos^{6} θ + \sin^{6} = \frac{1}{32} (20 + 12 \cos 4 θ) = \frac{1}{8} (5 + 3 \cos 4 θ)$

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