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 :
We have, \[x+\frac{1}{x}=2\cos \alpha \] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2=4{{\cos }^{2}}\alpha \]. \[\] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=4{{\cos }^{2}}\alpha -2\], \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=2(2{{\cos }^{2}}\alpha -1)=2\cos 2\alpha \] Similarly \[{{x}^{n}}+\frac{1}{{{x}^{n}}}=2\cos \,n\alpha \].
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