在三角形ABC中,C=60°,若a2=b2+1/2c2,求sin(A-B)的值
问题描述：
在三角形ABC中,C=60°,若a2=b2+1/2c2,求sin(A-B)的值
答
正弦定理:a/sinA=b/sinB=c/sinC
-b?=c?/2,∴sin?A-sin?B=sin?C/2=(√3/2)?/2=3/8
sin?A=(1-cos2A)/2,sin?B=(1-cos2B)/2
∴sin?A-sin?B=(1-cos2A)/2-(1-cos2B)/2=(cos2B-cos2A)/2=3/8
∴cos2B-cos2A=3/4
cos2B-cos2A=cos[(A+B)-(A-B)]-cos[(A+B)+(A-B)]
=cos(A+B)cos(A-B)+sin(A+B)sin(A-B)-[cos(A+B)cos(A-B)-sin(A+B)sin(A-B)]
=2sin(A+B)sin(A-B)=2sin120°sin(A-B)=√3sin(A-B)=3/4
∴sin(A-B)=√3/4