已知f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(x)≠0,若f(π/2)=0,求f(π)和f(2π)的值
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已知f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(x)≠0,若f(π/2)=0,求f(π)和f(2π)的值已知f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(x)
已知f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(x)≠0,若f(π/2)=0,求f(π)和f(2π)的值
已知f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(x)≠0,若f(π/2)=0,求f(π)和f(2π)的值
已知f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(x)≠0,若f(π/2)=0,求f(π)和f(2π)的值
题目不对
f(x)≠0,f(π/2)=0
显然矛盾
但先不管这个
令x=y=0
f(0)+f(0)=2f(0)*f(0)
f(x)≠0
所以两边除以f(0)
f(0)=1
令x=y
则f(2x)+f(0)=2f(x)*f(x)
f(2x)=2[f(x)]²-1
令x=π/2
则f(π)=2*0²-1=-1
同理
f(2π)=2*(-1)²-1=1
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