Question

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

Answer 1

[sin(A-B)]/[sin(A+B)}=(2c-b)/2c=(2sinC-sinB)/2sinC
sin(A+B)=sinC,就念枣弊有:
2sin(A-B)=2sinC-sinB=2sin(A+B)-sinB
sinB=2(sin(A+B)-sin(A-B))=2*2cosAsinB
得到:cosA=1/4
那岩磨么:cos(B+C)=-cosA=-1/4=2[cos(B+C/2)]^2-1
[cos(B+C/2)]^2=3/仔族8
所以cos[(B+C)/2] =根号6/4

Answer 2

[sin(A-B)]/[sin(A+B)}=(2c-b)/2c=(2sinC-sinB)/2sinC

2sin(A-B)=2sinC-sinB=2sin(A+B)-sinB
sinB=2(sin(A+B)-sin(A-B))=2*2cosAsinB
得棚羡神到:cosA=1/4
那么:cos(B+C)=-cosA=-1/4=2[cos(B+C/派游2)]^2-1
[cos(B+C/2)]^2=3/8
所链亏以cos[(B+C)/2] =√6/4