Question

△ABC中,a,b,c分别是角A,B,C所对的边,cos2B+3cos(A+C)+2=0,b=根号三,向量m=(根号三c,-c),n=(cosA,sinC)

△ABC中,a,b,c分别是角A,B,C所对的边,cos2B+3cos(A+C)+2=0,b=根号三,向量m=(根号三c,-c),n=(cosA,sinC),求m,n的取值范围。

Answer 1

cos(2B)+3cos(A+C)+2=cos(2B)-3cosB+2=2cosB^2-3cosB+1
=(cosB-1)(2cosB-1)=0，即：cosB=1/2或cosB=1(不和题意，舍去)
故：B=π/3
m=(sqrt(3)a,-c)，n=(cosA,sinC)，故：m·n=sqrt(3)a*cosA-c*sinC
而：a/sinA=c/sinC=b/sinB=2=2R，R是外接圆半径
即：a=2sinA，c=2sinC
故：m·n=sqrt(3)a*cosA-c*sinC=2sqrt(3)sinA*cosA-2sinC*sinC
=sqrt(3)sin(2A)-(1-cos(2C))=sqrt(3)sin(2A)+cos(2C)-1
=sqrt(3)sin(2A)+cos(4π/3-2A)-1
=sqrt(3)sin(2A)-cos(2A)/2-sqrt(3)sin(2A)/2-1
=sqrt(3)sin(2A)/2-cos(2A)/2-1
=sin(2A-π/6)-1，0<A<2π/3，故：-π/6<2A-π/6<7π/6
故：-1/2<sin(2A-π/6)≤1
故：-3/2<m·n≤0