155
$\begingroup$

Cosine is just a change in the argument of sine, and vice versa.

$$\sin(x+\pi/2)=\cos(x)$$ $$\cos(x-\pi/2)=\sin(x)$$

So why do we have both of them? Do they both exist simply for convenience in defining the other trig functions?

$\endgroup$
26
  • 37
    $\begingroup$ If you look at the $\cos$ and $\sin$ from a geometric point of view, it's quite clear why it's useful to have both of them. $\endgroup$
    – Asaf Karagila
    Aug 31, 2015 at 16:37
  • 49
    $\begingroup$ You really mean, "why do we name them both?" Because it is quite clear from your equation why one exists if the other does. $\endgroup$ Aug 31, 2015 at 16:39
  • 147
    $\begingroup$ Why do we have either $\sin$ or $\cos$, when we have the more fundamental $e^x$? $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}\qquad\qquad \cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$ $\endgroup$ Aug 31, 2015 at 16:58
  • 77
    $\begingroup$ @Zev: Why do we have $e^x$ when we have the more fundamental $\lim_{n\to\infty}(1+1/n)^x$? :-) $\endgroup$
    – Asaf Karagila
    Aug 31, 2015 at 19:43
  • 71
    $\begingroup$ 1 + 1 = 2 Why do we have both of them? 2s are just 1s added up. $\endgroup$
    – Plutor
    Sep 1, 2015 at 12:26

17 Answers 17

158
$\begingroup$

You seem to be asking why we name them both, rather than why they exist, since the very relationships you've written shows that if one exists, the other does, too.

Essentially, all mathematical notation and names, except for a very small subset, are for convenience/clarity/human communication. Math does not require that we name any function consistently across separate proofs, but it becomes much easier to communicate and think about things when we do.

I prefer the relationship:

$$\sin(x)=\cos(\pi/2-x)\\\cos(x)=\sin(\pi/2-x)$$ since this is symmetric and obviously geometric when $0\leq x\leq \pi/2$, and because this is also the relationship between "tangent" and "cotangent" and "secant" and "cosecant." It indicates a duality in these functions.

It is certainly something of a paradox that adding more names often simplifies our understanding. In particular, if you defined only one, it would give you a sense that one of these functions was "primary." There is a hint of that error even in the names "sine" and "cosine," which vaguely implies that "sine" is primary, but it would be particularly strong if we only defined "sine" and never defined "cosine." We would have a harder time grasping the duality that happens in trig functions.


If you actually must start with one function, most mathematicians wouldn't start with $\cos x$ or $\sin x,$ they'd start with the complex-valued function: $\operatorname{cis}(x)=\cos(x)+i\sin(x).$ This has the property $\operatorname{cis}(x)\operatorname{cis}(y)=\operatorname{cis}(x+y),$ and can also be written as $e^{ix}.$ You can define $\cos x$ and $\sin x$ in terms of $\operatorname{cis} x:$

$$\cos x=\frac{1}{2}\left(\operatorname{cis}(x)+\operatorname{cis}(-x)\right)\\ \sin x=\frac{1}{2i}\left(\operatorname{cis}(x)-\operatorname{cis}(-x)\right)$$


Addition: I recently heard, in a lecture that covered some of the accomplishments of the early Muslim world, that they acquired knowledge of the sine function from India, and developed the other trigonometric functions. So if you want someone to blame, we can blame them. 🤓

