Trigonometric Identities Question 60
Question: Given that $ \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} ) $ is equal to
Options:
A) 2
B) $ 2+4\sin \alpha $
C) $ 2-4\sin \alpha $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Given that $ \pi <\alpha <\frac{3\pi }{2}i.e.,\alpha $ is in third quadrant. Now, $ \sqrt{(4{{\sin }^{4}}\alpha +{{\sin }^{2}}2\alpha )}+4{{\cos }^{2}}( \frac{\pi }{4}-\frac{\alpha }{2} ) $ $ =\sqrt{(4{{\sin }^{4}}\alpha +4{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha )}+2.2{{\cos }^{2}}( \frac{\pi }{4}-\frac{\alpha }{2} ) $ $ =\sqrt{4{{\sin }^{2}}\alpha ({{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha )}+2[ 1+\cos ( \frac{\pi }{2}-\alpha ) ] $ $ =\pm 2\sin \alpha +2+2\sin \alpha $ On taking -ve, answer is 2 and on taking +ve, answer is $ 2+4\sin \alpha $ But $ \pi <\alpha <\frac{3\pi }{4}, $ Hence answer is $ 2-4\sin \alpha $ because $ \sin \alpha $ is $ -ve $ in third quadrant.
BETA
