Let $\varphi$ be the golden ratio. What is the supremum of $f(x)=\sin(x) + \sin(\varphi x)$?
My idea was that we can consider the sequence of functions $f_n(x) = \sin(x) + \sin(\frac{F_{n+1}}{F_n} x)$ for the Fibonacci numbers $F_n$, which converges to $f(x)$. The sequence $\{x_n\}_{n \geq 1}=F_n \pi + a_n$ satisfies $f_n(x_n) = \sin(a_n) + \sin(F_n a_n)$, so that if we choose the correct $a_n$, then $f_n(x_n) =2$ $\forall n$; which changes the problem to one with an integer instead of $\varphi$, but I'm not sure where to proceed from here.
Clearly the maximum is bounded above by $2$, as each $\sin$ has a maximum of $1$.
You get the maximum for the first function when $\sin x = 1$ or $x = 2 \pi k_1 + \pi/2$ for some integer $k_1$, and likewise when $\sin (\phi x) = 1$ or $\phi x = 2 \pi k_2 + \pi/2$. A little manipulation gives:
$$k_1 = \frac{k_2}{\phi} - \frac{\phi - 1}{4 \phi}$$
Because $\phi$ is not rational, one cannot solve this equation exactly for $k_1$ and $k_2$. However, you can approximate this arbitrarily closely by choosing $k_2$ extremely large (e.g., $10^{1000}$), and solving for $k_1$. In this way you can get as close to $f(x) = 2$ as you like.