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已知函数F(x)=Cosωx?sin(ωx%π6)+14(ω>0)的...

(1)f(x)=sinωx?cos(ωx+π6)=sinωx?(32cosωx-12sinωx)=32sinωxcosωx-12sin2ωx=34sin2ωx+14cos2ωx-14=12sin(2ωx+π6)-14∴f(x)=12sin(2ωx+π6)-14,∵函数f(x)图象的两相邻对称轴间的距离为π2.∴T=π,∴2π2ω=π,∴ω=1,∴ω的值1;(2)根...

(I)f(x)=4cosωx?sin(ωx-π6)+1=4cosωx(32sinωx?12cosωx)+1=3sin2ωx-cos2ωx=2sin(2ωx?π6).∵函数f(x)的最小正周期是π,∴2π2ω=π,解得ω=1.∴f(x)=2sin(2x?π6).∵2kπ?π2≤2x?π6≤2kπ+π2,解得kπ?π6≤x≤kπ+π3,k∈Z.∴f(x)的单调递增区间为[k...

f(x)=sin(2ωx-π/6)-4sin²ωx+2 =√3/2sin2ωx-1/2cos2ωx-4sin²ωx+2 =√3/2sin2ωx-1/2cos2ωx+2(cos2ωx-1)+2 =√3/2sin2ωx+(3/2)cos2ωx =√3sin(2ωx+φ),tanφ=√3→φ=π/3 ∵相邻两个交点的距离为π/2,最小正周期=π ∴2π/2ω=π→ω=1 解析式:f(x)=√3sin...

(Ⅰ)f(x)=2cosωx(sinωxcosπ6+cosωxsinπ6)+(cos2ωx-sin2ωx)(cos2ωx+sin2ωx)=3cosωxsinωx+cos2ωx+cos2ωx=32sin2ωx+cos2ωx+cos2ωx=32sin2ωx+32cos2ωx+12=3sin(2ωx+π3)+12,∵T=π,∴ω=1,则f(x)=3sin(2x+π3)+12;(Ⅱ)∵B为三角形锐角...

f(x)=4cos(ωx-π6)sinωx-cos(2ωx+π)=4(32cosωx+12sinωx)sinωx+cos2ωx=23cosωxsinωx+2sin2ωx+cos2ωx-sin2ωx=3sin2ωx+1,∵-1≤sin2ωx≤1,所以函数y=f(x)的值域是[1?3,1+3](II)因y=sinx在每个区间[2kπ?π2,2kπ+π2],k∈z上为增函数,令2k...

函数f(x)=sin(ωx+π6)+sin(ωx?π6)+cosωx=sinωx?cosπ6+cosωx?sinπ6+sinωx?cosπ6?cosωx?sinπ6+cosωx=3sinωx+cosωx=2sin(ωx+π6)由ω>0且函数f(x)在[?π2,π2]上是增函数,可得π2ω+π6≤π2解得ω≤23故ω的取值范围是(0,23]故答案为:(0,23]

∵f(x)=sinωx+cos(ωx+π6)(ω>0)=sinωx+cosωxcosπ6-sinωxsinπ6=sinωx+32cosωx-12sinωx=12sinωx+32cosωx=sin(ωx+π3)由图象上相邻两条对称轴间的距离为π,知周期T=2π∴ω=1,f(x)=sin(x+π3)令12π+2kπ≤x+13π≤32π+2kπ可得π6+2kπ≤x≤7π6+2kπ结合选...

(1)∵f(x)=23sinωxcosωx+1?2sin2ωx(ω>0)∴利用三角函数的降次公式,得f(x)=3sin(2ωx)+cos(2ωx)=2sin(2ωx+π6)∵函数f(x)的最小正周期为T=2π2ω=π∴2ω=2,可得函数f(x)的解析式为:y=2sin(2x+π6)令π2+2kπ<2x+π6<3π2+2kπ,得π6+kπ<...

(Ⅰ)f(x)=3sinωxcosωx+1+cos2ωx2+1=32sin 2ωx+12cos 2ωx+32=sin(2ωx+π6)+32.∵ω>0,∴T=2πω=π,∴ω=1.故f(x)=sin(2x+π6)+32.令2kπ?π2≤2x+π6≤2kπ+π2,解得:kπ?π3≤x≤kπ+π6.f(x)的单调递增区间为[kπ?π3,kπ+π6](k∈Z)(Ⅱ)∵0≤x≤π2,...

用和角公式变换 4cos(wx+π/6)sin(wx+π/6) =2√3cos*sin+2(cos)^2 =√3sin(2wx)+cos(2wx)+1 =2sin(2wx+π/6)+1 f(x)=2sin(2wx+π/6)+a+1 a=-1, 2π/2w=π,w=1

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