已知一兀/3<=x<=兀/4,f(x)=tan^2x十2tanx十2求f(x)的最值及相应的x值

Question

如题所述

Answer 1

利用公式cosαsinβ=[sin（α+β）-sin（α-β）]/2化简：
f(x)=4cosxsin(x+π/6)-1
=2sin(2x+π/6)-2sin(x-x-π/6)-1
=2sin(2x+π/6)

（1）f(x)的最小正周期T=2π/2=π

（2）在区间[-π/6，π/4]上，-π/6<=x<=π/4
所以：-π/6<=2x+π/6<=2π/3
所以：-1/2<=sin(2x+π/6)<=1
所以：-1<=f(x)<=2
所以：在区间[-π/6，π/4]上，f(x)的最大值为2，最小值为-1