JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Trigonometrical ratios of multiple and sub multiple angles

  • question_answer
    If \[\sin \theta +\cos \theta =x,\] then \[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta =\frac{1}{4}[4-3{{({{x}^{2}}-1)}^{2}}]\] for

    A) All real x

    B) \[{{x}^{2}}\le 2\]

    C) \[{{x}^{2}}\ge 2\]

    D) None of these

    Correct Answer: B

    Solution :

    On squaring the given relation \[\sin 2\theta ={{x}^{2}}-1\le 1\Rightarrow {{x}^{2}}\le 2\] or \[-\sqrt{2}\le x\le \sqrt{2}\] \[[\because \,\,\sin 2\theta \le 1]\] Now \[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta \]        \[={{({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )}^{3}}-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta ({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )\] \[=1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta =1-\frac{3}{4}{{\sin }^{2}}2\theta \] \[=1-\frac{3}{4}{{({{x}^{2}}-1)}^{2}}=\frac{1}{4}\{4-3{{({{x}^{2}}-1)}^{2}}\}\] Thus the given result will hold true only when \[{{x}^{2}}\le 2\]and not for all real values of x.

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