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?
 A: There is a small distinction to the statement $a \leq b \; \implies \; a^2 \leq b^2$ and that is that this statement holds only when $a, b \geq 0$. However if $a, b \leq 0$ the opposite is true $a \leq b \; \implies \; a^2 \geq b^2$. So you need to check the signs of the terms you are squaring in an inequality. 
A: It would be better to say
As $0\le|{\cos\theta}|\le1$,
$0^2\le|{\cos\theta}|^2\le1^2$
$0\le{\cos^2\theta}\le1$
A: The square function is not growing, so it does not preserve the inequality direction! $$-5 < -2 < -1$$ but $$(-5)^2 > (-2)^2 > (-1)^2$$
Additionally the square function is not monotonic, so it does not preserve inequalities at all! $$-1 < 1$$ but $$(-1)^2 = (1)^2$$
Squaring preserves inequality only if both sides of inequality are non-negative and so they belong to a part of the domain, where the square function is strictly growing.
A: You can not deduce that if $a\leq b$ for $a,b\in \mathbb{R}$, then $a^2\leq b^2$. For example, for all $a\neq b$, both negative, such inequality is false. However, you can state your original inequality as
$$
0\leq |\cos x|\leq 1,
$$
from which you can deduce
$$
0\leq |\cos^2 x|=\cos^2 x\leq 1,
$$
since the inequality $a^2\leq b^2$ does hold in the case of $0\leq a\leq b$.
A: Squaring does not preserve inequality since it is a 2-to-1 function over $\mathbb{R}$, hence nonmonotone. Mind you a monotone decreasing function would reverse the inequality there would still be a valid inequality pointing the opposite way.
A: 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 $
