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As an example, the function $\cos(\theta)$ represents the ratio of, relative to the angle $\theta$, the adjacent-side and hypotenuse of a right triangle. Strictly speaking, $\theta$ is measured in any units we want - it's the interpretation/definition of the $\cos(\theta)$ function relative to our chosen "units" for $\theta$ that matters. That's why it's just a matter of economics when choosing between radians, degrees, or anything else.

Of course, when finding arcs on a circle, it's convenient to define the angle $\theta$ in terms of the ratio between the radius of a circle and its circumference, $1/2\pi$ (e.g. when you've covered an angle $\pi$ in these units along the circumference of a circle, you've traveled $\pi/(2\pi)=1/2$ of the circumference). That way, calculating arc-lengths becomes simple multiplication by the angle $\theta$ (this obviously isn't true for degrees!).

However, I don't know how to interepret the power (Taylor/Maclaurin) series for trig functions like $\cos(\theta)$ and $\sin(\theta)$.

$$\cos(\theta)=1-\frac{\theta^2}{2}+\cdots,\,\,\,\,\sin(\theta)=\theta-\frac{\theta^3}{6}+\cdots$$

Why must we use radians in the above series representation? Why don't we use for the "units" of $\theta$, for example, the fraction of the entire circle that it covers (e.g. $1/4$ instead of $\pi/2$, $1$ instead of $2\pi$, etc.)? That would seem more natural and "unitless".

Arturo don Juan
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  • The fundamental reason we use $\pi$ as part of the measurement of angle is that the circumference and area of the unit circle are $2\pi$ and $\pi$ respectively. – Arkady Mar 07 '16 at 06:44
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    If you want to measure angles in terms of full turns instead of radians you are welcome to do so! Just substitute $\theta=2\pi r$ everywhere in the Taylor expansions. There will be powers of $\pi$ everywhere, but the formulas work all the same. Using radians keeps the coefficients as simple as they can be. The circle geometry just plays out the way it does. Basically because their derivatives are easier, when we measure in radians. See this thread for some nice arguments. – Jyrki Lahtonen Mar 07 '16 at 06:49
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    @Jake That's nice and all, but why in the $\cos(\theta)$ expansion? On the left we have an abstract representation, and on the right we have concrete numbers. How does one make the logical path from left to right? – Arturo don Juan Mar 07 '16 at 06:52
  • If you start with the power series definition, then the usual route is to show that there is a smallest $T>$ such that $\cos T = 0$, and we define $\pi = 2T$. Then we show periodicity and all the usual properties. – copper.hat Mar 07 '16 at 06:54
  • @JyrkiLahtonen I don't think I get you too well. We can't use any measurement we want because the right-hand side is a bunch of numbers (dimensionless), so we'll get different answers for different numbers, which doesn't agree with the "abstract/general representation" interpretation of $\cos(\theta)$. – Arturo don Juan Mar 07 '16 at 06:55
  • @copper.hat See my comment above. How can we define a specific set of units that are constrained? We can't define T to be anything else than an odd-multiple of $\pi/2$, because then the Taylor series wouldn't get the answer right. I'm guessing I'm just misunderstanding you though, so sorry if that's the case. – Arturo don Juan Mar 07 '16 at 06:58
  • Radians are defined by arc length/length of radius. So it is just a dimensionless number. Whatever units you measure the arc length and the radius with - feet, meters, furlongs,.... - will cancel out – Jyrki Lahtonen Mar 07 '16 at 06:58
  • @JyrkiLahtonen Yeah not those dimensions of course! Basically what I'm asking for is the reason for why we can't just plug in $\theta=90$ or $\theta=1/4$ in the power series of $\sin(\theta)$ and get exactly $1$ out of it. Like, when did the notion of arc-length, area, volume, etc., enter into the definition of $\theta$ in the abstract function $\cos(\theta)$? – Arturo don Juan Mar 07 '16 at 07:01
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    One word: derivatives. The coefficients of that power series come from higher order derivatives of the trig functions. And those derivatives are simple ($D\sin x=\cos x$ and $D\cos x=-\sin x$) only when we use radians. See the thread I linked to. If you are not familiar with derivatives, then study them a bit. – Jyrki Lahtonen Mar 07 '16 at 07:14
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    @ArturodonJuan: If you chose a different scaling for $\cos$, then there would be a different first zero $T'$ and a different value for $\pi'$. Just like when we use degrees, $\pi'=360 $ ($\neq 3.14 \cdots$). – copper.hat Mar 07 '16 at 07:18
  • Did any of you guys mentioned about $e^{i \pi} = -1$. These trigonometric function definitions are in great harmony with this exponential function, which gives "nice" results for radians. – crbah Mar 07 '16 at 07:37
  • Related: Why do we require radians in calculus?, Why does the derivative of sine only work for radians?, among many others. – Andrew D. Hwang Mar 07 '16 at 22:46
  • @ArturodonJuan: "On the left we have an abstract representation, and on the right we have concrete numbers. How does one make the logical path from left to right?" This answer shows a geometric way to do it, using curves known as involutes. – Blue Oct 11 '18 at 16:28

