函数y=sinxcos(x+π/4)+cosxsin(x+π/4)的最小正周期是

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函数y=sinxcos(x+π/4)+cosxsin(x+π/4)的最小正周期是
函数y=sinxcos(x+π/4)+cosxsin(x+π/4)的最小正周期是

函数y=sinxcos(x+π/4)+cosxsin(x+π/4)的最小正周期是
y=sinxcos(x+π/4)+cosxsin(x+π/4)
=sin(x+x+π/4)=sin(2x+π/4)
所以w=2,T=2π/w=π
所以其最小正周期为π
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y=sinxcos(x+π/4)+cosxsin(x+π/4)
=sin[x+(x+π/4)]
=sin(2x+π/4)
所以T=2π/2=π

是‘ Pai ’ 决定对!我们手机打不出这个符号,不好意思