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
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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).
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=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
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