Question

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

• A. $2$
• B. $-2$
• C. $0$
• D. $1$

Answer 1

The correct option is A $2$

$\sqrt{4{\sin }^4\theta +{\sin }^{2}2\theta }+4{\cos }^2(\frac{\pi }{4}-\frac{\theta }{2})$

$\theta$ lies in $3rd$ quadrant.

$\Rightarrow$ Put $\theta =\frac{3\pi }{2}$

$\Rightarrow \theta =\frac{3\pi }{2}$

$=\sqrt{4{\left(\sin \frac{3\pi }{2}\right)}^4+{(\sin \frac{3\pi }{2}\times 2)}^2}+4{\cos }^2(\frac{\pi }{4}-\frac{3\pi }{4})$

$=\sqrt{4{(-1)}^4+0}+4{\cos }^2(-\frac{\pi }{2})$

$=\sqrt{4}+0$

$=2$

Hence, the answer is $2.$