在△ABC中,sin(A-B)/sin(A+B)=2c-b/2c,求cos【(B+C)/2】

[sin(A-B)]/[sin(A+B)}=(2c-b)/2c=(2sinC-sinB)/2sinCsin(A+B)=sinC,就有:2sin(A-B)=2sinC-sinB=2sin(A+B)-sinBsinB=2(sin(A+B)-sin(A-B))=2*2cosAsinB得到:cosA=1/4那么:cos(B+C)=-cosA=-1/4=2[cos(B+C/2)]^2-1[cos...