Question

If $x\in (\pi ,\frac{3\pi }{2})$ then
$4{\cos }^2(\frac{\pi }{4}-\frac{x}{2})+\sqrt{4{\sin }^{4}x+{\sin }^{2}2x}$ is always equal to :-

• A. $-2$
• B. $1$
• C. $2$
• D. $0$

Answer 1

The correct option is C $2$

$formulaused$
$1+\cos x=2{\cos }^{2}x/2$
$\sin 2x=2\sin x\cos x$
$Given$
$4{\cos }^2(\frac{\pi }{4}-\frac{x}{2})+\sqrt{4{\sin }^{4}x+{\sin }^{2}2x}$
$=4{\cos }^2(\frac{\pi }{4}-\frac{x}{2})+\sqrt{4{\sin }^{4}x+{\left(2\sin x\cos x\right)}^2}$
$=4{\cos }^2(\frac{\pi }{4}-\frac{x}{2})+\sqrt{4{\sin }^{4}x+\left(4{\sin }^{2}x{\cos }^{2}x\right)}$
$=2\left(2{\cos }^2(\frac{\pi }{4}-\frac{x}{2})\right)+\sqrt{4{\sin }^{2}x({\sin }^{2}x+{\cos }^{2}x)}$
$=2(1+\cos (\frac{\pi }{2}-x))+\sqrt{4{\sin }^{2}x}$
$=2(1+\sin x)+2\left|\sin x\right|$
$asx\in (\pi ,\frac{3\pi }{2})$
$=2+2\sin x+2(-\sin x)$
$=2$