(2010•合肥模拟)已知a=(sin(ωx+ϕ) , 2) , b=(1&nb
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(2010•合肥模拟)已知
=(sin(ωx+ϕ) , 2) ,
=(1 , cos(ωx+ϕ))(ω>0 , 0<ϕ<
)
a |
b |
π |
4 |
![(2010•合肥模拟)已知a=(sin(ωx+ϕ) , 2) , b=(1&nb](/uploads/image/z/16852892-68-2.jpg?t=%EF%BC%882010%E2%80%A2%E5%90%88%E8%82%A5%E6%A8%A1%E6%8B%9F%EF%BC%89%E5%B7%B2%E7%9F%A5a%EF%BC%9D%28sin%28%CF%89x%2B%CF%95%29%26nbsp%3B%EF%BC%8C%26nbsp%3B2%29%26nbsp%3B%EF%BC%8C%26nbsp%3Bb%EF%BC%9D%281%26nb)
(1)f(x)=|
a|2−|
b|2-2=1-cos(2ωx+2ϕ).
由题意知 T=
2π
|2ω|=4 , ∴ω=
π
4,
又图象过A,则
3
2=1−cos(
π
2×1+2ϕ),sin2ϕ=
1
2,
又0<ϕ<
π
4 , ∴ϕ=
π
12,∴f(x)=1−cos(
π
2x+
π
6)
(2)由2kπ≤
π
2x+
π
6≤2kπ+π,得4k-
1
3≤x≤4k+
5
3(k∈Z),
∴递增区间为[4k−
1
3,4k+
5
3] (k∈Z).…(12分)
a|2−|
b|2-2=1-cos(2ωx+2ϕ).
由题意知 T=
2π
|2ω|=4 , ∴ω=
π
4,
又图象过A,则
3
2=1−cos(
π
2×1+2ϕ),sin2ϕ=
1
2,
又0<ϕ<
π
4 , ∴ϕ=
π
12,∴f(x)=1−cos(
π
2x+
π
6)
(2)由2kπ≤
π
2x+
π
6≤2kπ+π,得4k-
1
3≤x≤4k+
5
3(k∈Z),
∴递增区间为[4k−
1
3,4k+
5
3] (k∈Z).…(12分)
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