Prove $\sin x$ is uniformly continuous on $\mathbb R$ 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). 
 A: 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.
A: Let $\epsilon>0$ and $x,y\in \mathbb{R}$. We want
$$\left|f(x)-f(y)\right|<\epsilon\implies \left|\sin x-\sin y\right|<\epsilon\implies \left|2\cos\frac{x+y}2\sin\frac{x-y}2\right|$$
Because 
$$\left|2\cos\frac{x+y}2\sin\frac{x-y}2\right|\le 2\left|\sin\frac{x-y}2\right|$$
it suffices
$$2\left|\sin\frac{x-y}2\right|<\epsilon$$
when
$$\left|x-y\right|<\delta\implies \left|\frac{x-y}2\right|<\delta$$
SInce $\left|\sin x\right|\le \left|x\right|$,
$$2\left|\sin\frac{x-y}2\right|\le 2\left|\frac{x-y}2\right|<2\delta$$
Choosing $\delta=\frac{\epsilon}{2}>0$ will do the trick. Because $\delta$ doesn't depend on $x,y$, the continuity is uniform
A: By Mean Value Theorem, 
$$ |\sin{x}- \sin{y}| \leq |x-y| |\cos{\xi}| \leq |x-y|, \quad x\leq\xi \leq y.$$
Hence, you may choose $\epsilon=\delta$.
A: Since $\sin x$ is a periodic continuous function with a period $2\pi$, it suffices to prove that it is uniformly continuous on $[0, 2\pi]$. Since $[0, 2\pi]$ is compact, this follows from the well-known theorem.
