已知f(x)满足f(x+y)+f(x-y)=2f(x)*f(y)且f(0)≠2,若f(π/2)=0,求f(π)及f(2π)
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已知f(x)满足f(x+y)+f(x-y)=2f(x)*f(y)且f(0)≠2,若f(π/2)=0,求f(π)及f(2π)
已知f(x)满足f(x+y)+f(x-y)=2f(x)*f(y)且f(0)≠2,若f(π/2)=0,求f(π)及f(2π)
已知f(x)满足f(x+y)+f(x-y)=2f(x)*f(y)且f(0)≠2,若f(π/2)=0,求f(π)及f(2π)
f(x+y)+f(x-y)=2f(x)*f(y)
x=y=0
f(0)+f(0) =2[f(0)]^2
f(0)=0
x=y=π/2
f(π)+f(0)=2f(π/2).f(π/2)
f(π)=0
x=y =π
f(2π)+f(0)=2f(π)*f(π)
f(2π) = 0
令x、y均为0,所以f(0)+f(0)=2f(0)×f(0)所以f(0)^2-f(0)=0所以f(0)=1或f(0)=0,又f(0)≠0.所以f(0)=1设x=y,则f(2x)+f(0)=2f(x)f(0)=2f(x)所以f(2x)=2f(x)这似乎不能求出f(2π)
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