Trigonometric Identities Question 124
Question: If $ \sin \beta $ is the geometric mean between $ \sin \alpha $ and $ \cos \alpha , $ then $ \cos 2\beta $ is equal to
Options:
A) $ 2{{\sin }^{2}}( \frac{\pi }{4}-\alpha ) $
B) $ 2{{\cos }^{2}}( \frac{\pi }{4}-\alpha ) $
C) $ 2{{\cos }^{2}}( \frac{\pi }{4}+\alpha ) $
D) $ 2{{\sin }^{2}}( \frac{\pi }{4}+\alpha ) $
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Answer:
Correct Answer: C
Solution:
Since $ \sin \beta $ is G.M. between $ \sin \alpha $ and $ \cos \alpha $ .
$ \therefore {{\sin }^{2}}\beta =\sin \alpha \cos \alpha $ Now $ \cos 2\beta =1-2{{\sin }^{2}}\beta =1-2\sin \alpha \cos \alpha $ $ ={{(\cos \alpha -\sin \alpha )}^{2}}=2,{{( \frac{1}{\sqrt{2}}\cos \alpha -\frac{1}{\sqrt{2}}\sin \alpha )}^{2}} $ $ =2{{\sin }^{2}}( \frac{\pi }{4}-\alpha ) $ , which is given in (a). Also $ \cos 2\beta =2{{\cos }^{2}}{ \frac{\pi }{2}-( \frac{\pi }{4}-\alpha ) }=2{{\cos }^{2}}( \frac{\pi }{4}+\alpha ) $ , which is given in (c).
BETA
