JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Trigonometrical ratios of sum and difference of two and three angles

  • question_answer
    \[{{\cos }^{2}}48{}^\circ -{{\sin }^{2}}12{}^\circ =\]      [MNR 1977]

    A) \[\frac{\sqrt{5}-1}{4}\]

    B) \[\frac{\sqrt{5}+1}{8}\]

    C) \[\frac{\sqrt{3}-1}{4}\]

    D) \[\frac{\sqrt{3}+1}{2\sqrt{2}}\]

    Correct Answer: B

    Solution :

    \[{{\cos }^{2}}A-{{\sin }^{2}}B=\cos \,(A+B)\,.\,\cos \,(A-B)\] \[\therefore \,\,{{\cos }^{2}}{{48}^{o}}-{{\sin }^{2}}{{12}^{o}}=\cos \,\,{{60}^{o}}\,.\,\cos \,\,{{36}^{o}}\] \[=\frac{1}{2}\,\left( \frac{\sqrt{5}+1}{4} \right)=\frac{\sqrt{5}+1}{8}.\]


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