Introduction to Trigonometry by Geometric Construction

Posted 1 Jan 2001 at 20:45 UTC by Crutcher Share This

I got really tired of forgeting the Trig functions, and their equalities, so I wrote a short paper on them, that I think is not entirely valueless. A link is attached, as well as a summery.

So there I was a couple months back, wanting to write a continuous optimizer for vector space functions, when I realized that I had forgotten too much math. Several hundred dollars in textbooks and a lot of esspresso later, I am on track to have worked my way back up to vector calculus and linear optimization by the summer.

In school, I always remembered things best when I taught them to someone else, because I had to construct a self-contained set of information in order to pass it on. So, In re-learning the trig functions, I wrote an article over the last two days. It took that long because I wrote it in TeX, which I did not know.

Below is a castrated form of the article (We don't have MathML yet, and the graphs are done in LaTeX, sigh). I think that this is a much better approach than is usually used, though a little peer-review is wanted. The real one is available as a postscript document (With pictures!) from:

Trigonometry by Geometric Construction


Trigonometry is often introduced in terms of the graphs of its functions, I believe that this is a mistake in pedagogy. The functions were originally constructed geometrically, in order to solve geometric problems; and I believe that they are best understood in this form. This document represents an introduction to Trigonometry through geometric construction.

The Angle and The Circle

Given an angle, X, we shall construct the six trigonometric functions, sin(X), tan(X), sec(X), cos(X), cot(X), and csc(X) geometrically. To begin, we must inscribe a circle preed upon X's vertex. A unit circle is often used for this construction, but this masks some of the fundamental relationships between the functions, so we will use a circle of radius r, and expose those relationships.

The Sine and The Co-Sine

Sine meant, at one time (translated very poorly), a fold on a curve. In order to construct the sine and the complementary sine (the sine of X's complement, also called the co-sine), we draw a line segment connecting the intersection of one of X's rays with the closest point on the other ray. Recall from plane geometry that the intersection angle will be pi/2. The length of this new line segment is the sine of X, and we will designate it a. The length of the line segment connecting X's vertex and the point of intersection is the co-sine of X, and we will designate it b.

It is useful to note at this point that we have formed a right triangle, with sides a and, b, and hypotenuse r. The Pythagorean theorem yields:

   a**2 + b**2 = r**2

The functions sin(X) and cos(X) are defined to be the values of the sine and the co-sine of X upon a unit circle. To obtain these values from a and b, we must examine the relationship between the circle we have constructed, and the unit circle.

Notice at this point that we are dealing with two triangles of equal angles. The triangle on the unit circle is 1/r'th the size of the triangle on our constructed circle, so we can easily scale any length upon our constructed circle to its equivalent length on the unit circle by dividing by r. This allows us to now say that:

   sin(X) = a/r
   cos(X) = b/r

And we can also conclude that sin(X)**2 + cos(X)**2 = 1:

             a**2 + b**2 = r**2
   (1/r**2)(a**2 + b**2) = (1/r**2)(r**2)
   a**2/r**2 + b**2/r**2 = r**2/r**2
     (a/r)**2 + (b/r)**2 = 1
   sin(X)**2 + cos(X)**2 = 1

The Tangent and The Secant

Tangent meant, originally, touching; and secant meant cutting. To establish these values, we draw a line segment connecting the point of intersection of one of X's rays with the other ray, but this time, we begin our line segment with an angle of pi/2. Which ray we chose is irrelevant, but the math is easier if we start with the ray that we ended with last time. This new line segment is the tangent (touching the circle), and we will designate it j. The line segment connecting the point of intersection with the other ray and the vertex is the secant (cutting the circle), and we will designate it k.

