在△ABC中,cosA=1/3,求sin^2（（B+C)/2)*cos2A的值

网友回答

cosA=1/3, A=180-(B+C), cos(180-(B+C))=1/3
cos(B+C)=-1/3
sin^2((B+C)/2)*cos(2A)=((1-cos(B+C))/2)*(2(cosA)^2-1)
=(2(cosA)^2-1-2*cos(B+C)*(cosA)^2+cos(B+C))/2
=((2(cosA)^2-1)-(2*(cosA)^2-1)*cos(B+C))/2
=((2(cosA)^2-1)*(1-cos(B+C))/2
=(2*(1/3)^2-1)*(1+1/3))/2=-(14/27)