Question

在三角形ABC中,角A,B,C的对边长分别为a,b,c,若A-C=π/3,且a,b,c成等差数列，求sinB值等于

Answer 1

∵a,b,c成等差数列
∴a+c=2b
根据正弦定理：
∴sinA+sinC=2sinB=4sinB/2cosB/2
∵ C-A=π/3
∴C=A+π/3, B=π-A-C=2π/3-2A
∴sinA+sin(A+π/3)=4sinB/2cosB/2
∴sinA+sinAcosπ/3+cosAsinπ/3
=4sinB/2cosB/2
∴3/2sinA+√3/2cosA=4sinB/2cosB/2
∴√3sin(A+π/6)=4sin(π/3-A)cos(π/3-A)
∴√3sin(A+π/6)=4sin(A+π/6)cos(A+π/6)
∴cos(A+π/6)=√3/4
∴sin(A+π/6)=√[1-cos²(A+π/6)]=√13/4
∴sinB=sin(A+C)=sin(2A+π/3)
=2sin((A+π/6)cos(A+π/6)
=2×√3/4×√13/4=√39/8
（ 用和差化积更快些，但怕你没学）

追问

谢了，我们苗苗（谢老）教过我们了

追答

sinA+sinC=2sinB
2sin(A+C)/2×cos(A-C)/2=4sinB/2×cosB/2
∵(A+B)/2=π/2-C/2,C-A=π/3
∴cosB/2×cosπ/6=2sinB/2×cosB/2
∴sinB/2=√3/4
cosB/2=√13/4
sinB=2sinB/2×cosB/2=√39/8
简单多了

Answer 2

sinA+sinC=2sinB
2sin(A+C)/2×cos(A-C)/2=4sinB/2×cosB/2
(A+B)/2=π/2-C/2,C-A=π/3
cosB/2×cosπ/6=2sinB/2×cosB/2
sinB/2=√3/4
cosB/2=√13/4
sinB=√（1-（cosB）^2）=√39/8