I'm still struggling to understand why the derivative of sine only works for radians. I had always thought that radians and degrees were both arbitrary units of measurement, and just now I'm discovering that I've been wrong all along! I'm guessing that when you differentiate sine, the step that only works for radians is when you replace $\sin(dx)$ with just $dx$, because as $dx$ approaches $0$ then $\sin(dx)$ equals $dx$ because $\sin(\theta)$ equals $\theta$. But isn't the same true for degrees? As $dx$ approaches $\theta$ degrees then $\sin(dx \,\text{degrees})$ still approaches $0$. But I've come to the understanding that $\sin(dx \,\text{degrees})$ approaches $0$ almost $60$ times slower, so if $\sin(dx \,\text{radians})$ can be replaced with $dx$ then $\sin(dx \,\text{degrees})$ would have to be replaced with $(\pi/180)$ times $dx$ degrees.

But the question remains of why it works perfectly for radians. How do we know that we can replace $\sin(dx)$ with just $dx$ without any kind of conversion applied like we need for degrees? It's not good enough to just say that we can see that $\sin(dx)$ approaches $dx$ as $dx$ gets very small. Mathematically we can see that $\sin(.00001)$ is pretty darn close to $0.00001$ when we're using radians. But let's say we had a unit of measurement "sixths" where there are $6$ of them in a full circle, pretty close to radians. It would also look like $\sin(dx \,\text{sixths})$ approaches $dx$ when it gets very small, but we know we'd have to replace $\sin(dx \,\text{sixths})$ with $(\pi/3) \,dx$ sixths when differentiating. So how do we know that radians work out so magically, and why do they?

I've read the answers to this question and followed the links, and no, they don't answer my question.

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    $\begingroup$ I believe it has to do with the fact that an arc-length is given by $\theta r$ only if $\theta$ is measured in radians. There isn't anything special about this because we know (without radians) that the circumference of a circle is given by $2\pi r$. So if we define radians such that "$360^\circ$" is $2\pi$ then we get for the value of arc-lengths: $A_\theta = 2\pi r * \frac{\theta}{2\pi} = \theta r$ vs. degrees: $A_\theta = 2\pi r * \frac{\theta}{360^\circ} = \theta r * \frac{2\pi}{360^\circ}$. $\endgroup$ – Jared Jun 25 '15 at 23:48
  • $\begingroup$ food for thought plus.google.com/+TimothyGowers0/posts/hDLKkTnDgi8 $\endgroup$ – hHhh Jun 26 '15 at 0:10
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    $\begingroup$ Degrees and radians are not "arbitrary units of measure" because angles are unitless. As pseudo-units, "rad" is the value 1 (unitless) and "deg"/degree-sign is the value $\frac{\pi}{180}$. $\endgroup$ – R.. GitHub STOP HELPING ICE Jun 26 '15 at 3:02
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    $\begingroup$ I'm going to add this as a general comment. This question is very similar to the following question: Why did Euler use e to represent complex numbers? Most of the answers ignored the definition of $e$ (including the top answer) and I see that most of the answers here ignore the definition of trigonometric functions in a similar fashion--specifically the definition of $\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1$ and why it is that this is the true. $\endgroup$ – Jared Jun 26 '15 at 7:49
  • $\begingroup$ Related: Why do we require radians in calculus?. $\endgroup$ – Andrew D. Hwang Jun 26 '15 at 20:46

17 Answers 17


It seems to me that the best answer thus far is Simon S's. Others have hinted at the important property:

$$ \lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1 $$

Some have simply stated it's important with little reason as to why it's important (specifically in regards to your question about the derivative of $\sin(x)$ equaling the $\cos(x)$). Simon S's answer explained why that limit is important for the derivative. However, what I find lacking is why it is that the limit equals what it equals and what would it equal if we decided to use degrees instead of radians.

At this point, I want to acknowledge that my answer is essentially the same as Simon S's except that I am going to go into gruesome detail.

Before I go into this, there is absolutely nothing wrong with using degrees over radians. It will change what the definition of the derivative of the trigonometric functions are, but it won't change any of our math--it just introduces a tedious factor we always have to carry around.

