A) \[2{{\sin }^{2}}\left( \frac{\pi }{4}-\alpha \right)\]
B) \[2{{\cos }^{2}}\left( \frac{\pi }{4}-\alpha \right)\]
C) \[2{{\cos }^{2}}\left( \frac{\pi }{4}+\alpha \right)\]
D) \[2{{\sin }^{2}}\left( \frac{\pi }{4}+\alpha \right)\]
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 \[\cos 2\beta =1-2{{\sin }^{2}}\beta =1-2\sin \alpha \cos \alpha \] \[={{(\cos \alpha -\sin \alpha )}^{2}}=2\,{{\left( \frac{1}{\sqrt{2}}\cos \alpha -\frac{1}{\sqrt{2}}\sin \alpha \right)}^{2}}\] \[=2{{\sin }^{2}}\left( \frac{\pi }{4}-\alpha \right)\], which is given in (a). Also \[\cos 2\beta =2{{\cos }^{2}}\left\{ \frac{\pi }{2}-\left( \frac{\pi }{4}-\alpha \right) \right\}=2{{\cos }^{2}}\left( \frac{\pi }{4}+\alpha \right)\], which is given in (c).
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