A) \[{{2}^{n}}\cos \alpha \]
B) \[{{2}^{n}}\cos n\alpha \]
C) \[2i\sin n\alpha \]
D) \[2\cos \,n\alpha \]
Correct Answer: D
Solution :
Given that \[x+\frac{1}{x}=2\cos \alpha \] \[\Rightarrow \] \[{{x}^{2}}-2\cos \alpha +1=0\] \[x=\frac{+2\cos \pm \sqrt{4{{\cos }^{2}}\alpha -4}}{2}\] \[x=\cos \alpha +i\sin \alpha \] Now, \[{{x}^{n}}={{(\cos \alpha +i\sin \alpha )}^{n}}=\cos n\,\alpha +i\sin \,n\,\alpha \] and \[\frac{1}{{{x}^{n}}}={{(\cos \alpha -i\sin \alpha )}^{n}}=\cos n\,\alpha -i\sin \,n\,\alpha \] \[\therefore \,\,{{x}^{n}}+\frac{1}{{{x}^{n}}}=\cos \,n\,\alpha +i\sin \,n\,\alpha \] \[+\,\cos \,n\,\alpha -i\sin \,n\,\alpha \] \[=2\cos \,n\,\alpha \]
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