$-\sin x\geq-1 \ $ AND $ \ -\sin x\leq1$ but $\frac1{\sin x}\leq-1 \ $ OR $ \ \frac1{\sin x}\geq1$

Question

I just answered a simple question but that reminded me of a confusion that was lost in times but now revived again.

So, in the answer post above (in simple form) we had something like: $$-1\leq\sin x\leq1$$ Now consider the following $2$ "transformations"/"operations": $$1. \ 1\geq-\sin x\geq -1$$ $$2. \ -1\geq\frac1{\sin x}\geq 1$$

So, basically for seemingly similar "operations", we have 2 different results:

1. $-\sin x\geq-1 \ $ AND $ \ -\sin x\leq1$
2. $\frac1{\sin x}\leq-1 \ $ OR $ \ \frac1{\sin x}\geq1$

Why in one case we take And and OR in another?

Answer 1

The reciprocal function $t \mapsto 1/t$ is a decreasing functions on the two separate intervals $(-\infty,0)$ and $(0,\infty$). So, you cannot take reciprocals and reverse inequality sign if your functions not entirely belongs to $(-\infty,0)$ or $(0,\infty)$. For example, it is true that $-2 < 3$ but it is false that $1/3<-1/2$.

So you must distinguish two cases: $0< \sin x \le 1$ or $-1 \le \sin x < 0$. In both cases, we can take reciprocals and get $\frac{1}{\sin x} \le -1$ or $\frac{1}{\sin x} \ge 1$. This is where the "or" come from.

Answer 2

$\frac{1}{\sin(x)}$ is surely not bounded between $0$ and $1$.

You are falling into a pitfall since the inverse function is undefined in $0$, you actually have to apply it twice to both inequalities $-1 \leq \sin(x) \leq 0$ and $0 \leq \sin(x) \leq 1$ to obtain $-1 \geq \frac{1}{\sin(x)} \geq -\infty$ and $1 \geq \frac{1}{\sin(x)} \geq +\infty$ and thus $ \frac{1}{\sin(x)} \in (-\infty, -1] \cup [1, +\infty)$ .

You need to be careful when manipulating inequalities, you can only apply a function on the inequalities if it's well defined on the whole interval (also note that the inverse function is not a decreasing function on $\mathbb{R}^*$ which is basically what you tried to use).