CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    \[\sin \left( \frac{1}{2}{{\cos }^{-1}}\frac{4}{5} \right)\] is equal to :

    A)  \[-\frac{1}{\sqrt{10}}\]                

    B)  \[\frac{1}{\sqrt{10}}\]

    C)  \[-\frac{1}{10}\]                              

    D)  \[\frac{1}{10}\]

    Correct Answer: B

    Solution :

    We have \[\sin \left( \frac{1}{2}{{\cos }^{-1}}\frac{4}{5} \right)\] Now put \[\frac{4}{5}=\cos 2\theta \] \[\therefore \]  \[\sin \left( \frac{1}{2}\times 2\theta  \right)\] \[\sin \theta =\sqrt{\frac{1-\cos 2\theta }{2}}\]                 \[=\sqrt{\frac{1-\frac{4}{5}}{2}}=\sqrt{\frac{1}{5\times 2}}=\frac{1}{\sqrt{10}}\]


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