Physicist: Because calculus.

When you first start doing trigonometry the choice between radians, degrees, turns, or hexacontades is a matter of personal preference.  Most people use degrees because most other people use degrees (and other people seem pretty on the ball).  But when you get to calculus using radians is the most natural choice; anything else is just a headache waiting to happen.

To see why you have to get to know the unit circle.

The unit circle.  “Unit” means “1” and refers to the radius.

Start with a unit circle with a horizontal line through it and a radius (“a radius” means a line from the center to the edge somewhere).  The definition of sine and cosine of the angle between the radius an the horizontal line are in the picture above.  SOH CAH TOA is easy in this case because the hypotenuse is 1.

When you use radians you’re describing the angle by using the length of the arc it traces out on the edge of the unit circle.  The circumference of a circle or radius R is 2πR, so (since R=1 on the unit circle) the full circle is 2π radians around.  That is: 2π radians = 360 degrees.

You’ll notice that when the angle is very small (and measured in radians) the value of sin(θ) and the value of θ itself become very nearly equal.  Not too surprisingly, this is called the “small angle approximation” and it’s remarkably useful.

For small angles sin(θ)≈θ, but only when that angle is described in radians.

So for small values sin(θ)≈θ or $\frac{\sin(\theta)}{\theta}\approx 1$.

In fact, in the limit as the angle approaches zero they are equal, or in mathspeak: $\lim_{\theta\to0}\frac{\sin(\theta)}{\theta}= 1$.  When someone says “in the limit as ___ approaches ___” it means they’re about to talk about calculus (and true to form…).  All of the calculus around trig functions can be based on the fact that $\lim_{\theta\to0}\frac{\sin(\theta)}{\theta}= 1$.  For example, one of the more important things in the world (that’s not quite sarcasm) is the fact that $\frac{d}{dx}\left[\sin(x)\right] = \cos(x)$.

The derivative of a function is $\frac{d}{dx}f(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}{h}$, so:

$\begin{array}{ll} \frac{d}{dx}\left[\sin(x)\right]\\[2mm] = \lim_{h\to0} \frac{\sin(x+h)-\sin(x)}{h}\\[2mm] = \lim_{h\to0} \frac{\sin(x)\cos(h)+\sin(h)\cos(x)-\sin(x)}{h} & *\\[2mm] = \lim_{h\to0} \frac{\sin(x)\left(\cos(h)-1\right)+\sin(h)\cos(x)}{h} \\[2mm] = \lim_{h\to0} \frac{\cos(h)-1}{h}\sin(x)+\frac{\sin(h)}{h}\cos(x) \\[2mm] = \lim_{h\to0} \left[-\sin(h)\frac{\sin(h)}{h} \frac{1}{\cos(h)+1}\right]\sin(x)+\frac{\sin(h)}{h}\cos(x) & ** \\[2mm] = \left[-0\cdot 1 \cdot \frac{1}{1+1}\right]\sin(x)+\cos(x) \\[2mm] =\cos(x) \end{array}$

That doesn’t look like a big deal, but keep in mind that all of trigonometry is just a rehashing of sine.  For example, $\cos(x)=\sin\left(x-\frac{\pi}{2}\right)$ and $\tan(x)=\frac{\sin(x)}{\cos(x)}=\frac{\sin(x)}{\sin\left(x-\frac{\pi}{2}\right)}$.

If it weren’t for the fact that (when using radians) $\sin(x)\approx x$ we wouldn’t have $\frac{d}{dx}\left[\sin(x)\right] = \cos(x)$.

It’s not the end of the world if you try to do calculus with degrees (it’s close), it’s just that the result is multiplied by an inconvenient constant.  For example, if you’re using degrees: $\frac{d}{dx}\left[\sin(x)\right] = \frac{\pi}{180}\cos(x)$.  Same thing happens when you differentiate cosine or tangent or whatever.  It’s a lot easier to understand why if you look at a graph.

x, sin(x) in radians, and sin(x) in degrees.  Notice that when measured in radians sin(x)≈x for small x, and when using degrees sine is really stretched out.

