A) \[{{2}^{n}}\cos n\theta \]
B) \[{{2}^{n}}{{\cos }^{n}}\theta \]
C) \[2{{\cos }^{n}}\theta \]
D) \[2\cos n\theta \]
E) \[2\sin n\theta \]
Correct Answer: D
Solution :
\[{{x}^{2}}-2x\cos \theta +1=0\] \[\Rightarrow \] \[x=\frac{2\cos \theta \pm \sqrt{4{{\cos }^{2}}\theta -4}}{2}\] \[x=\cos \theta \pm i\sin \theta \] Take\[+\]sign, \[x=\cos \theta +i\sin \theta \] \[\Rightarrow \] \[{{x}^{n}}=\cos n\theta +i\sin n\theta \] \[\therefore \] \[{{x}^{n}}+\frac{1}{{{x}^{n}}}=\cos n\theta +i\sin n\theta +\cos n\theta -i\sin n\theta \] \[=2\cos n\theta \]
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