Question

已知函数f(x)=cosx·sin(x+π/3)-根号3cos²x+根号3/4,x∈R

已知函数f(x)=cosx·sin(x+π/3)-根号3cos²x+根号3/4,x∈R. (1)求f(x)在闭区间[-π/4,π/4]上的最大值和最小值; (2)若cosθ=3/5,θ∈(3/2π,2π),求f(θ)的值.

Answer 1

f(x)=cosx·sin(x+π/3)-√3cos²x+√3/4

=cosx(sinx·½+cosx·√3/2)-√3cos²x+√3/4

=¼sin2x-√3(cos2x+1)/4+√3/4

=¼sin2x-√3/4cos2x

=½sin(2x-π/3)
最小正周期=π
1.x∈[-π/4,π/4]
x=-π/12→2x-π/3=-π/2→f(x)取得最小值-½
x=π/4时→2x-π/3=π/6→f(x)取得最大值¼
2.cosθ=3/5→sinθ=-4/5
f(θ)=cosθ(sinθ·½+cosθ·√3/2)-√3cos²θ+√3/4

=⅗(-½·⅘+⅗√3/2)-√3·9/25+√3/4

=-24/100+18√3/100-36√3/100+25√3/100

=(7√3-24)/100