Question

求证:函数f(x)sin(x+θ)为偶函数的充要条件是θ=kπ+π/2(k∈Z)

φ=kπ+π/2(k∈Z) f(x)=sin(ωx+kπ+π/2) =coswx=cos(-wx)所以是充分条件 必要条件f(x)=f(-x) sin(ωx+φ)=sin(-ωx+φ) sin(ωx+φ)+sin(-ωx+φ)=0……………………………………这步怎么来的? 2sinφcoswx=0 sinφ=0 φ=kπ+π/2(k∈Z)

Answer 1

证明
函数f(x)=sin(x+θ)为偶函数,<===>
恒有f(x)=f(-x),即恒有
sin(x+θ)=sin(-x+θ).
<===>
恒有:
sin(θ+x)-sin(θ-x)=0
和差化积,可知,恒有2cosθsinx=0
<===>
恒有cosθ=0,
<===>
θ=2kπ±(π/2).
即θ=2kπ+(π/2)
或θ=2kπ-(π/2)=2kπ-[π-(π/2)]=(2k-1)π+(π/2)
合写就是:θ=kπ+(π/2)