Question

已知函数f(x)＝2√3sinx/3cosx/3-2sin²x/3 （1）求函数f(x)的值域

Answer 1

值域[-3,1],

Answer 2

解： f(x)=3sinx+4cosx =√(32+42)sin(x+φ)，(其中，tanφ=4/3) =5sin(x+φ) sin(x+φ)=-1时，f(x)取得最小值f(x)min=-5 此时，x=θ θ+φ=2kπ+ 3π/2，(k∈Z) θ=2kπ+ 3π/2-φ=(2k+2)π-(φ+π/2) sinθ=sin[(2k+2)π-(φ+π/2)]=-sin(φ+π/2)=-cosφ cosθ=cos[(2k+2)π-(φ+π/2)]=cos(φ+π/2)=-sinφ sin(2θ)=2sinθcosθ =2(-cosφ)(-sinφ) =2sinφcosφ =2sinφcosφ/(sin2φ+cos2φ) =2tanφ/(tan2φ +1) =2·(4/3)/[(4/3)2+1] =24/25