在三角形ABC中,a，b，c分别是对边的长度，设a+c=2b，sin2B=39^1/2\8,求A-C的值

a+c=2b
a=2RsinA
b=2RsinB
c=2RsinC

2sin[(A+C)/2]cos[(A-C)/2]
=sin[(A+C)/2+(A-C)/2]+sin[(A+C)/2-(A-C)/2]
=sinA+sinC
=2sinB
=2sin(A+C)

sin[(A+C)/2]cos[(A-C)/2]=sin(A+C)=2sin[(A+C)/2]cos[(A+C)/2]
cos[(A-C)/2]=2cos[(A+C)/2]=2sin(B/2)

cos(A-C)=2{cos[(A-C)/2]}^2-1=2[2sin(B/2) ]^2-1
=8[sin(B/2)]^2-4+3
=－4cosB+3

sin2B=√39/8
2(cosB)^2-1=cos2B＝5/8
2(cosB)^2-1=cos2B＝-5/8

cos(A-C)=-√3+3(舍)
cos(A-C)=＋√3+3(舍)
cos(A-C)=＋√13+3(舍)
cos(A-C)=-√13+3

A-C＝+-arccos(-√13+3)

其实也蛮难得：
a+c=2b
a=2RsinA
b=2RsinB
c=2RsinC

2sin[(A+C)/2]cos[(A-C)/2]
=sin[(A+C)/2+(A-C)/2]+sin[(A+C)/2-(A-C)/2]
=sinA+sinC
=2sinB
=2sin(A+C)

sin[(A+C)/2]cos[(A-C)/2]=sin(A+C)=2sin[(A+C)/2]cos[(A+C)/2]
cos[(A-C)/2]=2cos[(A+C)/2]=2sin(B/2)

cos(A-C)=2{cos[(A-C)/2]}^2-1=2[2sin(B/2) ]^2-1