求三角函数的值域①y=2sinx(-π/3`≤x≤2π/3) ②y=-2cosx+1(-π/3`≤x≤2π/3)

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求三角函数的值域①y=2sinx(-π/3`≤x≤2π/3)②y=-2cosx+1(-π/3`≤x≤2π/3)求三角函数的值域①y=2sinx(-π/3`≤x≤2π/3)②y=-2cosx+1(-π/

求三角函数的值域①y=2sinx(-π/3`≤x≤2π/3) ②y=-2cosx+1(-π/3`≤x≤2π/3)
求三角函数的值域①y=2sinx(-π/3`≤x≤2π/3) ②y=-2cosx+1(-π/3`≤x≤2π/3)

求三角函数的值域①y=2sinx(-π/3`≤x≤2π/3) ②y=-2cosx+1(-π/3`≤x≤2π/3)
解(1)由-π/3`≤x≤2π/3
知sin(-π/3)≤sinx≤sinπ/2
即-√3/2≤sinx≤1
即-√3≤2sinx≤2
即-√3≤y≤2
故函数y=2sinx(-π/3`≤x≤2π/3)
的值域为[-√3,2].
(2)由-π/3`≤x≤2π/3
知cos2π/3≤cosx≤cos0
即-1/2≤cosx≤1
即1≥-2cosx≥-2
即2≥-2cosx+1≥-1
即2≥y≥-1
故函数的值域为[-1,2].