# Are there 3 trig functions or are there 6 trig functions?

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?

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There are a lot of trig functions. – Lucian Apr 19 '14 at 18:49
There is only one trig function: $e^z$. The rest is clear. – Norbert Apr 19 '14 at 18:51
Also, all of their inverses. – imallett Apr 19 '14 at 20:20
There are 33 trigonometric functions: theonion.com/articles/… – KCd Apr 20 '14 at 16:20
Often the term "sinusoid" is used to denote "a trig function, sin or cos, doesn't matter which one". "Sinusoid" is a good candidate for the basis of all the other trig functions. – DanielV Apr 20 '14 at 17:55

It depends on how you look at it I guess, but:

$$\cot(x) = \frac{1}{\tan(x)}$$

$$\csc(x) = \frac{1}{\sin(x)}$$

$$\sec(x) = \frac{1}{\cos(x)}$$

So the three "extra" functions your friend told you about are just derived from the three you know. But if that's the rule, then two of the ones you know,

$$\cos(x) = \sin\left(\frac{\pi}{2} - x\right)$$

$$\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{\sin{x}}{\sin\left(\frac{\pi}{2} - x\right)}$$

are also just derived functions. Hence we would say there is only one trigonometric function, for example $\sin{x}$.

(As others have mentioned, this statement works even counting hyperbolic functions, because of properties like $\cosh(x) = \cos(ix)$ and so on, or using $e^{\theta} = \cos{\theta} + i\sin{\theta}$. But since you don't appear to be at this level of math yet, I won't go into detail about that.)

Bottom line: We only need one trigonometric function, but for practical reasons, there are more.

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Actually even counting the hyperbolic functions, the answer is still 1. Euler's formula (mindblowing when I first learned that one, such a surprising relationship) and the fact that cosh(x) = cos(ix),etc. are enough. Probably best to specify the minimal function as e^z with z being a complex number. – Voo Apr 19 '14 at 20:47
@Voo Thank you, I've fixed it. – naslundx Apr 19 '14 at 20:50
@naslundx No, I think you missed his point; the only true trig function is $\exp(i x)$. – AJMansfield Apr 20 '14 at 2:35
Is there any particular reason cosecant goes with sin and secant goes with cosine? It seems like it would make more sense for cosecant to go with cosine and secant go with sine. – Cole Johnson Apr 20 '14 at 21:45
I'm pretty sure everybody just blew this kid's mind with all of the comments relating to $e^z$. I'm pretty sure that a lot of this is just confusing him/her. – TheBluegrassMathematician Apr 21 '14 at 2:00

I strongly recommend this blog post by John Cook: as he says, the calculator answer is 3, the college textbook answer is 6, and the historical answer is at least 12

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That is a great link for this question. – Mark Bennet Apr 19 '14 at 17:52
Great link, didn't know that we used to consider 12. Another great example of the saying about perfection not being reached when nothing can be added anymore, but when nothing can be removed any longer. From 12 functions down to a single one that covers all of them with one elegant relationship. – Voo Apr 19 '14 at 20:53
The bottom line answer always seems to be "it depends." Great link, btw. – J.R. Apr 20 '14 at 9:53

The three sides $a,b,c$ of a right-angled triangle can make nine ratios $$\frac aa, \frac ab, \frac ac, \frac ba, \frac bb, \frac bc, \frac ca, \frac cb, \frac cc$$

Three of these ratios are trivially equal to $1$, and the other six all have names (depending where the right-angle is).

If $r$ is a ratio, then so is $\frac 1r$, so the ratios come in pairs. So sometimes people think of three basic ratios sine, cosine and tangent, because the others are their reciprocals, and these three are sufficient because you can divide by them if necessary.

As it happens when it comes to calculus (particularly integration) it turns out that the other ratios come in handy, so even though the names are not much used in early work, they are useful to know later.

Since tan=sin/cos and in a right-angled triangle $\sin^2+\cos^2=1$ we can write everything in terms of the sine function if we like (and don't mind taking square roots).

As it happens one of the most useful ways of writing these functions in terms of a single function is to use $t=\tan \frac {\theta}2$ when we get the nice expressions $$\sin\theta=\frac {2t}{1+t^2}, \cos \theta=\frac {1-t^2}{1+t^2}$$

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I recall using your $t$ parameterization when I was asked to give a parameterization of a circle, and I was feeling a little cheeky. I think it's nice that you can parameterize a circle using only rational functions (although you have to throw in $t=\infty$). – Mario Carneiro Apr 20 '14 at 15:59
@MarioCarneiro The substitution, which links also links integration with algebraic geometry, is generally attributed to Weierstrass. – Mark Bennet Apr 20 '14 at 22:46

To be a little silly, there is only one "trig" function. By Euler's formula, $e^{i \theta} = \cos \theta + i \sin \theta$, for real valued angles $\theta$. This means that $$e^{-i\theta} + e^{i \theta} = (\cos \theta + i \sin \theta) + (\cos (-\theta) + i \sin (-\theta)) = 2 \cos \theta$$ since $\cos \theta$ is even and $\sin \theta$ is odd.

