TITLE: Squaring a trigonometric inequality
QUESTION [8 upvotes]: 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?

REPLY [2 votes]: 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.