Why do we require radians in calculus?

Question

I think this is just something I've grown used to but can't remember any proof.

When differentiating and integrating with trigonometric functions, we require angles to be taken in radians. Why does it work then and only then?

Degrees aren't really numbers that you can do calculations with. Radians, on the other hand, are — Anonymous Computer, Mar 21 '14 at 08:40
A nice discussion: http://teachingcalculus.wordpress.com/2012/10/12/951/ — Siminore, Mar 21 '14 at 08:48
I read somewhere on MO, "Degrees are from the watcher's perspective, radians from the walker's". — geodude, Mar 21 '14 at 08:49
To define degrees you have to pull the number $360$ as an artificial unit out of a hat, whereas $2\pi$ is already there. — Christian Blatter, Mar 21 '14 at 09:18
@HansLundmark That is a very good argument for using radians. To expand a little, in certain industrial applications, like solar trackers, the preferred unit is milliradians, which is equivalent to mm of deviation per metre. This is a much more convenient unit than degrees. — Level River St, Mar 21 '14 at 14:23
@naslundx It means that if you walk on a circle of radius 1, radians are exactly how much you walk. On a general circle, radians are exactly the ratio of how much you walk, by the radius. — geodude, Mar 21 '14 at 16:51
@user18921 Please contribute with your answer then, would be very appreciated. — naslundx, Mar 22 '14 at 11:49
@goblin Spivak's Calculus mentions why you should use radians over degrees and builds the sin and cosine functions up from the unit circle. Supposedly there's two clear problems; one is that while it might be clear what 45 and 90 degrees are, what about $\sqrt{2}$ degrees? The other is that 360 is a fairly arbitrary choice, and supposedly one would not expect anything elegant to come from it. At least clear enough is that you'll have extra factors of $\pi/180$ appearing all over the place, once you relate the degrees to radians and know calculus for usual trig function in radians — snulty, Mar 04 '16 at 15:26

score 36 · Accepted Answer · answered Mar 21 '14 at 11:44

36

Radians make it possible to relate a linear measure and an angle measure. A unit circle is a circle whose radius is one unit. The one unit radius is the same as one unit along the circumference. Wrap a number line counter-clockwise around a unit circle starting with zero at (1, 0). The length of the arc subtended by the central angle becomes the radian measure of the angle.

From Why Radians? | Teaching Calculus

We are therefore comparing like with like the length of a radius and and the length of an arc subtended by an angle $L = R \cdot \theta$ where $L$ is the arc length, $R$ is the radius and $\theta$ is the angle measured in radians.

We could of course do calculus in degrees but we would have to introduce awkward scaling factors.

The degree has no direct link to a circle but was chosen arbitrarily as a unit to measure angles: Presumably its $360^o$ because 360 divides nicely by a lot of numbers.

answered Mar 21 '14 at 11:44

Warren Hill

3,092

5

I prefer this answer because it gets at the heart of why radians are unitless, not just that they're good because they are. – Almo Mar 21 '14 at 15:00
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I read that "360" comes from how a base 60 numbering system was used in babylon – Joe the Person Mar 21 '14 at 22:17
Excellent answer, this is exactly what I was looking for. – naslundx Mar 21 '14 at 23:55
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360 was also used in China or India (or both?) in part because there are roughly 360 days in a year. – R R Mar 22 '14 at 00:11
There's also gradians where $2\pi_r = 360_d = 400_g$ – Cole Tobin Mar 22 '14 at 04:14
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@Alom: But being unitless is not the point. Degrees are also unitless since they are a scalar multiple of a radian. The point is that one wants to avoid a multiplicative constant in derivatives, just like the exponential function $e^x$ is selected for this property. – Marc van Leeuwen Mar 22 '14 at 10:31

Andrew D. Hwang · Answer 2 · 2014-03-21T23:14:49.117

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To make commenters' points explicit, the "degrees-mode trig functions" functions $\cos^\circ$ and $\sin^\circ$ satisfy the awkward identities $$ (\cos^\circ)' = -\frac{\pi}{180} \sin^\circ,\qquad (\sin^\circ)' = \frac{\pi}{180} \cos^\circ, $$ with all that implies about every formula involving the derivative or antiderivative of a trig function (reduction formulas for the integral of a power of a trig function, power series representations, etc., etc.).

