Proving Trigonometric “Definitions”

Question

The expression trigonometric “definitions” refers here, rather narrowly, to statements expressing stable relations between the sides of right triangle.

Thus, for instance, the traditional definition of sine supposes that one has a demonstration that the ratios between opposite side and hypothenuses are independent of the size of the right triangle and are dependent instead on the amplitude of the angles of the right triangle.

My question is the following: How these stable relations, simply assumed by most trigonometric “definitions,” can actually be demonstrated?

A very similar question has been posted here but I remain unsatisfied by most answers — which seem to convoke either unnecessary complex mathematical objects or simplistic historical accounts.

I guess, what I am looking for, is something like a geometrical proof — but am open to others!

Answer 1

For proving that the sine function is well-defined, one uses two theorems from Euclidean geometry combined with a tiny bit of algebra. The two theorems are:

Theorem 1: In any triangle, the sum of the angles equals $\pi$.

I don't actually care about the numerical value of the sum, so perhaps one can state Theorem 1 more classically: the sum of the angles of any triangle is equal to the sum of two right angles. In any case, all that I'll use is that the sums of the angles of any two triangles are equal.

Theorem 2 (The "Angle-Angle-Angle" theorem): For any two triangles $\triangle ABC$ and $\triangle A'B'C'$, if $\angle ABC = \angle A'B'C'$ are congruent, and if $\angle BCA = \angle B'C'A'$ are congruent, and if $\angle CAB = \angle C'A'B'$ are congruent, then the triangles are similar. In more detail, this means that we have equality of ratios $$\text{Length}(\overline{AB}) \bigm/ \text{Length}(\overline{A'B'}) = \text{Length}(\overline{BC}) \bigm/ \text{Length}(\overline{B'C'}) = \text{Length}(\overline{CA}) \bigm/ \text{Length}(\overline{C'A'}) $$

So now let's consider two right triangles $\triangle ABC$ and $\triangle A'B'C'$, such that the $\angle ABC$ and $\angle A'B'C'$ are right angles. It follows that $\angle ABC = \angle A'B'C'$.

Suppose also that $\angle CAB = \angle C'A'B'$. By applying Theorem 1, it follows that $\angle BCA = \angle B'C'A'$. The hypotheses of Theorem 2 have therefore been verified, so its conclusions are true. From the equation $$\text{Length}(\overline{BC}) \bigm/ \text{Length}(\overline{B'C'}) = \text{Length}(\overline{CA}) \bigm/ \text{Length}(\overline{C'A'}) $$ we deduce, by a tiny bit of algebra, that $$\text{Length}(\overline{BC}) \bigm/ \text{Length}(\overline{CA}) = \text{Length}(\overline{B'C'}) \bigm/ \text{Length}(\overline{C'A'}) $$ In words, this says that if in triangle $ABC$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, and in triangle $A'B'C'$ we divide the length of the side opposite angle $A$ by the length of the hypotenuse, we get the same number. That number is the sine of the angle $A$.

This proves that the sine of an angle is well-defined no matter what right triangle we use for its calculation.

Answer 2

You should look up similarity. In particular, similarity of triangles, and especially right triangles.

It is because any two right triangles containing the same angle are similar that the functions depend only on the angle in question.

As for why right triangles. Why not, say, isosceles triangles, since these also allow one to define a certain trig function of the vertex angle uniquely, especially as they are laterally symmetric? Well, there's no reason why one couldn't do that; indeed, the first compiler of trig tables compiled them for isosceles triangles, and called them chords of the vertex angle. However, one can argue that that would not be as fundamental as using right triangles for the simple reason that whereas not all triangles can be divided into two isosceles triangles, all can always be partitioned into two right triangles. This keeps things simpler. NB. What was called the chord of $x$ is now double the sine of $x/2.$