If $\sin A + \sin 2 A = x, \cos A + \cos 2 A = y$ then $\cos A =$

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\frac{1}{2} (x^{2} + y^{2} - 2)

x^{2} + y^{2} + 2

\frac{1}{2} (x^{2} + y^{2} + 2)

2 (x^{2} + y^{2} + 2)

answer is B.

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Detailed Solution

\sin A + \sin 2 A = x

$\Rightarrow x^{2} = {[\sin A + \sin 2 A]}^{2}$

$\Rightarrow x^{2} = \sin^{2} A + \sin^{2} 2 A + 2 \sin A . \sin 2 A$

$y = \cos A + \cos 2 A$

$\Rightarrow y^{2} = \cos^{2} A + \cos^{2} 2 A + 2 \cos A . \cos 2 A$

$x^{2} + y^{2} = [\sin^{2} A + \cos^{2} A] + [\cos^{2} 2 A + \sin^{2} 2 A] + 2 \sin A . \sin 2 A + 2 \cos A . \cos 2 A$

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

$= 2 + 4 \sin^{2} A . \cos A + 4 \cos^{3} A - 2 \cos A$

$= 2 + 4 \cos A [\sin^{2} A + \cos^{2} A] - 2 \cos A$

$= 2 + 4 \cos A - 2 \cos A$

$\Rightarrow x^{2} + y^{2} = 2 + 2 \cos A$

$\Rightarrow x^{2} + y^{2} - 2 = 2 \cos A$

$\Rightarrow \frac{1}{2} [x^{2} + y^{2} - 2] = \cos A$

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