$\endgroup$
18
  • 23
    $\begingroup$ In my opinion, cosine seems primary. Also, why didn't people define $\sec:=\frac1\sin$? I have this extra "co" to deal with whenever I invert the value and it's so annoying… $\endgroup$ Aug 31, 2015 at 18:55
  • 38
    $\begingroup$ It does bug me that $\sec$ is related to $\cos$ and $\csc$ is related to $\sin$, because the "co" switches places. However, geometrically it makes sense: things without "co" (sine, tangent and secant) are to do with the sine length in the unit circle, which is vertical; meanwhile things with the "co" are to do with the cosine length, which is horizontal. $\endgroup$
    – Will R
    Aug 31, 2015 at 19:12
  • 8
    $\begingroup$ @columbus8myhw if on the unit circle, you draw the vertical line tangent to it at $(1,0)$, then take its intersection with angle ray (the other along the positive $x$-axis) then... (1) the height above the x-axis is the tangent of the angle, while (2) the length segment that cuts across the circle along the ray to the vertical line is the secant of the angle. Well, it might not satisfy you as an optimal naming convention, but that's why it was defined that way. $\endgroup$
    – Stan Liou
    Aug 31, 2015 at 19:13
  • 12
    $\begingroup$ Alternatively, $\sin,\sec,\tan$ are all increasing on $(0,\pi/2)$. But yeah, it bugged me when I took trig way back when. $\endgroup$ Aug 31, 2015 at 19:14
  • 2
    $\begingroup$ @Ovi Trying to remember what I was thinking two years ago: When you write them out in term of Euler's formula, cosine is $\dfrac{e^{i\theta}+e^{-i\theta}}2$ and sine is $\dfrac{e^{i\theta}-e^{-i\theta}}{2i}$; doesn't the former look a lot more primary and clean? In addition, $\cos(n\theta)$ can be written as a polynomial in terms of $\cos(\theta)$, while $\sin(n\theta)$ cannot be written as a polynomial of $\sin(\theta)$. $\endgroup$ Jun 9, 2017 at 22:42
112
$\begingroup$

$$ \sin^2 \alpha + \cos^2 \alpha =1, $$ $$ \sin^2 \alpha + \sin^2 \left( \alpha + \frac{\pi}{2} \right) =1. $$ Which one do you prefer?

Post scriptum: of course this is not a deep answer, but I think that sometimes mathematicians prefer elegance to logical "economy".

$\endgroup$
1
  • 38
    $\begingroup$ You should make a "PS" to explain what the "post scriptum" means $\endgroup$
    – Ooker
    Sep 6, 2015 at 12:25
34
$\begingroup$

The question is ridiculous. (I'm afraid to say.)

It's equivalent to asking "Why are there words for both North and South?"

(Of course you could just use "negative North" at all times, if desired.)

In my opinion, one would only ask the question at hand, if, one has rather naively "just noticed" - let's put it that way - that sine and cosine are complementary.

Note too that, indeed, in English co-sine is simply "sine" ... with the appropriate prefix!! Just as you'd expect.

Again to make analogy, one could ask questions such as "why do we label both matter and antimatter!" or "Why label both up and down?"

It's clear, traditional, and expected in languages that there are matching terms for complementary qualities {rather than, let us say "minimalistically," using only the one and then the negative of it} ...

heaven and hell, paradis et enfer.

$\endgroup$
7
  • 10
    $\begingroup$ A double plus good answer if ever there was one. $\endgroup$
    – T. Kiley
    Sep 3, 2015 at 10:54
  • 3
    $\begingroup$ If anything, I think it's more like asking "Why are there words for both North and East" - almost exactly so. $\endgroup$
    – user81060
    Sep 3, 2015 at 12:38
  • 1
    $\begingroup$ I had exactly the same thought--this is like asking why we have words for "up" and "down" when you could just use one word all the time. (I'm going not down this mountain right now!) Sine is opposite over hypotenuse, and cosine is adjacent over hypotenuse. May as well ask why a triangle has three sides! $\endgroup$
    – ErikE
    Sep 4, 2015 at 0:47
  • 3
    $\begingroup$ Coming back to this answer 5 years later, and I still don't like it. $\endgroup$
    – Tdonut
    Aug 4, 2020 at 14:51
  • 2
    $\begingroup$ This has a lot of upvotes for an answerless rant about the question. $\endgroup$
    – user
    Nov 26, 2020 at 4:15
20
$\begingroup$

I think it's conceptually cleaner to have both $\cos$ and $\sin$ as distinct notations, for the following reason: in my opinion, it's a bit of a "coincidence" that each of the functions $\{\cos,\sin\}$ can be described as translations of the other.

