Question

已知函数fx=sin²x-cos²x-2√3sinxcosx求f(2π/3)

已知函数f(x)=√3(sin²x-cos²x)-2sinxcosx
①求f(x)的最小正周期②设x∈［-π/3,π/3］求f(x)值域和单调增区间

Answer 1

f(x)=-√3cos2x-sin2x=-2[(√3/2)cos2x+(1/2)sin2x]=-2cos(2x-π/6)
最小T=2π/2=π
x=π/12,f(π/12)=-2
x=π/3,f(π/3)=0
-2≤f(x)≤0
-π/3≤x≤π/12单调递减
π/12≤x≤π/3 单调增加