I have learned about the correspondence of radians and degrees so 360° degrees equals $2\pi$ radians. Now we mostly use radians (integrals and so on)

My question: Is it just mathematical convention that radians are much more used in higher maths than degrees or do radians have some intrinsic advantage over degrees?

For me personally it doesn't matter if I write $\cos(360°)$ or $\cos(2\pi)$. Both equals 1, so why bother with two conventions?

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    $\begingroup$ The intrinsic advantage is that the limit $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$ holds only if we use radians. This is very important in calculus. $\endgroup$ – MathematicsStudent1122 May 24 '16 at 8:42
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    $\begingroup$ Well. For a unit circle the actual length of an arc spanning $n\deg = \pi/180$ units. Wheres as the actual length of the arc spanning $n$ radians is $n$ units. $\endgroup$ – fleablood May 24 '16 at 8:44
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    $\begingroup$ One advantage of using radians is that you don't have to bother writing that little circle that is the symbol for degrees. $\endgroup$ – bof May 24 '16 at 8:45
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    $\begingroup$ A better question is why degrees are used elsewhere. $\endgroup$ – anomaly May 24 '16 at 16:57
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    $\begingroup$ An advanced alien race would know about radians, but surely not degrees, which are an artifact here on Earth inherited from the Babylonian fascination with the number 60. Our alien friends might have inherited their own strange system of angle measurement, perhaps related to the number on tentacles sprouting from their foreheads. But, being advanced, they would have moved on to radians, or blozzletons, as they might call them. $\endgroup$ – zhw. May 24 '16 at 21:38

16 Answers 16


The reasons are mostly the same as the fact that we usually use base $e$ exponentiation and logarithm. Radians are simply the natural units for measuring angles.

  • The length of a circle segment is $x\cdot r$, where $x$ is the measure and $r$ is the radius, instead of $x\cdot r\cdot \pi/180$.
  • The power series for sine is simply $\sin(x)=\sum_{i=0}^\infty(-1)^i{x}^{2i+1}/(2i+1)!$, not $\sin(x)=\sum_{i=0}^\infty(-1)^i(x\cdot \pi/180)^{2i+1}/(2i+1)!$.
  • The differential equation $\sin$ (and $\cos$) satisfies is $f+f''=0$, not $f+f''\pi^2/(180)^2=0$.
  • $\sin'=\cos$, not $\cos\cdot 180/\pi$.

You could add more and more to the list, but I think the point is clear.

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    $\begingroup$ @Soke The real reason to use $\tau$ over $\pi$ in my opinion is that it would make the introduction to trigonometry (beyond right-angled triangles) a lot easier for those poor 17-year-olds (or however old they are where you're from). It would make math lessons a smidgen less frightening and complicated at no cost of factual accuracy or anything like that. The real cost is the rewriting and reprint of all the school books, plus teachers who don't see the point and teach $\pi$ anyways, which is even worse because now they're contradicting the book. $\endgroup$ – Arthur May 25 '16 at 19:09
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    $\begingroup$ @Soke, if you used $\tau$ you'd get random $\frac 12$'s everywhere instead. The only real benefit with $\tau$ is that it is easier to write. And while we are speaking of it, how come $\tau = 2\pi$ when $\pi$ has two legs, but $\tau$ only one? $\endgroup$ – Ennar May 25 '16 at 19:11
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    $\begingroup$ @Ennar Meh, I think $C = \tau r$ and $A = \frac{1}{2} \tau r^2$ is better. I would hardly call the $1/2$ in $\frac{1}{2} x^2$ to be a "random" half. $\endgroup$ – MCT May 25 '16 at 19:48
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    $\begingroup$ @Soke, it is as random as $2$'s with using $\pi$. I'm not convinced. On completely unrelated note, fraktur should probably disappear altogether... $\endgroup$ – Ennar May 25 '16 at 19:57
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    $\begingroup$ @Soke, again, $\frac 12$ comes from antidifferentiation of $x$ as clearly as $1$ comes from antidifferentiation of $2x$. In my opinion, the whole $\pi$/$\tau$ discussion brings no mathematical benefit and lies into "popular mathematics" sphere. $\endgroup$ – Ennar May 25 '16 at 20:25

As I teach my trigonometry students: "Degrees are useless."

