CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    The real part of\[{{\left[ 1+\cos \left( \frac{\pi }{5} \right)+i\sin \left( \frac{\pi }{5} \right) \right]}^{-1}}\]is:

    A)  \[1\]                                    

    B)  \[\frac{1}{2}\]

    C)  \[\frac{1}{2}\cos \left( \frac{\pi }{10} \right)\]   

    D)         \[\frac{1}{2}\cos \left( \frac{\pi }{5} \right)\]

    E)  \[\frac{1}{2}\sec \left( \frac{\pi }{10} \right)\]

    Correct Answer: B

    Solution :

     \[{{\left[ 1+\cos \frac{\pi }{5}+i\sin \frac{\pi }{5} \right]}^{-1}}\]\[=\frac{1}{2{{\cos }^{2}}\frac{\pi }{10}+2i\sin \frac{\pi }{10}.\cos \frac{\pi }{10}}\] \[=\frac{1}{2\cos \frac{\pi }{10}}{{\left[ \cos \frac{\pi }{10}+i\sin \frac{\pi }{10} \right]}^{-1}}\] \[\therefore \]Real part of\[{{\left[ 1+\cos \frac{\pi }{5}+isin\frac{\pi }{10} \right]}^{-1}}\]is\[\frac{1}{2}\].


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