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已知w>0,0<φ<π,(π/4,0),(3π/4,0)是函数F(x)=sin(w...

已知函数f(x)=Asin(wx+φ)(A>0,w>0,|φ|0,w>0,|φ|T=4π==>w=2π/T=1/2 ∴f(x)=2sin(1/2x+φ) ∵f(x)图像关于直线x=π/3对称, ∴f(x)在Y轴右侧第一个对称轴为:wx+φ=π/2==>x=(π-2φ)/(2w) 令(π-2φ)/(2w)=π/3==>(π-2φ)=π/3==>φ=π/3 ∴f(x)=2sin(1/2x+π/3) (2...

f(x)=a·tan(ωx+φ)? 相邻零点距离5π/6-π/6=2π/3为最小正周期 ω=π/(2π/3)=3/2 ω·π/6+φ=0 φ=-π/4 ∴f(x)=a·tan(3x/2-π/4) f(0)=a·tan(3x/2-π/4)=-a=-3 ∴a=3 f(x)解析式:f(x)=3tan(3x/2-π/4) (2)3tan(3x/2-π/4)≥√3 tan(3x/2-π/4)≥√3/3=tan(kπ+π/6) ∴k...

f(x)=(向量a+向量b)*(向量a-向量b) =a^2-b^2 =[sin(wx+p)]^2+4-{1+[cos(wx+p)]^2} =3-cos(2wx+2p), (1)y=f(x)的图像过点M(1,7/2),且相邻对称轴之间的距离为2,w>0,0

cosωx - sinωx = sqrt(2) (cosωx cosπ/4 - sinωx sinπ/4) = sqrt(2) cos(ωx + π/4) 该函数在 2kπ

不能啊 即sinwx最小是-1 此时wx=2kπ-π/2 w>0 -wπ/3

已知函数f(x)=Asin(x+φ)(a>0,0cosβ=12/13==>sinβ=5/13 f(α-β)=sin(α-β+π/2)=cos(α-β)=cosαcosβ+sinαsinβ =36/65+20/65=56/65

sin增 则2kπ-π/2

x∈[0,1], u=wx -wπ/8(w>0)的值域是[-wπ/8,w-wπ/8], y=sinu恰有三个最高点, ∴-wπ/8

解: ∵ 0<a<四分之π<β<四分之三π ∴ 3π/4

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