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It seems to me that radians have lots of very special properties that allow us to do maths with trigonometric functions. When I first came across radians, I was led to believe that they were designed purely to make finding areas and arc lengths of sectors more easy. But then I discovered that:

$$\frac{d}{dx}\sin x=\cos x$$ Clearly this only works when using radians. If you used degrees and found the gradient of $\sin x$ at $x=0$ you would get that gradient $=\frac{\pi}{180}^{\circ}$. So my question is: why does differentiating $\sin x$ give you $cosx$ when you are using radians? What makes radians so special in this respect?


marked as duplicate by N. F. Taussig, Andrew D. Hwang, A.P., Jonas Meyer, Johanna Apr 28 '15 at 17:03

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  • $\begingroup$ Heuristically, you can think of using radians as the most natural form of sin and cosine. The Taylor expansions are literally defined in radians, because its the most natural way. Changing to degrees results in a change of variables, because degrees aren't natural. They are a really bad approximation of a calendar year ;) $\endgroup$ – Zach466920 Apr 28 '15 at 15:34
  • $\begingroup$ Radians are a natural measure of an angle. If you take a unit circle, the the angle in radians of a point is the distance measured on the unit circle (not the straight line distance) between $(1,0)$ and the point in question (in an anti clockwise direction). $\endgroup$ – copper.hat Apr 28 '15 at 15:38
  • $\begingroup$ I'll have a look at the linked question. I really wanted to know specifically why we need radians when we differentiate and intergrate, as in a proof or demonstration of that fact. $\endgroup$ – bnosnehpets Apr 28 '15 at 15:39

No matter which unit for angles you use, the derivative of the sine will be some multiple of the cosine.

The radian measure is chosen as the angle measure that makes the proportionality constant $1$. If it was a different angle measure that did this we would be working with that as the natural angle measure instead of radians.

Geometrically think of the sine of a very small angle, that is, the ratio between short side and the hypotenuse in a very thin right triangle. We can scale everything so the hypotenuse has length 1; then the sine is simply the length of the short side.

So if we want $\sin'(0)=1\cdot\cos(0) = 1$ we had better choose our angle measure such that the measure of a small angle is close to the length of that short side.

Now, if we draw a circular arc next to the short side with the hypotenuse as the radius, until it reaches the extension of the long leg of the triangle, this will almost coincide with the short side -- and the thinner the triangle is, the better does it match.

And "arc length of a unit circle" is clearly additive on angles, so it will work as an angle measure. Therefore it is the angle measure that makes the derivative of the sine at 0 unity.

  • $\begingroup$ Ok, so its to do with the fact that radians is strictly a measure of arc length? $\endgroup$ – bnosnehpets Apr 28 '15 at 15:44
  • $\begingroup$ @user75473: It depends on what you're asking. I'm supposing that you take it as given that radians are defined as arc length, and are asking why that definition happens to produce an angle measure with nice differential properties. If you have a different definition of "radian", the answer won't necessarily involve geometry. $\endgroup$ – Henning Makholm Apr 28 '15 at 15:49
  • $\begingroup$ Yes, that is what I'm asking. By using arc length we are comparing like for like when we compare the length of the short side to the length of the arc. That answers my question, thanks. $\endgroup$ – bnosnehpets Apr 28 '15 at 15:54
  • $\begingroup$ but we essentially are finding the ratio of the arc length to the radius, so the angle in radians is a ratio not a length $\endgroup$ – danimal Apr 28 '15 at 15:59
  • $\begingroup$ Ok, but if we are using a unit circle, then the small side can also be considered a ratio. So aren't we still equating equal things? $\endgroup$ – bnosnehpets Apr 28 '15 at 16:10

to prove the differential of sin(x) is achieved using the fact that $$\lim_{x \rightarrow 0} \frac{sin(x)}{x} = 1$$ to prove that we say that for a small enough absoulte(x) $$sin(x)\leq x \leq tan(x)$$ this is proved using a bit (complicated) geometry, that is only true assuming x is in radians. enter image description here

KH is sin(x)

the arc KA is x

and LA is tan(x)

this however is not a proof, the full proof is a bit more complicated.

  • $\begingroup$ We can think of radian measure as the arc length of a unit circle. By the same token, we can think of degree measure as the arc length of a circle of radius $\frac{180}{\pi}$. The limit of the ratio of to height of point $K$ to the arc length (in degrees) becomes $$\lim_{x \rightarrow 0} \frac{\frac{180}{\pi}\sin(x)}{x} = 1$$ implying that $$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = \frac{\pi}{180}$$ This is why (where $x$ is in degrees) $$\frac{d}{dx}\sin x = \frac{\pi}{180}\cos x$$ $\endgroup$ – John Joy Apr 28 '15 at 16:20

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