在三角形ABC中,已知cos[(A-B)/2]=2sin(B/2),证明三边abc成等差数列
问题描述：
在三角形ABC中,已知cos[(A-B)/2]=2sin(B/2),证明三边abc成等差数列
答
应该是cos((A-C)/2)=2sin(B/2)2sin(B/2)cos(B/2)=cos(B/2)cos((A-C)/2)sinB=sin(π-B/2)cos((A-C)/2)=sin((A+C)/2)cos((A-C)/2)=(sinA+sinC)/22sinB=sinA+sinCa/sinA=b/sinB=c/sinC∴2b=a+ca,b,c成等差数列