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Why do radians *work* in calculus and higher maths? 
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#1
Jan2312, 01:54 PM

P: 6

Hello, I searched the forum and couldn't find this topic, so I'll try and describe my question.
I want to know why using radians work and while they are 'natural'. I think the key equation is... lim x→0 (sin x) / x = 1 ...because it makes the derivatives of trigonometric functions relatively simple. I'm not looking for a proof of the above limit  I know it can be proved various ways. What I want to know is why is works for radians. What property of radians satisfies this? Why does sin x ≈ x for small values? (I was thinking it might have to do with the Taylor series, but that relies on differentiating the function, so it can't be assumed.) I'm looking for an answer analogous to why the derivative of e^x = e^x (e.g. if you look at it as an infinite series, when you differentiate, the first time disappears, and each subsequent term becomes the one before it) This has been bugging me for some time. Please ask if you need any more clarification about what it is I'm asking Thanks 


#2
Jan2312, 02:34 PM

P: 1,395

most proofs start with sin(x) < x < tan(x), this means that in this picure, where OA = OB = 1, and there's a circular arc from A to B sin(x) = BC < length of arc AB < tan(x) = AD The length of the arc AB is defined as x radians. An angle of x radians means, that the ratio between the length of th arc and the radius is x, and here the radius is 1. 


#3
Jan2412, 10:24 AM

P: 475

In the days before calculators using radians made a lot of sense in many calculations involving rotation you could keep things as a proportion of 2 Pi and found at the end you'd have pi on both sides and just cancel it out.



#4
Jan2412, 12:21 PM

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Why do radians *work* in calculus and higher maths?
Here's another point in mathematics variables do not have to have units if a problem on a test defines [itex]f(x)= x^2[/itex] and asks you to find f(2), you would not need to ask if that is "2 feet" or "2 meters", or "2 kilograms". Similarly, if ff is defined as [itex]f(x)= sin(x)[/itex], and we are asked to find f(2), we should not have to ask "2 what" but to be correct, you had better put your calculator in "radian mode"! Why radians? In order to treat sine and cosine as functions, we have to define them for all real numbers. One way that is typically done, in precalculus and calculus, is to use the "unit circle": Construct the graph of [itex]x^2+ y^2= 1[/itex] on an x, y coordinate system. If t> 0, start at the point (1, 0) and measure counterclockwise around the circumference of the circle a distance t (If t< 0, measure clockwise). cos(t) and sin(t) are defined as the x and y values of the ending point.
Notice that the variable t, here, is NOT an angle at all! It is, rather, a distance around the circle. And, since this is a circle of radius 1, the total circumference is [itex]2\pi[/itex]. Of course, engineers tend to always think of sine and cosine as depending on angles so just to keep them happy, we 


#5
Jan2412, 12:57 PM

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Sometimes I wonder how it would work if math did try to use units. You'd end up with fun questions like "what is the sine of a second anyway?"



#6
Jan2512, 04:37 AM

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#7
Jan2812, 04:26 AM

P: 2

> I want to know why using radians work and while they are 'natural'.
If you convert (or start with) your angle measurements in radians then it turns out that: d (sin x) / dx = cos x d (cos x) / dx = sin x d (tan x) / dx = sec^2 x etc. You can work out the derivatives using degrees (we did it from 1st principles in my Alevel maths course about 22 years ago!) but you end up with "annoying constants to remember" that just cancel out if you use radian angle measure instead of degrees. So, ultimately, I think radians are more natural because they make the various results easier to remember! I can't find the relevant results to hand but I'll check later. 


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