若函数f(x)=sin(wx φ)(其中w>0)在(0,π 3)上单调递增,且

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+阿发)+cos(wx+阿发)=√2[√2∕2sin(wx+阿发)+√2∕2cos(wx+阿发)]（提取√2,根号2）=√2[cos45°sin(wx+阿发)+sin45°cos

&=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.

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

(1)sin(wx+π/6)=sinwxcosπ/6+coswxsinπ/6sin(wx-π/6)=sinwxcosπ/6-coswxsinπ/6f(x)=sin(wx+π/6)+sin(wx-π/6

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)=根号3sin(wx+a)-cos(wx+a)当a+π/3=kπ时f(x)为偶函数,而0<a<π,则a+π/3=πf(x)=2coswx,而函数y=f(x)图象的两相邻对称轴间

(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

（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+

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

已知函数f（x）＝sin（wx+∮）（w＞0.0＜∮＜派）为偶函数,其图像上相邻的一个最高点和一个最低点之间的距离为√（4+派的平方）.求f（x）的解析式解析：∵函数f(x)＝sin(wx+∮)（w＞

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

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/

(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(-