(2010•河南三模)已知函数f(x)=sin(3π4−x)−3cos(x+π4),x∈R,则f(x)是( )
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(2010•河南三模)已知函数f(x)=sin(
−x)−
cos(x+
),x∈R
3π |
4 |
3 |
π |
4 |
![(2010•河南三模)已知函数f(x)=sin(3π4−x)−3cos(x+π4),x∈R,则f(x)是( )](/uploads/image/z/588663-63-3.jpg?t=%EF%BC%882010%E2%80%A2%E6%B2%B3%E5%8D%97%E4%B8%89%E6%A8%A1%EF%BC%89%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%EF%BC%9Dsin%283%CF%804%E2%88%92x%29%E2%88%923cos%28x%2B%CF%804%29%EF%BC%8Cx%E2%88%88R%EF%BC%8C%E5%88%99f%EF%BC%88x%EF%BC%89%E6%98%AF%EF%BC%88%E3%80%80%E3%80%80%EF%BC%89)
f(x)=sin(
3π
4−x)−
3cos(x+
π
4)
=sin[π-(x+
π
4)]-
3cos(x+
π
4)
=sin(x+
π
4)-
3cos(x+
π
4)
=2[
1
2sin(x+
π
4)-
3
2cos(x+
π
4)]
=2sin[(x+
π
4)-
π
3]
=2sin(x-
π
12),
∵x∈R,∴x-
π
12∈R,
∴-1≤sin(x-
π
12)≤1,
则f(x)的最大值为2;
∵ω=1,∴周期T=
2π
1=2π;
当x-
π
12=kπ(k∈Z)时,f(x)图象关于某一点对称,
∴当k=0,求出x=
π
12,即f(x)图象关于x=
π
12对称,
故选B
3π
4−x)−
3cos(x+
π
4)
=sin[π-(x+
π
4)]-
3cos(x+
π
4)
=sin(x+
π
4)-
3cos(x+
π
4)
=2[
1
2sin(x+
π
4)-
3
2cos(x+
π
4)]
=2sin[(x+
π
4)-
π
3]
=2sin(x-
π
12),
∵x∈R,∴x-
π
12∈R,
∴-1≤sin(x-
π
12)≤1,
则f(x)的最大值为2;
∵ω=1,∴周期T=
2π
1=2π;
当x-
π
12=kπ(k∈Z)时,f(x)图象关于某一点对称,
∴当k=0,求出x=
π
12,即f(x)图象关于x=
π
12对称,
故选B
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