Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My question might sound dumb, but I don't really see why the graphics of arccos(x) and sec(x) are different, because as far as I know arccos is the inverse cosine function (cos(x)^-1) and sec equals 1/cos (source Wolfram|Alpha). Then why aren't they equal?

Thanks in advance.

share|improve this question
5  
That's the difference between $\cos^{-1}(x)$ and $\cos(x)^{-1}$. The former is an inverse for the composition law, the latter for the pointwise product. Notations are confusing with trigonometric functions, though. –  1015 May 15 '13 at 23:40
    
Others have explained the inconsistency of this notation. I'll just mention that my favorite calculator, and many others, have the $sin^{-1}$ key mere millimeters from the $x^{-1}$ key... –  User58220 May 16 '13 at 1:22
1  
So now I know that cos^−1(x) is the inverse of what the cos function does and not the multiplicative inverse of cos(x). cos^−1(x) is in practice a whole different function. Thank you all. –  user30632 May 16 '13 at 2:15
    
And now you are ready for: "the derivative of the inverse is the inverse of the derivative" :-) –  Francesco May 16 '13 at 8:56

5 Answers 5

up vote 6 down vote accepted

Well, it'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.

share|improve this answer
    
Thank you for the explanation, though I knew the formal definition of functions, it help me to understand where I was confusing things. –  user30632 May 16 '13 at 2:10

$\arccos x$ 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.

share|improve this answer
    
I wonder why the convention isn't simply $\cos_n\:x \equiv (\cos\:x)^n$, sub- rather than superscript; that might not be as self-explanatory but at least it would be unambiguous, and no less convenient. –  leftaroundabout May 16 '13 at 8:25
1  
My impression is that $arccos$ predates $\cos^{-1}$, but I've no reference. –  egreg May 16 '13 at 8:29

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 $$\sec(x)\cdot \cos(x)=1\quad\text{when }\cos(x)\neq 0.$$

share|improve this answer
    
Yes, the notation was really my main problem. –  user30632 May 16 '13 at 2:17

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!

enter image description here

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.