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

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)=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...

多一个sin(2x+π/6)吧! f(x)=sin(2x+π/6)+2cos²x =sin2xcosπ/6+cos2xsinπ/6+1+cos2x =√3/2sin2x+3/2cos2x+1 =√3(1/2sin2x+√3/2cos2x)+1 =√3sin(2x+π/3)+1 函数最小正周期T=2π/2=π

由题意得,f(x)=32cos2x-12sin2x+sin2x=32cos2x+12sin2x=sin(2x+π3),(1)f(x)的最小正周期T=2π2=π;(2)由x∈[0,π2]得,2x+π3∈[π3,4π3],当2x+π3=π2时,此时x=π12,函数f(x)取到最大值1,当2x+π3=4π3时,此时x=π2,函数f(x)取到最...

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...

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...

(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)...

把函数f(x)=sin2x+cos2x= 2 sin(2x+ π 4 )的图象向左平移 π 6 个单位,得到函数y=g(x)= 2 sin[2(x+ π 6 )+ π 4 ]= 2 sin(2x+ 7π 12 )=- 2 cos(2x+ π 12 )的图象,令2x+ π 12 =kπ,k∈z,求得x= kπ 2 - π 24 ,故函数g(x)的对称轴...

(1)f(x)=12(3-cos2x)-32[1-cos(2x-π2)]=32sin2x-12cos2x=sin(2x-π6),令-π2+2kπ≤2x-π6≤π2+2kπ,k∈Z,得到kπ-π6≤x≤kπ+π3,k∈Z,则函数f(x)的单调递增区间[kπ-π6,kπ+π3],k∈Z;(2)由f(B)=1,得到sin(2B-π6)=1,∴2B-π6=π2,即B=...

f(x)=sin(π/6-2x)+cos2x =sinπ/6cos2x-cosπ/6sin2x+cos2x =1/2cos2x-✔3/2sin2x+cos2x =-✔3/2sin2x+3/2cos2x =-✔3(sin2xcosπ/3-cos2xsinπ/3) =-✔3sin(2x-π/3) 振幅:✔3 最小正周期:2π/2=π

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