I am going to use this geometric proof as a way to make sense of the limit above:

Image taken from [https://proofwiki.org/wiki/Limit_of_Sine_of_X_over_X/Geometric_Proof]

There is only one part of the proof that will change if we decide to use degrees as opposed to radians and that is when we find the area of the sector subtended by $\theta$. When we use radians we get: $A_{AB} = \pi 1^2 * \frac{\theta}{2\pi} = \frac{\theta}{2}$--just as they found in the given proof. If however, we use degrees then we will get: $A_{AB} = \theta * \frac{\pi}{360}$. Now this changes their initial inequality which the rest of the proof relies on:

$$ \frac{1}{2}\sin(\theta) \leq \frac{\pi \theta}{360} \leq \frac{1}{2}\tan(\theta) $$

(the others don't change because the sine and tangents equal the same thing regardless of whether or not we use radians or degrees--with a proper, trigonometric definition of each, of course).

We still proceed in the same way (I'm going to be less formal and not worry about the absolute values--although we should technically). We divide everything by $\sin(\theta)$ which since I'm only worrying about the first quadrant won't change the directions of the inequalities:

$$ \frac{1}{2} \leq \frac{\pi \theta}{360 \sin(\theta)} \leq \frac{1}{2\cos(\theta)}\\ \frac{360}{2\pi} \leq \frac{\theta}{\sin(\theta)} \leq \frac{360}{2\pi \cos(\theta)} \\ \frac{\pi}{180} \geq \frac{\sin(\theta)}{\theta} \geq \frac{\pi}{180}\frac{1}{\cos(\theta)} $$

When we plug in $\theta = 0$ (whether radians or degrees) we get $\cos(0) = 1$ and thus we use the squeeze theorem to show that:

$$ \frac{\pi}{180} \leq \lim_{\theta \rightarrow 0} \frac{\sin(\theta)}{\theta} \leq \frac{\pi}{180} $$

Therefore, if we use degrees, then:

$$ \lim_{\theta \rightarrow 0} \frac{\sin(\theta)}{\theta} = \frac{\pi}{180} $$

Going back to Simon S's answer, this gives, as the definition of the derivative for $\sin(x)$:

$$ \lim_{h \rightarrow 0} \frac{\sin(x + h) - \sin(x)}{h}\\ \lim_{h \rightarrow 0} \frac{\sin(x)\cos(h) + \sin(h)\cos(x) - \sin(x)}{h} \\ \lim_{h \rightarrow 0} \frac{\sin(h)\cos(x) + \sin(x)(\cos(h) - 1)}{h} $$

This may be a little sloppy, but when $h = 0$ $\cos(h) - 1 = 1 - 1 = 0$, so we can drop the second part and are left with:

*Actually this is extremely sloppy, at this point I would refer back to Simon S's Answer

$$ \lim_{h \rightarrow 0} \frac{\sin(h)}{h}\cos(x) = \cos(x)\lim_{h \rightarrow 0} \frac{\sin(h)}{h} $$

Using our above result we find the following:

$$ \frac{d}{dx}\sin(x) = \frac{\pi}{180}\cos(x) $$

This is what the derivative of $\sin(x)$ is when we use degrees! And yes, this will work fine in a Taylor series where we plug in degrees for the polynomial as opposed to radians (although the Taylor series will look different!).

And hopefully you already realize that this is what the derivative of $\sin(x)$ is when we use degrees, because if we accept that we must use radians, then we must convert our degrees to radians:

$$ \sin(x^\circ) = \sin\left(\frac{\pi}{180}x\right) $$

Now using the chain rule we get:

$$ \frac{d}{dx}\sin(x^\circ) = \frac{\pi}{180}\cos(x^\circ) $$

So the question isn't really why does it only work with radians--it works just as well with degrees except that we get a different definition of the derivative. The reason we prefer radians to degrees is that radians doesn't require this extra factor of $\frac{\pi}{180}$ every single time we differentiate a trigonometric function.


Radians, unlike degrees, are not arbitrary in an important sense.

The circumference of a unit circle is $2\pi$; an arc of the unit circle subtended by an angle of $\theta$ radians has arc length of $\theta$.

With these 'natural' units, the trigonometric functions behave in a certain way. Particularly important is $$\lim_{x\to 0} \frac{\sin x}{x} = 1 \quad\quad - (*)$$

Now study the derivative of $\sin$ at $x = a$:

$$\lim_{x \to a} \frac{\sin x - \sin a}{x-a} = \lim_{x \to a}\left( \frac{\sin\left(\frac{x-a}{2}\right)}{(x-a)/2}\cdot \cos\left(\frac{x+a}{2}\right)\right)$$

This limit is equal to $$\cos a$$ precisely because of the limit $(*)$. And $(*)$ is quite different in degrees.

