Question

已知△ABC的面积为S，满足1≤S≤√3，且向量AC乘以向量CB=-2,∠ACB=θ

Answer 1

S=|AC|*|BC|*sinC/2=CA*CB*tanA/2=tanA
∵1<=S<=√3
∴1<=tanA<=√3,
∴π/4<=θ<=π/3
1)f(θ)=sinθ*√2/2-cosθ*√2/2+2√2sin2θ-cosθ*√2/2+sinθ*√2/2-2
=√2sinθ-√2cosθ+2√2sin2θ-2=2sin(θ-π/4)+2√2cos(2θ-π/2)-2
=2√2-4√2sin(θ-π/4)^2+2sin(θ-π/4)(t=sin(θ-π/4))
=-4√2t^2+2t+2√2,开口向下,对称轴t=√2/4
π/4<=θ<=π/3,∴0<=t<=(√6-√2)/4<√2/4
∴最大值在θ=π/3,t=(√6-√2)/4处取得,为-4√2(8-4√3)/16+2(√6-√2)/4+2√2=(3√6-√2)/2
2)|m|=|n|=1,mn=sin2Acos2B+cos2Asin2B=sin(2A+2B)=sin(2π-2C)=-sin2C
∵π/4<=C<=π/3,∴π/2<=2C<=2π/3,
∴√3/2<=sin2C<=1
|2m-3n|^2=4+9-12mn=13+12sinC
∴13+6√3<=|2m-3n|^2<=13+12
∴√(13+6√3)<=|2m-3n|<=5