My preferred definition of these two functions is the following: first, position yourself at $(1,0)$ on the unit circle. Then start walking anticlockwise at unit speed. It follows that if the time elapsed is $t$, your $x$-coordinate will be $\cos t$ and your $y$-coordinate will be $\sin t$.

But notice that, in general, this way of producing pairs of functions (namely: positioning yourself on a curve and walking at unit speed, and then projecting onto the $x$ and $y$ coordinate axes) won't usually result in a pair of functions such that each can be defined as a translation of the other. The ability to define $\cos$ and $\sin$ in terms of each other is specifically a quirk of circles centered at the origin. In some sense, it's kind of a coincidence.

By the way, this viewpoint explains why $\cos^2 t + \sin^2 t = 1$; it's because you're walking on the curve defined by $x^2+y^2= 1$. This also explains why $\cos(0) = 1$ and $\sin(0)= 0$; it's because we positioned ourselves at $(1,0)$ to begin with. It also explains why $(\cos' t)^2+(\sin't)^2= 1$; it's because you're walking at unit speed. And finally, this explains why $\sin'(0)>0;$. It's because we chose to start walking anticlockwise. I'm pretty sure these four conditions (listed below for your convenience) completely characterize the ordered pair ($\cos,\sin$) among all ordered pairs of differentiable functions $\mathbb{R} \rightarrow \mathbb{R}$.

  1. $\cos^2 t + \sin^2 t = 1$
  2. $\cos(0) = 1, \sin(0) = 0$
  3. $(\cos't)^2+(\sin't)^2 = 1$
  4. $\sin'(0) > 0$
$\endgroup$
7
  • 1
    $\begingroup$ Those conditions aren't complete. Be nice and provide a proof with the correct conditions :) (easy counterexample, stick with cosine but instead of sine use negative sine) $\endgroup$ Aug 31, 2015 at 19:31
  • 9
    $\begingroup$ In some sense, it's a kind of coincidence, but in another sense it isn't; the reason we care at all about parameterizing circles is precisely that they're symmetric, and the ability to express $\sin$ and $\cos$ in terms of each other is a consequence of that symmetry... $\endgroup$
    – Micah
    Aug 31, 2015 at 19:35
  • 1
    $\begingroup$ @Chan-HoSuh, thanks for the counterexamples; I've added a fourth condition to eliminate the problem. Honestly, I really have no idea how to prove this. Analysis is a bit of a blind spot for me at the moment. Feel free to edit your own proof into the question. I've community-wikified it. $\endgroup$ Aug 31, 2015 at 19:35
  • 1
    $\begingroup$ @goblin,Of course your conditions work, since they translate (in words) to "$(\cos(x),\sin(x))$ is the arclength parameterization of the unit circle, starting at $(0,1)$ and moving in the positive direction", which is basically the definition of sine and cosine used most commonly in calculus courses. It is not too hard to show, inductively, that all orders of derivatives at $0$ are what they are supposed to be, by differentiating implicitly. Try it out! So this definition agrees with the power series definition, at least. $\endgroup$ Aug 31, 2015 at 21:27
  • 2
    $\begingroup$ @Chan-HoSuh, I'm not following. Note that $\cos^2 t+\sin^2t=1$ forces us to remain on unit circle centered at the origin. $\endgroup$ Sep 1, 2015 at 4:50
12
$\begingroup$

Suppose you're walking counterclockwise around the unit circle, starting at the point with coordinates $(1, 0)$. When you've walked a distance of $\theta$, your coordinates are $(\cos \theta, \sin \theta)$.

To me, that's why it's natural to name both cosine and sine: it takes two coordinates to describe a point in the plane, and both of those coordinates deserve names.

