Inequality; when we switch signs? From basic to complicated Let's say I have
$$a > b$$
Now we know that the inequality sign will switch if:


*

*We take the reciprocal of both sides

*Multiply $-1$ to both sides.
But, are there more? What if we do more "complicated" things to it? 
Will it change if I apply inverse trig on it? For instance is it still true then if 
$$\arccos(a) > \arccos(b)$$
Note that $a,b \in\mathbb R$.
So what I really want to ask is, are there any other operations that would make you change the inequality signs?
 A: *

*Multiplication by negative numbers flips inequalities. Multiplication by positive numbers respects inequalities.
If $a\lt b$ and $c\lt 0$, then $ac\gt bc$; if $d\gt 0$, then $ad\lt bd$. 

*You can deduce the case of reciprocals (which is more complicated than what you wrote) from this. 
If $0\lt a\lt b$, then multiplying by $\frac{1}{ab}$ (which is positive) gives $\frac{a}{ab} \lt \frac{b}{ab}$, which yields $\frac{1}{b}\lt \frac{1}{a}$; i.e., a "flipped inequality".
If $a\lt b \lt 0$, then multiplying by $\frac{1}{ab}$ (which is also positive, since $a$ and $b$ are both negative) again yields $\frac{1}{b}\lt \frac{1}{a}$.
If $a\lt 0 \lt b$, then multiplying by $\frac{1}{ab}$ (which is negative) yields
$$\frac{a}{ab}\gt \frac{b}{ab}$$
or $\frac{1}{b}\gt \frac{1}{a}$; that is, the inequality is not "flipped."

*A function $f$ is strictly increasing if $a\lt b$, both in the domain of $f$, implies $f(a)\lt f(b)$. Examples of increasing functions are $f(x)=x^3$, $f(x)=\arctan(x)$, $f(x)=e^x$, $f(x)=\log(x)$. A function $f$ is increasing if $a\lt b$, both in the domain of $f$, implies $f(a)\leq f(b)$.
Symmetrically, a function $f$ is strictly decreasing if $a\lt b$ and both in the domain imply $f(a)\gt f(b)$; and decreasing if $a\lt b$ and both in the domain implies $f(a)\geq f(b)$.
You can also deduce most of the case of reciprocals from this: $f(x)=\frac{1}{x}$ is strictly decreasing on the positive reals, so if $0\lt a \lt b$, then $f(a)\gt f(b)$; $g(x)=\frac{1}{x}$ is strictly decreasing on the negative numbers, so if $a\lt b\lt 0$, then $g(a)\gt g(b)$. But the case $a\lt 0\lt b$ has to be dealt with separately anyway.
A: It's not entirely true that when you take the reciprocals of both sides, you change "$\lt$" to "$\gt$" and vice-versa.  If $a$ and $b$ are both positive or if they are both negative, and $a<b$, then $1/a>1/b$.  But if one is positive and the other negative, and $a<b$ (which means $a$ must be the one that's negative) then $1/a<1/b$; the direction doesn't get reversed.
Generally if $a<b$ and $g$ is a strictly decreasing function, then $g(a)>g(b)$.  That's actually the definition of the concept of "strictly decreasing function".  So the fact that when you multiply by a negative number, you invert the inequality relation, is the same as saying that multiplication by a negative number is a strictly decreasing function.  For example, if $g(x) = -5x$ for all values of $x$, then $g$ is a strictly decreasing function.  If $g(x)=1/x$, then the restriction of $g$ to the positive numbers is a strictly decreasing function, and the restriction of $g$ to the negative numbers is a strictly decreasing function, but $g$, over its whole domain, is not a strictly decreasing function.
$\arccos$ is a strictly decreasing function: as a number increases from $-1$ to $1$, its arccosine decreases, i.e. if $a<b$, then $\arccos a>\arccos b$.
One thing that will tell you that a function is strictly decreasing is that it's derivative is everywhere negative and its domain has no gaps.  The reciprocal function has an everywhere negative derivative, but its domain has a gap at $0$.