2 Answers2

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Think of radians as having a special identity multiplication property.

$radians\cdot radians=radians$. The 'unit' is unaffected unlike other units.

In fact, I learned in physics the other day that $radians\cdot meters=meters$, which means that 'radians' aren't even really units.

This allows the Taylor series for trig functions to make sense in terms of units.

Since degrees have the same property, it is simply how we derive the Taylor series of our trigonometric functions that determines that we use radians over degrees.

It has to do with derivatives, which makes a well defined Taylor series for our trig functions if we use radians.

If we use degrees, then the Taylor series for the trig functions are

$$\cos(\theta)=1-\frac{\left(\frac{90}{\pi}\right)^2\theta^2}{3!}+\frac{\left(\frac{90}{\pi}\right)^4\theta^4}{5!}-\dots$$

Of course, this makes sense and all, but the reason it is like this is because $\frac d{d\theta}\cos(\theta)=-\frac{90}{\pi}\sin(\theta)$, if $\theta$ is given in degrees. This results from simple chain rule, which you will learn about in Calculus I.

Simply Beautiful Art
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To go off what Simply Beautiful Art said,

This works because radians multiplied by a meters is still meters. This works for any translational distance. Why? Take the instance where we are moving $\theta_{\text{radians}}$ radians around a circle of radius $r$ meters. Let us take look at the case where $\theta_{\text{radians}} = \pi$. This is the same as going around half of the circle's circumference. Since $$\text{circumference} = 2\pi r$$ we know $$\frac{\text{circumference}}{2} = \pi r$$ Clearly, $\frac{\text{circumference}}{2}$ is in meters, even though it is the product of radians and meters. Thus, radians isn't a unit. Radians are the ratio of circle arc length to circle radius.

This is not true for degrees, however. If we take the product of degrees and meters, we get $180^{\circ} r$. Note how we write a $^\circ$, where we don't write a unit for radians. This is because degrees is a unit, not a ratio between distances. You might wonder why this is important. If you modify degrees so you multiply the raw number of degrees by some scalar, you can still get meters when multiplying by meters. You are right! However, we can't treat it the same as radians as this is like saying 180 dollars is equal to 180 pennies. They are different units. One is a ratio, and one is not. Thus, we actually need to cancel the degrees out by multiplying by $\frac{1}{1} = \frac{\text{ratio}}{\text{degrees}} = \frac{\text{radians}}{\text{degrees}})$.

Given an angles in degrees, we know the proportion we go around a circle is $\theta_\text{degrees} \cdot \frac{1}{360^{\circ}}$. Thus,

$$\text{circumference} = \left(\theta_\text{degrees} \cdot\frac{1}{360^{\circ}}\right)2\pi r = \frac{\theta_\text{degrees}}{180^{\circ}} \pi r$$

Thus, we need to multiply $180^{\circ} r$ by $\frac{\pi r}{180^{\circ}}$ to get a value in meters.

This is the same reason the power series is in radians. If we wanted it in degrees, we would need to factor in $\frac{\pi r}{180^{\circ}}$ when applying The Chain Rule.

andrewgazelka
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    Radians are as much an unit as degrees. – José Carlos Santos Oct 11 '18 at 16:47
  • @JoséCarlosSantos is my base reasoning wrong, though? I mean, aren't radians essentially ratios of (arclength) distance given a distance? – andrewgazelka Oct 11 '18 at 16:50
  • It is not correct to say that if you multiply radians by meters you get something totally different from what you get if you multiply degrees by meters. After all $1^\circ$ is equal to $\frac\pi{180}$ radians. It's as if you said that dividing meters by seconds is totally different than dividing meters by hours. – José Carlos Santos Oct 11 '18 at 17:02