Solve: $f(x+\frac{1}{y}) + f(x-\frac{1}{y}) = 2f(x).f(\frac{1}{y})$ Here i have one functional equation:
If $$f(x+\frac{1}{y}) + f(x-\frac{1}{y}) = 2f(x)\cdot f(\tfrac{1}{y})\text{ for all x},y\in\mathbb{R}-{0}$$ and $f(0) = \frac{1}{2}$ , then find the value of $f(4)$.
As noted in the comments, $f(x)$ is symmetric.  Set $a=f(1), b=f(2), c=f(3), d=f(4)$.  Then we have the following four equations:

$$d+f(0)=2b^2$$
$$d+b=2ac$$
$$c+a=2ab$$
$$b+f(0)=2a^2$$

This has two solutions according to alpha; in both solutions $d=-\frac{1}{2}$.  Finding them by hand doesn't seem too difficult.
$f(x)=\frac{1}{2}\cos x$ satisfies the condition.