If $\pi<\alpha<\frac{3\pi}{2}$ then the expression $\sqrt{4\sin^4\alpha+\sin^2 2\alpha}+4\cos^2(\frac{\pi}{4}-\frac{\alpha}{2})$

Question

For $\sqrt {4\sin^4\alpha+\sin^2 2\alpha}$

=$\sqrt {4\sin^2\alpha + 4\sin^2\alpha \cos^2\alpha}$

=$2\sin\alpha$

Then for $4\cos^2(\frac{\pi}{4}-\frac{\alpha}{2})$

$=2+2\sin\alpha$

Adding them we get $$2+4\sin\alpha$$

My questions is why isn’t it $2-4\sin\alpha$? I get that it may have have something to do with the calculation, but I still would like help with that.

Thanks!

The right answer is $2+4\sin\alpha$

Answer 1

We note that in $\pi < \alpha < \frac{3\pi}{2}$, the value of $\sin \alpha$ is negative.

Now, $$\sqrt{4\sin^4 \alpha + \sin^2 2\alpha}$$ $$= \sqrt{4\sin^2 \alpha[\sin^2 \alpha + \cos^2 \alpha]} $$ $$ = |2 \sin \alpha| = - 2\sin \alpha$$

Now, you have correctly simplified the value of $4\cos^2(\frac{\pi}{4} - \frac{\alpha}{2})$ as $2 + 2\sin \alpha$.

As a result, the answer must simply be $2$.