Added: Regarding Yves Daoust's comment, I read the question, "Why does it work [if angles are taken in radians] and only then?", as asking, "Why do the derivative formulas for $\sin$ and $\cos$ take their familiar form when (and only when) $\sin$ and $\cos$ are $2\pi$-periodic (rather than $360$-periodic)?" If this interpretation is correct, and if one accepts that one full turn of a circle is both $360$ units of one type (degrees) and $2\pi$ of another (radians), then the above formulas are equivalent to $\sin' = \cos$ and $\cos' = -\sin$, and (I believe) do justify "why" we use the $2\pi$-periodic functions $\cos$ and $\sin$ in calculus rather than $\cos^\circ$ and $\sin^\circ$.

Of course, it's possible naslundx was asking "why" in a deeper sense, i.e., for precise definitions of "cosine and sine in radians mode" and a proof that $\cos' = -\sin$ and $\sin' = \cos$ for these functions.

To address this possibility: In my view, it's most convenient to define cosine and sine analytically (i.e., not to define them geometrically), as solutions of the second-order initial-value problems \begin{align*} \cos'' + \cos &= 0 & \cos 0 &= 1 & \cos' 0 = 0, \\ \sin'' + \sin &= 0 & \sin 0 &= 0 & \sin' 0 = 1. \end{align*} (To say the least, not everyone shares this view!) From these ODEs, it's easy to establish the characterization: $$ y'' + y = 0,\quad y(0) = a,\ y'(0) = b\quad\text{iff}\quad y = a\cos + b\sin. $$ One quickly gets $\cos' = -\sin$ and $\sin' = \cos$, the angle-sum formulas, power series representations, and periodicity (obtaining an analytic definition of $\pi$). After this, it's trivial to see $\mathbf{x}(\theta) = (\cos \theta, \sin \theta)$ is a unit-speed parametrization of the unit circle (its velocity $\mathbf{x}'(\theta) = (\sin\theta, -\cos\theta)$ is obviously a unit vector). Consequently, $\theta$ may be viewed as defining a numerical measurement of "angle" coinciding with "arc length along the unit circle", and $2\pi$ units of this measure equals one full turn.

edited Mar 21 '14 at 23:14

answered Mar 21 '14 at 10:36

Andrew D. Hwang

78,195

5

Does this really explain why it "works" with radians ? – Mar 21 '14 at 11:07
If radians aren't really units, why do we have things like "rad/s" in physics? – Lazar Ljubenović Mar 21 '14 at 11:51
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@LazarLjubenović for convenience and transparency. Just like we use $N\cdot m$ for torque instead of $J$, which is equivalent. – Ruslan Mar 21 '14 at 11:57
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Ruslan, that's begging the question. If you treat angles as a dimension, then torque is N m / rad while energy is still N m. There are valid reasons but it isn't as clear-cut as it's sometimes made out to be. – Doug McClean Mar 21 '14 at 12:13
@Dustan Levenstein: yes, it seems first thoughts remain the best and sin'(0 degrees) is pi/180. I'll retract my other comment. – Tom Collinge Mar 21 '14 at 15:27
In Physics, radians are unitless, but sometimes still mentioned. Just like counts/s or something like that. I think it is good to make it explicit. – Martin Ueding Mar 21 '14 at 23:28
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@Ruslan: No, how can you say that on a Mathematics site! There is a world of difference between *N⋅m* and *J*; namely that one measures the magnitude of a pseudo-vector, and the other the magnitude of a vector. – Pieter Geerkens Mar 22 '14 at 02:42
@DougMcClean Torque is $\vec r\times\vec F$, thus $\text{m}\cdot\text{N}$, where would $\text{rad}^{-1}$ come from? – Ruslan Mar 22 '14 at 05:11
@PieterGeerkens units don't know what a quantity measures, they still coincide for torque and energy (and energy is scalar not vector). What makes these quantities different in nature is not the units, it's how they change under transformation of coordinates. – Ruslan Mar 22 '14 at 05:14
Very nice addition, but why would you define them analytically in the way you have done here, i.e. why should we want $\cos'' + cos = 0$ and so on? – naslundx Mar 22 '14 at 12:38
@naslundx: There are multiple ways to address this "why", depending what you mean: How would one think to make these definitions? What geometric/physical/analytic properties of (co)sine do these definitions express? Do these definitions make the technical development simpler than other (e.g., geometric) definitions? Complete answers could be quite lengthy.... :) – Andrew D. Hwang Mar 22 '14 at 13:51
@Ruslan How to get it into that equation is actually an interesting question. It pretty naturally arises in all the others (if you exert a torque through an angle you get an energy, so the dimensions of torque * angle = energry, etc). Not having a good answer is one reason to prefer not treating angles as a dimension, but that's not self-referential in the way that the first explanation was. An internally consistent but unsatisfying answer is to say that torque "just is" 1/rad * r x F, this could be the case with dimensionless angles and we just can't "see" it. – Doug McClean Mar 23 '14 at 02:40

