$\displaystyle \sup_{t\in \mathbb{R}}\left(\sin(\sqrt 2 \ t)+\sin(t)\right)=2$

Question

How to prove that $$\sup_{t\in \mathbb{R}}\left(\sin(\sqrt 2 \ t)+\sin(t)\right)=2$$ I know that $\displaystyle \sup_{t\in \mathbb{R}}\left(\sin(\sqrt 2 \ t)+\sin(t)\right)\leq 2$. I also know that the supremum is not reached on some number $t_0$, otherwise we would have $\sqrt 2 \in \mathbb{Q}$.

Answer 1

Hint: For any irrational number $c$ the set of all numbers of the form $nc+m$ where $n$ and $m$ are integers is dense in the real line. Using this show that we can find integers $n,m$ with $|2n\pi+\frac {\pi} 2-\sqrt 2 (2m\pi+\frac {\pi} 2)| <\epsilon$. Since $|\sin t -\sin s| \leq |t-s|$ you can find $t$ such that both $\sin t$ and $\sin \sqrt 2 t$ are close to $1$.