You want to know the length of a circular arc? It's $r \theta$ where $r$ is the radius of the circle and $\theta$ is the angle it subtends in radians. If you use degrees, you get ridiculous answers.

You want to know the area of a sector? It's $\frac{1}{2} r^2 \theta$, with $r$ and $\theta$ as above. Again, if you use degrees, you get ridiculous results.

To really understand this, move on to calculus and study arc length. The arc length of the graph of the circle gives radian results. Or, look at the power series expansion of the circular trigonometric functions: if you use radians, everything works with small coefficients; if you use degrees, extra powers of $\frac{\pi}{180}$ scatter around.

What are degrees any good for? Dividing circles into even numbers of parts. That's it. If you want to actually calculate something, degrees are useless.

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    $\begingroup$ @MonK: historical reasons. The Sumerian or Babylonian astronomers who invented degrees knew some geometry of course, but they didn't have power series and differential equations, and so they didn't do so much calculation with angles as we do now. They were mostly interested in expressing position along an arc, so like Eric says they wanted an even division of the circle into units they could handle, preferably as integers. And it's only been 3000 years since then, which isn't long enough to persuade the general public to revisit the decision ;-) $\endgroup$ – Steve Jessop May 24 '16 at 10:28
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    $\begingroup$ "Degrees are useless." - except in the real world. I am building a CNC machine. Lots of trigonometry. No angle sensor, rotary encoder or any other method of measuring angle directly reads radians. (the most common digital angle measurement is in multiples of 1/(2^n) rotations.) $\endgroup$ – hildred May 24 '16 at 11:59
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    $\begingroup$ -1 for saying "degrees are useless". As an engineer, I can tell you that NO ONE ever describes an angle in a design in radians. While the mathematical side is defined in radians, degrees are what people use for historical reasons. In addition, the convenience of writing $90^o$ vs $\pi/2$ is not to be ignored. $\endgroup$ – FundThmCalculus May 24 '16 at 13:17
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    $\begingroup$ From an engineering perspective, I think the key staying power for degrees is the $90^\circ$ association that @FundThmCalculus mentioned. It's very common to want to specify exact right angles, which don't have a nice decimal representation in radians. For practical construction considerations, there are lots of tools that nominally yield 90 degree relationships, so it would require special conventions if using decimal angle values to specify "nominally a right angle" even though other angles are otherwise specified to some fixed degree of precision. $\endgroup$ – Dan Bryant May 24 '16 at 15:12
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    $\begingroup$ Would it not be better to wait until your students graduate before you tell them that degrees are useless ;) $\endgroup$ – user_of_math May 24 '16 at 17:32

Radians naturally arise when you look at some circles (note that they are a dimensionless unit). On the contrary, full circle being $360^\circ$ is due to some dude dividing the circle to as many pieces as there are days in the year (for some historical reason this resulted in $360$).

Why people think in radians then? My personal guess is that the reason is simply that mathematicians prefer to work with things that are somehow intrinsic to the object in consideration.

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    $\begingroup$ The reason might be that 360 as a lot of divisors (24 exactly) so it can be cut easily. (Same reason for the 24 hours day and 60 minutes hour) $\endgroup$ – luxcem May 24 '16 at 12:57
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    $\begingroup$ For your information, the Babylonians liked the sexagesimal (base-60) system: $6*60=360$ degrees in a circle, 60 minutes/hour, 60 seconds/minute, etc. $\endgroup$ – FundThmCalculus May 24 '16 at 13:18
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    $\begingroup$ @FundThmCalculus - Which is of course itself a bi-product of us using a decimal numbering system in the first place. There's nothing special about base-10 other than that's how many fingers our species is typically born with. Things would be much easier if we were born with one less finger on each hand, then we'd all be using base-8, which is a natural extension of binary, and that at least is fundamentally significant. Base-10 is just arbitrary, really. $\endgroup$ – Darrel Hoffman May 24 '16 at 18:42
  • $\begingroup$ There's not really an answer to the question in here. $\endgroup$ – djechlin May 24 '16 at 21:07
  • $\begingroup$ In my opinion, this is the most important point: It is a dimensionless unit (it is not a real unit like degrees, it is merely just a scalar). You do not have to mess with extra multiplications when using radians. $\endgroup$ – rexkogitans May 25 '16 at 8:14

radians are the natural unit of measure for angles. it's no anthropocentric convention. aliens on the planet Zog that do calculus and solve physics problems will also be understanding the naturalness of describing angles in radians.

