已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(0)不等于0,若f(π /2)=0,求f(π )和f(2π).
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已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(0)不等于0,若f(π/2)=0,求f(π)和f(2π).已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),
已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(0)不等于0,若f(π /2)=0,求f(π )和f(2π).
已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(0)不等于0,若f(π /2)=0,求f(π )和f(2π).
已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(0)不等于0,若f(π /2)=0,求f(π )和f(2π).
f(x+y)+f(x-y)=2f(x)f(y),
令x=y=0
则有;
f(0)+f(0)=2f(0)^2
2f(0)=2f(0)^2
f(0)不等于0
f(0)=1
令x=y=pai/2
则有:
f(pai)+f(0)=2f(pai/2)f(pai/2)
f(pai)=0-f(0)
=0-1
=-1
令x=y=pai
f(2pai)+f(0)=2f(pai)f(pai)
f(2pai)=2f(pai)^2-f(0)
=2(-1)^2-1
=1
令x=1,y=0,由f(x+y)+f(x-y)=2f(x)f(y),2f(1)=2f(1)f(0),f(0)不等于0,所以f(0)=1.
令x=y,f(2x)+f(0)=2f(x)f(x)
f(2x)=2f(x)f(x)-1
f(π )=2f(π /2)f(π /2)-1=-1
f(2π)=2f(π )f(π )-1=1
f(pi)=-1,f(2pi)=1
令x=y=0,得f(o)=1
令x=y=pi/2,得f(pi)=-1
令x=y=pi,得f(2pi)=1
这么厉害~ 我现在还是不太掌握这种抽象函数啊……
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