Why is differentiation and integration impossible with the angle expressed in degrees? Why the angle have to be expressed in radians?

I have wondered about this my whole... time doing those differentiations and integrations(Hi)

- October 14th 2008, 05:20 AMssadiWhy differentiation and integration is impossible with angle expressed in degrees?
Why is differentiation and integration impossible with the angle expressed in degrees? Why the angle have to be expressed in radians?

I have wondered about this my whole... time doing those differentiations and integrations(Hi) - October 14th 2008, 05:29 AMmr fantastic
- October 14th 2008, 05:37 AMssadi
- October 14th 2008, 05:39 AMmr fantastic
- October 14th 2008, 05:49 AMssadi
- October 14th 2008, 06:01 AMPlato
- October 14th 2008, 08:14 AMssadi
I know the definition of radian. But that does not help much.

I found the explanation in wikipedia

Somebody out there must understand the stuff better, all it succeded to do was leaving my eyes sore. :D

But I really wanna understand...

Why is the derivative of sin(x) only cos(x) when x is measured in radians? Algebra man 18:56, 18 March 2007 (UTC)

To avoid all ambiguity, let sinrad denote the sine function whose argument is in radians, and likewise for cosrad, so d/dxsinrad(x) = cosrad(x). Define sinscaleda(x) = sinrad(2πx/a), and again likewise for cosscaled. So, for example, sinscaled360(x) gives the sine of angle x measured in degrees. Now, by the chain rule, d/dx sinscaleda(x) = d/dx sinrad(2πx/a) = cosrad(2πx/a) d/dx(2πx/a) = (2π/a) cosrad(2πx/a) = (2π/a) cosscaleda(x), This only equals cosscaleda(x) if a = 2π. I hope this qualifies as a "reason". --LambiamTalk 19:25, 18 March 2007 (UTC) The working out of the derivative of sin(x) requires the sine of small angles rule which states that sin(x)~x only when x is in radians. Alexs letterbox 06:41, 19 March 2007 (UTC) Which is quick to show informally. Define a radian as 1/(2π) of a circle. Start from a point x = r on the positive x axis, and draw a very short arc, moving through θ radians counter-clockwise. Then draw a line from the top of that arc to the origin. This gives an almost-triangle, except that the short side is an arc, not a line. For a short arc that's negligible, though, so we can pretend that the arc is a line, and that our shape is a triangle. (It's actually a sector of a circle, like a slice of cake.) The long sides are both the same length r, because the shape is a sector of a circle. The arc has a length of θr. (If it were the entire circle, then θ would be 2π, so we would have the familiar formula for the circumference of a circle, 2πr.) The angle θ was very small, and the angles of a triangle should sum to 180 degrees; this means that the other two angles are almost ninety each, so we have a right triangle. The sine of θ is equal to opposite/hypoteneuse = θr/r = θ. If we were using different angle units, and not radians, then the formula for the arc length would have been different. The first answer sort of begs the question, I think. I hope this is more convincing. —The preceding unsigned comment was added by 71.192.58.216 (talk) 10:03, 19 March 2007 (UTC). All the information is in the preceeding answers; I shall attempt a synthesis.

- The derivative of a function
*f*at a point*x*is the ratio of a small displacement in input from*x*divided into the resulting small displacement in output from*f*(*x*). - Therefore sin(2
*x*) will have a derivative twice as large as sin(*x*), for example; and in general the derivative will depend on the units of input. (This is true for any non-constant function.) - At
*x*= 0, cos(*x*) is exactly 1 no matter what input units we use. - Therefore we can have only one possible choice of input unit for the sine function if its derivative is to match.
- Radian measure (arc length) works, as shown by a geometric argument.
- Therefore radian measure is the unique measure permitting the derivative of sin(
*x*) to be cos(*x').*

This line of reasoning does not prove that the correspondence holds for all values of*x*, but it does show that*if*it is to hold at all, we must use radians. (I assume that anyone who could handle the general proof wouldn't be asking this question.) --KSmrqT 23:36, 19 March 2007 (UTC) - The derivative of a function