Question

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]

Answer 1

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】