A geometry problem by Shivam Hinduja

First, we look at $4\sin^{4} \theta+\sin^{2} 2\theta$ .

Using the double-angle identities, we see that

$4\sin^{4} \theta = (1-\cos 2\theta)^{2} = 1-2\cos 2\theta+\cos^{2} 2\theta$

and thus

$4\sin^{4} \theta+\sin^{2} 2\theta = 2-2\cos 2\theta = 4\sin^{2} \theta$

Taking the square root immediately gives $2|\sin \theta|$ .

Using the double-angle identities yet again,

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

Finally, we combine them:

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

If $\theta$ is in the third quadrant, $\sin \theta$ must be negative and so $2|\sin \theta|$ and $2\sin \theta$ cancel each other out, leaving

$2|\sin \theta|+2+2\sin \theta = \boxed{2}$

I just putted theta as 240 degrees