as mentioned previous, the angle, expressed in radians, is the amount of circular arc swept by the angle divided by the radial arm with both lengths expressed in the same units. so radians are dimensionless. they're just a number. no units.

if you have a wheel of radius $r$ on a slip-free surface, the distance the wheel moves on the surface $x$ is equal to the angle turned, $\theta$ (in radians) times the radius of the circle of the outer rim of the wheel, $r$.

$$ x = r \cdot \theta $$

if the wheel (of radius $r$) is spinning at a rate of $\omega = \frac{\text{d} \theta}{\text{d} t}$, the speed that the wheel moves relative to the surface is $$v = \frac{\text{d} x}{\text{d} t} = r \cdot \omega \ .$$

if the angle was measured in any other manner, there would be constants of proportionality necessary for those equations to be true but those constants of proportionality are equal to 1 (and go away) if the angle is measured in radians.

radians are as natural to angles as $e \ \approx \ $ 2.718281828... is natural as a base for logarithms and exponentials in calculus. Euler's equation

$$ e^{i \theta} = \cos(\theta) + i \sin(\theta) $$

would need more nasty constants of proportionality if the base was not $e$ or the angle $\theta$ was not in radians.

so it's the opposite of human convention that in calculus the measure of angles are expressed in terms of radians.

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    $\begingroup$ It's such a great honor meeting someone who has visited planet Zog and worked with their scientists! Pray tell, good sir, kind sir, did they reveal to you their secrets of interstellar travel? $\endgroup$ – Asaf Karagila May 25 '16 at 4:40
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    $\begingroup$ i never said i visited Zog. but i am still quite sure that any aliens that do calculus understand the concept of the dimensionless measure of angle in radians. $\endgroup$ – robert bristow-johnson May 25 '16 at 4:46
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    $\begingroup$ It's very easy being certain that what you do is the universal way. If history taught us any lessons, is that whenever you're sure that the way you do it "is the reasonable way, and anyone should understand it", you're almost guaranteed to be wrong about it (not about how you do it, rather about how others are doing it). I only assumed, in that case, that confidence came from visiting unearthly planets. $\endgroup$ – Asaf Karagila May 25 '16 at 4:49
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    $\begingroup$ well, i would say that the concept of minimum is universal. and that mathematical truth (that means the axioms and theorems) are universal. the aliens that do calculus will discover, as did humans, that the minimally complicated expression of angles in analytical geometry is in radians. the arc length divided by the radial arm. no scaling numbers that any aliens or humans have pulled outa their ass. that is minimal complexity. sure we (or they) could do calculus with a base other than $e$ and angles of some other measure, but there would be scaling constants that would be added to the math. $\endgroup$ – robert bristow-johnson May 25 '16 at 4:58
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    $\begingroup$ i don't know if aliens will canonize the value $\pi \approx$ 3.14159... or instead $\tau = 2 \pi \approx$, 6.28318... but i am pretty sure it will be one of the two. and, if they do calculus, they will understand $e \approx$ 2.71828... as the natural base of logarithms and exponentials. those numbers, in whatever numbering system they use, will be "special" to the mathematicians and scientists on the planet Zog. $\endgroup$ – robert bristow-johnson May 25 '16 at 5:32

Degrees are a mistake of history (not speaking of minutes and seconds). Division in four quadrants of ninety degrees is quite arbitrary and inconvenient, but for one thing: it allows an easy representation of the remarkable angles, $30°$ and $45°$. In this respect, it is a little better than the $4\times100$ subdivisions in grades.

As explained by many others, radians are a natural unit as they avoid a constant factor when taking derivatives and make the formulas for the arc length or sector area the simplest.

IMO, the opportunity to use an interesting alternative is gone forever: the revolution. Counting angles in revolutions is pretty convenient as you can handle them modulo $1$, i.e. just using the decomposition in integer and fractional part.

