Question

If ${}^{\prime }{\theta }^{\prime }$ is in the III quadrant then $\sqrt{4{\sin }^4\theta }+4{\cos }^2(\frac{\pi }{4}-\frac{\theta }{2})=$

• A. $2$
• B. $-2$
• C. $0$
• D. $1$

Answer 1

The correct option is A $2$

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

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

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

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

$2{\sin }^2\theta +4{(\cos \frac{\theta }{2}+\sin \frac{\theta }{2})}^2{\left(\frac{1}{\sqrt{2}}\right)}^2$

$2{\sin }^2\theta +2(1+\sin \theta )$

$2{\sin }^2\theta +2+2\sin \theta$

$2(1+{\sin }^2\theta +\sin \theta )$

In third quadrant $\sin \theta$ is negative

$\Rightarrow 2(1+1-1)=2$