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已知函数F(x)=(A+2Cos2x)Cos(2x+θ)为奇函数,且F(π4)=0,其中A∈R,θ∈...

(1)f【4/π】=-(a+1)sinθ=0,∵θ∈(0,π). ∴sinθ≠0,∴a+1=0,即a=-1 ∵f(x)为奇函数,∴f(0)=(a+2)cosθ=0,∴cosθ=0,θ=2/π.(2)由(1)知f(x)=(-1+2cos2x)cos(2x+2/π)=cos2x(-sin2x)=-2/1sin4x,∴f(4/a)=-2/1 sinα=- 5/2,∴sinα=5/4 ∵α∈(2/π,π),∴cosα=厂125/16=5/3,∴sin(α+3/π)=sinαcos3/π+cosαsin3/π=10/4-3厂3.望采纳

解:(1) f(π/4)=0=-(a+1)sinθ ∵θ∈(0,π). ∴sinθ≠0,∴a+1=0,即a=-1 ∵f(x)为奇函数,∴f(0)=(a+2)cosθ=0,∴cosθ=0,θ=π/2

∵f(x)=(a+2casx)cos(2x+θ)为奇函数∴f(-x)=-f(x),即(a+2casx)cos(-2x+θ))=-(a+2casx)cos(2x+θ)∴cos(-2x+θ)=-cos(2x+θ)cos(-2x+θ)+cos(2x+θ)=02cosθcos2x=0cosθ=0∵θ∈(0,π).∴θ=π/2, π/2+θ∈(π/2,3π/2)∵f(π/4)=0∴(a+2casπ/4)cos(π/2+θ)=0∵ π/2+θ∈(π/2,3π/2)∴cos(π/2+θ)≠0∴a+2cosπ/4=0∴a=-1

f(x)=2cos(2x+π/4)+2T=π函数最大值是4,最小值是0(2)x不变y增大两倍得到y=2sinxy=2sinx 向上平移2个单位得到y=2sinx+2函数向左平移π/2个单位得到y=2cosx+2函数y不变x变为原来的1/2,得到y=2cos2x+2函数向左平移π/8个单位得到y=2cos(2x+π/4)+2

因为函数f(x)=2cos(2x+)是奇函数,所以=kπ+π2,函数在(0,π4)上是增函数,所以f(x)=2sin2x,所以=π2.故选A.

1)f(x)=2cos^2x+cos(2x+π/3)-1 =cos2x+cos2xcosπ/3-sin2xsinπ百/3 =cos2x+1/2cos2x-√度3/2sin2x =3/2cos2x-√3/2sin2x =√3(cos2xcosπ/6-sin2xsinπ/6) =√3cos(2x+π/6) 所以知最小值周期T=2π/w=2π/2=π 因为当道(2x+π/6)∈(2kπ-π,回2kπ)答

(1)f(x)=a(sin2X+cos2X)=√2asin(2X+π/4) f(π/4)=√2asin(3π/4)=a=-1 ∴a=-1 (2)f(x)= -√2sin(2X+π/4) 当2X+π/4=3π/2+2kπ, k∈Z 时 即{X / X=5π/8+kπ k∈Z } f(x)max=√2 (3)f(x)= -√2sin(2X+π/4) 当2X+π/4∈[-π/2+2kπ,π/2+2kπ], k∈Z时 函数f(x)的单调递增 ∴单调递增区间为[-3π/8+kπ,π/8+kπ], k∈Z

解:(1):f(派/4)=0,且f(x)是奇函数,所以:f(-派/4)=-f(派/4)=0.f(0)=0,即:-(a+1)sinfi=0,(a+1)sinfi=0,(a+2)cosfi=0.解得:a=-1,fi=派/2;(2):f(x)=(-1+2cos^2x)(cos(2x+派/2))=-sin2xcos2x=-1/2sin4x.f(a/4)=-2/5,即:sina=4/5,cosa=正负3/5.sin(a+派/3)==1/2sina+根号3/2cosa=(4+3倍根号3)/10或(4-3倍根号3)/10

f(x)=1+cos2x+根号3sin2x+a =2sin(2x+π/6)+a+1 1、若f(x)max=2,则sin(2x+π/6)=1,即2+a+1=2,得a=-1 2、正弦的单调减区间在第二和第三象限, 所以2x+π/6∈〔2kπ+π/2,2kπ+3π/2] 所以x∈〔kπ+π/6,kπ+2π/3〕为单调减区间

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