  • $\begingroup$ the rotation is indeed a great unit to measure angle in particularly for cnc applications. look for absolute rotary encoder which does exactly that (although math with grey code is a barrel of something unprintable) $\endgroup$ – hildred May 25 '16 at 1:56
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    $\begingroup$ Revolutions are implicitly used with frequency (typically in Hz, rpm) while radians are used with angular frequency. $\endgroup$ – user332714 May 25 '16 at 13:43

it's because calculus would be really annoying using degrees.

$$ (\sin x)' = \cos x. $$ in radians but not in degrees.

Also, the related fact $$ \cos x = \frac{e^{ix}+e^{-ix}}{2} $$ holds in radians.


Everybody's talking about radians vs degrees. What about radians vs turns, a turn being $2\pi$ radians? I'm asking because it seems logical to think of a measure of angle as the fraction of a full circle, also I end up putting so many $2\pi$'s everywhere (mostly in sound processing) that it makes me wonder why not go the other way and use turns. The worst it can do is require you to add $\tau^{-1}$ here and there. I wonder why it's not an option discussed more often, if the world was to divide itself in two camps you'd think it would be between pro-radians and pro-turns.

In computer programming using turns instead of radians also has the advantage that integers become perfect for storing angles, they give you fixed angular precision and wrap around perfectly when they overflow.


The simple reason is that radians incorporate pi as part of the ratio which tends to be more convenient for arbitrary calculations and more complex mathematical functions.

For strictly practical manufacturing type applications degrees are often preferred because they are easier to visualise and subdivide.

In engineering radians tend to be used for kinematic calculations where you are dealing with lots of trig functions as it effectively skips a step in approximating for pi and angular velocities are generally quoted in radians per second. For example if you know the angular speed of a disk in radians per second finding the tangential speed at the circumference is a simple multiplication by the radius without any messing about with irrational constants.

However angular dimensions are generally quoted in degrees as this gives more convenient units and you are mostly working with integers or a small number of decimal places.


Radians tell you the arc length. If you have 60° you then have a bit of work to figure out the arc length: Al = n°/360 * 2πr but if you have π/3 radians, you know that the arc length is π/3 radii or π/3 * r


I don't know if this answer is good or not.

Reason 1

Lets say you want to measure the distance of a very distant star from earth like in the below image.

Consider that small circle earth and the big one, some distant star.

Say you observe the star from two different points on earth.

Now you can find the angle $\alpha$ and $\beta$ approximately and you can also find the distance between those two places, say $l$.

now you know $\alpha$ and $\beta$ and you can find $\theta$.

Thus, the distance of that star from earth would be $\frac{l}{\theta}$, considering $\theta$ is in radians.

not the most important use of radians but nevertheless a use of it.

Reason 2

Lets say i defined the area of that red rectangle $1 z$. Then i can defined area of any polygon (i took polygons only we can take any closed curve) as $x z$ where $x$ is some real number.

now what would you prefer, having area in some relation with the side or $x z$, where $z$ is just some randomly chosen area ?

This is the difference between radians and degrees.

  • $\begingroup$ The skinny triangle rule which is a cool application in astronomy relies on the fact that $\sin\theta\approx\theta$ when $\theta$ is a small angle. The formula would thus b $180/\pi\cdot l/\theta$ if the angle were measured in degrees. This goes along with the other answers showing that using radians results in trig functions with nicer properties (e.g. no need to the $\pi/180$ factor). $\endgroup$ – jdods May 27 '16 at 0:43
  • $\begingroup$ @jdods Absolutely. I just showed the practical use of radians in favour of degrees. $\endgroup$ – A---B May 27 '16 at 0:48

(Note: See edits below original post.)

Here is another reason why:

  • Radians are unitless (a.k.a. dimensionless).

This means that "$2\pi$ radians" actually equals $2\pi \ $ (the number) without any need to signify a unit of measurement. With $360^\circ$, it is absolutely necessary that we include the "$^\circ$" symbol to signify that it is in the unit of "degrees".

This is important because in dimensional analysis, you can insert radian measurements everywhere without affecting the overall units.

