A) \[-1\]
B) 0
C) \[\sqrt{2}\]
D) \[-\sqrt{2}\]
Correct Answer: D
Solution :
Since, \[A.M.\ge G.M.\] \[\therefore \]\[\frac{{{\sin }^{4}}\alpha +4{{\cos }^{4}}\beta +1+1}{4}\] \[\ge {{({{\sin }^{4}}\alpha .4{{\cos }^{4}}\beta .1.1)}^{1/4}}\] \[\Rightarrow \]\[{{\sin }^{4}}\alpha +4{{\cos }^{4}}\beta +2\ge 4\sqrt{2}\sin \alpha \cos \beta \] But according to question, we have \[{{\sin }^{4}}\alpha +4{{\cos }^{4}}\beta +2=4\sqrt{2}\sin \alpha \cos \beta \] \[\Rightarrow \]\[A.M.=G.M.\Rightarrow si{{n}^{4}}\alpha =1=4{{\cos }^{4}}\beta \] \[\Rightarrow \]\[sin\alpha =1,\cos \beta =\pm \frac{1}{\sqrt{2}}\] \[\therefore \]\[sin\alpha =1,and\,\sin \beta =\frac{1}{\sqrt{2}}\]as\[\alpha ,\beta \in [0,\pi ]\] Now,\[\cos (\alpha +\beta )-\cos (\alpha -\beta )=-2sin\alpha \beta \] \[=-2.1.\frac{1}{\sqrt{2}}=-\sqrt{2}\]
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