Note that we have now constructed another right triangle, this one of sides j, and r, and hypotenuse k. The Pythagorean theorem yields:

   j**2 + r**2 = k**2

The functions tan(X) and sec(X) are defined to be the values of the tangent and secant of X on a unit circle. Scaling by 1/r, we can now say:

   tan(X) = j/r
   sec(X) = k/r

And we can also conclude that tan(X)**2 + 1 = sec(X)**2:

             j**2 + r**2 = k**2
   (1/r**2)(j**2 + r**2) = (1/r**2)(k**2)
   j**2/r**2 + r**2/r**2 = k**2/r**2
            (j/r)**2 + 1 = (k/r)**2
           tan(X)**2 + 1 = sec(X)**2

If we wish to represent tan(X) and sec(X) in terms of a, b and r, we need only note that the new triangle and the first triangle also share angles, and we can easily establish a proportion of r/b between them. So, substituting for j and k, and simplifying, we obtain tan(X) = a/b and sec(X) = r/b:

        j = a(r/b)
   tan(X) = j/r
          = a(r/b)/r
   tan(X) = a/b
        k = r(r/b)
   sec(X) = k/r
          = k(r/b)/r
   sec(X) = r/b

The Co-Tangent and The Co-Secant

We now seek to construct the complementary tangent (the co-tangent), and the complementary secant (the co-secant). In constructing the co-tangent and the co-secant, we draw a line segment, beginning on the circle, which lies tangent to the circle, parallel to one of X's rays, and terminating at the intersection with the other ray. This new line segment is the co-tangent, and we will designate it p. The line segment from the point of intersection of this new segment with X's ray and X's vertex is the co-secant, and we will designate it q.

And look, we've constructed yet another right triangle, this one with sides p, and r, and hypotenuse q. The Pythagorean theorem yields:

   p**2 + r**2 = q**2

The functions cot(X) and csc(X) are defined to be the values of the co-tangent and co-secant of X upon a unit circle. Using the 1/r proportion from before, we can say:

   cot(X) = p/r
   csc(X) = q/r

And we can also conclude cot(X)**2 + 1 = csc(X)**2:

             p**2 + r**2 = q**2
   (1/r**2)(p**2 + r**2) = (1/r**2)(q**2)
   p**2/r**2 + r**2/r**2 = q**2/r**2
            (p/r)**2 + 1 = (q/r)**2
           cot(X)**2 + 1 = csc(X)**2

If we wish to represent cot(X) and csc(X) in terms of a, b, and r, we must once again establish a proportion between the first triangle and this new triangle; which in this case is a factor of r/a. Substituting for p and q, and simplifying, we obtain cot(X) = b/a and csc(X) = r/a:

        p = b(r/a)
   cot(X) = p/r
          = b(r/a)/r
   cot(X) = b/a
        q = r(r/a)
   csc(X) = q/r
          = r(r/a)/r
   csc(X) = r/a

preed == centered, damn s/center/pre/g, posted 1 Jan 2001 at 20:49 UTC by Crutcher » (Journeyer)


Complex numbers help a lot, posted 2 Jan 2001 at 01:50 UTC by Bram » (Master)

Generally when confronted with a trig problem I rederive everything using complex numbers - think of a point in the plane as representing x + yi, with x being the horizontal component and y being the vertical component. (remember i is the square root of -1).

Now, viewing theta as the angle between a line from the origin to (x, y) to the x-axis, cos(theta) is x/sqrt(x*x + y*y) and sin(theta) is y/sqrt(x*x + y*y)

The magic formula you can now use to rederive everything in trig is -

e^(i*theta) = cos(theta) + i*sin(theta)

More well known, (and, strangely, widely considered more 'elegant',) is the special case

e^(i*pi) = -1

But that still leaves the other funcs., posted 2 Jan 2001 at 02:12 UTC by Crutcher » (Journeyer)

And a cohesive set to hang your mind on. Having a derivation that is:

  • Graphical,
  • Named, and
  • Mathematical
performs the magic in your head of cross relating a concept into multiple sections of your mind. The article is really much better in TeX form, with its nice figures. I really cant wait for MathML and SVG to land fully.

Now do spherical trig?, posted 2 Jan 2001 at 02:22 UTC by jwalther » (Journeyer)

Wow. That cheat sheet at the end especially was great. I found the motivation for csc and cot functions could use some improvement, but seeing the diagrams, and greek letters all together made it really nice.