Clearly when using degrees the slope (derivative) of sine at zero is not 1, it’s much smaller (it’s 2π/360 in fact).  If you don’t want any weird extra constants, then you need to use radians.  But if you don’t mind them, then you be you.  You can certainly use degrees or whatever, but you need to be careful with all those extra 2π/360’s.

* This is a trigonometric identity.

** That isn’t obvious:

$\begin{array}{ll} \frac{\cos(h)-1}{h} \\[2mm] = \frac{\cos(h)-1}{h}\frac{\cos(h)+1}{\cos(h)+1} \\[2mm] = \frac{\left(\cos^2(h)-1\right)}{h\left(\cos(h)+1\right)} \\[2mm] = \frac{-\sin^2(h)}{h\left(\cos(h)+1\right)} \\[2mm] = -\sin(h)\frac{\sin(h)}{h} \frac{1}{\cos(h)+1} \\[2mm] \end{array}$

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### 8 Responses to Q: Why radians?

1. Charles Nicholas says:

I just had to comment on this. In High School (1961) my first teacher taught the study of trigons. I loved it and made all A’s. Second half semester the main teacher came back and taught “identities”. No mention of the unit circle or why you would use the unit circle. Made all F’s and was clueless. When I taught Trig as a graduate student I made sure BOTH views were taught and why electrical engineers liked the unit circle while the Forestry students loved trigons.

2. Anthony Rose says:

*Really* useful – thank you so much!

Why radians? I understand that the first commentator tried to explain but most probably lost you in the first paragraph if not sooner.
Radians connect algebra to trigonometry to geometry like nothing else can.
One radian measures the angle when the radius of a circle is traced out on its edge. Radians relate right back to Pi. (3.14………….)
The first thing to look at to get a closer view of this is a math object called the “Spiral of Squares”
This spiral is easy to construct but it reveals some relationships that are very hard to explain. For instance when you begin this spiral and measure the distance between the coils you start out with a measurement that is very close to Pi. The further out you go on this spiral the closer this measurement gets to Pi.
The next one is just as, if not more mysterious than the first. When you measure the angles between the whole number intervals on the spiral you will find that the angles between them approach 2 Pi radians.

4. Yanis Zidelmal says:

Radians are not only convenient, in some cases they are the only correct choice.
Let me give you an example. We were told at school that the value of pi was first
approximated by drawing a regular polygon inscribed in a circle and dividing its perimeter by the radius. The more sides this polygon had, the closer to pi this value was.
So when the number of sides tends towards infinity, the perimeter of the polygon tends towards that of the circle.
here are the calculations:
let the natural number n denote the number of sides of the polygon and the positive real number r the radius of the circle (angles are taken in radians) the perimeter of the polygon is then given by the formula :
p = 2nr * sin(pi/n).
Let’s then find the limit of p as n approches positive infinity (to resolve the indeterminate form, we define N as 1/n and we use the definition of derivatives.)
Depending on the unit we use, the limit l we find is different:
l = 2r*pi which is a basic geometry rule, but when we use degrees the result is totally different (360r).
I can’t really explain why this problem isn’t unit agnostic as I first thought, though.

5. Zidelmal Yanis says:

IGNORE THE ABOVE STATEMENT. You get 2r*pi in both cases. I got them wrong because when I used degrees I omitted multiplying by pi/180. this formula is indeed unit-agnostic.

6. Usha Kotelawala says:

Beautifully written piece. I especially like many of your comments in parentheses. My only suggestion would be that you consider increasing font size on the equations – that font size in fractions is beginning to seem smaller 🙂 and I do slowly verify as I read.

I came to this question considering trigonometry for a population of non-STEM majors. You make excellent points and I’m not a person who would want those additional constants although somehow the 180 seems less acceptable than pi – is this esthetics?

My sense is that random collection of college graduate subway riders would struggle with correctly describing what a radian is a year after taking math but they may be able to talk a bit about degrees. I wonder if I should consider the idea six pizza slices and a tiny extra piece?

Thanks.

7. Scott Halvorson says:

Because *of* calculus. We need to give grammatically incorrect Internet memes a quick and very painful death. Let’s not perpetuate them any longer.

8. 1/(Scott Halvorson) says:

So prescription. Much wrong.