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    $\begingroup$ But that doesn't answer my question. I'm more interested in why $\lim_{x\to 0} \frac{\sin x}{x} = 1$, and it looks like your answer just assumed that to be true. (I've since found proof here: oregonstate.edu/instruct/mth251/cq/Stage4/Lesson/sinProof.html) What is $(*)$? $\endgroup$ – Kyle Delaney Jun 27 '15 at 16:59
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    $\begingroup$ If your question was how to prove the limit I have labeled $(*)$, then it's always best to ask it explicitly: it has been addressed a few times on this site, e.g., math.stackexchange.com/questions/75130/… Based on your original question, I wanted to point out was that the derivative works out nicely precisely because $(*)$ lies at the heart of the limit of the difference quotient used to define the derivative. It looks like 25+ people agreed with a reading of your question where that was a point worth making. $\endgroup$ – Simon S Jun 27 '15 at 17:29
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    $\begingroup$ I'm sorry. Sometimes you need to know the answer to a question in order to know exactly how to ask it. I didn't know exactly how to convey my question and so it was a difficult question to ask. I was fumbling around in the dark here. But in these cases it seems like a lot of people jump on an interpretation that's easier to answer. $\endgroup$ – Kyle Delaney Jun 27 '15 at 18:07
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    $\begingroup$ This is still the best answer imo, my answer only copies it and it wasn't until this answer that I realized why that limit (that many other answers mentioned) was so important. $\endgroup$ – Jared Jun 28 '15 at 1:16
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    $\begingroup$ @KyleDelaney That's true, but Simon S did not merely state that the given limit is 1 (granted he assumed this without proof)! He showed why it is that if that limit equals 1 that the derivative of the sine is the cosine. As I understand it, that was your original question (sort of). Now, as is usually the case, the "true" answer to your question lies in the details. Nevertheless my answer would not have been possible (by me anyway) without Simon S's illumination of the problem (and an illumination that I find lacking in almost all of the other answers). $\endgroup$ – Jared Jun 30 '15 at 21:10

It works out precisely because $$ \lim_{x \to 0} \frac{\sin x}{x} = 1,$$ which in turns happens precisely because we've chosen our angle to be the same as the arclength around a unit circle (and for small angles, the arc is essentially a straight line).

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    $\begingroup$ I have to say that I'm flabbergasted as to why this has so many upvotes. Any high school Calculus student can tell you that the given limit equals $1$, but merely stating it's that value shows no real understanding of why it's that value. And I see no connection given as to why it has to be radians to get this value for the limit. My question would be: why is it that choosing an "angle to be the same as the arclength around a unit circle" proves this identity? Doesn't matter what unit you choose for your angle, small angles always result in an "arc is essentially a straight line". $\endgroup$ – Jared Jun 26 '15 at 7:38
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    $\begingroup$ The value statement isn't that the limit is $1$. It's that the limit is $1$ because the angle is the arclength of a unit circle. This is the defining attribute of radians. $\endgroup$ – davidlowryduda Jun 26 '15 at 13:56
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    $\begingroup$ But why is it $1$ because $2\pi$ radians is the arclength of a unit circle? That was my question. I don't see the connection. $\endgroup$ – Jared Jun 27 '15 at 3:56
  • $\begingroup$ @Jared: the connection is the usual geometric proof that you gave in your answer. $\endgroup$ – Martin Argerami Jun 27 '15 at 11:39

It might help to forget, just briefly, that sine and cosine have anything to do with triangles. Instead, let's look at them in this light:

Let the function $s(x)$ be the unique function with the property that $s''(x) = -s(x)$ such that $s(0) = 0$ and $s'(0) = 1$. So it's a function that, when differentiated twice, gives the same function with a change of sign, and the line $y = x$ is the tangent to the function at $x = 0$.

Similarly, let $c(x)$ be the unique function that $c''(x) = -c(x)$, $c(0) = 1$ and $c'(0) = 0$.

These are perfectly useful functions. They provide a basis for equations of simple harmonic motion, they're related to each other (because you can show that $s'(x) = c(x)$), they're related to the exponential function via the complex numbers, and so forth.