In fact, I was so impressed, would you consider doing spherical trig? I'd love to see a treatment of it the same as the treatment you just gave regular trig.

Thanks for providing the Postscript version, it was much more readable. Have you considered using latex2html? It generates png images, but I can deal with that. Ghostscripts on the fly rendering gets irritating.

Yes, the co-functions need work., posted 2 Jan 2001 at 03:02 UTC by Crutcher » (Journeyer)

Yes, the co-functions need some more work, but fully exploring the complement's relationship wasn't something I realized needed doing until today. I haven't done it yet, because I didn't want to fool with TeX pictures today, but I will do it soon.

The proper approach is to construct the sine, then construct the complementary sine, after constructing the complementary angle. Then the cotangent and cosecant can easily be introduced, and be obvious.

But I'll have to rewrite it to introduce the tangent and the cotangent simultaneously, and the secant and cosecant simultaneously, which will be a pain, but probably worth the effort. To do it clean, I'll end up adding a least 2 pages (3 figures and some text/equations).

As for hyperbolic trig, I have to figure out how those functions are constructed before I write a paper on them.

Complex numbers and trig identities, posted 2 Jan 2001 at 03:51 UTC by Pseudonym » (Journeyer)

Bram: You are so right. It's amazing what you can derive using this method.

For example, if you ever forget how to expand cos(A+B), simply take the real part of the complex exponent:

= Re(exp(i*(A+B))
= Re(exp(i*A) * exp(i*B))
= Re((cos A + i * sin A) * (cos B + i * sin B))
= Re(cos A cos B + i * (sin A cos B + cos A sin B) - sin A sin B)
= cos A cos B - sin A sin B

Taking the imaginary part instead gives you:

sin(A+B) = sin A cos B + cos A sin B

Have fun coming up with your own identities...

MathML/SVG, posted 2 Jan 2001 at 17:01 UTC by sab39 » (Master)

Crutcher, have you considered providing a MathML/SVG version anyway? Some of us are using browsers that can handle them :)

Ah, but advogato can not., posted 3 Jan 2001 at 05:49 UTC by Crutcher » (Journeyer)

And so I'm not using SVG or MathML

spherical trig II, posted 3 Jan 2001 at 08:23 UTC by jwalther » (Journeyer)

I believe the treatment of spherical trig would need to start off with 2 pages or so introducing the hyperbola, its salient use, and then the constructions would probably go from there. Btw, even if you don't do spherical trig, I look forward to seeing your new, updated, improved version of the trig derivations. It has a lot of potential; don't let it succumb to bloat. Its simplicity and brevity are its strong points.

Excellent, posted 3 Jan 2001 at 14:01 UTC by damo » (Apprentice)

Thank you, Crutcher - that looks very good and you might even want to email a few universities about it (or stick a note on a relevant newsgroup). I think it would be of particular use to A-level maths or first year physics undergrads in the UK (I could have made good use of it back then) and presumably to similar people in other countries.

Only negative comment is that you have written origianly in section 3 where you meant to write originally! Besides that, top notch!

Okay, I've finnished re-writting it., posted 4 Jan 2001 at 05:23 UTC by Crutcher » (Journeyer)

Its still at the same location, but I've basicaly touched everything, changed the intro order, etc. I'm very pleased with it, and its only 7pgs :)

Trigonometry *Without* Geometry is Simpler, posted 4 Jan 2001 at 09:47 UTC by moshez » (Master)

If you want to work formally with Trigo by geometry, you need to get into some crufty parts of measure theory. What's wrong with just working straight off the power series?

cos x = sigma -1^i*x^(2i)/(2i)!

oo sin x = sigma -1^i*x^(2i-1)/(2i-1)! 1

You can easily prove that sin^2+cos^2=1 (simple power series manipulations). You can prove that cos' = -sin, and cos(0)=1. You can prove cos has a zero (because it just keeps going down for a long while...), and define the infimum of zeros as half-pi. Then you can prove all the periodic qualities you need, and all the identities you need, with simple calculations of the power series. And, what's more, you get for free the fact that cos+sin=e^i, and so the natural extension of cos and sin to the complex numbers.