E.g. consider the gravitational acceleration constant: $$g=9.8 \ \text{m}/\text{s}^2. $$

If we, for whatever odd reason, needed to multiply $g$ by $\pi$ radians, the units of measurement are not affected:

$$\pi \text{ radians} \times g = 9.8\pi \ \text{m}/\text{s}^2. $$

If you felt compelled to do so, you could include the declaration of radians as a unit of measurement used:

$$\pi \text{ radians} \times g = 9.8\pi \ \text{m $\cdot$ rad}/\text{s}^2, $$

but it isn't necessary.

Of course, one might argue that this isn't a reason why we use radians, but it is simply a result of using radians. That may be like a chicken and the egg question though. However, I bet some historical research could answer the question.

Edits: Details added to avoid possible confusion.

After some feedback, here is an addendum. I was using the term "unitless" to mean "identical to pure quantity". This is not technically incorrect, per se, but is not a rigorous concept.

1) An angle $\theta$ is a dimensionless physical extent as opposed to distance $d$ which has dimensions of length.

2) Units of measurement of angles are therefore all dimensionless. However, measurements of length will always have dimension.

3) Radians are a unitless measure of angle, however, degrees are not. I.e. measurements in radians are identical to pure quantity, but this is not true for measurements in degrees. (Details below.)

Due to the way our system of mathematics has been historically constructed, measurements in radians are identical to pure quantity and therefore, the declaration of radians as the unit of measurement is redundant. (Of course, it is helpful sometimes.) For this reason, radians are unitless as they are identical to pure quantity. I.e., in the same way that Euler's Constant $e$ requires no unit attached to it (unless it represents a specific dimensional quantity), $1 \ \texttt{rad}$ does not actually need the text "$\texttt{rad}$" attached to it as it is identical to the number $1$.

$$\theta \ \texttt{rad} = \theta = \frac{\text{length of arc on a circle subtended by angle $\theta$}}{\text{radius of the circle}}$$ which has dimensions $$[\theta \ \texttt{rad}] = [\theta] = 1 = \frac{[\text{length}]}{[\text{length}]}$$ and units $$[\theta \ \texttt{rad}] = [\theta] = 1 = C \frac{[\texttt{u}_{\text{arc}}]}{[\texttt{u}_{\text{radius}}]}$$ where $\texttt{u}_x$ are the units used to measure $x$ and the conversion factor is $C[\texttt{u}_{\text{arc}}]=[\texttt{u}_{\text{radius}}].$ For example, if we measure the arc in $\texttt{cm}$ and the radius in $\texttt{m}$, then we get that $1 \ \texttt{rad} = 100 \ \texttt{cm}/\texttt{m} = 1$.

However, if we were to use degrees, then $$\theta^\circ = \theta \ \texttt{degrees} \cdot 1 = \theta \ \texttt{degrees} \cdot \frac{\pi \ \texttt{radians}}{180 \ \texttt{degrees}} = \theta\frac{\pi}{180} \ \texttt{radians} = \theta\frac{\pi}{180}.$$

Hence, $\texttt{degrees}$ are not unitless, as converting them to pure quantity requires a conversion factor.

We could have constructed our system of mathematics such that $\texttt{degrees}$ were unitless instead though. However, that would lead to more complicated formulae as noted by others in comments and answers.

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    $\begingroup$ I think this is entirely missing the point. As you mention in the last paragraph, radians are only "unitless" because we've decided to make them the standard unit of angle. $\endgroup$ – Eric Wofsey May 25 '16 at 4:19
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    $\begingroup$ That's not correct. They are unitless because they are a ratio of distances. $\endgroup$ – jdods May 25 '16 at 5:51
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    $\begingroup$ Careful: "Dimensionless" is not the same as "unitless". Radians are dimensionless units, as are degrees. -- en.wikipedia.org/wiki/Dimensionless_quantity#Properties $\endgroup$ – Daniel R. Collins May 25 '16 at 6:16
  • $\begingroup$ Maybe I was incorrect with the language. I refer, perhaps incorrectly, to meters and inches add two units of measurement. The point is that degrees need to be declared as the unit, radians do not. As far as OP's question is concerned, this seems to be an appropriate point. The exposition on calculus etc is included in other answers. $\endgroup$ – jdods May 25 '16 at 7:04
  • $\begingroup$ @DanielR.Collins, I did a little looking around and it seems normal to use "unitless" they way I have. Do you have any "authoritative" references that discuss this? Otherwise, this seems like a semantic quibble. However, the comments are welcome and add good content. Cheers. $\endgroup$ – jdods May 25 '16 at 7:26

As others noted, radians are a natural choice in mathematics. However, the same reason Babylonians chose 360 as the number of degrees in a circle (nice subdivisions of the whole circle) makes 360 a better choice for a full circle in numeric applications intended for graphics than $2\pi$: every different floating point format has a different numerical approximation of $2\pi$, and it is a major nuisance if $\pi/2$ may not be representable "exactly" when changing precision and/or be not exactly a third of $3\pi/2$ and/or be inequal to $2\pi-\pi/2$ even while keeping full precision.

