Okay I was given this problem. I concluded that it's false, since if we move the sin(x) to the left side of the equation, we will have a periodic expression that and no matter what the function f does it will remain periodic.. Am I missing something? and how would you prove it?
That is completely right. $f(\cos x) - \sin(x)$ is a $2\pi$-periodic function (no matter what $f$ is), and $\operatorname{sign}(x)$ is not periodic.
It would also suffice to set $x= -\pi$ and $x= \pi$ to obtain a contradiction: $$ f(\cos(-\pi)) = \sin(-\pi) + \operatorname{sign}(-\pi) \implies f(-1) = 0 + (-1) \\ f(\cos(\pi)) = \sin(\pi) + \operatorname{sign}(\pi) \implies f(-1) = 0 + 1 \\ $$