Ben Grossmann · Answer 3 · 2014-03-21T11:48:03.250

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It really comes down to the following limit: $$ \lim_{x\to 0} \frac{\sin(x)}{x} = 1 $$ Or in other words, "$\sin x \approx x$ for small $x$". As a consequence, we have $$ \frac{d}{dx}\sin x = \cos x, \qquad \frac{d}{dx}\cos x = -\sin x $$ For any other choice of angular unit, these derivatives require some sort of coefficient (such as $\pi/180$). In this sense, radians are the "natural" unit for an angle, as far as calculus is concerned.

edited Mar 21 '14 at 11:48

answered Mar 21 '14 at 10:57

Ben Grossmann

225,327

1

I am afraid that this is a circular argument, as the given limit is computed by the l'Hospital rule or Taylor expansion. – Mar 21 '14 at 11:02
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@YvesDaoust that is very much not the case. This limit is computed using the squeeze theorem, and used to find the derivative. In fact, notice that $$ \lim_{x \to 0} \frac{\sin x}{x} = \lim_{h \to 0} \frac{\sin(0 + h) - \sin(0)}{h} $$ – Ben Grossmann Mar 21 '14 at 11:10
How do you establish the bounding functions (presumably approximations $x$ and $x-x^3/6$ of $\sin x$ ??) to apply the squeeze theorem ? – Mar 21 '14 at 11:15
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@YvesDaoust see this answer – Ben Grossmann Mar 21 '14 at 11:28
I agree with the geometric answer. But we need to admit equivalence of the geometric and analytic concepts. – Mar 21 '14 at 14:37
@YvesDaoust Don't we "need to admit" such an equivalence when we define $\sin$ and $\cos$ in the first place? – Dustan Levenstein Mar 21 '14 at 15:22
@Dustan: we can do with just considering the analytic definitions $\frac {e^{ix}+e^{-ix}}{2}$ and $\frac {e^{ix}-e^{-ix}}{2i}$ that imply the $2\pi$ period. – Mar 21 '14 at 15:35
@YvesDaoust those are (at least traditionally) not the definitions of $\sin$ and $\cos$. Rather, this is an equality that emerges from the definitions of the trigonometric functions and the unrelated definition of $e$. – Ben Grossmann Mar 21 '14 at 15:41
@YvesDaoust Okay, as soon as you change all high school curricula to reflect this definition, then we'll talk. :) – Dustan Levenstein Mar 21 '14 at 15:42
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@YvesDaoust all properties of the analytic functions $\sin(x)$ and $\cos(x)$ can be derived from the following properties: $\lim_{x\to0}\frac{\sin x}{x} = 1$, $\sin(a + b) = \sin(a)\cos(b) + \cos(b)\sin(a)$, and $\cos(x) = \sin(\pi/2-x)$. – Ben Grossmann Mar 21 '14 at 15:46
Yep, the first (and third) relations are valid for angles in radians, this is what makes it a circular argument. – Mar 21 '14 at 15:54
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@YvesDaoust nothing about this is a circular argument. I am claiming that if we define sin and cos to be the geometric functions on angles measured in radians, then all the nice analytic properties of sin and cos can be derived. If we chose a different unit for angles, we would get different analytic properties. Each of the properties I've given can be derived from the geometric definitions of sin and cos. – Ben Grossmann Mar 21 '14 at 16:09
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The function $(\cos, \sin): \mathbb R \to S^1$ can be defined as a parametrization of the circle by arclength, starting at $(1, 0)$ for $t=0$, and moving counterclockwise. Of course you need a little calculus to make sense of this definition, but nothing circular (other than, you know, the circle itself). – Dustan Levenstein Mar 21 '14 at 16:46

score 6 · Answer 4 · 2014-04-28T07:15:50.283

Assume that the formula $\sin'_r(x)=\cos_r(x)$ is true for some angular unit, let "$r$". For another angular unit, let "$d$", there is a conversion factor, let $\lambda_{d\rightarrow r}$, and we can write:

$$\sin_d'(x)=\sin_r'(\lambda_{d\rightarrow r} x)=\lambda_{d\rightarrow r}\cos_r(\lambda_{d\rightarrow r} x)=\lambda_{d\rightarrow r}\cos_d(x).$$

So the derivation formula can only be simple ($\lambda=1$) for the specific angular unit $r$, which we use to call radians.

But how do we know how much is a radian ?