And, as it so happens, they are exactly the sine and cosine functions defined in radians. Just like we choose the value of $e$ so that $\frac{d}{dx} e^x = e^x$, choosing to work in radians gives us those nice derivatives.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Daniel Fischer Jun 29 '15 at 18:40
  • $\begingroup$ Wow, that is so cool. Is this definition identical to the usual definition of the sin and cos functions? $\endgroup$ – Algo Aug 2 '20 at 12:28
  • $\begingroup$ They are provably the same functions, although the proof works a lot better if you work with complex numbers (because then you can also use the fact that $e^{ix} = \cos x + i \sin x$ to simplify most of the work). $\endgroup$ – ConMan Aug 2 '20 at 23:21

I believe your confusion stems from a misperception of "sine" and "angle". We tend to think of the units we measure an angle in (degrees, or radians, or grads, or whatever) as being independent of the sine function. In truth, this is not so: the sine of an angle measured in degrees, and the sine of an angle measured in radians are two different functions.

As such, it is perfectly reasonable to suspect that they have differing rates of change (derivatives).

In fact, it is only the function "sine of an angle measured in radians" that has a slope (rate of change) of $1$ at the origin: to see this, suppose we consider the function:

$f(x) = \sin(ax)$, where $x$ is measured in radians, and $a$ is some unit conversion factor (for degrees we have: $a = \frac{180}{\pi}$).

Taking the derivative, we have (by the chain rule):

$f'(x) = \cos(ax)\cdot a$, and thus: $f'(0) = \cos(0)\cdot a = a$, which equals $1$ only when $a = 1$.

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    $\begingroup$ That doesn't answer my question. I know that sine of degrees and sine of radians have differing rates of change, and I know why they have differing rates of change. You've used the derivative of sine in your explanation without explaining why that derivative is what it is, which is more what my question was about. (I've since found proof here: oregonstate.edu/instruct/mth251/cq/Stage4/Lesson/sinProof.html) $\endgroup$ – Kyle Delaney Jun 27 '15 at 17:16
  • $\begingroup$ In finding the derivative of $\sin$ straight from the definition, we're led to the limit as $h \to 0$ of $\frac{\sin h}{h}$. What this ratio is, as we approach $0$ is going to depend vitally on "how we measure" $h$. With a function, its graph is "$x$ versus $y$", and if we stretch or compress the $x$, we get a different "shape" for the curve the graph forms. In fact, for any function, we have $f'(ax) = a\cdot f'(x)$, the "scaling factor" comes out in front. Choosing radian measure is "the right scaling factor for our units". That's the why of it, but of course, there's more in the "how". $\endgroup$ – David Wheeler Jun 27 '15 at 17:27

Mathematicians generally don't consider the sine to be a function of a physical quantity with a unit at all. They just consider it as a function mapping real numbers to real numbers, and often it is actually defined as the solution1 to the differential equation $\frac{\mathrm{d}^2}{\mathrm{d}x^2}\sin(x) = -\sin(x)$ with $\sin(0)=(0)$ and $\frac{\mathrm{d}\sin(x)}{\mathrm{d}x}|_{x=0} = 1$. (Or, equivalently, by the corresponding Taylor series.)

You can then go on to derive all the familiar geometrical properties from that, but it'll only work when you equate those real-number argument with radian angles. One way to show it is to determine the zeroes of the sine (easy from the Taylor series), which are at $0, \pi, 2\pi \ldots$ and obviously not at e.g. $0, 180, 360\ldots$.

1It can be proven that this exists and is unique, by Lipschitz continuity.

  • $\begingroup$ One can also do the following: define $\arcsin$ first by using the integral $\int_0^x \sqrt{1-t^2}\ dt$ for $x \in [-1,1]$, then show this function is injective on the range given, take its inverse, and extend periodically. However, the way calculus is usually taught means thus delaying introduction of trigonometric functions until after these "more advanced" topics are covered. This is logically sound, but goes against the traditional curriculum development. $\endgroup$ – David Wheeler Jun 27 '15 at 17:39

We can only find $ \frac{\mathrm{d}}{\mathrm{d}x}\sin(x) $ if we know $ \lim_{h\to0}\frac{\sin(h)}{h} $ . In radians, this limit is $1$ but in other systems of measuring angles, messy factors come out; namely $\frac{180}{\pi}$ for degrees.. This is why you differentiate $\sin(x)$ when $x$ is in radians or change degrees to radians before you do so.

