Question

If $\pi <\alpha <\frac{3\pi }{2},$ then the expression $\sqrt{4{\sin }^4\alpha +{\sin }^{2}2\alpha }+4{\cos }^2\left(\frac{\pi }{4}-\frac{\alpha }{2}\right)$ is equal to

• A. $2-4\sin \alpha$
• B. $2+4\sin \alpha$
• C. 2
• D. none of these.

Answer 1

Answer: (C)

$\sqrt{4{\sin }^4\alpha +{\sin }^{2}2\alpha }+4{\cos }^2\left(\frac{\pi }{4}-\frac{\alpha }{2}\right)$
$=\sqrt{4{\sin }^4\alpha +4{\sin }^2\alpha {\cos }^2\alpha }+2\left\{1+\cos \left(\frac{\pi }{2}-\alpha \right)\right\}$
$\left(\because 2{\cos }^2\theta =1+\cos 2\theta \right)$
$=\sqrt{4{\sin }^2\alpha }+2+2\sin \alpha$
$=2\left|\sin \alpha \right|+2+2\sin \alpha$
$=2\left(-\sin \alpha \right)+2+2\sin \alpha =2$
$\left(\because \pi <\alpha <\frac{3\pi }{2},\text{therefore}\left|\sin \alpha \right|=-\sin \alpha \right)$