Using $\sin_r'(x)=\cos_r(x)$ (and in turn $\cos_r'(x)=-\sin_r(x)$) allows to derive various Taylor-McLaurin series expansions, among which that of the arc tangent, and eventually leads to the Gregory-Leibnitz formula. This defines the constant $\pi$ and shows that an eighth of a turn (angle of the isosceles right triangle) is $\frac{\pi}{4}$ radians, equivalent to 45 degrees (by definition of the degrees).

In the end, $\lambda_{d\rightarrow r}=\frac{\pi}{180}$ and $\sin_d'(x)=\frac{\pi}{180}\cos_d(x)$.

Can we be sure that $sin'(r) = cos(r)$ (where $\lambda = 1$) actually has a solution, or do we just try and see that it actually does? — naslundx, Mar 22 '14 at 22:40
Actually, this is the way to define the $sin$ and $cos$ functions analytically. Starting from $dy/dx=y$ in complex numbers, we integrate $dy/y=dx$ and define the logarithm $ln(y)=x$ and its inverse, the exponential $y=e^x$ (such that $e^x:=e^{Re(x)}(\cos Im(x)+j.\sin Im(x)$). This defines both $e$, the base of natural logarithms, and $2\pi$, the "base" of "natural" trigonometric functions. — , Mar 23 '14 at 09:41

score 4 · Answer 5 · answered Mar 21 '14 at 11:18

4

Simple answer: radian is not really a unit, it's an absence of one. Degree, on the other hand, is. Working with dimensionful quantities in calculus is the last thing you would want to do (unless you're into kinky things) :)

answered Mar 21 '14 at 11:18

user132181

2,726

As someone pointed out, degrees are also unitless since they are just a scalar multiple of radians. – naslundx Mar 22 '14 at 23:15
@naslundx It's more accurate to say that degrees and radians are units (angular measures) but that angles are dimensionless. – ryang Oct 28 '21 at 05:25

namsap · Answer 6 · 2014-03-21T10:54:02.480

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In Calculus, sine and cosine are defined via the exponential function, meaning that $ \cos x = \mathrm{Re}\{e^{ix}\}$ and $ \sin x = \mathrm{Im}\{e^{ix}\}$ and as you know, $e^0 = e^{i2\pi} = 1$ which means that $360^{\circ}$ which is the full circle corresponds to $2\pi$. For further reference see here.

edited Mar 21 '14 at 10:54

answered Mar 21 '14 at 10:47

namsap

910

That's not the original definition though. – naslundx Mar 23 '14 at 08:54
I know of two possible definitions: using a right-angled triangle or, what I thought of, using the unit circle. If you think of the unit circle as $e^{iz}, z=0..2\pi$ in the complex plane, sine and cosine can exactly be defined like I did. Then, connection between degree and radian become clear. – namsap Mar 23 '14 at 21:16

score 2 · Answer 7 · answered Mar 21 '14 at 15:08

It is similar to why SI units are used for scientific calculations.

Many people will notice some quantity which they want to measure, for example length or angle. They will come up with repeatable ways of measuring these, designed to be easy to perform. For example the cubit and the foot are based on body parts. Degrees were chosen because they make division easy for many numbers (this is closely related to our measurement of time). The Babylonians used such number systems extensively.

After a long time, some pattern may be spotted. For example, we can calculate the energy used to move something by multiplying the force applied over the distance travelled. This lets us say things like "Energy in calories is proportional to a force in pounds multiplied by a distance in feet", but it doesn't give us an equation. To turn this into an equation requires a constant of proportionality, which will usually be difficult to remember because the units were chosen arbitrarily.

Later on, some bright spark will create a new unit, defined by these patterns. For example, in SI units the energy in Joules is equal to the force in Newtons multiplied by the distance in metres. The constant is 1 by design.

The same is true of radians. Degrees get horrible numbers everywhere because 360 divisions is arbitrary. We can make the equations nicer by changing the number of divisions to a pattern, like Tau (the circumference of a circle / its radius). A system with Tau divisions, instead of 360, makes all of the equations nicer, just like SI units do. That system is Radians.

Note that Tau = 2 * Pi, since Pi is circumference / diameter and diameter = 2 * radius.

