3. Given the function $f(x)$ satisfies: $f(1)=\frac{1}{4}, 4 f(x) f(y)=f(x+y)+f(x-y)(x, y \in \mathbf{R})$, then $f(2019)=(\quad)$
A: $\frac{1}{2}$
B: $\quad-\frac{1}{2}$
C: $\frac{1}{4}$
D: $\quad-\frac{1}{4}$
Answer B.
Analysis Taking $x=1, y=0$, we get $f(0)=\frac{1}{2}$;
Taking $x=1, y=1$, we get $4 f^{2}(1)=f(2)+f(0)$, so $f(2)=-\frac{1}{4}$;
Taking $x=2, y=1$, we get $4 f(1) f(2)=f(3)+f(1)$, so $f(3)=-\frac{1}{2}$;
Taking $x=n, y=1$, we have $f(n)=f(n+1)+f(n-1)$;
Similarly, $f(n+1)=f(n+2)+f(n)$, combining these, we get $f(n+2)=-f(n-1)$, hence $f(n+6)=f(n)$. Therefore, the period is 6, so $f(2019)=f(336 \times 6+3)=f(3)=-\frac{1}{2}$.