若π/4≤x≤π/3,则函数y=[2sin(x+π/6)]/cosx的值域
问题描述：
若π/4≤x≤π/3,则函数y=[2sin(x+π/6)]/cosx的值域
答
y=2(sinxcosπ/6+cosxsinπ/6)/cosx
=2(1/2*sinx+√3/2cosx)/cosx
=(sinx+√3cosx)/cosx
=sinx/cosx+√3
=tanx+√3
π/4≤x≤π/3
tanπ/4≤tanx≤tanπ/3
1≤tanx≤√3
所以值域[1+√3,2√3]