f(x)＝2sin(wx+φ)

f(x)=sinwxcosPai/3+coswxsinPai/3-coswxcosPai/6+sinwxsinPai/6+coswx=sinwx+coswx=根号2sin(wx+Pai/4)T=2Pa

f(x)=sin(wx+φ)-cos(wx+φ)=√2[√2/2sin(wx+φ)-√2/2cos(wx+φ)]=√2sin(wx+φ-π/4)∵函数y=f(x)图像的两相邻对线轴的距离为π/2.∴f

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

(1)f(x)=sin²ωx+2√3sin(ωx+π/4)cos(ωx-π/4)-cos²ωx-√3=2√3·√2/2(sinωx+cosωx)·√2/2(sinωx+cosωx)

(1)f(x)=根号3sin(wx+φ)-cos(wx+φ)=2Sin(wx+φ-π/6)由于是偶函数,即f(x)=f(-x）即2Sin(wx+φ-π/6)=2Sin(-wx+φ-π/6)即Sinwx

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

就以前两个为条件T=2π/w=πw=2f(x)=sin(2x+φ)sinx的对称轴就是取最值的地方即sin(2x+φ)=±12x+φ=kπ+π/2x=-π/6所以φ=kπ+5π/6由φ范围,取k=-1

（1)f(x)=根号3sin(wx+φ)-cos(wx+φ)=2Sin(wx+φ-π/6)由于是偶函数,即f(x)=f(-x）即2Sin(wx+φ-π/6)=2Sin(-wx+φ-π/6)即Sinwx

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+

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/

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

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

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

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)=sin(wx+φ){w>0,|φ|0,|φ|φ=π/6∴f(x)=sin(2x+π/6)2x+π/6=2kπ==>x=kπ-π/12；2x+π/6=2kπ+π==>x=kπ+5π/12；点对

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

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

由1,3作为条件,可以得到2,由2,3作为条件,可以得到1,由1,3得到2,证明：由3可知w=2或-2,设定w=2时,由1可以得到2*π/12+t=kπ/2,k为不等于0的整数.得到t=kπ/2-π/

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)=sin(wx+φ)+cos(wx+φ)=√2sin（x+π/4+φ）而f（x）是偶函数,故f（x）关于x=0对称所以π/4+φ=kπ+π/2（k是整数）而,|φ|