TITLE: Are there 3 trig functions or are there 6 trig functions?
QUESTION [17 upvotes]: In my algebra class we are being taught that there are only the 3 basic trig functions (cosine, sine, and tangent). But my friend who is 2 math grade levels ahead of me is saying that there is 6 trig functions (cosine, sine, tangent, cotangent, secant, and cosecant). Does anyone know why we are being taught differently and which one of us is correct?

REPLY [3 votes]: There is a very strong argument that there are 2 important functions whose behavior you need to have a fair understanding of for trigonometry, and then a somewhat arbitrary collection of auxiliary functions who find various uses but whose properties are completely determined by these two.  Let me go over some of this argument to elaborate.
Why not 1?
First, it's important to understand the argument that there may only really be one important function:
$$e^x$$
Euler showed that the function $e^{ix}$ traversed a circle in the imaginary plane and that taking the imaginary and real parts of this gave:
$$e^{ix} = \cos(x) + i\sin(x)$$
$$Re\{e^{ix}\} = \cos(x)$$
$$Im\{e^{ix}\} = \sin(x)$$
So it seems possible to simply define $e^x$ and you are good.  Of course, this is absolutely true, but you also need to have $Re:\mathbb{C} \mapsto \mathbb{R}$ and $Im:\mathbb{C} \mapsto \mathbb{R}$ defined for this approach.
And the tricky thing here is that those two functions have a lot of the same nice properties as $\cos$ and $\sin$!  They are projection functions that cannot be defined in purely arithmetic terms, and that projection definition applies to both axes of the complex plane symmetrically producing two functions in the same manner.  It's very hard to argue that the functions Re and Im are somehow simpler and easier to ignore as irrelevant parts of the definition from $e^x$ when many of the key reasons for looking at a pair of functions over a single function gets lost in these details.
So one may try a different approach: there is still a functional relationship between $\cos(x)$ and $\sin(x)$, namely
$$\cos^2 (x) + \sin^2 (x) = 1$$
so aren't they "really" still a single function expressed in different ways?
The problem here is that we've introduced another essential "choice of 2" by squaring.  The square root is a multivalued function, and at any particular $x$ we really only mean for one of those branches to be taken.  How do we describe that?
To do so, we have to express 

a particular value to be taken by the "other" function at some given point
a continuity requirement
a differentiability requirement

