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

问题描述:

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

f(x+0)+f(x-0)=2f(x)f(0)
=2f(x)
f(0)=1
f(pai/2+pai/2)+f(pai/2-pai/2)=2f(pai/2)f(pai/2)
f(pai)+1=0
f(pai)=-1
f(pai+pai)+f(0)=2f(pai)f(pai)=2
f(2pai)=1