已知函数f(x)满足f(0)=1,且对x,y∈R,恒有f(x+y)+f(x-y)=2f(x)*cosy.求f(x)

Question

已知函数f(x)满足f(0)=1,且对x,y∈R,恒有f(x+y)+f(x-y)=2f(x)*cosy.求f(x)

Answer 1

当x=0,f(x+y)+f(x-y)=2f(x)*cosy为f(y)+f(-y)=2cosy；
当x+y=0,f(x+y)+f(x-y)=2f(x)*cosy为1+f(-2y)=2f(-y)*cosy;
当x-y=0,f(x+y)+f(x-y)=2f(x)*cosy为f(2y)+1=2f(y)*cosy;
由第二和第三式子看出,f（y）是偶函数,即f(y)=f(-y),将其带入第一个式子,得到f(y)=cosy；所以f(x)=cosx