TITLE: Prove $\sin x$ is uniformly continuous on $\mathbb R$
QUESTION [24 upvotes]: How do I prove $\sin x$ is uniformly continuous on $\mathbb R$ with delta and epsilon? 
I proved geometrically that $\sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|\sin x_1 - \sin x_2|\le|\sin x_1|+|\sin x_2|<|x_1|+|x_2|$$ 
But this doesn't help me much finding a delta... 
Thanks for any help!
P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation  (because they haven't been regorously proven).

REPLY [4 votes]: There is an elementary geometric way to do this.  Let $x$ and $y$ be real numbers; for now, assume $x, y\in(-\pi, \pi]$.  Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle.  Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle.  $|\sin(x) - \sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$.  Hence, in this case
$$|\sin(x) - \sin(y) | \le |x - y|.$$
This gives us uniform continuity on $(-\pi, \pi]$, so by periodciity the sine function is uniformly continuous on the entire line.