If theta is in the third quadrant,then sqrt{4 | vedclass

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(B) Given expression: $E = \sqrt{4 \sin^4 	heta + \sin^2 2 	heta} + 4 \cos^2 \left(\frac{\pi}{4} - \frac{	heta}{2}ight)$.First,simplify the term

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$2$

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(B) Given expression: $E = \sqrt{4 \sin^4 	heta + \sin^2 2 	heta} + 4 \cos^2 \left(\frac{\pi}{4} - \frac{	heta}{2}ight)$.First,simplify the term inside the square root:$4 \sin^4 	heta + \sin^2 2 	heta = 4 \sin^4 	heta + (2 \sin 	heta \cos 	heta)^2 = 4 \sin^4 	heta + 4 \sin^2 	heta \cos^2 	heta = 4 \sin^2 	heta (\sin^2 	heta + \cos^2 	heta) = 4 \sin^2 	heta$.So,$\sqrt{4 \sin^2 	heta} = 2 |\sin 	heta|$.Since $	heta$ is in the third quadrant,$\sin 	heta Next,simplify the second term:$4 \cos^2 \left(\frac{\pi}{4} - \frac{	heta}{2}ight) = 2 \left[2 \cos^2 \left(\frac{\pi}{4} - \frac{	heta}{2}ight)ight] = 2 \left[1 + \cos \left(\frac{\pi}{2} - 	hetaight)ight] = 2 (1 + \sin 	heta) = 2 + 2 \sin 	heta$.Adding both parts:$E = -2 \sin 	heta + 2 + 2 \sin 	heta = 2$.

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

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