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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    If\[1+cos\text{ }x=k,\]where\[x\]is acute, then\[\sin \frac{x}{2}\]is:

    A)  \[\sqrt{\frac{1-k}{2}}\]

    B)         \[\sqrt{2-k}\]

    C)  \[\sqrt{\frac{2+k}{2}}\]

    D)         \[\sqrt{\frac{2-k}{2}}\]

    E)  \[\sqrt{\frac{k}{2}}\]

    Correct Answer: E

    Solution :

    Given that, \[1+\cos x=k\] \[\Rightarrow \]               \[2{{\sin }^{2}}(x/2)=k\] \[\Rightarrow \]               \[\sin (x/2)=\sqrt{\frac{k}{2}}\]


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