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Let $x+y=3-\cos 4 \theta$ and $x-y=4 \sin 2 \theta$ then the greatest of $x y$ is
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$x=\frac{3-\cos 4 \theta+4 \sin 2 \theta}{2}$
$=\frac{3-\left(1-\sin ^{2} 2 \theta\right)+4 \sin 2 \theta}{2}=(1+\sin 2 \theta)^{2}$
$y=\frac{3-\cos 4 \theta-4 \sin 2 \theta}{2}$
$=\frac{3-\left(1-\sin ^{2} 2 \theta\right)+4 \sin 2 \theta}{2}=(1-\sin 2 \theta)^{2}$
$\therefore x y=\left(1-\sin ^{2} 2 \theta\right)^{2}=\cos ^{4} 2 \theta \leq 1$
$=\frac{3-\left(1-\sin ^{2} 2 \theta\right)+4 \sin 2 \theta}{2}=(1+\sin 2 \theta)^{2}$
$y=\frac{3-\cos 4 \theta-4 \sin 2 \theta}{2}$
$=\frac{3-\left(1-\sin ^{2} 2 \theta\right)+4 \sin 2 \theta}{2}=(1-\sin 2 \theta)^{2}$
$\therefore x y=\left(1-\sin ^{2} 2 \theta\right)^{2}=\cos ^{4} 2 \theta \leq 1$
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