To summarise, you can differentiate in any system of measuring angles but it'll be messy if you don't use radians.


Geometrically, the radian unit is the only intrinsic unit, since it depends on no arbitrary choice, and is the same whatever the radius of the circle.

There is another reason which leads to the fundamental limit $\,\dfrac{\sin x}x\to 1$ when $x\to 0$: it is based on the fact that the area of a circular sector of radius $r$, with angle $x$ is equal to $$\frac12r^2x=\frac12 r\cdot rx=\frac12\,\text{radius}\cdot\text{arclength}$$ exactly the same of the formula for the area of a triangle: $\dfrac12\,\text{height}\cdot\text{basis}$. And indeed, you can consider a circular sector as a sort of curvilinear triangle. Geometric proofs for the fundamental limit stem from this.

  • $\begingroup$ "is the same whatever the radius of the circle." Every unit that could be applied to angles is the same whatever the radius of the circle.I understand that radians serve a specific purpose, but so do degrees. It still seems subjective to me to say that radians depend on no arbitrary choice. $\endgroup$ – Kyle Delaney Jun 30 '15 at 15:58
  • $\begingroup$ It is intrinsic, depending only on this geometric object – the circle, not on having a basis 60 numeration system. $\endgroup$ – Bernard Jun 30 '15 at 16:16
  • $\begingroup$ You could make lots of angle units that don't depend on any base. What if instead of 2pi, a full circle was pi? What if a full circle was 1, so then an angle measure was literally just represented as a fraction of a full circle? What if the angle was represented as a fraction of a right angle, so then a full circle was 4? Radians are just as arbitrary as any of those. $\endgroup$ – Kyle Delaney Jul 1 '15 at 18:05

Unlike any other unit for angles, radians really are the most natural choice. Let me quickly sketch how '1 rad' can be defined in the first place.

Let's start with the exponential function, defined via series: $$\exp(x) = \sum_{n=0}^\infty{\frac{x^n}{n!}}.$$ Differentiating it clearly yields $\exp'=\exp$. Also, one can show that $$\exp(a+b) = \exp(a)\exp(b)$$ for any $a$ and $b$ (even complex).

The sine and cosine functions are now defined as the real and imaginary parts of the exponential function with an imaginary argument: $$\exp(ix) =: \cos(x) + i \sin(x).$$ Note that the sine and cosine functions fulfill the differentiation rules as we know them ($\sin'=\cos$ and $\cos'=-\sin'$), but so far we did not yet talk about angles.

For this, first note that $|\exp(ix)|^2 = \exp(ix) \times \overline{\exp(ix)} = \exp(ix) \exp(-ix) = \exp(0) = 1$. Thus $\exp(ix)$ parametrizes the unit circle. Even more so, it does it with a constant velocity (as required by $\exp(a+b)=\exp(a)\exp(b)$).

At this point, we may now link the geometric concept of an angle to the argument $x$ to the function $\exp(ix)$. That's basically what an angle is: it is some measure on how far we have advanced on the parametrization of the unit circle. If we define the angle formed between $\exp(1i)$ and the real axis to be 1 rad, we effectively identify the angle to the argument of the sine and cosine functions as defined above.

Conclusion / TL;DR

In the end, it comes down to this: sine and cosine functions do a priori not coincide with the geometric functions as we know them, that give us relations between sides and angles in a triangle. They are analytical functions defined via series. It is only after some thought that they can be identified to their homonyms from geometry. It is in this step that we need to specify the concept of an angle, and it is only for radians that we can just plug in the numbers right away. Any other units will require some conversion first.


For convenience, let's measure time in seconds and distance in meters, so the unit circle has a radius of one meter.

Look at a point moving around the unit circle along the path $x(t) = (\cos t, \sin t)$. The usual trig functions are scaled so that the point will make a full circle every $2\pi$ seconds, which means it's moving at a speed of $1$ meter per second. In other words, the point's velocity $x'(t)$ is a unit vector.

Position and velocity for the usual trig functions.

Knowing this, it's not hard to figure out that $x'(t)$ is just $x(t)$ rotated a quarter turn counterclockwise. Now we have two ways of figuring out $x'(t)$: by taking the derivatives of the components of $x(t)$, and by rotating $x(t)$ a quarter turn counterclockwise. The fact that those two descriptions give you the same vector $x'(t)$ is precisely the fact that $$(\cos' t, \sin' t) = (-\sin t, \cos t).$$

Now, what if we rescaled our trig functions so that the point made a full circle every $T$ seconds instead of every $2\pi$ seconds? Let's call the rescaled trig functions $\cos_T$ and $\sin_T$. When $T = 360$, we get the trig functions for angles measured in degrees.