$\endgroup$
1
  • $\begingroup$ While not as "deep" as the other answers; I think this is the best. You can take any curve in the plane or on the earth and ask for the coordinates (x,y) and have two functions x(t),y(t) . If you constrain the curve to be a circle (constant radius) then the coordinates reflect the symmetry of your chosen curve and are thus degenerate in the questioners sense. $\endgroup$
    – rrogers
    Sep 2, 2015 at 17:19
11
$\begingroup$

Mathematics is about making definitions that are as elegant and nice as possible, and allow your conclusions to easily or straightforwardly follow. In this case, it actually turns out the most elegant function, and the most helpful to understanding, is not cosine or sine. It is $$ \textbf{cis}(x) = e^{ix} = \cos x + i \sin x. $$ Of course, this is a complex-valued function--it takes in a real number $x$ and returns the point on the unit circle at angle $x$. But all of the properties you know about sine and cosine can be derived from this.

But more importantly, we see that $\cos$ and $\sin$ naturally arise in tandem, one as the real part and one as the imaginary part of $\text{cis}(x)$. So it makes little sense to define only one of the two without defining both.

Why do we define $\cos$ and $\sin$ at all, if we could just use $\text{cis}$? Because we often work with real numbers, and it is nice to have names for the real number functions arising from the complex exponential so that we don't have to insert imaginary numbers into real-number expressions that eventually cancel out. Also for historical reasons.

$\endgroup$
0
5
$\begingroup$

I would suggest two reasons why we have both sin and cos.

First, historically, those who originally developed trigonometry started down the path of defining both functions, and then discovered various relationships between the two functions. Rather than redefine the work they did, we retain the functions as separate entities.

Second, for simplicity. The equation for a circular arc for example, can be more simply written as $x=\cos(t)$, $y=\sin(t)$, than as $x=\cos(t)$, $y=\cos(t-\pi/2)$.

If we are really trying to economize on functions, we could rewrite the tangent function $\tan(x)=\sin(x)/\cos(x)$ as $\cos(x-\pi/2)/\cos(x)$, but I think it would be more difficult to talk about the function this way, or to describe some of its remarkable properties, such as $\tan'(x)=\sec^2(x)$ or $\tan^2(x)+1=\sec^2(x)$.

From these two identities we can deduce: $\tan'(x) = \tan^2(x)+1$, which means $\tan(x)$ is a solution to the differential equation: $y' = y^2+ 1$.

I'll leave it to you to rewrite this derivation using $\cos(x)$ only.   :)

$\endgroup$
4
$\begingroup$

Consider:

  • Why do we say up and down instead of just up, since down is just reversed up?
  • Why is liquid measured in liters when you could just use cubic decimeters?

Because each unit or word is appropriate to the thing it's measuring.

A triangle has three sides. In a right triangle, each side's length divided by another side's length gives a ratio that reliably relates to one of the (non-right) angles of the triangle. You'll notice that the permutations of 3 items taken 2 at a time is 6, which just happens to be how many basic trigonometric functions there are (not counting their inverses):

opposite / hypotenuse : sine        hypotenuse / opposite : cosecant
adjacent / hypotenuse : cosine      hypotenuse / adjacent : secant
opposite / adjacent   : tangent       adjacent / opposite : cotangent

Which of these ratios is not useful?

The fact that all of these can be determined from the others is irrelevant. By the same logic as your question, why use more than just one of the six names?

While they all describe the same thing, a triangle, they do it from useful, even if different, perspectives.

$\endgroup$
0
3
$\begingroup$

Because $-1$ has no root in the reals.

Both sine and cosine can be defined as solutions to the differential equation $$ f''(x) = -1 \cdot f(x). $$ Which one you get depends only on the boundary condition.

Now, if the second derivative is so useful, then surely the first derivative should matter too! It's tempting to write $$ f'(x) = \sqrt{-1} \cdot f(x). $$ and indeed it kind of works: it gives the complex function $\backslash t \mapsto e^{it}$. But for the applications the trigonometric functions were invented for – geometry, physics... – complex numbers can't really be used, as in, you can't measure a complex quantity.

