函数f x2sin(wx φ)(w>0,0

函数f(x)=sin(ωx+φ)(w>0,0≤φ≤π)是R上的偶函数∴f(-x)＝f(x)→sin(-wx+φ)=sin(wx+φ)→-sinωxcosφ=sinωxcosφsinωx不恒等于0,∴c

T=2π／3=2π／ω,∴ω=3.∵最小值为﹣2,∴A＝2.将﹙5π／9,0﹚代入函数,可得：2sin(5π／9×3＋φ）=0,解得：φ=kπ－5π／3.∵φ的绝对值＜π,∴﹣π＜φ＜π,即：∵﹣π＜

y=2sin(2x+π/6)f(0)=1f(π/2)=-11

T=π=2π/w-->w=2最高点的纵坐标为3/2-->A=3/2对称轴方程是x=π/6-->因为sin函数的对称轴在π/2+kπ,上,所以φ=-π/6+kπ+π/2--->φ=π/3y=1.5sin

f(x)=sin(wx+φ)+cos(wx+φ)=√2sin(wx+φ+π/4)T=2π/w=πw=2f(x)=√2sin(2x+φ+π/4)f(-x)=f(x),所以f(-π/8)=f(π/8)si

A=22sinφ=√3φ=π/3w*(5π/6)+π/3=π或w(5π/6)+π/3=2πw=4/5w=2f(x)=2sin【(4/5)x+π/3】f(x)=2sin(2x+π/3)-π/2

f(x)=(√3)sin(ωx+φ)-cos(ωx+φ)=2{[(√3)/2]sin(ωx+φ)-(1/2)cos(ωx+φ)}=2[sin(π/3)sin(ωx+φ)-cos(π/3)cos(ωx+

根据图象求出函数的周期,再求出ω的值,根据周期设出M和N的坐标,利用向量的坐标运算求出A的值,即求出A•ω的值．望采纳,谢谢

有图得A=3两个相邻的零点π/3,5π/6相距半个周期∴T/2=5π/6-π/3=π/2,∴T=π由2π/w=π,得w=2f(x)=3sin(2x+φ)当x=(π/3+5π/6)/2=7π/12时,函

A=(最大值-最小值)/2=2,b=(最大值+最小值)/2=1,周期T：T/2=2π/3-π/6=π/2,T=π,W=2,所以,f(x)=2sin(2x+φ)+1,代入最大值点(π/6,3),化简,s

（1）可以看出周期是π则w=2π/T=2至于φ,当x=0时f（x）=sinφ=-1则根据题意,φ=-π/2（2）f（x）=sin（2x-π/2）=-cos2xg（x）=2√2f（x/2）f(x/2-π

画图可以画出分数线的形式,但是慢了点,我这幅图看得懂吧?求采纳

(a)T=7pai/12-(-pai/12)=8pai/12=2pai/3w=2pai/(2pai/3)=3y=Asin(3x+Q)=Asin(3(x+q))y=Asin3x向左移了pai/12所以,

求函数函数y=Asin(wx+φ))(A≠0,w>0)的单调区间解析：∵函数y=Asin(wx+φ))(A≠0,w>0)单调增区间：2kπ-π/2

周期是（x0+π/2-x0）*2=π所以T=2π/w=πw=2最大值为3所以A=3f(x)=3sin(2x+φ)f(0)=3sinφ=3/2|φ|

f(x)=√3sin(wx+φ/2)*cos(wx+φ/2)+sin^2(wx+φ/2)=(√3/2)sin(2wx+φ)+(1/2)[1-cos(2wx+φ)]=sin(2wx+φ-π/6)+1/2

函数f(x)=sin(wx+φ)(w>0,|φ|0,|φ|φ=2π/3f(x)=sin(2x-2π/3+φ)=-sin2x==>φ=-π/3∵|φ|x=kπ+5π/122x-π/3=2kπ-π/2==

f(x)=√2sin(8（x/4+π/2)+φ）因为加了个π/2所以变成了cos所以变成偶函数

(1)f(x)=√3sin(wx+φ)-cos(wx+φ)=2sin(wx+φ-π/6)相邻对称轴间的距离为π/2,最小正周期为π所以w=2π/π=2又知f(0)=2sin(φ-π/6)=00

f(x)=√2sin(wx+φ+π/4)2π/w=πw=2f(x)=√2sin(2x+φ+π/4)f(-x)=√2sin(-2x+φ+π/4)f(x)=f(-x)sin(2x+φ+π/4)=sin(-