SI units are up to humans to choose and define as we will. Pi defines itself outside of a human's choice. — R R, Mar 22 '14 at 00:17
The vast majority of SI units are not up to humans, since there are only a few degrees of freedom in the system. If humans define the metre, second, Coulomb and kilogram, then we 'automatically' get definitions for Newtons, litres, Joules, Watts, Hertz, Bequerels, Amperes, Volts, etc. without any choice. The same happens with radians, except there are no degrees of freedom; as you say, "Pi defines itself". — Warbo, Mar 24 '14 at 09:35
@ R R, Not true pi comes from the diameter not the radius and therefore not fundamental. Warbo stated $\tau = 2\pi$, but this is a little off, it should be $\pi := \tau/2$ and the rest follows. — John Nicholson, Mar 17 '15 at 01:51

score 2 · Answer 8 · answered Mar 22 '14 at 11:31

Another perspective, in my opinion the most "proper" one:

degree is just a named numerical constant that equals ${}^\circ=\frac{\pi}{180}$. So when you read $180^\circ$ you are actually multiplying by that constant! In a similar fashion, conversion into degrees just means that you divide & multiply by a degree and carry out the division, leaving the multiplication unevaluated. In that sense, the degree is just like SI prefixes (kilo, mega, etc.). It just makes the notation more convenient, but it ultimately has no deeper meaning.

The trig functions themselves are, as all other functions, defined for pure number arguments. For instance, through power series $\sin x=\sum\frac{(-1)^n x^{2n+1}}{(2n+1)!}$, and there are no degrees implied in these formulas, it doesn't even imply an angle. It is just a function, it takes any numerical argument.

"Angles" may also appear in formulas outside trig functions (also common in physics). In that case, you cannot survive simply by saying that you used special degree-versions of the trigonometric functions. For instance, the function $${\rm sinc}(x)=\frac{\sin x}{x} $$ take both $x$ as pure numbers.

Aaron Meyerowitz · Answer 9 · 2015-05-11T12:03:13.833

Here is a rationale admittedly guided by where we want to end up, which is not to say that it isn't natural. Warm-up: From the origin I have a unit vector $(x,y)$ and a second unit vector $(x',y')$ perpendicular to the first and counterclockwise from it. Use geometry to show that $y'=x$ and $x'=-y$ hint: From perpendicularity we have slopes $y'/x'=-x/y.$

OK. We want to analyze uniform circular motion so we fix a time unit and distance unit. A particle moves with variable position vector $(x,y)$ and associated velocity vector $(x',y'). $ Fix the center at the origin and the two vectors to have length $1$. (aka "move counterclockwise about the unit circle at a linear velocity of $1$ unit/sec".) Then we have$x,y,x',y'=\cos{\theta},\sin{\theta},\cos'{\theta},\sin'{\theta}$ where $\theta$ is the angle moved. The derivative identities? We did that in the warm-up. (And you thought the primes were just decorations , tricky!) Q What is the angular velocity? A: 1 rad/sec.

An optional digression: if we are trying to develop calculus from first principles then there seems an issue of "circularity" in that the linear velocity involved accumulating arc length and that already requires a fair amount of calculus . A solution (see Apostol, I think) is to instead accumulate sector area.

score 0 · Answer 10 · answered Sep 23 '15 at 21:03

0

\begin{align} \lim_{x\to0} \frac {\sin x} x & = (\text{some constant that is not 0}) \\[12pt] \frac d {dx} \sin x & = \Big( (\text{some constant that is not 0}) \cdot \cos x \Big) \end{align}

The "constants" are equal to $1$ if, but only if, radians are used.

answered Sep 23 '15 at 21:03

Michael Hardy

1

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Sure, but why? – naslundx Sep 24 '15 at 15:51
@naslundx : Here is one explanation: http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1/75149#75149 – Michael Hardy Sep 24 '15 at 16:59

Manuel Ocaña · Answer 11 · 2022-03-23T13:46:02.000

You can't remember any proof because 'we require radians in calculus' is not a logically well formulated statement, and only those have proofs. The problem with using degrees in your calculations is that when you integrate or differentiate a function you probably left many logical gaps, which you fill with geometric intuition, which is sometimes misleading.

Differentiating and integrating is a mathematical operation that has nothing to do with the warping you do to the input and/or output of your function so that it takes and returns a range of 'nice' values. This 'warping' MUST be made after all calculations are done, with the only purpose of formatting the data. The problem is that you have $g_1(f(g_2(x)))$ instead of $f(x)$.

Remember: You MUST always first have the analytic definition, then derive the 'geometric interpretation' which is usually just a bunch of analityc properties. Therefore angles must be first defined analytically. The easiest way do it is by the anlytic definitions of sine and cosine, or of the complex exponential if you like. In no part of this process nobody mentions the word 'degree' or 'radian', even though the functions are perfectly differentiable and integrable.

I hope my answer helped.

Why do we require radians in calculus?

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