To stay in the reals, we just leave the definition a second-order differential equation, but also give the first derivative a name of its own.

$\endgroup$
2
  • $\begingroup$ Mh, isn't that a counter-argument ? The first order equation has a single solution $Ze^{it}$, which corresponds to a sinusoid with indeterminate phase and amplitude, like $A\cdot\sin(t;\phi)$ or $A\cdot \sin_\phi(t)$. $\endgroup$
    – user65203
    Sep 2, 2015 at 10:50
  • 1
    $\begingroup$ @YvesDaoust: but the first-order equation can't even be phrased without complex numbers. And though complex numbers are of course a greatly useful concept, there are good reasons to also know how to live without them. $\endgroup$ Sep 2, 2015 at 11:01
3
$\begingroup$

Before calculators and computers we had tables of values trig functions. Having all the trig functions: $\sin, \cos, \sec, \csc, \tan, \cot$ made doing geometric calculations easier. You didn't have to first convert a cos into a sin before looking it up in the tables. It saved a step.

The $\sec$ and $\csc$ came into mathematical fashion when calculations for sailing ships across oceans were easier using them.

You can read more about the history of trig functions over thousands of years at http://www-history.mcs.st-and.ac.uk/HistTopics/Trigonometric_functions.html

$\endgroup$
2
  • $\begingroup$ I'd like to add a reference to the diagram on wikipedia: upload.wikimedia.org/wikipedia/commons/9/9d/Circle-trig6.svg -- A dozen different named and highly redundant trig functions. Many have fallen out of use, but they all were named at one point. $\endgroup$
    – zebediah49
    Sep 3, 2015 at 17:55
  • 2
    $\begingroup$ Nice diagram! I can imagine the midshipmen on 18th century sailing ship learning all these trig functions to do their spherical geometry calculations to figure out where the ship was today. :-) $\endgroup$ Sep 4, 2015 at 13:20
2
$\begingroup$

They must both have names in order to preserve the symmetry when writing down relationships: in complex arithmetics ($e^{ix}=\cos x+i\sin x$) or in calculus ($\sin' x=\cos x$), we see that they are two parts of the same coin, so if you only keep the name of one, expressions become lopsided and just plain ugly. You could just define one (and for the purpose of computing it numerically, you can certainly use just one to compute both), but in math, you want elegance.

Speaking of, a more pertinent question would be, why do $\csc$ and $sec$ exist. They have no real use except for creating confusion: they are fractions, don't naturally arise from linear relationships (sin/cos are components of complex numbers, describe rotation, solutions to common differential equations) and, most importantly, they make otherwise simple expressions look unrecognizable and unreadable.

$\endgroup$
2
  • $\begingroup$ Secants were of major interest to the maritime schools and unsolved problems for many years with regards to secants made it a major topic of curiosity probably. $\endgroup$
    – user507974
    Sep 3, 2015 at 5:50
  • $\begingroup$ @user507974 of course, there are problems when you need to divide by a cosine. But mathematically, there is no elegance or structure that would make defining sec=1/cos better instead of worse. There are just twice as many relationships between functions to remember. It's sort of like invention of grads as angular units for the sake of "convenience". $\endgroup$
    – orion
    Sep 3, 2015 at 7:06
2
$\begingroup$

This is an elaboration on Vectornaut's answer and hopefully more elementary than most others; actually two viewpoints. The explanations are kept elementary in order to emphasize the origin of calculations sin() and cos() and subsequent relations like the one in the question are based long after the original terms and are reflections of underlying symetries that people needed and wanted.

