Why can we say $0\leq\sin^2(x)\leq 1$? I often see instructors write:
$0\leq\sin^2(x)\leq 1$
Why is this valid?
Isn't it supposed to be between $1$ and $-1$?
 A: Because
$\sin^2(x) \ge 0$,
$\cos^2(x) \ge 0$,
and
$\sin^2(x)+\cos^2(x)
= 1
$.
A: Edit: The original question asked why $0 < \sin^2(x) < 1$ for $x \not= 0$ but has now been changed to something else.
The note in the question is not correct: if $x = \pi/2$, then $\sin^2 x = 1$, and if $x = \pi$, then $\sin^2 x = 0$, so $0 < \sin^2 x < 1$ fails to hold true in those cases.
What is true, through, is that $0 \leq \sin^2 x \leq 1$ for all $x$ (including $x = 0$): since the square of any real number is non-negative, you have $0 \leq \sin^2 x$. On the other hand, if a real number $y$ satisfies $0 \leq y \leq 1$, then $y^2 \leq 1$ (since, for instance, $y \mapsto y^2$ is increasing on $[0, \infty)$, and $0^2 = 0$, $1^2 = 1$). Recalling that $\sin x \leq 1$ for all $x$, this implies that also $\sin^2 x \leq 1$ for all $x$.
A: If you plot the to graphs it would be easier for you to see the boundaries.

A: $x \in \mathbb R$ then $x^2 \ge 0$.
If $|x| \le 1$ then $x^2 = |x|^2 = |x||x| \le |x|*1 = |x| \le 1$.
So as $-1 \le \sin x \le 1$, it follows $(\sin x)^2 \le 1$.  And as $(\sin x)^2 \ge 0$, $0 \le \sin^2 x \le 1$.
