急急急!!一道数学题,在线等,20分钟内有加分!帮帮忙!

在△ABC中：
sin(A-B)/sin(A+B)
=(sinAcosB-sinBcosA)/sin(A+B)
=[(sinAcosB+sinBcosA)-2sinBcosA]/sin(A+B)
=[sin(A+B)-2sinBcosA]/sin(A+B)
=1-2sinBcosA/sin(A+B)
=1-2sinBcosA/sinC

又∵根据正弦定理：b/sinB=c/sinC
∴(2c-b)/2c
=1-b/2c
=1-sinB/2sinC
又∵sin(A-B)/sin(A+B)=(2c-b)/2c
∴1-2sinBcosA/sinC=1-sinB/2sinC
即2sinBcosA/sinC=sinB/2sinC
解得：cosA=1/4
即sinA=√(1-cos²A)=√15/4

sin(A-B)/sin(A+B)=sin(A+B-2B)/sin(A+B)=sin(C-2B)/sinC=(c-b/2)/cB=pi/6 C=pi/2 A=pi/3

sin(A-B)/sin(A+B)=2c-b/2c,sin(A-B)/sinc=2sinc-sinb/2sinc, 2sin(A-B)=2sin(A+B)-sinb,
2sin(A-B)-2sin(A+B)=-sinb,2(sinacosb-cosasinb-sinacosb-cosasina)=-4cosasinb=-sinb
cosa=1/4,sina=根号15/4

先展开 然后利用正弦定理和余弦定理分别替换sina sinb cosa cosb 就可以求出cosa 然后就可以得到sina了

后面是(2c-b)/w2c对吗？