Functional Equation Involving Trigonometric Functions: $f(\sin x +\sin y)=f(x+y)$

Question

Find all functions $f : \mathbb{R} \to \mathbb{R} $ such that $f(\sin x +\sin y)=f(x+y)$. My guess is that the function is constant, I've found that:

1. The function is even
2. $f(\sin x + \sin y)=f(\cos x + \cos y)$
3. The function is periodic with period $\pi$

I don't really know what else to do.

Answer 1

hint

If $y=0$, then

$$f(x)=f(\sin(x))$$ and $$f(\sin(x))=f(\sin(\sin(x)))=...=f(0)$$ if continuity at $0$.

Answer 2

$$f(\sin(x)+\sin(y))=f(x+y)\iff f\left(2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})\right)=f(x+y)$$ Making $x+y=\pi$ we have $$f(2(\cos(\frac{x-y}{2}))=f(\pi)=f(2(\cos(\frac{\pi}{2}-y))=f(2\sin(y))$$ The range of the function $g(y)=2\sin(y)$ is equal to $[-2,2]$ equal to the range of the periodic function $h(x,y)=\sin(x)+\sin(y)$.

The function $f(x)$ is in fact constantly equal to $f(\pi)$.