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** A Right Triangle **

The geometry of trigonometry is about the relationship of the parts of a triangle to each other and to derived or related points, lines, and angles. The trigonometric functions are so generally useful that their basic relations are often lost in the memorizing of formulas and their algebraic manipulations.

The point of trigonometry from the geometric view is to deal with arbitrary triangles, but the definitions of trigonometric functions are based on right triangles.

Given an acute angle *θ*, a right triangle with one acute angle equal to *θ* can be constructed.

We can call the vertex the angle *θ* point A, the vertex of the right angle the point C, and the vertex of the other angle point B.

Such a triangle is shown in blue in the figure. Yes this figure was constructed on the SVG grid, but it is easy (perhaps easier) to create such a figure on plain paper with compass and straightedge given any acute angle and choosing any convenient length *b*. For the time being *θ* is restricted to being an acute angle. In general trigonometric functions are not restricted to acute angles. although some are not defined for every angle.

It should not be news to anyone that a right triangle has, besides the right angle two acute angles. The angle we are given we are calling *θ*, but other than it must be an acute angle, it is entirely arbitrary (some choices of *θ*, such as near a right angle or very tiny, would make a sketch awkward to construct and difficult to understand, but in principle it could be made given a sufficiently large paper,compass, and straightedge).

The side of the triangle opposite the right triangle is called the hypotenuse (labelled *c* here). The trigonometry functions relate other things to the hypotenuse of this triangle, the triangle with an acute angle equal to *θ*.

Nothing has been said about how long the sides of {$\triangle {\color{blue}{ABC}} $} might be. That is because for present purposes, it does not matter. All right triangles with one acute angle equal to *θ* are similar, so their corresponding parts are proportional. Because trigonometry is about comparing so many things to the hypotenuse (*c*), when trigonometry is taught it is customary to assign the value of 1 to the length of the hypotenuse. It might as well be 1 (or another number), but 1 has the advantage of simplifying the math since it is easy to divide or multiply by 1 and to take the root of 1. It has the disadvantage of disguising some relationships, so you end up memorizing and manipulating trigonometric expressions without learning what many of them mean.

In the following, *hypotenuse* always means, unless explicitly stated otherwise, the hypotenuse (*c*) of the right triangle with an acute angle *θ* ({$\triangle {\color{blue}{ABC} }$}.) This needs to be made clear because we will be comparing *c* with some things not actually in {$\triangle {\color{blue}{ABC}}$}, and those things may belong to other triangles.

When reference is made to the adjacent side, that means the side of the triange adjacent to *θ* that is not the hypotenuse. In the figure, the adjacent side is labelled *b*. There is only one side opposite *θ*, and the opposite side is labelled *a*

Like many other things mathematical, the word *sine* comes to us from Arabic, in this case, mistranslated Arabic. The original Arabic word (*jiba*) meant cord (of a circle) or bowstring. But it was mistaken for another Arabic word (*jaib*) which was translated into Latin as *sinus.* What we call sine is actually a relationship involving half of a cord, but it is too late the patch up the language now.

The sine of an angle is defined as the ratio of the side opposite (*a*) the angle *θ* to the hypotenuse (*c*) of a right triangle as constructed above.

As labeled above:

{$$ \sin{\color{blue}{\theta}} = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{\color{blue}{\overline{BC}}}{\color{blue}{\overline{AB}}} = \frac {\color{blue}{a}} {\color{blue}{c}} $$}

The *co* of *cosign* stands for *compliment.* *Cosign* (and the other *co-* terms) is frequently hyphenated in old texts.

The cosine of an angle is defined as the ratio of the side adjacent (*b*) to the angle *θ* to the hypotenuse (*c*) of a right triangle as constructed above.