PostScript, PDF, and METAFONT/METAPOST use trigonometric functions based on 360°/circle. Similarly the Cairo graphical library.


Radians are in some sense the “natural” units in which to measure angles. For a circle of radius $r$, an angle of $A$ radians will subtend an arc on that circle of length $rA$.

The use of degrees or grads is just a change of units for measuring angles, but if one uses units other than radians, one must always carry around conversion factors like $180/\pi$ all over the place. This pain goes beyond simple geometry since the relationship of arc-length to angle permeates so much of math, including the circular functions (sin, cosine) and other trig. functions, etc. The further one goes in math the more radians become natural. So much so that I now naturally answer “$\pi/2$” when I am asked the measure of a right angle — doesn’t everyone?

  • $\begingroup$ No, I don't. I answer $pi/2$ when asked the measure of a right angle. $\endgroup$ – BudgieJane May 24 '16 at 20:26
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    $\begingroup$ @BudgieJane: I answer $\tau/4$ ;) $\endgroup$ – TRiG May 24 '16 at 20:29
  • $\begingroup$ Really proves the point that it's much easier to remember a right angle in degrees. $\endgroup$ – djechlin May 24 '16 at 21:08
  • $\begingroup$ So why not "diametrians"? $\endgroup$ – Random832 May 25 '16 at 3:07
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    $\begingroup$ "diametrians"? a new one to me. Thanks. Maybe that will be "natural" for future generations in that 2 \pi can get replace by \tau with no leading constant. Alas, my brain is already too ossified to internalize that innovation completely. As the old quote goes, academic revolutions occur apace with caskets. That said, both are objectively more "natural" than degrees or grads. $\endgroup$ – John Cobb May 25 '16 at 21:07

The answer is simple, it's a distance measure. When you move in a straight line you use inches or metres, in circles it is radians.
If you are at Disneyland and ask how far it was to Anaheim Stadium [go, Angels!] and I tell you that from my house it's about 45º, you are probably not going to be happy.
You want the distance traveled, at 1 mile out from my house, from one point to another. This distance is pi/4 * 1 mi = about 0.8 miles. Enjoy the game.

  • $\begingroup$ You said "in circles it is radians" but gave the final answer of "0.8 miles." So which is it? $\endgroup$ – djechlin May 24 '16 at 21:10
  • $\begingroup$ It's r * pi / 4 radians. The radius of the radians is 1 mile and if you traverse it, you will travel .8 miles in an arc. $\endgroup$ – Engineer May 24 '16 at 23:25
  • $\begingroup$ Let me put it another way. The radian measurement is pi / 4 times the radius. Google says this is 0.78539816339; call it 0.8. You multiply that by the radius of the circle. My circle is 1 mile around me, and the two places I mentioned are on that circle. From my viewpoint, they are 45º apart. My friend needs to traverse the arc for 45º, which will mean he has to walk 0.8 miles to get to the arena. If I change the example and say those places were 2 miles away, then 45º for me would mean he has to walk 1.56 miles (2 * pi / 4). $\endgroup$ – Engineer May 27 '16 at 2:24
  • $\begingroup$ In private pilot ground school we use degrees. With degrees, it is easy to find the approximate sine of an angle, by sin(x)=x/60 ( the "clock rule"). However, that rule breaks down beyond 33 degrees. For 33 to 55 degrees, subtract a correction factor, (x-33)/250. All of this is easy to calculate in your head without resort to an E6-B computer (circular slide rule). I guess I am on a little project to refine these approximations. We could memorize a few sines and do linear interpolation. Two digits of accuracy is all we need for airplane navigation. $\endgroup$ – richard1941 Jun 19 '16 at 2:05
  • $\begingroup$ Well a 30 degree angle is pi/6. Pi is about 3.1, so it's 31/60 or about .52. $\endgroup$ – Engineer Jun 20 '16 at 6:25

It is natural to define $$ \sin x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}\tag1 $$ rather than $$ \sin x=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\left (\frac {2\pi x}{360}\right)^{2n+1}\tag2. $$

The function defined by (1) has many nice properties, for example it is the solution of the ODE: $$f''(x)=-f (x), \quad f (0)=0,f'(0)=1. $$

The same reason is the background of a special role of the Euler number $e $.