So it's not just one function that must be described but a function + a relation with sufficient ambient data to describe the other function.  So this isn't a very natural way to argue for one function...
But!  What about the coup de grace, the secret weapon that proves once a for all that there really is only one function here and that there is a simple relationship that avoids all the ambiguities that the other approaches bring:
$$\sin(x) = \cos(x-\frac{\pi}{2})$$
?? That seems unambiguous and clearly defines one function in terms of the other.  Right?  You don't need any additional data beyond the relations.
Well... of course it does.  All of the methods described so far are mostly unambiguous because we have a clear understanding of the data that is left unspoken.  But there has always been an argument that the unspoken data is complex enough that we shouldn't ignore it in a discussion of the "natural number of trigonometric functions".  So what is the unspoken data here?
The obvious one is the use of $\frac{\pi}{2}$.  How do we define this?  Clearly, $\tau=2\pi$ is the natural period of one function, and the function has symmetries with crossing zero every $\pi$ and each section between zeroes having reflection symmetries at it's halfway point and odd symmetry over the zeroes.  So to get this constant as a natural part of the function, you have to be willing to accept even and odd symmetric relations of the one function as being natural based on what you use to define that function.  Most definitions do not have this property (such as the taylor expansion), but the ones that do (such as circle projections) seem to all define the functions $\sin$ and $\cos$ together.
But another issue with this kind of definition is that it manipulates the arguments of the function, not the function externally.  There is, though, an ambient manipulation that shifts functions:
$$e^{\alpha D} f(x) = f(x+\alpha)$$
where
$$D = \frac{d}{dx}$$
is the differential operator.  So this kind of definition, when viewed under the (admittedly restricted) view of external manipulation requires looking at one of $\sin,\cos$ and also $e^x$ applied to operators.
So why 2?
Let's assume some of the above makes some kind of case.  Why is any of this arguing for two functions being natural and not "one function + one point" or 3 functions or 10?
Look at the definition in terms of circle projections.  Even though the choice of basis is arbitrary, and even though one base may be transformed to the other, this is still a two dimensional system and there are two bases to project on.  The relationship between the two projection operators is very much the relationship between the two trigonometric functions.  The shift relation mentioned above is really a representation of a rotation transformation that translates one basis to the other.  
The relation $\cos^2(x)+\sin^2(x)=1$ is really just a representation of the invariant distance of a point on a circle from the origin center.  We don't just study two points and say we have studied a circle, because that misses all of the other relationships between other points on the circle.  Similarly, we don't study just a single trigonometric function, because it is the relationship between the functions that expresses the relationships and symmetries between projections to the two bases across all points of the circle.
And these symmetries of projection are also seen in the series expansion.  Take the series:
$$e^{ix} = \sum_{j=0}^{\infty} \frac{(i x)^j}{(1)_j}$$
and project this on to the even and odd elements of the expansion basis $x^j$.  This gives:
$$\cos(x) = \sum_{j=0}^{\infty} \frac{(-1)^j (x)^{2j}}{(1)_{2j}}$$
$$i\sin(x) = i\sum_{j=0}^{\infty} \frac{(-1)^j (x)^{2j+1}}{(1)_{2j+1}}$$
This operation of projecting a series onto certain subbases of the expansion is known as multisection.  The particular operator that projects from series expanded over $x^j$ and only selects those elements with $j \equiv m \pmod{n}$ can be written as:
$$ f = \sum_{j=0}^{\infty} f_j x^j$$
$$M_{m,n} f = \sum_{j=0}^{\infty} f_{nj+m} x^{nj+m}$$
and have very useful and important properties.  Note that with $n=2$ this selects out the even and odd components of the function.  This is one of the big "points" of trigonometry, in particular in it's application to Fourier series.  With n=2, there are two such projection operators.
Now let's step back from that approach for a second and look at another way to define these functions.
$$D^2 y = -y$$
The solution space of this differential equation is TWO dimensional.  We can choose as a basis $(\cos(x),\sin(x))$ or even $(e^{ix}, e^{-ix})$ or linear transforms of these, but we always must choose two to have the complete solution space.  (NOTE: this is another reason why $e^{ix}$ is insufficient.  Even with the definitions $\cos(x)=\frac{e^{ix} + e^{-ix}}{2}$ and $\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$ we have to appeal to a second function that again is not externally definable through arithmetic operations (and again the relation between them, the reflection operator this time, is fundamentally related to the symmetries that are being studied).
Bringing this all together, this diffy-que can be studied with the series projection operator mentioned above.  This operator has the interesting property:
$$ D M_{m,n} = M_{m-1,n} D$$
so differentiation shifts projections.  This is precisely the differential property we see with the two basic trigonometric functions, that bases are rotated.
So... the point here is that studying two functions is kind of the point of trigonometry: it is studying the symmetries of projection, reflection, and rotation.  We can express those relationships with two functions.  They completely specify the solution space.
And a little more
And if we take this approach, you get a lot more for free.  There is a very classical theorem on the multisection operator that says:
$$M_{m,n} f(x) = \frac{1}{n} \sum_{j=0}^{n-1} \zeta_n^{-mj} f(\zeta_n^j x)$$
where $\zeta_n$ is an $n^{th}$ root of unity.  With this theorem, you immediately get the form of the trigonometric functions in terms of $e^{ix}$ and $e^{-ix}$ given above.  You also get the product and sum formulas for the trigonometric functions almost immediately as well.  Not only that, but it is clear that this can be extended to higher order multisections and you still get very simple and natural functions that explore the cubic symmetries or septic symmetries or any higher $n^{th}$ order symmetries, with simple product and sum formulas.  In this way, the extension of the exploration of exponential symmetries has a very elegant extension in higher order trigonometries.  See "Beyond Sin and Cos" by Muldoon and Ungar for more on the history of this approach, which goes back at least to Vincenzo Riccatti in 1757, but has been forgotten and rediscovered many times since.  This approach shows that there are always n functions in the $n^{th}$ order theory that participate in the standard relations and give the full solution space to $D^n y = -y$.
As for tangent and others, I could really care less.  Their relations are natural and, importantly, arithmetical consequences of the bases argued for here.  Although they are often a distraction, they can usefully be used in things like deriving special relativistic velocity addition laws and a number of interesting finite and infinite sums, so it's good to have familiarity with them.  But really, all of that can be simply (and arithmetically) expressed in terms of the original bases above without effort.