A) \[2\cos 3(\beta -\alpha )\]
B) \[2\cos 3(\beta +\alpha )\]
C) \[2\sin 3(\beta -\alpha )\]
D) \[2\sin 3(\beta +\alpha )\]
Correct Answer: C
Solution :
\[\because \]\[x+\frac{1}{x}=2\sin \alpha ,\,\,y+\frac{1}{y}=2\cos \beta \] \[\therefore \]\[{{x}^{2}}-2\sin \alpha \,x+1=0,\,\,{{y}^{2}}-2\cos \beta y+1=0\] \[\Rightarrow \] \[x=\frac{2\sin \alpha \pm \sqrt{4{{\sin }^{2}}\alpha -4}}{2}\] \[\Rightarrow \] \[x=\sin \alpha \pm i\cos \alpha \] Similarly, \[\therefore \]\[xy=(\sin \alpha \pm i\cos \alpha )(\cos \beta \mp i\sin \beta )\] \[y=\cos \beta \mp i\sin \beta \] \[=\sin (\beta -\alpha )\pm i\cos (\beta -\alpha )\] and \[\frac{1}{xy}=\sin (\beta -\alpha )\mp i\cos (\beta -\alpha )\] \[\therefore \] \[{{(xy)}^{3}}+\frac{1}{{{(xy)}^{3}}}=\sin 3(\beta -\alpha )\] \[\pm i\cos 3(\beta -\alpha )+\sin 3(\beta -\alpha )\] \[\mp i\cos 3(\beta -\alpha )\] \[=2\sin 3(\beta -\alpha )\]
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