I don't know what the real reason is. I see one possible good reason to use radians. Some people define $\cos$ and $\sin$ of any number of degrees in terms of a circle. Other people define $\cos$ and $\sin$ purely as a function from $\mathbb{R}$ to $\mathbb{R}$ where the operand of $\cos$ or $\sin$ is just a real number with no units according to the differential equations

  • $\cos(0) = 1$
  • $\sin(0) = 0$
  • $\forall x \in \mathbb{R}\cos'(x) = -\sin(x)$
  • $\forall x \in \mathbb{R}\sin'(x) = \cos(x)$

A radian is defined to be $\frac{180}{\pi}°$. It can be shown that $\forall x \in \mathbb{R}$, $\cos(x \text{ radians})$ using the first definition of $\cos$ is equal to $\cos(x)$ using the second definition of $\cos$ and $\sin(x \text{ radians})$ using the first definition of $\sin$ is equal to $\sin(x)$ using the second definition of $\sin$.

Radians are also used in one of the formulae for centripetal acceleration. The angular frequency of a uniformly rotating object denoted $\omega$ is the number of revolutions per amount of time divided by $2\pi$. That formula is $|a| = \omega^2r$. Finally when you take the Taylor series of $\cos$ and $\sin$ in radians, it's not that hard to compute that $\forall x \in \mathbb{R}\cos(x) = \sum_{i = 0}^\infty\frac{(-1)^ix^{2i}}{(2i)!}$ and $\forall x \in \mathbb{R}\sin(x) = \sum_{i = 0}^\infty\frac{(-1)^ix^{2i + 1}}{(2i + 1)!}$. That makes it so easy to compute given a number its $\cos$ and $\sin$ in radians because the $\pi$ does not appear anywhere in the Taylor series' of $\cos$ and $\sin$ in radians.

The author of the question https://academia.stackexchange.com/questions/78068/is-there-a-place-in-academia-for-someone-who-compulsively-solves-every-problem-o wants to solve every problem on their own. There might be some people like that in a job who refuse to accpet that the first 16 digits of $\pi$ are what they were told they are and use the value of $\pi$ to build something when they cannot afford to get it more than a tiny bit off until they figure out its value to a high enough accuracy on their own because people sometimes make a mistake and spread false rumours in research. In fact, the Taylor series of $\sin$ can be used to calculate $\pi$. $\pi$ is $6 \times \sin^{-1}(\frac{1}{2})$. Given that $\sin^{-1}$ has a power series centered around 0 that has a nonzero radius of convergence and is equal to $\sin^{-1}$ in its radius of convergence, there is an obvious way to use the rule of composition for a power series and the fact that $\forall x \in [-\frac{\pi}{2}, \frac{\pi}{2}], \sin(\sin^{-1}(x)) = x$ to compute the power series for $\sin^{-1}(x)$. Dropping the assumption that $\sin^{-1}$ has a power series with a nonzero radius of convergence that is equal to $\sin^{-1}$ in the radius of convergence, if we then see that that series has a nonzero radius of convergence, then for all numbers in the radius of convergence, if you first apply that power series then apply $\sin$, you will get back the original number because of the rule for composition of two power series'. From this, we can deduce that that power series is $\sin^{-1}$ in its radius of convergence. If its radius of convergence is more than $\frac{1}{2}$, that can be used to calculate $\pi$. I will not however give the power series of $\sin^{-1}$ here because in my opinion, it's better when people who want to find out the power series for $\sin^{-1}$ have to either figure it out on their own or do without knowing. Then maybe research groups that want to will be able to actually get people who for real are able to solve that problem on their own.


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