三角函数√3sinxcosx，X给了范围(-π/6,π/6)，怎么求值域？

Question

如题所述

Answer 1

√3sinxcosx
=(√3/2)sin(2x)
x∈(-π/6,π/6)
所以
2x∈(-π/3,π/3)
则 sin(2x)∈(-√3/2,√3/2)
所以
值域为 (-3/4,3/4)

Answer 2

y=√3sinxcosx=√3/2sin(2x)
∵x∈(-π/6,π/6)
∴2x∈(-π/3,π/3)
y在(-π/6,π/6)单调增加
ymin=√3/2sin(-π/3)=-3/4
ymin=√3/2sin(π/3)=3/4
y∈(-3/4,3/4)

Answer 3

y=(√3/2)sin2x
-π/6<x<π/6;
-π/3<2x<π/3;
-√3/2<sin2x<√3/2
-3/4<√3/2)sin2x<3/4
值域为(-3/4,3/4)