A) \[\cos 2n\theta \]
B) \[sin2n\theta \]
C) 0
D) R - {0}
Correct Answer: C
Solution :
[c] : From given relation, we have \[x=\cos \theta \pm i\sin \theta \] Take \[x=\cos \theta +i\sin \theta \] \[\therefore \]\[{{x}^{n}}=\cos \theta +i\sin n\theta \][By De-Moivres theorem] and \[1/{{x}^{n}}=\operatorname{cosn}\theta +i\sin n\theta \] \[\therefore \]\[{{x}^{n}}+(1/{{x}^{n}})=2cosn\theta \] \[\therefore \]\[{{x}^{2n}}-2{{x}^{n}}cos\,n\theta +1=0\]
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