JEE Main & Advanced Sample Paper JEE Main - Mock Test - 6

  • question_answer
    If\[{{x}^{2}}-2x\cos \theta +1=0\]then,\[{{x}^{2n}}-2{{x}^{n}}\operatorname{cosn}\theta +1\]is equal to

    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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