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设函数F(x)=sin(ωx%π/6)%Cosωx(ω>0),且F(x)的图像...

∵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)

f(-x)=f(x).sin( -ωx+ψ)+cos(-ωx+ψ)=sin( ωx+ψ)+cos(ωx+ψ).-sinωx*cosψ+cosωx*sinψ+cosωx*cosψ+sinωx*sinψ=sinωx*cosψ+cosωx*sinψ+cosωx*cosψ-sinωx*sinψ化简得sinψ=cosψ→ψ=四分之π+kπ,k属于Z

∵函数f(x)=sinωx+cosωx(ω≠0)= 2 sin(ωx+π 4 ),对任意实数x都有f(π 6 +x)=f(π 6 ?x),故函数的图象关于直线x=π 6 对称,故有ω?π 6 +π 4 =kπ+π 2 ,k∈z,∴ω=6k+3 2 .令ω=3 2 ,则f(π 3 ?π ω )= 2 sin[ω?(π 3 ?π ω )+π 4 ]= 2 sin(-π 4 )=-1,故选A.

sin(wx+π/6)+cos(wx+π/6) =根2[根2/2(sin(wx+π/6)+cos(wx+π/6))]=根2[sin(wx+π/6+π/4)] 可以看出x的系数为w 函数y=f(x)图像的两相邻对称轴间距离为π/2,即D=π/2 T=2D=πT=2π/w=πw=2所以原式=根2[sin(2x

f'(x)=ωcos(ωx+π/6)最大值为3,ω>0,则ω=3f(x)=sin(3x+π/6)-1sin(x)的对称轴在x=(2n+1)π/2处,则f(x)的对称轴在3x+π/6=(2n+1)π/2处,n为整数.选项A附和要求

f(x)=sin(ωx+π/6)+sin(ωx-π/6)-2cos^2(ωx/2), =sinωxcosπ/6+cosωx sinπ/6+sinωxcosπ/6-cosωx sinπ/6-1+cosωx =√3sinωx+cosωx-1 =2sin(ωx+π/6)-12sin(ωx+π/6)-1=-1sin(ωx+π/6)=0在一个π内有且仅有两个不同的交点,即周期为πω=2单调增区间:2kπ-π/2 评论0 0 0

f(x)=sin(ωx+π/3)cos(ωx-π/6) =sin(wx+π/3)sin(wx+π/3) =1-cos^2(wx+π/3) =1-(1+cos(2wx+2π/3) =cos(2wx+2π/3) T=2π/2w=π w=1 (1)f(x)=cos(2x+2π/3) (2) 2x+2π/3属于 【π+2kπ,2π+k

f(x)=cos(wx)cosπ/6+sin(wx)sinπ/6+sin(wx)=√3/2*cos(wx)+3/2*sin(wx)=√3[1/2*cos(wx)+√3/2*sin(wx)]=√3*[sinπ/6*coswx+cosπ/6*sinwx]=√3sin(wx+π/6)T=2π/w=πw=2f(x)=√3sin(2x+π/6)∵x∈[0,π/2]∴2x∈[0,π],2x+π/6∈[π/6,7π/6]sin(2x+π/6)∈[-1/2,1]f(x)∈[-√3/2,√3]f(x)的最大值为√3

(Ⅰ)f(x)=sin(ωx-π6)-2cos2ω2x+1=sinωxcosπ6-cosωxsinπ6-21+cosωx2+1=32sinωx-32cosωx=3(12sinωx32cosωx)=3sin(ωxπ3).∵函数f(x)的最大值为3,以题意,函数f(x)的最小正周期为π

1、f(x)=1/2*sin2ωx+√3*(1+cos2ωx)/2+a=1/2*sin2ωx+√3/2*cos2ωx+√3/2+a=sin2ωxcosπ/3+cos2ωxsinπ/3+√3/2+a=sin(2ωx+π/3)+√3/2+a从原点到第一个最高点距离是T/4所以T/4=π/6T=2π/2ω=2π/3ω=3/22、f

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