Question

请教老师~设函数f(x)=sin(wx+φ)+cos(wx+φ)(w>0,|φ|<π/2)的最小正周期为π，且f(-x)=f(x)则

A f(x)在（0，π/2)单调递减。
B f(x)在（π/4, 3π/4)单调递减。
C f(x)在(0,π/2）单调递增。
D f(x)在（π/4,3π/4)单调递增
刚开始的解是
f(x)=sin(wx+φ)+cos(wx+φ)
=√2sin(wx+φ+π/4)这步怎么理解？？

Answer 1

asinx+bcosx=√(a²+b²)sin(x+φ)
∴f(x)=sin(wx+φ)+cos(wx+φ)=√2sin(wx+φ+π/4)
∵T=π
∴w=2
∵f(-x)=f(x)
∴函数是偶函数
∴φ+π/4=kπ+π/2
∴φ=kπ+π/4
∵-π/2＜φ＜π/2
∴φ=π/4
∴f(x)=)=√2scos2x
∴在区间[-π/2+kπ，kπ](k∈Z)是单调递增
在区间[kπ，kπ+π/2](k∈Z)是单调递减
∴选A

Answer 2

f(x)=sin(wx+φ)+cos(wx+φ)=√2sin(wx+φ+π/4) ,最小正周期为π，w=2,且f(-x)=f(x)则偶函数，对称于y轴，起点坐标（-π/4,0) , -(φ+π/4) /w=-π/4 ,φ=π/4
y=√2sin(2x+π/2)
A f(x)在（0，π/2)单调递减。

Answer 3

f(x)=sin(wx+φ)+cos(wx+φ)
=√2*(√2/2)*sin(wx+φ)+√2*(√2/2)*cos(wx+φ)
=√2sin(wx+φ)cosπ/4+√2cos(wx+φ)sinπ/4
=√2[sin(wx+φ)cosπ/4+cos(wx+φ)sinπ/4]
=√2sin(wx+φ+π/4)