Question

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

• A. $ 2 - 4 \, \sin \, \alpha $
• B. $ 2 + 4 \, \sin \, \alpha $
• C. 2
• D. none of these.

Answer 1

Answer: (C)

$\sqrt{4 \sin^{4} \alpha + \sin^{2} 2\alpha} + 4 \cos^{2} \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) $
$= \sqrt{4 \sin^{4} \alpha + 4 \sin^{2} \alpha \cos^{2} \alpha} + 2\left\{1+\cos\left( \frac{\pi}{2} -\alpha\right)\right\}$
$ \left(\because 2 \cos^{2} \theta = 1 + \cos2\theta\right)$
$ = \sqrt{4\sin^{2} \alpha} + 2 + 2 \sin \alpha $
$ = 2 \left|\sin\alpha\right| + 2 + 2 \sin\alpha$
$ = 2\left(-\sin\alpha\right) + 2 + 2 \sin\alpha = 2$
$ \left(\because \pi< \alpha < \frac{3\pi}{2} , \text{therefore} \left|\sin\alpha\right| =- \sin\alpha\right)$