Question

If $\theta$ lies in the third quadrant, then the expression $\sqrt{(4{\sin }^4\theta +{\sin }^{2}2\theta )}+4{\cos }^2(\frac{1}{4}\pi -\frac{1}{2}\theta )$ equals $2.$

• A. True
• B. False

Answer 1

The correct option is A True

$True$.

We have

$\surd \{4{\sin }^4\theta +{\sin }^{2}2\theta \}=\surd \left\{4{\sin }^2\theta ({\sin }^2\theta +{\cos }^2\theta )\right\}$

$\sqrt{\left(4{\sin }^2\theta \right)}=2\left|\sin \theta \right|$.

But since $\theta$ lies in the third quadrant, we have $\sin \theta <0$ and $|\sin \theta |=-\sin \theta$.

Hence $\surd 4{\sin }^4\theta +{\sin }^{2}2\theta =-2\sin \theta$.

and $4{\cos }^2(\frac{\pi }{4}-\frac{\theta }{2})=2[1+\cos (\frac{\pi }{2}-\theta \left)\right]=2+2\sin \theta$.

Hence the given expression

$=-2\sin \theta +2+2\sin \theta =2$.

[Note that $|x|=x$ if $x\ge 0$ and $=-x$ if $x\le 0$