Aubrey Hendricks

## Answered question

2022-01-25

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

### Answer & Explanation

rakije2v

Beginner2022-01-26Added 12 answers

We note that in $\pi <\alpha <\frac{3\pi }{2}$, the value of sinα is negative.
Now,
$\sqrt{4{\mathrm{sin}}^{4}\alpha +{\mathrm{sin}}^{2}2\alpha }$
$=\sqrt{4{\mathrm{sin}}^{2}\alpha \left[{\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha \right]}$
$=|2\mathrm{sin}\alpha |=-2\mathrm{sin}\alpha$
Now, you have correctly simplified the value of $4{\mathrm{cos}}^{2}\left(\frac{\pi }{4}-\frac{\alpha }{2}\right)$ as $2+2\mathrm{sin}\alpha$
As a result, the answer must simply be 2

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