已知函数fa=Asin(x π 4 gx=tanx

2sin^2x=1-cos2x2sinxcosx=sin2xf(x)=a-acos2x-a√3sin2x+b=a+b-2a(cos2xcosπ/3+sin2xsinπ/3)=a+b-2acos(2x-

f(5π/12)=Asin(5π/12+π/4)=Asin(2π/3)=A*√3/2,(√为根号）=3/2A=√3f（θ）+f（-θ）=3/2√3sin(θ+π/4)+√3sin(-θ+π/4)=3/

应仔细审题∵函数y=Asin(ωx+φ)(x∈R,A,ω>0,-π/20,∴A≠-√2,所以不讨论A=-√2你画的y=-√2sin(π/8x-3π/4)图像是错的,再者φ-π/2

3=f(2011)=asin(2011π+α)+bcos(2011π+β)+4=-asinα-bcosβ+4∴asinα+bcosβ=1∴f(2012)=asin(2012π+α)+bcos(2012

函数f(x)=Asin(X+φ)(A>0)在x=π/2处取得最小值,即sin（π/2+φ）=-1所以φ=-π.f(3π/4-X)=Asin(3π/4-X-π)=Asin(-x-π/4).所以它既不是奇

(1)A=1T/4=π/12+π/6得到T=πw=2所以f（x）=sin（2x+φ）f（π/12）=0故π/6+φ=2kπ+π/2（k是整数）而-π/2

∵0≤x≤π2，∴π6≤2x+π6≤7π6，∴-12≤sin（2x+π6）≤1．①当a＞0时，2asin（2x+π6）∈[-a，2a]，得2asin（2x+π6）+a+b∈[b，3a+b]∴b＝−53

f(x+pai/6)=Asin(2x+pai/3+a)=Acos(pai/6-a-2x)pai/6-a=2kpai,pai/6-a=2kpai+paif(x)=Asin(2x+pai/6-2kpai)

最小正周期是T=2π/(π/3)=6.设S点坐标为(4,0),则三角形QRS为含π/6的直角三角形,RS=√3QS=√3A=3,A=√3.

已知函数f(x)=-2asin(2x+π/6)+2a+b,x∈[π/4,3π/4】,是否存在常数a,b∈Q,其中Q为有理数集,使得f(x)值域为【-√3,√3-1】,若存在,请求出a,b不存在,请说明

最大值是3,则A=3.函数周期是π,则2π/w=π,w=2.f(x)=3sin(2x+α)当x=π/6时f(x)取得最大值3,则3=3sin(π/3+α),π/3+α=π/2,α=π/6.∴f(x)=

已知函数f(x)=Asin(wx+a)(A>0,w>0,-π/20,w>0,-π/2π/3+a=π/2==>a=π/6∴f(x)=3sin(2x+π/6)单调增区间：2kπ-π/2x0=0==>2x0

最大值,最小值的中间量为2所以n=2最大值-最小值=4所以振幅=4/2=2T=π/2=2π/ww=4y=2sin(4x+φ）+2对称轴x=π/3所以sin(4π/3+φ）=±1φ=π/6再问：y=2s

0≤x≤π-π/4≤x-π/4≤3π/4sin(x-π/4)∈【-√2/2,1】ab=3最小值√2a*1+a+b=2------>a=-1/(√2+1)=1-√2a=1-√2,b=3

【解】函数f(x)和φ(x)的最小正周期之和是3π/2,则2π/k+π/k=3π/2,k=2.由f(π/2)=φ(π/2)可得,asin(π+π/3)=btan(π-π/3),-√3a/2=-√3b,

f(x)=-sin^2x+2asin(x-π/2)=-(1-cos^2x)+2a(-cosx)=cos^2x-2acosx-1=(cosx-a)^2-a^2-1由于-1

(1)因为最大值为2+m说明A=2,最大最小值之间的最小距离为π/2,所以W=1所以在x取（-π/4,π/6）时,f(x)最大=f（π/12）=2+mf（x）最小=f（-π/4）=-1+m所以m=2(

(1)∵当Asin(wx+π/4)=1时得最大值A∴A=2最小正周期=2π/w=8w=π/4f(x)=2sin(πx/4+π/4)(2)f(2)=2sin(π*2/4+π/4)=2sin(π/2+π/

已知函数f(x)=Asin(wx+4/π)(其中x属于R,A>0w>0)的最大值为2最小正周期为8(1)求函数f(x)的解析式、(2)若函数f(x)图像上的两点P.Q的坐标依次为2,4,O为坐标原点,