已知△ABC中,|AC|=√3,∠ABC=5π/6,∠BAC=θ,记f（θ）=向量AB*向量BC,求f(θ)的值域

Question

已知△ABC中,|AC|=√3,∠ABC=5π/6,∠BAC=θ,记f（θ）=向量AB*向量BC,求f(θ)的值域

Answer 1

题目不好做呀,t表示theta：
AB·BC=|AB|*|BC|*cos(π-B)=accos(π/6)=sqrt(3)ac/2
b^2=a^2+c^2-2accos(5π/6)=a^2+c^2+sqrt(3)ac=3
即：ac=sqrt(3)-(a^2+c^2)/sqrt(3)
即：f(t)=sqrt(3)ac/2=3/2-(a^2+c^2)/2
a/sinA=c/sinC=b/sinB=2sqrt(3),故：a=2sqrt(3)sint
c=2sqrt(3)sinC=2sqrt(3)sin(π/6-t)
故：f(t)=3/2-(12sint^2+12sin(π/6-t)^2)/2
=3/2-3(1-cos(2t)+1-cos(π/3-2t))
=-9/2+3(cos(2t)+cos(π/3-2t))
=3*2cos(π/6)cos(2t-π/6)-9/2
=3sqrt(3)cos(2t-π/6)-9/2
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