Question

Let $x+y=3-\cos 4\theta$ and $x-y=4\sin 2\theta$ then the greatest of $xy$ is

• A. 34
• B. 1
• C. $\frac{1}{2}$
• D. 2

Answer 1

Answer: (B)

$x=\frac{3-\cos 4\theta +4\sin 2\theta }{2}$
$=\frac{3-\left(1-{\sin }^{2}2\theta \right)+4\sin 2\theta }{2}=(1+\sin 2\theta {)}^2$
$y=\frac{3-\cos 4\theta -4\sin 2\theta }{2}$
$=\frac{3-\left(1-{\sin }^{2}2\theta \right)+4\sin 2\theta }{2}=(1-\sin 2\theta {)}^2$
$\therefore xy={\left(1-{\sin }^{2}2\theta \right)}^2={\cos }^{4}2\theta \le 1$