Calculation vs Application, posted 4 Jan 2001 at 16:01 UTC by Crutcher » (Journeyer)

Yes, there are some interesting shortcuts available for caluclating the plane trig functions, but even knowing those methods, you still need to understand their geometric application to get any use out of them outside of the specialized fields that provide the shortcuts.

I basicaly do not think that an introduction to trig should begin:

first, using the complex plane ...

Btw, I had enough fun doing this that I am going to do hyperbolic trig by construction over the weekend, and maybe even get to spherical trig by construction later on.

YAY!, posted 5 Jan 2001 at 03:55 UTC by jwalther » (Journeyer)

Nuff said :-)

Some thoughts, posted 5 Jan 2001 at 06:58 UTC by inri » (Apprentice)

Disclaimer: I'm a first year PhD student in mathematics. That said, I haven't used the geometric meanings of the trig functions in years and have little intuition for them.

Crutcher: There is an error in the cheatsheet currently: the last line should read:

tan * cot = sin * csc = cos * sec = 1 (times instead of divide)

That said, I find this is an excellent elementary exposition of the elementary trig functions; I had never learned where tangent and secant came from or why the co-s were named as they are -- I only knew them as defined in terms of sine and cosine, and I just had to memorize them. Your article fills a gap in my education. Could you give a bibliography? What language do sine, secant and tangent come from?

You might also want to note that the complement of the complement of an angle is the original angle, so the co-cosine is just sine.

For the hyperbolic functions, I think these are constructed by analogy with the standard trig functions as follows: the unit circle x^2 + y^2 = 1 can be parametrized (with unit speed) by f(x) = (cos(x), sin(x)); the corresponding parametrization of the unit hyperbola x^2 - y^2 = 1 is g(x) = (cosh(x),sinh(x)) (note that in order for this to have unit speed, you need a different notion of length than the standard, so maybe this isn't too helpful) (formally you can say that these functions are the unique isometric immersion of the real line into the unit circle/hyperbola sending 0 to (1,0) and the positive tangent vector at 0 to the positive y direction) -- you then contrust tanh, coth, sech, csch via the identities tanh = sinh/cosh, sech = 1/cosh etc.

moshez: Yes, if you want to work formally, you should define exp(x) = 1 + x + x^2/2! + ... + x^n/n! + ... and define cos(x) = (e^ix + e^-ix)/2, sin(x) = (e^ix - e^-ix)/2i; you then get that these are entire analytic functions, the relation e^ix = cos x + i*sin x, the derivatives of exp, cos and sin, can prove that cos has a zero (and can thus define the constant Pi abstractly as twice the first zero of cosine), show that they are periodic, etc., etc. However, this gives no intuition as to where they come from or to why you should care. A priori they are just arbitrary functions with no geometric meaning. For example one can define the hyperbolic trig functions formally as sinh(x) = (e^x - e^-x)/2, cosh(x)= (e^x - e^-x)/2 (as my calculus text did) and have no idea why you should care, or their meaning. I still find it kinda mysterious that exp (constructed as an exponent/power series) and sin/cos (constructed geometrically) are related -- I basically chalk it up to the relations between the derivatives, which can (in the case of sin/cos) be derived geometrically. (incidentally the corresponding e^x identity is e^x = cosh x + sinh x)

Crutcher: You might consider including an additional section (at a higher level of abstraction than the previous ones) mentioning the above interpretation/definition of sin/cos, sinh/cosh, which would give a certain completeness; also, the expression of sinh/cosh in terms of exp is very concrete and useful -- one can prove it by noting that cosh^2 - sinh^2 = 1 (i.e., (cosh x, sinh x) is on the unit parabola) (though this leaves the question of speed -- if you want, this is a good opportunity to introduce pseudo-riemannian metrics as in relativity, but that's likely far, far beyond the scope of this paper).

