f(x)=sin(π/6-x)-cosx
=sinπ/6cosx-cosπ/6sinx-cosx=cosx*1/2-√3/2*sinx-cosx=-1/2cosx-√3/2*sinx=-（sin(π/6+x)）∵x∈[0,π/2] → （x+π/6）∈[π/6,2π/3] 又∵函数sin(π/6+x)在[π/6,π/2] 为增；函数sin(π/6+x)在[π/2,2π/3] 为减，且1/2=sin(π/6）＜sin(2π/3)＜sin(π/2）=1∴函数sin(π/6+x)的值域为[1/2,1] → -（sin(π/6+x)）的值域为[-1,-1/2]∴函数f(x)=sin(π/6-x)-cosx的值域为[-1,-...
|
