已知函数f(x)=2sin(wx φ)对任意x都有f(3分之π+x)

&=π/2,w=2.f(x)=sin(2x+π/2)=cos2x,偶函数,关于点M(3π/4,0)对称,且在[0,π/2]上是单调递减函数.

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

已知函数f(x)=sin(wx+π/3)(w>0)若f(π/6)=f(π/2),且f(x)在区间(π/6,π/2)内有最大值,无最小值,则w的最小值为?解析：∵函数f(x)=sin(wx+π/3)(w

向量a⊥向量bcosφ=sinφφ=pi/4周期为piw=2f(x)=sin(2x+pi/4)g(x)=sin(2(x-6)+pi/4）=sin(2x-12+pi/4)-pi/26-3pi/8

f(x)=sinwxcospi/6+coswxsinpi/6+sinwxcospi/6-coswxsinpi/6-coswx-1=根3sinwx-coswx-1=2sin(wx-pi/6)-1所以值域

相邻的两条对称轴的距离=T/2(T为最小值周期)T/2=π/2周期T=2π/2w=πw=1x=-π/4F(-π/4)=2根号3sin(-π/2+@)=0@=π/2所以F(x)=2√3sin(2x+π/

∵sin^2wx=1-cos^2wx=1-(cos2wx+1)/2=1/2-(cos2wx)/2根号3sinwx*sin(wx+派/2)=根号3sinwx*coswx=(根号3sin2wx)/2∴f(

解由图象上的两个相邻的最高点和最低点的距离为2根号2,即T=2*2√2=4√2由T=2π/w=4√2即w=√2π/4即f(x)=sin(√2π/4x+φ),由图象过点(2,1/2),即sin(√2π/

它是先得出：pi/2〈=ωx

2π/w=6π所以w=1/3x/3+φ=π/2+2kπ或x/3+φ=-π/2+2kπ（k属于z)φ=π/3+2kπ或φ=-5π/6+2kπ又-π

（1）∵F(x)的图像经过点（π/6,2)代入：2=2sin(π/6w+π/6）由于sin（kπ+π/2）=0,且0

已知函数f(x)=根号3sin(wx+φ)++(w>0,-π/2x=(π-2φ)/4=π/3==>φ=-π/6∴f(x)=√3sin(2x-π/6)(2)解析：设f(a/2)=√3/4,(π/6＜a＜

【参考答案】w=3/2由函数图象知,当x=2π/3时,取得最大值；当x=0时,取得最小值.∴函数半个周期是2π/3-0=2π/3函数最小正周期是2×(2π/3)=4π/3即T=2π/w=4π/3∴w=

1：(sinwx)^2+√3sinwxsin(wx+π\2)=(sinwx)^2+√3sinwxcoswx=2[(sinwx)^2+(√3\2)sin2wx]\2=[2(sinwx)^2+√3sin2

f(x)=根号3sinwxcoswx+sin^2wx=[（根号3）/2]sin2wx+(1-cos2wx)/2=sin(2wx-兀/6)-1/2(1)T=2兀/(2w)=兀w=1f(x)=sin(2x

f(x)=f(-x)sin(x+w)+sqrt(3)*cos(x-w)=sin(w-x)+sqrt(3)*cos(x+w)2(cos30度cos(x+w)-sin30度sin(x+w))=2(cos3

T=π/2*2=π即2π/w=π∴w=2f(0)=2sinφ=根号3sinφ=√3/2∴φ=π/3

由已知可得,T/2=7π/8-3π/8=π/2,所以T=π=2π/w,则w=2.当f(3π/8)=2sin(3π/4+φ)=2,必有f(7π/8)=2sin(7π/4+φ)=-2sin(3π/4+φ)

f(x)=sinwx·sin(π/2-ф)-sin(π/2+wx)sin(π+ф)=sinwx*cosф+coswxsinф=sin(wx+ф)是R上的偶函数∴ф=π/2+kπ,k∈整数0