求下列函数的最小正周期,值域,单调区间(1)y=sin(x+π/3) (2)y=2sin(1/2x-π/4) (3)y=cos(2x-π/6)

求下列函数的最小正周期,值域,单调区间(1)y=sin(x+π/3) (2)y=2sin(1/2x-π/4) (3)y=cos(2x-π/6)
(4)y=2cos(x-π/3) (5)y=sin(2x-π/6)+3 (6)y=3tan(2x-π/6)要详细过程~
数学人气:622 ℃时间:2019-10-08 10:20:02
优质解答
(1) y = sin(x + π/3)
最小正周期:T = 2π
值域:y ∈ [-1 ,1]
单调增区间:2kπ - π/2 ≤ x + π/3 ≤ 2kπ + π/2 ,x ∈[2kπ - 5π/6 ,2kπ + π/6]
单调减区间:2kπ + π/2 ≤ x + π/3 ≤ 2kπ + 3π/2 ,x ∈[2kπ + π/6 ,2kπ + 7π/6]
(2) y = 2sin(1/2x - π/4)
最小正周期:T = 2π/(1/2) = 4π
值域:y ∈ [-2 ,2]
单调增区间:2kπ - π/2 ≤ x/2 - π/4 ≤ 2kπ + π/2 ,x ∈[4kπ - π/2 ,4kπ + 3π/2]
单调减区间:2kπ + π/2 ≤ x/2 - π/4 ≤ 2kπ + 3π/2 ,x ∈[4kπ + 3π/2 ,4kπ + 7π/2]
(3) y=cos(2x - π/6)
最小正周期:T = 2π/2 = π
值域:y ∈ [-1 ,1]
单调减区间:2kπ ≤ 2x - π/6 ≤ 2kπ + π ,x ∈[kπ + π/12 ,kπ + 7π/12]
单调增区间:2kπ + π ≤ 2x - π/6 ≤ 2kπ + 2π ,x ∈[kπ + 7π/12 ,kπ + 13π/12]
(4) y=2cos(x - π/3)
最小正周期:T = 2π
值域:y ∈ [-2 ,2]
单调减区间:2kπ ≤ x - π/3 ≤ 2kπ + π ,x ∈[2kπ + π/3 ,2kπ + 4π/3]
单调增区间:2kπ + π ≤ x - π/3 ≤ 2kπ + 2π ,x ∈[2kπ + 4π/3 ,2kπ + 7π/3]
(5) y=sin(2x - π/6) + 3
最小正周期:T = 2π/2 = π
值域:y ∈ [-1 ,1]
单调增区间:2kπ - π/2 ≤ 2x - π/6 ≤ 2kπ + π/2 ,x ∈[2kπ - π/6 ,2kπ + π/3]
单调减区间:2kπ + π/2 ≤ 2x - π/6 ≤ 2kπ + 3π/2 ,x ∈[2kπ + π/3 ,2kπ + 5π/6]
(6) y = 3tan(2x - π/6)
最小正周期:T = π/2
值域:y ∈ (-∞ ,+∞)
单调增区间:kπ - π/2 < 2x - π/6 < kπ + π/2 ,x ∈(kπ/2 - π/6 ,kπ/2 + 2π/6)
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