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已知F(x)=12sin2x?Cosπ6+12Cos2xsinπ6(1)函数F...

f(x)=sin(2x-π6)+2cos2x-1=32sin2x-12cos2x+cos2x=32sin2x+12cos2x=sin(2x+π6)由-π2+2kπ≤2x+π6≤π2+2kπ(k∈Z)得:-π3+kπ≤x≤π6+kπ(k∈Z)由π2+2kπ≤2x+π6≤3π2+2kπ(k∈Z)得:π6+kπ≤x≤2π3+kπ(k∈Z)故f(x)的单调递增区间是[-π3+kπ,π6+kπ]...

f(x)=sin(5π/6-2x)-2sin(x-π/4)cos(x+3π/4) = sin[π-(5π/6-2x)] - { sin[(x-π/4)+(x+3π/4)] + sin [(x-π/4)-(x+3π/4)] } = sin(2x-π/6) - { sin(2x+π/2) - sin(-π) = sin(2x-π/6) - cos(2x) = sin2xcosπ/6-cos2xsinπ/6-cos2x = √3/2sin2x-3/2cos...

f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2cosx)-cos2x-1/4, =√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4, =√3/4sin2x+1/4(1+cos2x)-cos2x-1/4, =√3/4sin2x-3/4cos2x =√3/2(1/2sin2x-√3/2cos2x) =√3/2sin(2x-π/3) 2x-π/3∈[2kπ-π/2,2kπ+π/2]单...

f(x)=[cos(x+π/12)]^2=[1+cos{2(2x+π/6)}]÷2 (x=5π/12对称轴)代入g(x)=1+1/2sin2x=5/4(2)h(x)=[cos(x+π/12)]^2+1+1/2sin2x=3/2+1/4sin2x+√3/4cos2x=2/3+1/2sin(π/3+2x)至于值域自己会球吧

1、 f(x)=sinx(sinxcosπ/6+cosxsinπ/6) =√3/2*sin²x+1/2*sinxcosx =√3/2*(1-cos2x)/2+1/2*(sin2x)/2 =√3/4+1/2*(1/2*sin2x-√3/2*cos2x) =√3/4+(sin2xcosπ/3-cos2xsinπ/3) =√3/4+sin(2x-π/3) 所以T=2π/2=π 2、 0≤x≤π/2 -π/3≤2x-π/3≤2π/3 -√3...

(I)∵函数f(x)=sin(7π6?2x)+2cos2x?1=sin7π6cos2x-cos7π6sin2x+cos2x=32sin2x+12cos2x=sin(2x+π6).故函数f(x)的周期为T=2π2=π.再令 2kπ-π2≤2x+π6≤2kπ+π2,k∈z,求得 kπ-π3≤x≤kπ+π6,k∈z,故单调递增区间为[kπ-π3,kπ+π6],k∈z.(II)...

f(x)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

见图 解:(I)f(x)==sin2x+cos2x=sin(2x+). 令 2kπ-≤(2x+)≤2kπ+,可得 kπ-≤x≤kπ+,k∈z. 即f(x)的单调递增区间为[kπ-,kπ+],k∈z. (II)在△ABC中,由,可得sin(2A+)=,∵<2A+<2π+, ∴<2A+= 或,∴A= (或A=0 舍去). ∵b,a,c成...

f(x)=2cosxcos(x?π6)?3sin2x+sinxcosx=2cosx(32cosx+12sinx)-3sin2x+sinxcosx=3(cos2x-sin2x)+2sinxcosx=3cos2x+sin2x=2(32cos2x+12sin2x)=2sin(2x+π3),(1)∵ω=2,∴T=2π2=π;(2)∵f(x)=1,即2sin(2x+π3)=1,∴sin(2x+π3)=12,...

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