My question might sound dumb, but I don't really see why the graphics of $\arccos(x)$ and $\sec(x)$ are different.
As far as I know $\arccos(x)$ is the inverse cosine function $(\cos(x)^{-1})$ and $\sec(x)$ = $\frac{1}{\cos(x)}$ (source Wolfram|Alpha).
So why aren't they equal?
Thanks in advance.
On
The notation $\cos^{-1}(x)$ means the same thing as $\arccos(x)$. If you wanted to talk about $\sec(x)$, which is $1/\cos(x)$, you would write $(\cos(x))^{-1}$.
Yes, this is very confusing - it's inconsistent with the common use of $$\cos^n(x)$$ to mean $(\cos(x))^n$ when $n\geq 2$; it is just a historical accident that notational conventions turned out this way. All the more reason to just never ever write "$\cos^{-1}(x)$" and use one of the options that is unambiguous: $\arccos(x)$ if you want to say something like $$\cos(\arccos(x))=x,$$ and either $\sec(x)$ or $\dfrac{1}{\cos(x)}$ if you want to say something like $$\sec(x)\cdot \cos(x)=1\quad\text{when }\cos(x)\neq 0.$$
On
The notation $\cos^{-1}(x)$ is very confusing. it does not mean $(\cos(x))^{-1}=1/\cos(x)$, but rather $\arccos(x)$, which is quite a different function!
On
$\arccos x$ means the angle $\alpha$ in the interval $[0,\pi]$ such that
$$ \cos\alpha=x $$
while
$$ \sec x=\frac{1}{\cos x} $$
So $\arccos0=\pi/2$, while $\sec0=1$.
I believe you're misled by the $\cos^{-1}$ notation somebody uses for $\arccos$.
From a "function theory" point of view, the notation $\cos^2 x$ is inappropriate, because it should mean $\cos(\cos x)$; however, $\cos^2x=(\cos x)^2$ has been used for centuries because it's practical and in many formulas you need the cosine squared, while the cosine of the cosine is rarely needed.
Mixing the two notations, that is using $\cos^{-1}$ for the "inverse function of the cosine" is, at the least, confusing.
On
The word "inverse" is context-sensitive. There are additive, multiplicative and compositional inverses. The $\arccos$ function is a compositional inverse. The secant function is a multiplicative inverse to the cosine function, which is defined off the zeroes of the cosine function.
On
One of them, $\arccos$, is the inverse function to $\cos$. It reverses the operation. The other, $\sec$, is $1\over \cos$, so it raises the results of the operation to the power of $-1$.
Think of a simpler function, $x^2$, to see how different they are.
a) The inverse function would be square root. If $y=x^2$, then $x = \pm\sqrt{y}$.
b) But $x^2$ raised to $-1$ would be $1\over x^2$.
If, say, $x=2$, example a) would be $\pm 1.4142$... but example b) would give you $0.25$
That's a problem of notation and probably a lack of definitions. We define $\sec x$ as the multiplicative inverse of $\cos x$, in other words, fixed $a \in \mathbb{R}$, $\sec a$ is the number such that $\sec a \cos a = 1$. Now $\arccos x$ is a little different thing: it's the inverse function of $\cos x$.
I don't know if you've learned this but the formal definition of a function is that of a collection of ordered pairs. In other words, since a function from a set $A$ to a set $B$ should be a rule assigning for each $a \in A$ some $b \in B$ we can simply define a function as the set of all ordered pairs of elements in $a$ together with related elements in $b$. However, we require the additional property that if $(a,b) \in f$ and if $(a,c)\in f$ then $b = c$ and this is just the formal way to state the "vertical line rule". Since the second element in each pair is unique we give it a name: if $(a,b) \in f$ then $b = f(a)$. Also to state starting and ending sets we write functions from $A$ to $B$ as $f: A \to B$.
Now, if you have a function you have a collection of ordered pairs right? So, you can create a new set of ordered pairs by reversing the pairs. So if $f : A \to B$ is a function from $A$ to $B$ we define the inverse $f^{-1}$ by the property that $(a,b) \in f^{-1}$ when $(b,a)\in f$. Now it's not at all clear when $f^{-1}$ is a function. Just to show you that consider the following function that maps naturals to naturals:
$$f = \{(1,2), (3,2), (4,1)\}\subset \mathbb{N}\times \mathbb{N}$$
This is a function by our definition. Now the inverse is $f^{-1} = \{(2,1), (2,3), (1,4)\}$, now this isn't a function because $(2,1)\in f^{-1}$ and $(2,3)\in f^{-1}$. So $f^{-1}$ will be a function if the original function also satisfies $f(x) = f(y)$ implying $x = y$. This kind of function is called one-one, and so if $f$ is one-one, $f^{-1}$ will be a function called then inverse function.
Also, if $f: \mathbb{R} \to \mathbb{R}$ has an inverse function $f^{-1}:\mathbb{R} \to \mathbb{R}$ then $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. So $\arccos$ is defined precisely this way: fixing one interval where $\cos$ is one-one, you define $\arccos$ in that interval by the property that $\arccos x$ is the number $y$ such that $\cos y = x,$ in other words, it returns you the value of angle whose cosine is $x$.
Just a reference to finish: you can find treatments like this in books like Spivak's Calculus or Apostol's Calculus Vol. 1. I hope the way I exposed this helps you a little. Good luck!
EDIT: The problem of notation I've mentioned and forgot talking about is that both the multiplicative inverse and the inverse function are in some contexts denoted by $\cos^{-1}$ and this usually happens to all trigonometric functions. So to avoid confusion, I recommend writing $\arccos$, $\arcsin$ and so on for the inverse functions.