Since the point is making a full circle every $T$ seconds, it's moving at a speed of $\tfrac{2\pi}{T}$ meters per second.

Position and velocity for time-rescaled trig functions.

You can calculate the derivatives of the rescaled trig functions using the same reasoning as before, and you get $$(\cos_T' t, \sin_T' t) = \tfrac{2\pi}{T}(-\sin_T t, \cos_T t).$$ So this is why setting $T = 2\pi$ gives derivatives with no extra factors: it's the scaling that makes $x(t) = (\cos_T t, \sin_T t)$ a unit-speed path.

I can't resist mentioning one more thing. If you look at a point moving up the unit hyperboloid along the path $x(t) = (\cosh t, \sinh t)$, you'll see it doesn't move at unit speed. What's up with that?

Well, if you change your point of view, you can see that $x(t)$ really is a unit-speed path: not through $2$-dimensional space, but through $(1+1)$-dimensional spacetime! In spacetime, of course, you have to be careful about the word "speed": when I say $x(t)$ is unit-speed, I mean that $x'(t)$ always has a Minkowski norm of $1$.


Try thinking of both degrees and radians as measures of arc length, where the radii differ between the two measurements. enter image description here Notice how $\sin \theta < \theta < A$ where $$ \theta: \text{the angle measured in radians}\\ A:\text{the angle measured in degrees} $$ In fact, $A = \frac{180}{\pi}\theta\approx 57.3\theta$ So the angle measured in degrees is always at least as large as 57.3 times the sine of the angle, no matter how small (i.e. close to zero) they become.


Well we pick the unit radians so it has this property. That is why the circle is split up into 2pi radians this is actually slightly inconvenient and is why when we don't need calculus we typically use degrees.


First, note that when one speaks of "units" for an angle, what you are doing is utilizing a mapping from real numbers to angles, and in the case of trig functions, you really want to consider oriented angles.

What is meant by "$\pi/2$ radians" is that the radian mapping sends $\pi/2$ to a certain oriented angle, and "$\pi/2$ radians" is really a name for that angle. But actually this radian mapping is incorporated into how the sine function is computed. What actually goes into the sine function as input is $\pi/2$, which is a real number and has no units. In other words, the derivative of sine is unitless as the ratio of two unitless quantities.

  • $\begingroup$ You can call it unitless, but is it dimensionless? An angle is a different dimension from a length or a ratio, right? Since a sine is a ratio but x is an angle, x and y use two different dimensions. And that would mean the derivative of sine is a ratio divided by an angle, right? $\endgroup$ – Kyle Delaney Jun 27 '15 at 18:03
  • $\begingroup$ Not sure what you mean by "dimension" but think of it this way. A unit is a parametrization of the space of angles by real numbers. ideally sine should be a function on angle space directly. But it isn't. The convention is to define sine as a composition of the radian parametrization and a function that acts on angle space. This way the usual sine function is a mapping from real to real, not angle to real. $\endgroup$ – Chan-Ho Suh Jun 27 '15 at 20:08
  • $\begingroup$ Length is a dimension. Feet and meters are units of that dimension. Mass is a dimension. Pounds and grams are units of that dimension. owlnet.rice.edu/~labgroup/pdf/Dimensions_units.htm $\endgroup$ – Kyle Delaney Jun 29 '15 at 0:11
  • $\begingroup$ According to your reference, 1) angles are dimensionless and 2) dimensionless quantities do not have units. It goes on further to say ratios and angles may have certain "titles" associated to them like Mach or mol (in the case of ratios) and radians or degrees (in the case of angles). I don't see how this counters anything I said in my answer or comment. I tend to think that this discussion on dimension is kind of a red herring as far as your original question goes. $\endgroup$ – Chan-Ho Suh Jun 29 '15 at 0:45

Assume you only know the equation of the (trigonometric) circle, $x^2+y^2=1$.

If you want to travel a length $l$ from the point $(1,0)$, you need to compute the integral

$$l=\int_{t_0}^{t_1}\sqrt{\dot x^2(t)+\dot y^2(t)}\,dt$$ for some parameterization.

