Question

已知函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y),且f(0)≠0,f(π/2)=0,求f(0),f(2π)的值

Answer 1

f(1+0)+f(1-0)=2f(1)f(0)
2f(1)=2f(1)f(0)
f(0)=1

f(x+y)+f(x-y)=2f(x)f(y)
f(π/2+π/2)+f(π/2-π/2)=2f(π/2)f(π/2)

f(π）+f(0)=0
则f(π)=-1

f(π+π）+f(π-π)=2f(π)f(π)
f(2π)+f(0)=4f(π)
f(2π)=-4-1=-5

Answer 2

令x=y=0
则f(0)+f(0)=2f(0)*f(0),且f(0)≠0
所以f(0)=1

令x=y=π/2
则f(π)+f(0)=2f(π/2)*f(π/2)=0
f(π)=-f(0)=-1

令x=y=π
则f(2π)+f(0)=2f(π)*f(π)=2
f(π)=2-1=1

Answer 3

x=y=0时 f(0)+2f(0).f(0) f(0)=1
x=y=π/2 时 f(π)+f(0)=2f(π/2).f(π/2) f(π)=-1
x=y=π 时 f(2π)＋f(0)=2f(π)．f(π) f(2π)=1