3

Expanding question and question2, I was wondering why the geometric definition of trigonometric functions is well-defined.

Background

In my mind, the sine function $\mathrm{sin}(\theta): \mathbb{R} \to \mathbb{R}$ is defined as a composition of functions $$ \mathbb{R} \to \left [ 0, 2\pi \right ) \to \mathbb{S}^1 \to \mathbb{R} $$ defined by $$ \theta \mapsto \theta \bmod{2\pi} \mapsto \text{point $(x,y)$ corresponding to angle $\theta$} \mapsto \frac{y}{\sqrt{x^2+y^2}} $$ Here $\mathbb{S}^1$ is the unit circle.

It seems clear that $\mathbb{R} \to [0, 2\pi)$ and $\mathbb{S}^1 \to \mathbb{R}$ are well-defined. So we just need to show that $[0, 2\pi) \to \mathbb{R}^2$ is well-defined. When we first learn sine function, one "constructs" a right triangle with one angle $\theta$. By similarity of triangles, one can "construct" the right triangle to have hypotenuse of length 1. By constructing the right triangle in a standard way (one end of the hypotenuse is the origin, and one side lies on the $x$-axis), the other end of the hypotenuse is $(x,y) \in \mathbb{S}^1$.

Question

How could one "rigorously" construct a triangle with an angle $\theta$? To rephrase, given a real number $\theta \in [0, 2\pi)$, how can I associate bijectively a point in $\mathbb{S}^1$ so that the composition of functions above is the usual sine function that we know without trigonometric functions?

Of course, I can use the real and the imaginary part of $e^{i\theta}$ to associate a point on the unit circle, but I thought it would defeat the original purpose of defining sine "geometrically".

libofmath
  • 438
  • Is it okay to just say: starting at the origin, draw a line with angle $\theta$ until you hit the unit circle. $\sin \theta$ is the height of that line. I believe this is constructible in standard Euclidean geometry if $\theta$ is constructible. – Xodarap Apr 21 '21 at 01:17
  • @Xodarap I guess I should have been more clear with the question, but given a real number $\theta$, what line would we draw without cosine and sine? I somehow feel like trig functions and the definition of angle are somewhat circular when defined geometrically. – libofmath Apr 21 '21 at 01:30
  • The usual unit-circle definition as I recall it is that $\theta$ is the length of an arc of the circle. By the way, there is no requirement for a "mod $2\pi$" function; we are allowed to have an arc longer than $2\pi$. It will simply wrap around as many times as needed. And if $\theta$ is negative we simply take the arc in the opposite direction. – David K Apr 21 '21 at 03:39
  • In fact, looking at the linked questions, I see that the use of arc lengths was spelled out several times in various ways, so I'm wondering why there is still a question. Did you understand the use of "length of an arc" in this question, and can you explain why you are not satisfied with this answer? – David K Apr 21 '21 at 03:54
  • @DavidK: First, quoting from the answer: "The length is the smallest upper bound of the set of all lengths of polygonal paths moving in one direction along the arc." I understood it as the supremum of all polygonal path where polygonal path is a finite collection of lines where the two end points of them are on the circle. This makes sense. I still don't know given such arc-length, how would you pick a point in $\mathbb{S}^1$? (which I intuitively think of it as constructing the right triangle) – libofmath Apr 21 '21 at 04:33
  • Maybe it is enough. Should the argument be something like: for $\theta \in [0, 2\pi)$, there exists a point $(x,y) \in \mathbb{S}^1$ such that the arc-length defined by $(1,0)$ and $(x,y)$ is $\theta$? Not completely sure how to show existence. With this definition, can we show that the function $\mathbb{S}^1 \to \mathbb{R}$ defined by $(x,y) \mapsto \mathrm{arc}\text{-}\mathrm{length}((x,y) \text{ and } (0,1))$ is continuous? – libofmath Apr 21 '21 at 04:40
  • 1
    For existence we might need something like the construction of the real numbers themselves, that is, you assume that between the set of all points that are too close too $(1,0)$ along the circle and the set that are too far from $(1,0)$ there is a point that is at exactly the arc distance $\theta$. At some point during all this I think we have to make some assumption that allows the construction of points that are not at rational coordinates and are cannot be constructed by compass and straightedge, because almost all the points on the arc are like that. – David K Apr 21 '21 at 11:34

0 Answers0