Given that pi

Question

(A) Given $\pi < \alpha < \frac{3\pi}{2},$ $\alpha$ lies in the third quadrant.Simplify the expression: $\sqrt{4\sin^4 \alpha + \sin^2 2\alpha} + 4\co

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given $\pi Simplify the expression: $\sqrt{4\sin^4 \alpha + \sin^2 2\alpha} + 4\cos^2 \left( \frac{\pi}{4} - \frac{\alpha}{2} \right)$$= \sqrt{4\sin^4 \alpha + 4\sin^2 \alpha \cos^2 \alpha} + 2 \left[ 2\cos^2 \left( \frac{\pi}{4} - \frac{\alpha}{2} \right) \right]$$= \sqrt{4\sin^2 \alpha (\sin^2 \alpha + \cos^2 \alpha)} + 2 \left[ 1 + \cos \left( \frac{\pi}{2} - \alpha \right) \right]$$= \sqrt{4\sin^2 \alpha} + 2(1 + \sin \alpha) = 2|\sin \alpha| + 2 + 2\sin \alpha$Since $\pi Expression $= 2(-\sin \alpha) + 2 + 2\sin \alpha = 2$.

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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