As labeled above:

{$$ \cos{\color{blue}{\theta}} = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{\color{blue}{\overline{AC}}}{\color{blue}{\overline{AB}}} = \frac {\color{blue}{b}} {\color{blue}{c}} $$}

The other acute angle in the right triangle is the complement of *θ*, and *b* is the side that is opposite the complement of *θ*, so *cosine* is the sine of the angle that is the complement of *θ* or 'the compliment sine' which is shortened to *cosine.*

Now in unit-circle teaching of trigonometry (where *c* is given the value of 1, the next thing you get is:

{$$ \tan{\color{blue}{\theta}} = \frac{\text{opposite side}}{\text{adjacent side}} = \frac {\sin{\color{blue}{\theta}}}{\cos{\color{blue}{\theta}}}$$}

Whoa! This is what prevented me from become a great mathematician
and physicist. The equation is perfectly true, but by setting *c* equal to 1, and doing some algebra in the background what tangent is about was completely obscured. They drew the tangent line, but why it would be sine divided by the cosine was simply handwaving magic, some kind of cosmic coincidence. Accept it and move on (a thing entirely foreign to my nature). So I was stuck.

So here is the real dish:

*Tangent* here means a segment of a tangent line. In plain geometry, a tangent line is a line that touches a circle (other curve) at only one point no matter how far it is extended in either direction. It is also perpendicular to the radius of the circle at that point. In this case we mean only a segment of the tangent line.

There are at least two ways to represent the tangent segment geometrically. This is one of them:

Construct the perpendicular to {$ \color{blue}{\overline{AB}} $} at point *B*. Because the angle *θ* was restricted to acute angles, there must be a point where the extension of the perpendicular constructed intersects with the extension of {$ \color{blue}{\overline{AC}} $}. The point of intersection will be called *T*.

For present purposes we will call {$ \color{red}{\overline{\color{blue}{B}T}} $} the tangent (short for tangent segment).

We define {$ \tan \theta $} thus:

{$$ \tan\color{blue}{\theta} = \frac{\text{tangent}}{\text{hypotenuse}} $$}

(Just a gentle reminder here: *hypotenuse* means the hypotenuse (*c*) of the original triangle unless otherwise specified.)

Now

{$$ \angle \color{blue}{CB}\color{red}{T} = \color{blue}{\theta} $$}

because they are both complements of {$ \angle \color{blue}{ABC}.$}

{$$ \triangle \color{blue}{ACB} \sim \triangle \color{blue}{CB}\color{red}{T} $$}

since we know they each have two angles that are equal to angles in the other, the right angles and *θ* = {$ \angle \color{blue}{CB}\color{red}{T} $}.

From {$ \triangle \color{blue}{ACB} $}

{$$ \color{blue}{a} = \color{blue}{c} \sin\color{blue}{\theta} \tag{1}$$}

From {$ \triangle \color{blue}{BC}\color{red}{T} $}

{$$ \color{blue}{a} = \overline{\color{blue}{B}\color{red}{T}} \cos \color{blue}{\theta} $$}

In other words, *a* is the side opposite *θ* in the original triangle, but *a* is the side adjacent to the angle that is equal to *θ* in the new triangle.

In this case {$ \overline{\color{blue}{B}\color{red}{T}} $} is the hypotenuse of {$ \triangle \color{blue}{BC}\color{red}{T} $}; it just is not the hypotenuse we usually mean by *hypotenuse.*

So

{$$ \frac{\color{blue}{a}}{\cos\color{blue}{\theta}} = \overline{\color{blue}{B}\color{red}{T}} $$}

Substituting by (1)

{$$ \frac{\color{blue}{c} \sin\color{blue}{\theta}}{\cos\color{blue}{\theta}} = \overline{\color{blue}{B}\color{red}{T}} $$}

Dividing by *c*

{$$ \frac{ \sin\color{blue}{\theta}}{\cos\color{blue}{\theta}} = \frac{\overline{\color{blue}{B}\color{red}{T}}} {\color{blue}{c}} = \frac{\text{tangent}}{\text{hypotenuse}} = \tan\color{blue}{\theta} $$}

So unit-circle people may have their hearts in the right place, but they just do not express themselves well.

