6. Let $f(x), g(x)$ be two functions defined on $(-\infty,+\infty)$, for any real numbers $x, y$, satisfying $f(x+y) +$ $f(x-y)=2 f(x) \cdot g(y)$. If $f(0)=0$, but $f(x)$ is not identically zero, then
A. $f(x), g(x)$ are both odd functions
B. $f(x), g(x)$ are both even functions
C. $f(x)$ is an even function, $g(x)$ is an odd function
D. $f(x)$ is an odd function, $g(x)$ is an even function
6. D Let $x=0, f(y)+f(-y)=0, f(-y)=-f(y)$, so $f(y)$ is an odd function. Also, by substituting $-y$ for $y$, we have $f(x-y)+f(x+y)=2 f(x) g(-y)$, thus $g(-y)=g(y)$.