Squaring a trigonometric inequality A very, very basic question.
We know
$$-1 \leq \cos x \leq 1$$
However, if we square all sides we obtain
$$1 \leq \cos^2(x) \leq 1$$
which is only true for some $x$.
The result desired is
$$0 \leq \cos^2(x) \leq 1$$
Which is quite easily obvious anyway. 
So, what rule of inequalities am I forgetting?

since $a>b \ \space  \forall a,b \in R$
It is not necessary
...
$a^2>b^2  \  \  \forall  a , b \in  R$
let  $ a=-1  b=-4 $ 
$ a>b   \ But  \  b^2>a^2 $
