Question

If $\theta \in \left[\pi ,\frac{3\pi }{2}\right]$, then $\sqrt{4{sin}^4\theta +{sin}^{2}2\theta }+{4cos}^2(\frac{\pi }{4}-\frac{\theta }{2})=$

• A. 2
• B. -2
• C. 0
• D. 1

Answer 1

The correct option is A 2

Given $\theta \in \left[\pi ,\frac{3\pi }{2}\right]⟹\sin \theta <0⟹|\sin \theta |=-\sin \theta$

$\sqrt{4{\sin }^4\theta +{\sin }^{2}2\theta }=\sqrt{4{\sin }^4\theta +4{\sin }^2\theta {\cos }^2\theta }=2\sqrt{{\sin }^2\theta ({\sin }^2\theta +{\cos }^2\theta )}=2|\sin \theta |=-2\sin \theta$

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

$=2(\cos \frac{\theta }{2}+\sin \frac{\theta }{2}{)}^2$

$=2({\cos }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}+2\sin \frac{\theta }{2}\cos \frac{\theta }{2})$

$=2+2\sin \theta$

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