Let us choose $x=\dfrac1{\sqrt{1+t^2}},y=\dfrac t{\sqrt{1+t^2}}$, then


We can evaluate the integral by the Taylor development,


If we take $T=1$, we cover an eighth of a turn. Then if we define

$$\pi=4\sum_{k=0}^\infty\frac{(-1)^k}{2k+1},$$ ($4$ being chosen instead of $8$ for obscure historical reasons), the length of a full turn is $2\pi$.

We have now established a relation between $l, x, y$, via $t$. Let us denote $\cos(l):=x(l)$ and $\sin(l):=y(l)$. Obviously, $$\lim_{t\to0}\frac{\sin(l)}l=\lim_{t\to0}\frac{\frac t{\sqrt{1+t^2}}}{\sum_{k=0}^\infty\frac{t(-t^2)^k}{2k+1}}=1.$$

This makes it an attractive unit to measure the angles. With the same defintions, we could show that $\sin'(l)=\cos(l)$.

  • $\begingroup$ What do you mean by obscure historical reasons? Can you clarify this point please? $\endgroup$ – explogx Apr 3 '20 at 14:48
  • $\begingroup$ Also, how do you know that $l = \int_0^t \frac{dt}{1 + t^2}$ is one eighth of the unit circle, since you need the arctan function to solve this integral for $\frac{\pi}{4}$ $\endgroup$ – explogx Apr 3 '20 at 15:06
  • $\begingroup$ @user631896: by exploiting symmetries, we can cover a full circle with eight similar integrals. There is no "arc tangent" in this development. $\endgroup$ – Yves Daoust Apr 3 '20 at 16:18
  • $\begingroup$ You proof seems to make use of the Leibniz formula for $\pi$ which is not trivially deduced from your integral $\endgroup$ – explogx Apr 3 '20 at 17:16
  • $\begingroup$ @eigenslacker: the expansion of $1/(t^2+1)$ is an elementary geometric progression. Allow me to call trivial the term-wise integration of an entire series. And the formula defines $\pi$ rather than uses known facts about it. By the way, who is Leibnitz ? I don't see him among the contributors. $\endgroup$ – Yves Daoust Apr 3 '20 at 19:19

The short answer: because the length of the circumference of a circle is $2\pi r$, not $360r$.


A different explanation is the following:

When writing the sin function as a taylor series (which approximates the function with polynomials) you find that in first order (meaning small errors for small values of x) sin(x) = x + O(x^{3})

That means, that for small values of x (ca. 5 degrees) you can replace sin with x. This approximation is used often in the physics of harmonic oscillations (pendulums and the like). It is not relevant if you measure the angle in degrees or radians.


Simply put: Because a radian is defined as the unit of measurement that makes $\sin(dx)\approx dx$.

As you have realized, for any unit of measurement you define as the basis of $\sin$, you'll have $\sin(dx)\approx \alpha\ dx$ for some $\alpha$. There is a specific unit of measurement for which $\alpha=1$. Call this unit a radian, and you're done.

Well, not really done because there are a lot of connections you still need to prove. But that's a basic idea that underlies much of math: You can define things however you like, but you can only define each once. If you want several different things to be true about a concept, you have to take one as the definition and prove the rest.

  • $\begingroup$ Will the downvoter care to explain? I tried in this answer to get to the core of the issue. There are technical issues such as the proof that the radian I defined is the same as the radian defined as $1/(2\pi)$ of a circle - which are important, but other answers have addressed that and it didn't seem like what the OP truly wondered about. $\endgroup$ – Meni Rosenfeld Jul 27 '16 at 11:32
  • $\begingroup$ π/2 isn't approximately 1 though $\endgroup$ – Kyle Delaney Jan 8 at 21:46
  • $\begingroup$ @KyleDelaney: $dx$ is meant to represent a very small value. (It's the symbol for a differential). I didn't say that $\sin(a)\approx a$ for every value of $a$; this is true only for $a \approx 0$. $\pi/2$ isn't close to 0, so it doesn't apply to it. Of course, the rigorous way to write this is $\lim_{a\to0}\frac{\sin(a)}{a}=1$ (or, for a more general definition for the unit, $\lim_{a\to0}\frac{\sin(a)}{a}=\alpha$). $\endgroup$ – Meni Rosenfeld Jan 17 at 18:00

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