Question

已知向量m=（sinB，1-cosB），且与向量n=（2，0）所成角为60度，其中A，B，C为三角形内角

1.求角B大小
2.求sinA+sinC的取值范围
要有过程，谢谢

Answer 1

m*n=2sinB
|m|²=sin²B+(1-cosB)²=2-2cosB=2(1-cosB)=4sin²(B/2)
|m|=2sin(B/2)
|n|=2
cos60
=m*n/(|m|*|n|)
=2sinB/4sin(B/2)
=4sin(B/2)cos(B/2)/4sin(B/2)
=cos(B/2)=1/2
B/2=60,B=120度

sinA+sinC
=(1/2)sin(A+C/2)cos(A-C/2)
=(1/2)*sin(60/2)cos(A-C/2)
=cos(A-C/2)
取值范围是：(根号3)/2<sinA+sinC<=1