JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • question_answer
    Direction: Let ABC be a triangle, R be the circumradius of the triangle. Also, given \[{{R}^{2}}=\frac{1}{8}\,({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\],  then
    Which of the following is true?

    A)  \[\sum{\cos \,2A=-1}\]          

    B)  \[\sum{\cos \,2A=1}\]

    C)  \[\sum{\sin \,2A=1}\]

    D)  \[\sum{\sin \,2A=-1}\]

    Correct Answer: A

    Solution :

     By Sine rule, \[\sum{{{a}^{2}}=4{{R}^{2}}\sum{{{\sin }^{2}}A}}\] \[\Rightarrow \]            \[4{{R}^{2}}\sum{{{\sin }^{2}}a=8{{R}^{2}}}\] \[(\because \,\,\,{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=8{{R}^{2}})\] \[\Rightarrow \]            \[2\sum{{{\sin }^{2}}}A=4\] \[\Rightarrow \]            \[2\sum{(1-\cos 2A)=4}\] \[\Rightarrow \]            \[\sum{\cos \,2A=-1}\]

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