1.y=sin(1/2x+π/6),x属于[0,π/3] 2.y=-cos(3x-π/3),x属于[-π/3,π/3]

题目

1.y=sin(1/2x+π/6),x属于[0,π/3] 2.y=-cos(3x-π/3),x属于[-π/3,π/3]

答案

1
y=sin(1/2x+π/6)
x属于[0,π/3]
π/6≤1/2x+π/6≤π/3
1/2 ≤ sin(1/2x+π/6) ≤ 根号3/2
值域【1/2,根号3/2】
2
y=-cos(3x-π/3)
x属于[-π/3,π/3]
-π-π/3 ≤ 3x-π/3 ≤ π- π/3
(3x-π/3)的变化范围涵盖了从--π-π/3 到π- π/3正好2π的区间
∴y=-cos(3x-π/3)值域为【-1,1】