Q:

Taylor expansion and trigonometric functions

I've seen a proof including this claim:
$$x-\frac{x^3}{6} \le \sin x$$
Now, for my understanding, the series $x-\frac{x^3}{6} +\frac{x^5}{120} -... $ is converging to $\sin x$ in an alternating way.  meaning, with each term, the sum of the terms is getting closer to $\sin x$, one time from the bottom and one time from the top.   

I'd be glad if you could expand about this behavior.   
Is it true only for $\sin x$ and $\cos x$? 

Update:
I want to make my question more clear (but maybe a more general one):  
Suppose you're given the first $n$ terms of the expansion of $\cos x$, $\sin x$ (maybe $e^x$ too). 
How can you infer if the summation of those terms is bigger or smaller than $f(x)?$
(where $f(x)$ can be $\sin x$, $\cos x$ etc..)  
For example, is there a pattern I can rely on? ( i.e. alternation)
Thanks. 

A:

You can prove these inequalities by using increasing functions.  Consider 
the inequality,
$$1-\frac{x^2}{2} < \cos x$$ we take the function
$f(x)=\cos x -1 +\frac{x^2}{2}$ then 
$f^{\prime}(x)= -\sin x + x $.  Now $0 < f^{\prime}(x)$ because $\sin x < x$ for $0<x$. this means that $f$ is increasing, thus
$$f(0) < f(x)$$ for $0<x$ therefore 
$$0 <\cos x -1 +\frac{x^2}{2}$$
Now try the same thing for 
$$g(x)=\sin x -x +\frac{x^3}{6}$$
You can continue the pattern at the next stage you get
$$\cos x < 1-\frac{x^2}{2}+\frac{x^4}{4!}$$ and 
$$\sin x < x-\frac{x^3}{3!}+\frac{x^5}{5!}$$
So the stages alternate between lower and upper bounds.