Trigonometric Identities Question 346
Question: $ 2,{{\sin }^{2}}\beta +4\cos ,(\alpha +\beta )\sin ,\alpha ,\sin ,\beta +\cos ,2,(\alpha +\beta )= $
[MNR 1993; IIT 1977]
Options:
A) $ \sin 2\alpha $
B) $ \cos 2\beta $
C) $ \cos 2\alpha $
D) $ \sin 2\beta $
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Answer:
Correct Answer: C
Solution:
$ \cos 2(\alpha +\beta )=2{{\cos }^{2}}(\alpha +\beta )-1,,2{{\sin }^{2}}\beta =1-\cos 2\beta $ L.H.S. $ =-\cos 2\beta +2\cos (\alpha +\beta ),[2\sin \alpha \sin \beta +\cos (\alpha +\beta )] $ $ =-\cos 2\beta +2\cos (\alpha +\beta )\cos (\alpha -\beta ) $ $ =-\cos 2\beta +(\cos 2\alpha +\cos 2\beta )=\cos 2\alpha $ .
BETA
