Show that $f(x)=\cos x$ is uniformly continuous in $\mathbb{R}$, directly from the definition 
Possible Duplicate:
Proving that cosine is uniformly continuous 


Show that $f(x)=\cos x$ is uniformly continuous in $\mathbb{R}$. Use the definition.

I don't know how to do this.
 A: Let $\epsilon>0$ and $x,y\in \mathbb{R}$. We want
$$\left|f(x)-f(y)\right|<\epsilon\implies \left|\cos x-\cos y\right|<\epsilon\implies \left|-2\sin \frac{x+y}2\sin\frac{x-y}2\right| < \epsilon$$
Because 
$$\left|-2\sin \frac{x+y}2\sin\frac{x-y}2\right|\le 2\left|\sin\frac{x-y}2\right|$$
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: Hint
$$\left| \cos(x)- \cos(y) \right| =2 \left| \sin( \frac{x-y}{2}) \right| \left| \sin(\frac{x+y}{2}) \right| \leq 2 \left| \sin( \frac{x-y}{2}) \right| $$
A: Hint The trigonometric identity
$$\cos(x)-\cos(y) = 2 \sin \left(\frac{x+y}{2}\right) \cdot \sin \left( \frac{x-y}{2} \right)$$
holds. Use $|\sin(z)| \leq 1$, $|\sin(z)| \leq |z|$ ($z \in \mathbb{R}$) to find an estimate for $|\cos(x)-\cos(y)|$.
Another possibility: Apply the mean-value theorem to $x \mapsto \cos(x)$.
(In both cases you obtain that $x \mapsto \cos(x)$ is Lipschitz-continuous which makes it pretty easy to prove the uniform continuity using the definition.)
A: You want to show that for each $\epsilon>0$ there exists a $\delta>0$ such that for each $x,y$ $$|x-y|<\delta\implies |\cos(x)-\cos(y)|<\epsilon$$
Now recall that the cosine has the sine as it derivative, so that for any $x,y$, there exists a $\mu$ such that
$$|\cos x-\cos y |=|\sin\mu||x-y|$$
But 
$$|\sin\mu|\leq 1$$ for any value of $\mu$, so that
$$|\cos x-\cos y |\leq |x-y|$$
for any choice of $x,y$. Thus, given $\epsilon >0$, choose $\delta =\epsilon /2$. This means that, whenever $|x-y|<\delta=\epsilon/2$, we'll have
$$|\cos x-\cos y |\leq |x-y|=\epsilon/2 <\epsilon$$
so that uniform continuity is achieved. 
Functions with the property that
$$|f(x)-f(y)|<\alpha |x-y|$$ are called Lipschitz continuous, and in particular, every Lipschitz continuous function is uniformly continuous, but not conversely.
