If $(1 + \cos θ - i \sin θ) (1 + \cos 2 θ + i \sin 2 θ) = x + i y$ then $x^{2} + y^{2} =$

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$16 \cos^{2} θ \sin^{2} (\frac{θ}{2})$

$16 \sin^{2} θ \cos^{2} (\frac{θ}{2})$

$16 \sin^{2} θ \sin^{2} (\frac{θ}{2})$

$16 \cos^{2} θ \cos^{2} (\frac{θ}{2})$

answer is D.

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Detailed Solution

$((1 + \cos θ) - i \sin θ) ((1 + \cos 2 θ) + i \sin 2 θ)$

$= (2 \cos^{2} \frac{θ}{2} - i 2 \sin \frac{θ}{2} \cos \frac{θ}{2}) (2 \cos^{2} θ + i 2 \sin θ \cos θ)$

$= 2 \cos \frac{θ}{2} 2 \cos θ [\cos \frac{θ}{2} - i \sin \frac{θ}{2}] [\cos θ + i \sin θ]$

$x + i y = 4 \cos \frac{θ}{2} \cos θ (\cos \frac{θ}{2} + i \sin \frac{θ}{2})$

Taking modulus both side and squaring

$∴ x^{2} + y^{2} = 16 \cos^{2} θ \cos^{2} \frac{θ}{2} (\cos^{2} \frac{θ}{2} + \sin^{2} \frac{θ}{2})$

$= 16 \cos^{2} θ \cos^{2} \frac{θ}{2}$

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