How do you prove arc length is greater than chord length?

Graphically, it's obvious that given two different points $a$ and $b$ on a circle of radius $r$, the linear distance (chord length) from $a$ to $b$ is less than the arc length from $a$ to $b$. How do you prove this?

I know (because I've proven it before) that if the angle between $a$ and $b$ is $\theta$ in radians, then the arc length $s$ is $s = r \theta$, and the chord length $d$ is $d = r * \sqrt{2 - 2\cos(\theta)}$.

The proof should be simple, but somehow it's not apparent to me how to prove $d < s$.

Solution 1:

The question is: What is arc length anyway? One possible definition is: For $\epsilon>0$, consider all point sequences $a=x_0,x_1,\ldots, x_n=b$ such that all $x_i$ are on the circle (or in any other set $S$ with respect to which we want to measure an arc length) and the distance $d(x_i,x_{i+1})$ between $x_i$ and $x_{i+1}$ is $<\epsilon$. Let $d_\epsilon(a,b)$ be the infimum of $d(x_0,x_1)+d(x_1,x_2)+\ldots +d(x_{n-1},x_n)$ over all such sequences. Then the arc length from $a$ to $b$ can be defined as $\lim_{\epsilon\to 0} d_\epsilon(a,b)$.

By the triangle inequality, $$d(x_0,x_1)+d(x_1,x_2)+\ldots +d(x_{n-1},x_n)\ge d(x_0,x_n)=d(a,b)$$ for all considered point sequences. Hence $d_\epsilon(a,b)\ge d(a,b)$ for all $\epsilon$, and hence the same inequality holds for the limit. To show that the arc length is in fact strictly greater, pick any point $c$ on the arc between $a$ and $b$, and note that all point sequences with step width $<\epsilon$ have an intermediate point $x_i$ close to $c$ (that is: with $d(x_i,c)<\epsilon$). Hence in the limit we obtain that the arc length from $a$ to $b$ is $\ge d(a,c)+d(c,b)>d(a,b)$, where the last strict inequality follows from the fact that $c$ is not on the straight line segment $ab$.

Solution 2:

You wish to prove that (note the missing $r$): $$r\theta >r\sqrt{2-2\cos\theta}$$ Now use the half angle identity, to get: $$r\sqrt{2-2\cos\theta}=2r \sin(\theta/2)$$ And since for $x>0$ we have $\sin x < x$, $$2r \sin(\theta/2)<r\theta$$