How may I prove that both are uniformly continuous?:
- $f(x,y)=\sin(x+y)$ for 1, I said, let $\epsilon > 0$ if $d((x,y),(u,v))<\lambda$:
$|f(x,y)-f(u,v)|<|f(x,y)|+|f(u,v)|=|\sin(x+y)|+|\sin(u,v)|<|x+y|+|u+v|$
But how may I continue from here?
2026-03-29 20:49:39.1774817379
Proving that $\sin(x+y)$ is uniformly continuous
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Note that\begin{align}|\sin(x+y)-\sin(u+v)|&\leqslant|\sin(x+y)-\sin(x+v)|+|\sin(x+v)-\sin(u+v)|\\&\leqslant|y-v|+|x-u|.\end{align}Can you take it from here?