(2014•南昌模拟)已知函数f(x)=sin2x+2sinx•sin(π2-x)+3sin2(3π2-x)
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(2014•南昌模拟)已知函数f(x)=sin2x+2sinx•sin(
π |
2 |
解 (1)f(x)=sin2x+2sinx•cosx+3cos2x=
sin2x+2sinxcosx+3cos2x
sin2x+cos2x=
tan2x+2tanx+3
tan2x+1=
17
5;
(2)f(x)=sin2x+2sinx•cosx+3cos2x=sin2x+cos2x+2=
2sin(2x+
π
4)+2,
∵ω=2,
∴f(x)的最小正周期为T=
2π
2=π;
由
π
2+2kπ≤2x+
π
4≤
3π
2+2kπ,k∈Z,解得:
π
8+kπ≤x≤
5π
8+kπ,k∈Z,
则f(x)的单调递减区间为[
π
8+kπ,
5π
8+kπ],k∈Z.
sin2x+2sinxcosx+3cos2x
sin2x+cos2x=
tan2x+2tanx+3
tan2x+1=
17
5;
(2)f(x)=sin2x+2sinx•cosx+3cos2x=sin2x+cos2x+2=
2sin(2x+
π
4)+2,
∵ω=2,
∴f(x)的最小正周期为T=
2π
2=π;
由
π
2+2kπ≤2x+
π
4≤
3π
2+2kπ,k∈Z,解得:
π
8+kπ≤x≤
5π
8+kπ,k∈Z,
则f(x)的单调递减区间为[
π
8+kπ,
5π
8+kπ],k∈Z.
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