f(x)=sin²(x-π/6)+sin²(x+π/6)
=(√3/2sinx-1/2cosx)²+(√3/2sinx+1/2cosx)²
=3/4sin²x-√3/2sinxcosx+1/4cos²x+3/4sin²x+√3/2sinxcosx+1/4cos²x
=3/2sin²x+1/2cos²x
=3/2sin²x+1/2-1/2sin²x
=sin²x+1/2
x∈[-π/3,π/6]
sinx∈[-√3/2,1/2]
sin²x∈[0,3/4]
sin²x+1/2∈[1/2,5/4]
值域是[1/2,5/4]把sin(x-兀/6)^2化为1-cos2(x-兀/6)/2可以做吗？？一样，都需要打开能写一下吗，麻烦了。。。f(x)=sin²(x-π/6)+sin²(x+π/6)=1/2[1-cos(2x-π3)]+1/2[1-cos(2x+π3)]=1-1/2cos(2x-π3)-1/2cos(2x+π3)=1-1/2(cos2xcosπ3+sin2xsinπ3）-1/2(cos2xcosπ3-sin2xsinπ3）=1-cos2xcosπ3=1-1/2cos2x点击“采纳为满意答案”,谢谢！