Consider any curve on a plane (or in space if you like the reasoning is the same. To embed it in a mathematical structure we can envision writing the curve paramatricized by s thusly (x(s),y(s). Now we use this technique to describe a circle with a parameterization of distance from the center and angle from x. We get: $$\left(r\cdot \cos(\theta\right),r\cdot \sin\left(\theta\right))$$ In the abstract the terms are simply names for calculating the (x,y) coordinates for a particular curve. The identity mentioned is a reflection of the symmetry of that particular curve. If I had defined a hyperbola I would still have equations for (x,y) but they wouldn't have that particular symmetry; and the question wouldn't have been asked since the OP would see that two different functions would be expected with different relationships. All of the other properties of sin(),cos() are inherited from the underlying symmetry.
The alternate explanation comes from dropping the "r" dependence. Although this seems it would be more abstract but historically it was first :) In ancient time people still had property boundaries and wanted to measure and design things remotely; say by drawing for communication. In order to layout property or build a temple some way of scaling had to be invented. Our ancestors found out the ratios of certain triangles were constant and therefore models or drawings could be scaled down; this can also be considered a symmetry. On areas of the earth that are relatively flat we have the ratios $$ \left(\cos(\theta\right),\sin\left(\theta\right))$$ (They were as smart as we think we are) Since they didn't know any of the properties to start with they simply called the ratios different names. We discovered a multitude of properties later; over 1000's of years.

I hope the readers will forgive the emphasis on symmetries but I am studying Lie Groups/Algebras and they seem to be advanced versions and descriptions of symmetries like the above.

$\endgroup$
2
  • $\begingroup$ Up-voted for the key point: "Since they didn't know any of the properties to start with they simply called the ratios different names. We discovered a multitude of properties later; over 1000's of years." $\endgroup$
    – RayInNoIL
    Sep 3, 2015 at 11:37
  • $\begingroup$ @RayatERISCorp Thanks, I really didn't expect any points for this but I thought the question showed a lack of mathematical "maturity". In this case I thought an elementary explanation in a historical context was appropriate. Hopefully this provides motivation and knowledge to encourage younger people; rather than making someone feel dumb about not having a deep background. $\endgroup$
    – rrogers
    Sep 3, 2015 at 17:59
1
$\begingroup$

Having both $sin(x)$ and $cos(x)$ allows you to uniquely identify a point on the unit circle. $sin(x)$ is insufficient since you would also have to know $x$ to deduce $sin(\frac{\pi}{2} + x)$, which in turn renders $sin(x)$ useless since $x$ already uniquely identifies said point.

$\endgroup$
0
$\begingroup$

Just as every individual has a name , different from others ; so should every function have a name different from others .
This enables us to remember different functions in short, without having to write the detailed mathematical expressions , representting these functions.

$\endgroup$
2
  • $\begingroup$ If every function had a name, then these names would need to be so long to be unique that it would be easier to “inline” the defining expression. (Or, really, it would just be impossible, because there are clearly uncountable-infinitely many functions.) Alternatively, if the names weren't unique, it would be extremely confusing (actually, that's a problem in particular physics has already). $\endgroup$ Sep 3, 2015 at 17:23
  • $\begingroup$ @leftaroundabot It is_particle_ physics... $\endgroup$
    – Sasha
    Sep 4, 2015 at 13:30
0
$\begingroup$

It is similar to saying that why do spoon and fork exist even thought both of them do somewhat similar jobs....

Again yes there is a way to write cosine in terms of sine but if we write everything in sine the question which you will get will not look friendly. The more simple something is more is the motivation for you to do it.

Mathematics is all about making complicated things simpler not simpler things complicated

$\endgroup$
0
$\begingroup$

In a right triangle, the three basic trig functions sine, secant, and tangent are related to their co-functions cosine, cosecant, and cotangent by having the same relation to the sides of the triangle, but from the point of view of the complementary angle. For example, in a triangle with angles 30, 60, and 90, the sine of 30 is the cosine of 60. In this sense, the cosine and sine are complements; really one thing and its complement, not two separate things.

$\endgroup$
0
$\begingroup$

I think that cos is just the shortcut for cosine, so it is somehow derived. You have vectors and covectors and all kinds of other duaities in math.

Unlike the others here, I don't really think we need two different names -- and indeed, we don't have them.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .