The two most commonly used measures for angles are degrees and radians. There are 360 degrees in a full circle (a right angle is 90 degrees), and radians in a full circle (there are radians in a right angle), so there are about 57 degrees in a radian.

Students typically learn about degrees before they learn about radians, which brings up the question: Why learn about radians if degrees are good enough for measuring angles? There are two reasons, but both are grounded in the fact that the radian is a *unitless* measure, which I’ll explain in the following paragraph.

The radian is defined to be (see the diagram) the ratio of the length of the arc of a circle (indicated by in the diagram) to the length of the radius of the circle (indicated by in the diagram), where **each length is measured in the same unit**. Therefore, when you divide by to get the radian measure of the angle, the units for the two lengths cancel, and you end up with a measure that has no units:

When is the entire circumference of the circle, the corresponding angle is that of the entire circle. Since the circumference of a circle is , the angle of a full circle is .

Unless there is a good reason, one is free to use any measure whatsoever for angles, such as degrees, radians, or one of the less common ones. There are two good reasons to use radians:

1. If you are working with the derivative of a trigonometric function, then it is preferable to use radian measure for angles, because then derivative formulas (and limit formulas) are easier. For example, using radians, the derivative formulas for sine and cosine are:

and

However, if degrees are used, the derivative formulas for sine and cosine are:

and

The latter pair of formulas are a pain because of the additional factors, so why wouldn’t we use the simpler first pair of formulas? It’s the laziness principle that is used so often in mathematics … we always do the easiest thing. Therefore, we use radians whenever we are dealing with derivatives of trigonometric functions. (The derivatives of the other trigonometric functions are similarly simplified if radians are used … work them out, if you wish!)

2. In describing rotational motion, one rearranges the definition of radian measure to relate the linear displacement to the angular displacement:

Differentiating both sides of this relation with respect to time (and assuming a constant radius), one obtains a relation between the linear velocity and the angular velocity for motion in a circle:

If is measured in radians, then is measured in radians per second, and then the speed comes out in a natural unit; for example, if is measured in metres, then comes out in m/s. However, if were measured in degrees, then the units for would be m·degrees/s, which is an unnatural unit, and hard to understand.

Differentiating the relation from the previous display with respect to time once more, we get a relation between the linear acceleration and angular acceleration for an object moving in a circle:

A similar comment about units is also relevant here: Measuring in radians/s^{2} means that if is measured in metres, then comes out in natural and easily understood units, m/s^{2}. But if is measured in degrees/s^{2}, or in revolutions/s^{2}, then the units for come out an awkward mess.

Finally, there is something arbitrary about degree measure (why 360? why not 100, or 1000, or some other round number?) that makes it aesthetically not pleasing. On the other hand, there is something aesthetically pleasing about the definition of radian measure, as it is a simple ratio. This makes it a *natural* measure, wouldn’t you say?

Which leads me to think about the use of the word “natural” in mathematics … it would be nice to discuss this another time.

p.s.: By the way, did you notice that the factors in the formulas for the derivatives of sine and cosine in degree measure are the same as the factor that is used to convert degrees to radians? This makes sense if you think about the conversion from degrees to radians as a scale change (i.e., a change of variable) and then use the chain rule for differentiation.

**Update**, 20 September 2012: Ted Burke notes in a comment to this post that Euler’s formula

is valid only when the angle is measured in radians. If the angle is measured in degrees, Euler’s formula would have to be written in the considerably uglier form

This is another situation in which radian measure is more convenient than degree measure.

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This a very nice post and came up when I posed the question to my colleagues (the type of basic question which you sometimes have to be old(er) to pose) ‘why radians?. I started thinking about the answer when I got lazy and did a google instead. I am teaching a class on ac circuits and phasors and realized that I wasnt sure (had forgotten?) why I had to use the radian setting on the calculator to plot the sine wave, i.e. why the sine formula is x(t) = sin( 2 PI f t)

Thanks, Richard! The word “phasor” brings back some fond memories. All the best in your teaching!

Thanks, a very good explanation! One other thing that I was reminded of while reading your post is that radian angles make Euler’s formula particularly elegant.

Thanks, Ted! And it’s a good point about Euler’s formula, which I’ll add to the post. All the best!

I agree that students would benefit from learning radian measure earlier. Think of how beautiful the interior angle sums for polynomials are in radians! Triangle: pi, Quadrilateral: 2pi, Pentagon: 3pi… My precal kids got so flustered that I ended up making radian scale protractors for them! You can check them out on my website, proradian.net.

Radian-scale protractors … very cool!

Reblogged this on jensilver.

Reblogged this on Singapore Maths Tuition and commented:

Why radian is preferred in Math? Very good explanation here…

Very nice explanation! So radians are used because it’s more beautiful!

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(0) equals 0. But isn’t the same true for degrees? As dx approaches 0 degrees then sin(dx degrees) still approaches 0. But I’ve come to the understanding that sin(dx degrees) approaches 0 almost 60 times slower, so if sin(dx radians) can be replaced with dx then sin(dx 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 sin gets very small. Mathematically we can see that sin(.00001) is pretty darn close to .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 sixths) approaches dx when it gets very small, but we know we’d have to replace sin(dx sixths) with pi/6 dx sixths when differentiating. So how do we know that radians work out so magically, and why do they?

These are good questions.

One answer to why radians work out so nicely is that you just do the calculation (of the derivative of the sine function) in arbitrary angle units, and then see which one comes out the simplest. As you suggest, you could use “sixths” or any other angle units. When you try this, you find that the derivative formula comes out simplest for angles measured in radians. This is motivation to use radian measure in calculus, but if you wished you could use any other angle measure you wish, provided that you accepted the specific derivative formula for the sign function that comes with that angle measure.

You can find the derivation of the derivative of the sine function in many places; search for a derivation involving a limit calculation. If you go through the details carefully, you’ll note that the limit of sine theta divided by theta is 1 only when theta is measured in radians. For other angle measures, this limit is not 1, and so the derivative formula must be multiplied by this factor.

OK, that’s one answer. Another answer is geometrical. Think of the slope of a line, which is defined as rise divided by run. If you change the units for the run, you will get a different value for the slope.

A derivative can be interpreted as the slope of the tangent line to the graph of the function in question. Using different angle measures amounts to changing the units of the run, which changes the slope. The units for the slope are (rise units)/(run units). Changing the angle measure changes the units of the run, but not the rise.

Does this make sense?

Yes, I understand all that about slopes, but that’s not really what my question was about. I understand why it would be different depending on which units you use. I just don’t understand how to mathematically prove that radians are those perfect units. I guess I need to look further into proving limits.

OK, I get your question now. There is a geometric argument that you might find compelling:

Draw a circle centred at the origin. Now draw a ray from the origin angled off into the first quadrant somewhere, intersecting the circle. The purpose of doing this is to convince ourselves that the limit of sin (dx) divided by dx as dx approaches 0 is 1.

Drop a vertical line segment from the point of intersection of the ray and the circle down to the x-axis. This produces a right triangle, formed from the vertical line segment (of length Y), the line segment along the x-axis (of length X) and the segment along the radius of the circle (of length R). Label the angle as dx. Then in the right triangle, sin (dx) = Y/R.

Note that the arc of the circle subtended by the angle is not very different in length from Y; furthermore, the approximation gets better and better as the angle gets smaller. Thus, in the limit as dx approaches 0, Y can be replaced by the arc of the circle subtended by the angle, which we can call S. Thus, as dx approaches 0, sin (dx) gets closer and closer to S/R.

But, if we measure the angle in radians, then S/R = dx, an exact relation that is the definition of radian measure. Thus, in the limit as dx approaches 0, sin (dx) approaches dx, and so sin (dx) divided by dx approaches 1.

If we used some other angle measure, then we would have to modify S/R = dx by some factor, and the same factor would appear in the derivative formula.

Okay, I get it now, thank you. And I also found a detailed proof here: http://oregonstate.edu/instruct/mth251/cq/Stage4/Lesson/sinProof.html

Hello Mr. Bushell,

If you can point to an error at a specific location in the post, then I’m happy to hear it. Otherwise take your verbal abuse elsewhere.