For $\pi<\alpha<\frac{3\pi}{2}$ what is the value of $\sqrt{4\sin^4\alpha + \sin^2 2\alpha} + 4\cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$?

Question

I solved part of it this way: $$\sqrt{4\sin^4\alpha + \sin^2 2\alpha} + 4\cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$$ $$= \sqrt{4\sin^4\alpha + 4\sin^2\alpha \cos^2\alpha} + 2\cos\left(\frac{\pi}{2} - \alpha\right) + 2$$ because $\sin 2\alpha = 2\sin\alpha\cos\alpha$ and $\cos2\alpha = \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha - 1$.

The expression then reduces to: $$\sqrt{4\sin^2\alpha(\sin^2 \alpha + \cos^2\alpha)} + 2\sin\alpha + 2$$ since $\cos\left(\frac{\pi}{2} - \alpha\right) = \sin\alpha$. Now, because $\sin^2 \alpha + \cos^2\alpha = 1$, the expression becomes: $$\sqrt{4\sin^2\alpha} + 2\sin\alpha + 2$$ Keep in mind that $\pi < \alpha < \frac{3\pi}{2}$. How do you proceed? I always seem to be left with extraneous sines of $\alpha$, while according to my textbook, the answer is $2$.

Answer 1

Now, since $\pi<\alpha<\frac{3\pi}{2}$, it's $$-2\sin\alpha+2\sin\alpha+2=2$$ because for $x<0$ we have: $$\sqrt{x^2}=-x.$$

Answer 2

Hint: For that range of $\alpha$, $\sin\alpha\leq 0).

Also note that for all real numbers $x$, one has that $\sqrt{x^2}=|x|$.