Recently I've computed the ratio between arc and corresponding chord in a circle of given radius $r$ and noticed that it doesn't tend to $1$ as the angle $x$ approaches $0$ as expected, but to $\pi/180$. I used the formulas of arc ($\pi×r×x/180$) and corresponding chord ($2×r×\sin(x/2)$) in terms of angle and radius, where the radius is fixed and the angle is variable. I used the fundamental limit $\lim_{x\to 0}x/\sin(x)=1$ and found that $$\lim_{x\to 0}\frac{\text {arc}}{\text{chord}}=\lim_{x\to 0}\frac{\pi×r×x/180}{2r×\sin(x/2)}=\lim_{x\to 0}( \frac{\pi/180×(x/2)}{\sin(x/2)},$$ which tends to $\pi/180$.

Now, maybe it theoretically does make sense, but the good sense doesn't tell me the same thing. I mean, this whole thing I proved means that by making the angle (or the arc) close enough to $0$, we reach a point where the arc becomes smaller than the chord, which doesn't seem natural. Can someone explain me the link between the common sense and theoretical result, or whether I have a fault in interpretation or anything else? Thanks in advance! Edit: $x$ is in degrees.

  • $\begingroup$ The problem is most likely that (assuming no computational error) in mathematics the limit of anything involving angles is expressed in radians and not degrees. Convert your angles to radians and see if you get the expected limit now. $\endgroup$ – Allawonder Jan 8 '18 at 15:03

You must distinguish the sine function that assumes its argument in degrees rather than in radians, and

$$\lim_{d\to0}\frac{\sin_° d}d=\lim_{d\to0}\frac{\sin \dfrac{\pi d}{180}}d=\frac\pi{180}.$$

[$\sin_°$ denotes the sine applied to an angle in degrees.]

  • $\begingroup$ x is in degrees, but it is in degrees in both formulas, I don't get why you used it both in radians and degrees. Either I reffered to x as the angle in degrees or in radians, it wouldn't make sense if I did reffered to both versions as I was using the same angle value. $\endgroup$ – Anonymus Jan 8 '18 at 13:12
  • $\begingroup$ @Anonymus: which "both formulas" ? $\endgroup$ – Yves Daoust Jan 8 '18 at 14:26
  • $\begingroup$ @Anonymus: actually, the angle unit of $d$ doesn't matter at all, as $d$ is variable and tends to $0$. What matters is to understand the connection between the functions $\sin_°$ and $\sin$. $\endgroup$ – Yves Daoust Jan 8 '18 at 14:27

I presume you are using $x$ in degrees when you say that arc length is $r \cdot x \cdot \frac{\pi}{180}$. Lets be clear and rather write it as $x^\circ$.

The problem lies on the line where you say $ \lim_{x\to 0}\frac{\sin(x^\circ)}{x} = 1$. This is not true. The limit $\lim_{x\to 0}\frac{\sin(x)}{x} = 1$ if only angle $x$ is in radians.

Otherwise you may convert it into radians by multiplying with $\frac{\pi}{180}$:

$$\lim_{x\to 0}\frac{\sin(x^\circ)}{x} = \lim_{x\to 0}\frac{\sin(x \cdot \frac{\pi}{180})}{x} = \frac{\pi}{180}$$

  • $\begingroup$ If the angle is in degrees, why isn't in degrees at the denominator too? $\endgroup$ – Anonymus Jan 8 '18 at 13:16
  • $\begingroup$ @Anonymus Its the same value! If $f(x) = \frac{\sin(x^\circ)}{x}$ then $f(90) = \frac{\sin(90^\circ)}{90}$. Significance of degree/radian is only for trigonometric function. $\endgroup$ – samjoe Jan 8 '18 at 13:19
  • $\begingroup$ also note that $x^\circ = x\cdot \frac{\pi}{180} \text{radians}$ $\endgroup$ – samjoe Jan 8 '18 at 13:24
  • $\begingroup$ if x is in degrees, then shouldn't be the ratio $sin(x^∘)/x^∘$, and then transformed in radians as $sin(x*\pi/180)/(x*\pi/180)$ $\endgroup$ – Anonymus Jan 8 '18 at 13:51
  • $\begingroup$ @Anonymus As i said, $x^\circ$ has only significance for trignometric function. What is meant by $x^\circ$ notation is that interpret $x$ as degrees. If you don't like this, use Yves' notation (which is surely removes this confusion). $\endgroup$ – samjoe Jan 8 '18 at 17:27

Common sense and theoretical result are in agreement.

Working in radians presents no problem$$\lim_{x\to 0}\frac{arc}{chord}=\lim_{x\to 0}\frac{\pi\times r \times \frac{x}{\pi}}{2r\times \sin(\frac{x}{2})}=\lim_{x\to 0}\frac{r\times x}{2r\times \sin(\frac{x}{2})}=\lim_{x\to 0}\frac{\frac{x}{2}}{\sin(\frac{x}{2})}=\frac{1}{1}$$

When working in degrees, however, recall that $\pi=3.14159...$, the length of the semi-circumference of a circle measured in radii (radians), is equal to $180$, the same length measured in degrees. Hence$$3.14159...\neq180$$ but $$3.14159...radians = 180 degrees$$(Similarly, $1\neq1000$ but $1$ kilometer = $1000$ meters.)

We can work in radians or degrees but not both at once. Working in degrees we get finally$$\lim_{x\to 0}\frac{arc}{chord}=\frac{\pi}{180}=\frac{180}{180}=\frac{1}{1}$$ Arc and chord approach equality as $x$ tends to $0$.


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