• # 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)  $\sum{\sin \,2A=-1}$

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