The alternative construction of the tangent goes like this:

Extend the line containing {$ \overline{\color{blue}{AC}} $}. Copy the length of *c* to the extended line, starting at point *A*. The dotted black arc represents the path of the compass in making the copy.

The root of the word *secant* is Latin meaning "cutting." This is not much. Secant is more generally used to mean a line which intersects a curve in two places (that is a line containing a chord), but there are a lot of those.

The unit-circle definition of secant is:

{$$ \sec{\color{blue}{\theta}} = \frac{\text{hypotenuse}}{\text{adjacent side}} = \frac{1}{\cos{\color{blue}{\theta}}} $$}

The segment {$ \overline{\color{blue}{A}\color{red}{T}}$}, or {$ \overline{\color{blue}{A}\color{red}{T^\prime}}$} which is equivalent as we shall see, is the secant. See the connection between the definition and the line segment? Neither do I.

So, what is the ratio of the segment {$ \overline{\color{blue}{A}\color{red}{T}}$} to the hypotenuse of the original triangle, which is the segment we call *c* for short?

{$$ \frac{\text{secant}}{\text{hypotenuse}} = \frac {\overline{\color{blue}{A}\color{red}{T}}} {\color{blue}{c}} = ? $$}

Now we notice

{$$ \triangle \color{blue}{ACB} \sim \triangle \color{blue}{AB}\color{red}{T} $$}

since they have the angle *θ* in common and they each have a right angle, according to their construction.

In {$ \triangle \color{blue}{AB}\color{red}{T} $}, {$ \overline{\color{blue}{A}\color{red}{T}} $} is the hypotenuse and the leg adjacent to *θ* is *c*. So:

{$$ \frac{c}{\overline{AT}} = \cos\theta $$}

{$$ \overline{\color{blue}{A}\color{red}{T}} = \frac {\color{blue}{c}} {\cos\color{blue}{\theta}} $$}

Back to answer our ?

{$$ \frac{\text{secant}}{\text{hypotenuse}} = \frac{\overline{\color{blue}{A}\color{red}{T}}}{\color{blue}{c}} = \frac{\frac {\color{blue}{c}} {\cos\color{blue}{\theta}}}{\color{blue}{c}} $$}

Mutiplying through by *c*:

{$$ \overline{\color{blue}{A}\color{red}{T}} = \frac {\color{blue}{c}} {\cos\color{blue}{\theta}} $$}

Dividing through by *c*:

{$$ \frac{\text{secant}}{\text{hypotenuse}} = \frac{\overline{\color{blue}{A}\color{red}{T}}}{\color{blue}{c}} = \frac {1} {\cos\color{blue}{\theta}} $$}

Now

{$$ \sec\color{blue}{\theta} = \frac{1}{\cos\color{blue}{\theta}} $$}

Makes sense. It is also obviously a fact

{$$ \frac{1}{\cos\color{blue}{\theta}} = \frac{\text{hypotenuse}}{\text{adjacent side}} $$}

but this is not particularly germane to the understanding of secant.

The red dotted arc was not used in the construction of the alternate view of the tangent {$ \overline{\color{#F000F0}{V}\color{red}{T^\prime}} $}. The dotted red arc is just conjecture at this point. So we will demonstrate that {$ \overline{\color{#blue}{A}\color{red}{T^\prime}} $} is equal to the secant and also that the alternate view of the tangent ({$ \overline{\color{#F000F0}{V}\color{red}{T^\prime}} $}) is equal to the tangent {$ \overline{\color{blue}{B}\color{red}{T}} $}.

We observe:

{$$ \triangle \color{blue}{ACB} \sim \triangle \color{blue}{A}\color{#F000F0}{V}\color{red}{T^\prime} $$}

because both of them have the angle *θ* and a right angle.

*Graph:*

*Sources:*

*Elements of the Differential and Integral Calculus***Wikisource***Calculus Made Easy*by Silvanus P. Thompson**Project Gutenberg**- Online Etymology Dictionary

*Recommended:*

*Paul's Online Notes: Calculus I*while not a source of material here, sometimes helpful when stuck.

**Category:** Math

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