已知函数f(x)=sin(2x+π6)+2sin2(x+π6)−2cos2x+a−1(a∈R,a为常数)
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已知函数f(x)=sin(2x+
)+2sin
π |
6 |
(1)f(x)=sin(2x+
π
6)+2sin2(x+
π
6)−2cos2x+a−1
=sin(2x+
π
6)-cos(2x+
π
3)-2cos2x+a
=sin2x•
3
2+cos2x•
1
2-cos2x•
1
2+sin2x•
3
2-2×
1+cos2x
2+a
=
3sin2x-cos2x+a-1=2sin(2x-
π
6)+a-1.
故函数f(x)的最小正周期等于
2π
2=π.
(2)由2kπ-
π
2≤2x-
π
6≤2kπ+
π
2,k∈z,可得kπ-
π
6≤x≤kπ+
π
3,k∈z,
故函数f(x)的单调递增区间为[kπ-
π
6,kπ+
π
3],k∈z.
(3)若x∈[0,
π
2]时,有-
π
6)+2sin2(x+
π
6)−2cos2x+a−1
=sin(2x+
π
6)-cos(2x+
π
3)-2cos2x+a
=sin2x•
3
2+cos2x•
1
2-cos2x•
1
2+sin2x•
3
2-2×
1+cos2x
2+a
=
3sin2x-cos2x+a-1=2sin(2x-
π
6)+a-1.
故函数f(x)的最小正周期等于
2π
2=π.
(2)由2kπ-
π
2≤2x-
π
6≤2kπ+
π
2,k∈z,可得kπ-
π
6≤x≤kπ+
π
3,k∈z,
故函数f(x)的单调递增区间为[kπ-
π
6,kπ+
π
3],k∈z.
(3)若x∈[0,
π
2]时,有-
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