求救，高中数学题

Question

已知向量m=(sinB,1-cosB)， n=(2,0)，两向量所成角为 π/3，A,B,C为 △ABC内角，求 sinA+sinC取值
求sinA+sinC取值范围

Answer 1

解:由两向量所成角为 π/3
cosπ/3=(sinB,1-cosB).(2,0)除以向量m与n的模,通过运算可求B=2π/3
A+C=π/3,
sinA+sinC=sinA+sin(π/3-A)=sin(A+π/3),
A+π/3的范围是(π/3,2π/3)
所以,sinA+sinC取值为(根号3除以2.1]

Answer 2

2*√[sin²B+(1-cosB)²]*cosπ/3=√3sinB/3=2sinB sinB>0 1-cosB>0 sin²B+(1-cosB)²=4sin²B
1-cosB=√3sinB 1/2=1/2cosB+√3/2sinB=sin(B+π/6) π/6<B+π/6=5π/6
A+C=π-2/3π=π/3
sinA+sinC=sinA+sin(π/3-A)=1/2sinA+√3/2sinA=sin(A+π/3)
π/3<A+π/3<2π/3
√3/2<sin(A+π/3)<=1

Answer 3

m*n=2sinB，1m1=根号2-2cosB 1n1=2 cos=1/2 计算后得1-2cos2B+cosB=0
将cosB设为x 计算得1舍或-1/2,
A+C=60 sinA+sinC范围是 （根号3）/2 到1