Using something similar for $\sin \theta$, we can write $$\cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2},$$ $$\sin \theta = \frac{e^{i \theta} - e^{-i\theta}}{2i}.$$

From the identity $\tan \theta = \frac{\sin \theta} {\cos \theta}$, we can write

$$\tan \theta = \frac{e^{i \theta} - e^{-i\theta}}{i(e^{i \theta} + e^{-i \theta})}.$$

And the additional trig functions follow easily from the above and we write: $$\sec \theta = \frac{2}{e^{i \theta} + e^{-i \theta}},$$ $$\csc\theta = \frac{2i}{e^{i \theta} - e^{-i\theta}},$$ $$\cot \theta = \frac{i(e^{i \theta} + e^{-i\theta})}{e^{i \theta} - e^{-i \theta}}.$$

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I second this approach. I wish I knew about this when I was learning trig; it makes all of the ridiculous number of identities in trig boil down to variations on $e^{a+b}=e^ae^b$. – Mario Carneiro Apr 19 '14 at 21:32
Holy crap. Identities finally makes sense! I fail at trig for the same reason I fail a history or chemistry - too many things to memorize and can't be derived from first principles. – slebetman Apr 21 '14 at 2:26

You can say that there are 3 (or 2) basic (and I guess I should be careful with this term) trig functions. All the other functions you mentioned are just other names for different expressions of the same 3 functions: $$\sec x = \frac1{\cos x}$$ $$\csc x = \frac1{\sin x}$$ $$\cot x = \frac1{\tan x}$$ In my eyes, it's as if I just decided to create a function called Wambo: $$\operatorname{Wambo} x = \frac{\sin x + \cos x + \tan x}{\sum_{w=1, a=0, m=0, b=0, o=w^2}^{\infty}\sqrt{w^{ambo}}}$$ The point is, this wouldn't really be another "trig" function. It would be a derivative of the original ones.

However, technically speaking, even a Wambo would count as a trig function according to the definition (will be posted later in this post). And when you think about it, even the basic trig functions don't stand on their own, and they all define one another. I mean:

$$\tan x = \frac{\sin x}{\cos x}$$

You will learn later in trig that:

$$\sin x = \cos (90 - x)$$

And there are tons of others... They are all connected.

The Wikipedia article about this defines a trig function as any function that relates the angles of a triangle to the lengths of its sides. Surprisingly, it doesn't include Wambo. What it says regarding how many there are:

In modern usage, there are six basic trigonometric functions, tabulated here with equations that relate them to one another. Especially with the last four, these relations are often taken as the definitions of those functions, but one can define them equally well geometrically, or by other means, and then derive these relations.

The definition implies $\infty$, but the paragraph above says there are 6 (except, not really). Other places say there are 3, but I'd argue there are 2 because $\sin$ and $\cos$ are the only 2 unique relationships. But then again, so would $\sin$ and $\tan$. Or $\cos$ and $\tan$. I guess I'd just say 3 then. But then again, $\tan x = \frac{\sin x}{\cos x}$. But then again... stop.

When the zombie apocalypse arrives, there will only be one: Wambo. Until then, there's no real answer as to how many there are. Your friend's answer is just as acceptable as yours.

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You know, I think $\operatorname{Wambo}x=0$, since the sum diverges (note that $\sqrt{w^{ambo}}=1$ for each $w$ when $a=0$). So I don't think $\rm Wambo$ is going to take off as a new trig function. – Mario Carneiro Apr 20 '14 at 15:51
@MarioCarneiro Not quite. The sum may diverge, but it actually opens the rabbit hole to Narnia. – Shahar Apr 20 '14 at 20:30

If you want to get picky, there is only one, provided that $$\tan \alpha = \cfrac{\sin \alpha}{\cos \alpha}, \hspace{10pt} \cos \alpha = \sqrt{1 - \sin^2 \alpha},$$ thus allowing you to write $$\tan \alpha = \cfrac{\sin \alpha}{\sqrt{1 - \sin^2 \alpha}}.$$

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This appears to assume cosine is positive. – Mark S. Apr 19 '14 at 20:42
$\cos\alpha=\pm\sqrt{1-\sin^2\alpha}$ depending on where $\alpha$ is. – Hakim Apr 20 '14 at 16:40
Or just $cos(a) = sin(a+ pi/2)$ – ApproachingDarknessFish Apr 21 '14 at 6:58

There are more! The versine, the haversine, the covercosine, the exsecant.... This picture suggested by David H illustrates some of them.

Calling any of them "basic" is simply a pedagogical convention; it's not really an actual technical term.

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If I may make a suggestion, a visual aid such as this one would be a helpful addition to your answer. – David H Apr 21 '14 at 4:36

There are three that are the main ones: sinx, cosx and tanx. Then there are the reciprocals of these, 1/sinx is called cscx, 1/tanx is called cotx and 1/cosx is called secx. You can view these functions as the ratios of the sides of a right triangle in the rectangular coordinate system. If you go a certain angle, you can imagine that there is a point somewhere on the line and then taking the opposite side and dividing it by the hypotenuse (sinx) or any other combination. That's all these are, the ratios of the sides of right triangles.

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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.

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Actually there is hyperbolic, elliptical, cardioid, integral, cubic, quartic, and many more. It can be viewed as curve functions.

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Care to elaborate? It sounds like you want to call all named functions trigonometric, which is a stretch. – Mario Carneiro Apr 21 '14 at 2:01