Given that pi

Question

(A) Given $\pi < \alpha < \frac{3\pi}{2}$,$\alpha$ lies in the third quadrant,so $\sin \alpha < 0$.Let $E = \sqrt{4\sin^4 \alpha + \sin^2 2\alpha} + 4

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given $\pi Let $E = \sqrt{4\sin^4 \alpha + \sin^2 2\alpha} + 4\cos^2 \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$.Using $\sin 2\alpha = 2\sin \alpha \cos \alpha$,we have $\sin^2 2\alpha = 4\sin^2 \alpha \cos^2 \alpha$.$E = \sqrt{4\sin^4 \alpha + 4\sin^2 \alpha \cos^2 \alpha} + 2(1 + \cos(\frac{\pi}{2} - \alpha))$$E = \sqrt{4\sin^2 \alpha (\sin^2 \alpha + \cos^2 \alpha)} + 2(1 + \sin \alpha)$$E = \sqrt{4\sin^2 \alpha} + 2 + 2\sin \alpha$$E = 2|\sin \alpha| + 2 + 2\sin \alpha$Since $\alpha$ is in the third quadrant,$\sin \alpha $E = 2(-\sin \alpha) + 2 + 2\sin \alpha = 2$.

Given that $\pi < \alpha < \frac{3\pi}{2}$,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

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