x=cosθ+isinθ⇒1x=cosθ−isinθ2cosθ=x+1x⇒2isinθ=x−1x⇒(x+1x)4(x−1x)2=(24cos4θ)(22(−1)sin2θ)⇒−26cos4θsin2=(x2−1x2)2(x+1x)2⇒−26cos4θsin2=(x4+1x4−2)(x2+1x2+2)⇒−26cos4θsin2=x6+x2+2x4+1x2+1x6+2x4−2x2−2x2−4⇒−26cos4θsin2=2cos6θ+4cos4θ−2cos2θ−4cos4θsin2θ=132(2+cos2θ−2cos4θ−cos6θ)

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$\cos^{4} θ \sin^{2} θ =$

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$\frac{1}{32} (\cos 6 θ + 2 \cos 4 θ + \cos 2 θ + 2)$

$\frac{1}{32} (2 + \cos 2 θ - 2 \cos 4 θ - \cos 6 θ)$

$\frac{1}{32} (2 + \cos 2 θ + 2 \cos 4 θ - \cos 6 θ)$

$\frac{1}{32} (\cos 6 θ + 2 \cos 4 θ - \cos 2 θ - 2)$

answer is C.

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Detailed Solution

$x = \cos θ + i \sin θ \Rightarrow \frac{1}{x} = \cos θ - i \sin θ$

$2 \cos θ = x + \frac{1}{x} \Rightarrow 2 i \sin θ = x - \frac{1}{x}$

$\Rightarrow {(x + \frac{1}{x})}^{4} {(x - \frac{1}{x})}^{2} = (2^{4} \cos^{4} θ) (2^{2} (- 1) \sin^{2} θ)$

$\Rightarrow - 2^{6} \cos^{4} θ \sin^{2} = {(x^{2} - \frac{1}{x^{2}})}^{2} {(x + \frac{1}{x})}^{2}$

$\Rightarrow - 2^{6} \cos^{4} θ \sin^{2} = (x^{4} + \frac{1}{x^{4}} - 2) (x^{2} + \frac{1}{x^{2}} + 2)$

$\Rightarrow - 2^{6} \cos^{4} θ \sin^{2} = x^{6} + x^{2} + 2 x^{4} + \frac{1}{x^{2}} + \frac{1}{x^{6}} + \frac{2}{x^{4}} - 2 x^{2} - \frac{2}{x^{2}} - 4$