Why Use Radians instead of Degrees?
Date: 09/25/2001 at 23:33:55 From: Kanishka Joshi Subject: Radians? Hello everyone on the Math Forum team, For the past few days, I have been trying to figure out why we use radians instead of degrees. Some points that I have realised are as follows: 1. Radians have evolved from a circle. 2. A radian is a dimensionless quantity (this is probably the crux) 3. Radians provide simple formulae in terms of derivations of arc lengths and sector areas. But the problem still remains. For example, consider the Maclaurin's theorem. From the theorem we derive the Taylor series. Then suppose we have the series of sin(x), which goes as follows. sin(x) = x- (x^3)/3! + (x^5)/5! - ... Notice a strange thing here. To the right of this series are pure numbers, and suddenly to the left we get the answer in radians. Why is this happening? (Please send me the proof of Maclaurin's theorem, if possible).
Date: 09/26/2001 at 12:17:04 From: Doctor Rick Subject: Re: Radians? Hi, and welcome to Ask Dr. Math. All angle measures can be said to be dimensionless. A radian is the ratio of an arc length to a radius, and the ratio of two lengths is dimensionless. A degree is 180/pi radians, and the constant 180/pi is dimensionless, so a degree is also dimensionless. Trig functions are also by nature dimensionless. They can be defined as ratios of sides of a right triangle. Again, the ratio of two lengths is a dimensionless quantity (a "pure number"). Thus there is nothing surprising about the dimensional analysis of the Taylor series for sin(x). The variable x represents a (dimensionless) angle, so the Taylor series represents a dimensionless quantity, as does the sine of x. You can see that dimensional analysis isn't much use in investigating what is special about radians as a measure of angles. I consider the basis for the unique significance of radian measure to be this: The derivative of sin(x) is cos(x) (without any scale factor) when the angle is in radians. The MacLaurin series derives from this fact. Without going through the proof that the derivative of sin(x) is cos(x), we can see that the scale factor is 1 when the angle is measured in radians. Consider a unit circle and an angle near zero. | | *********** ***** | ***** *** | *** ** | ** ** | ** * | * * | * (x,y) * | 1 ---+ * | ------- * s * | ------ y|* ----*--------------------+-------------------+*---- * | x * * | * * | * * | * * | * ** | ** ** | ** *** | *** ***** | ***** *********** | | The sine of the small angle is y/1 = y. The radian measure of the angle is s/1 = s. For small angles, y is approximately equal to s; therefore the sine of the angle is approximately equal to the angle. sin(s) ~= s The slope (or derivative) of sin(s) with respect to theta is therefore close to 1 for small angles. If we know that the derivative of sin(s) is a*cos(s) for some scale factor a, then the sine of a small angle will be close to a*cos(0) = a. Therefore the scale factor must be 1 when the angle is measured in radians. If you used a different angle measure, such as degrees, the angle would not be equal to s; it would be some constant (such as 180/pi) times s. The derivative of sin(x) would then be that constant times cos(x). You can find discussions of the derivation of the MacLaurin series for sin(x) in our Dr. Math archives. Here is one such discussion: Power Series for Sine and Cosine http://mathforum.org/dr.math/problems/mark.10.12.00.html Does this satisfy your curiosity on this point? If you have more questions or thoughts, we'd be glad to talk about them. It's fun to discuss math with someone who is asking "why?" - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.