What is an Angle?

As many people know intuitively, the number $\pi \approx 3.14$ is defined to be the ratio of a circle’s circumference to its diameter. Equivalently, we can consider it to be $1/2$ the circumference of a circle with radius $1$ (i.e. a unit circle). But what is the circumference of a circle?

To answer that question, we first recall that a circle has cartesian equation $x^2 + y^2 = 1$ . For the top half of the circle, we have the equation $y = \sqrt{1-x^2} =: f(x)$ . What is the length of this curve?

Let’s partition the base $[-1, 1]$ into $n$ equal subintervals with $x$ -coordinates given by $x_k = -1 + 2k/n$ for $k = 0,1,\dots,n$ .

We can approximate the curve by considering the lengths $L_k$ of each secant line connecting $(x_k, f(x_k))$ to $(x_{k+1}, f(x_{k+1}))$ .

Denoting $\Delta x_k := x_{k+1} - x_k = 2/n$ and $\Delta y_k := f(x_{k+1}) - f(x_{k})$ , each $L_k$ can be computed using Pythagoras’ theorem via

$\displaystyle L_k := \sqrt{{\Delta x_k}^2 + {\Delta y_k}^2} = \sqrt{1 + \left( \frac{\Delta y_k}{\Delta x_k} \right)^2 } \Delta x_k$

By definition,

$\displaystyle \lim_{n \to \infty} \frac{\Delta y_k}{\Delta x_k} = \lim_{\Delta x_k\to 0} \frac{f(x_{k}+\Delta x_k) - f(x_{k})}{\Delta x_k} = f'(x_k).$

Therefore, the arc length is defined, whenever $f$ is continuously differentiable on $[a, b]$ , as

$\displaystyle \lim_{n \to \infty} \sum_{k=1}^n L_k = \lim_{n \to \infty} \sum_{k=1}^n \sqrt{1 + \left( \frac{\Delta y_k}{\Delta x_k} \right)^2 } \Delta x_k = \int_a^b \sqrt{1+ (f'(x))^2}\, \mathrm dx.$

Here is the formal definition for the arc length.

Definition 1. Let $f : [a, b] \to \mathbb R$ be continuously differentiable, perhaps except at the endpoints. Assuming the integral exists, the arc length of $f$ is defined by

$\displaystyle \int_a^b \sqrt{1+ (f'(x))^2}\, \mathrm dx.$

Setting $f(x) = \sqrt{1-x^2}$ and simplifying, one obtains the length of the upper-half of the unit circle, defined by $\pi$ , as follows.

Definition 2. $\displaystyle \pi := \int_{-1}^1 \frac{1}{ \sqrt{1-x^2} }\, \mathrm dx$ .

This is where we obtain $\pi$ as the angle corresponding to half of a turn, and $2\pi$ corresponding to a full turn. Strictly speaking, this is an improper integral, but we will explore this later on.

This actually opens up an interesting discussion on what an angle is. Intuitively, it should be measuring the “separation” between two lines. How do we formalise this rigorously?

Our definition of $\pi$ , really, is the technical formulation of an arc length. The Greek letter $\theta$ is usually used to denote an angle, and in that spirit, we will define the function

$\displaystyle \theta : [-1, 1] \to [0, \pi],\quad \theta(r) = \int_{r}^1 \frac{1}{ \sqrt{1-x^2} }\, \mathrm dx.$

We can verify that this function is continuous and strictly decreasing in $r$ , and therefore, has an inverse $\theta^{-1}$ . A principal angle in $[0,\pi]$ , therefore, is (rather anticlimactically), any real number of the form $\theta(r)$ . It is acute if $0 < \theta(r) < \pi/2$ and obtuse if $\pi/2 < \theta(r) < \pi$ .

Any other angle is therefore a real number of the form $k\pi + \theta(r)$ , where $k$ is an integer, and $\theta(r)$ is a principal angle. Observe that $\theta(1) = 0$ and $\theta(-1) = \pi$ . Using arguments involving symmetry, we could even establish that $\theta(0) = \pi/2$ .

In fact, the way we defined $\theta$ is intimately connected with the usual trigonometric functions. We will now formally define them here.

Definition 3. Let $\alpha$ be an acute angle. The cosine and sine of $\alpha$ respectively are defined by

$\displaystyle \cos(\alpha) := \theta^{-1}(\alpha),\quad \sin(\alpha) := \sqrt{1 - \cos^2(\alpha)}.$

Almost by construction, we obtain $\sin^2(\alpha) + \cos^2(\alpha) = 1$ . Furthermore, we recover the usual characterisations of $\sin(\alpha) = \text{(opposite)}/\text{(hypotenuse)}$ , $\cos(\alpha) = \text{(adjacent)}/\text{(hypotenuse)}$ , and $\cos(\alpha) = \sin(\pi/2-\alpha)$ (this yields $\cos(\alpha) = \sin(\pi/2 + \alpha)$ .

From this definition and its immediate corollaries, we can derive all relevant formulae for $\sin$ and $\cos$ , including that they have inverses on $[-\pi/2,\pi/2]$ and $[0,\pi]$ respectively. In fact, this definition confirms our suspicion that $\theta = \cos^{-1} \equiv \arccos$ , since $\cos^{-1}(1) = 0, \cos^{-1}(0) = \pi/2, \cos^{-1}(-1) = \pi$ .

Most importantly: at last, we have the foundational logical grounds for trigonometry, from which we define $\tan$ and the co-circular functions $\sec$ , $\csc$ , $\cot$ .

In the next few posts, we will finally prove that $\sin' = \cos$ . The plan initially was to do this in a discussion in real analysis, but it turns out that we have the necessary tools for the job.

—Joel Kindiak, 27 Oct 24, 1109H

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What is an Angle?

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