A) \[\sin \,\,2\alpha \]
B) \[\cos \,\,2\beta \]
C) \[\cos \,\,2\alpha \]
D) \[\sin \,\,2\beta \]
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. \[=-\cos 2\beta +2\cos (\alpha +\beta )\,[2\sin \alpha \sin \beta +\cos (\alpha +\beta )]\] \[=-\cos 2\beta +2\cos (\alpha +\beta )\cos (\alpha -\beta )\] \[=-\cos 2\beta +(\cos 2\alpha +\cos 2\beta )=\cos 2\alpha \].
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