d'Alembert's functional equation

Remarks

Example: $f(x) = \cos(2x)$ is a solution, by the product formula $2 \cos u \cos v = \cos(u + v) + \cos(u - v)$. Non-example: $f(x) = 2$ is not a solution, since the left side is $4$ and the right side is $8$. Setting $y = 0$ gives $2 f(x) = 2 f(x) f(0)$, so either $f = 0$ or $f(0) = 1$. If $f(0) = 1$, setting $x = 0$ gives $f(-y) = f(y)$, so every nonzero solution is even. The continuous solutions are $f = 0$, $f(x) = \cos(ax)$ and $f(x) = \cosh(ax)$ with $a \in \R$. The same equation is studied for complex-valued functions on groups, in the form $f(xy) + f(xy^{-1}) = 2 f(x) f(y)$.