有个结论:A、B、C共线,且λOA+αOB+βOC=0向量,则λ+α+β=0
由此可得 1+f(x)-√3sinxcosx-1/2-(sinx)^2=0
所以 f(x)=√3sinxcosx+(sinx)^2-1/2
=√3/2*sin2x+(1-cos2x)/2-1/2
=√3/2*sin2x-1/2*cos2x
=sin(2x-π/6)
有个结论:A、B、C共线,且λOA+αOB+βOC=0向量,则λ+α+β=0
由此可得 1+f(x)-√3sinxcosx-1/2-(sinx)^2=0
所以 f(x)=√3sinxcosx+(sinx)^2-1/2
=√3/2*sin2x+(1-cos2x)/2-1/2
=√3/2*sin2x-1/2*cos2x
=sin(2x-π/6)