设f(x)=cos(wx+φ)(w>0,0
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设f(x)=cos(wx+φ)(w>0,0设f(x)=cos(wx+φ)(w>0,0设f(x)=cos(wx+φ)(w>0,01.由于周期为π,故2π/w=π,因此w=2由于f(x)是R上的奇函数,因
设f(x)=cos(wx+φ)(w>0,0
设f(x)=cos(wx+φ)(w>0,0
设f(x)=cos(wx+φ)(w>0,0
1.由于周期为π,故2π/w=π,因此w=2
由于f(x)是R上的奇函数,因此f(-π/2)=-f(π/2),由于0
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