Again, thanks for a very informative article; could you also post the TeX source sometime?

anecdote: Yesterday, on our first day of classes for the new term, someone had written sec^2 + 1 = tan^2 on a side board in class; at the end of class, I got up and corrected this, to the thanks of my classmates who had been irritated by it all class long.

Addendum: Speed, posted 5 Jan 2001 at 07:29 UTC by inri » (Apprentice)

Sorry, my interpretation of hyperbolic functions was a little off-the-cuff, esp. my passing remarks about `unit speed' etc. Basically, if you have a path in the plane, given by f(x)=(p(x),q(x)), then the speed of f at a value x0 is given by p'(x0)^2 + q'(x0)^2 (where ' indicates derivative). In other words, the speed is the length of the tangent vector (p'(x),q'(x)) with the standard notion of length. One can put other notions of length on the plane (or other objects), giving different geometries. This can be summarized by the metric at a point, which is a 2x2 matrix M such that the speed of f at the point x is given by:

[p'(x) q'(x)]M[p'(x)]


In the case of the standard euclidean geometry, M is the identity matrix, so we get:

[p' q'][1 0][p'] = [p' q'][p'] = [p'^2 + q'^2]

       [0 1][q']          [q']

If we instead take

M=[-1 0]

  [0  1]

as a curvature, we get that the parametrization (cosh(x),sinh(x)) has speed -(sinh(x))^2+cosh(x)^2 = 1, as desired.

As a last note, note that giving the speed of a path at any given point allows us to define distance as the integral of speed over time, just as in single variable calculus (where a smooth function is the integral of its derivative). Sorry if this is long/off-topic etc. -- I figured that most Advogato readers would know calculus and some linear algebra, but not riemannian geometry, so that this would be a pleasent diversion.

Correction made, thanks., posted 5 Jan 2001 at 14:32 UTC by Crutcher » (Journeyer)

I've corrected the tan X / cot X error, and reposted. I've also posted the document as plane_circular_trig_construction.{tex,ps}, where it will live as I add more docs. The website is indexable, so have at.

More stuff..., posted 5 Jan 2001 at 15:17 UTC by Crutcher » (Journeyer)

For starters, tangent and secant I already new the defs for, but I checked to be sure before writing. They are both Latin.

However, sine I had no idea, and returned (after reading very much math returns) that it is a Latin mistranslation of an arabic word, jayb, as though the arabic meant "fold in a garment".

As for the biblio, I should add that, yes. I used Handbook of Mathematics by Bronshtein and Semendyayev. It contained a figure almost identical to the one I used, but only mentioned breifly a correspondence, and a few equalities, before jumping into standard trig. It is a very good reference, but sucks as an instructor.

Geometric Intuition?, posted 8 Jan 2001 at 08:20 UTC by moshez » (Master)

crutcher: Why would anything having to do with sin/cos has to do with geometry? Personally, I have never needed the geometry. Possibly, the only thing I needed is the following (formal?) fact:

For every complex z, s.t., |z|=1, there is a unique x, 0<=x<2*PI, s.t. z = e^{i*x}.

That fact (which can be easily proved formally) says that x->(cos x, sin x) is the universal cover of the circle by the plane. Then, using basic identities, you can *define* the angle of a vector by the unique x which maps to its normalization

Of course, I'm a western mathematician, the Former USSR school has a whole different view of mathematics. Biased as I am, I think that their view of math makes it terribly ugly...

Baffled about so much praise, posted 9 Jan 2001 at 00:33 UTC by mvw » (Journeyer)

I don't want to sound arrogant, but what you wrote is rather basic stuff that everyone should have learned in high school mathematics or physics class (when treating circular motion and oscillations), perhaps not the origin of words, but the formulas are standard.

In university you might rather get this geometric introduction in an experimental mechanics lecture. The mathematics at university is more analytic (involving limit proceesses) or algebraic than geometric today.

Perhaps one will stumble in some historic lecture on greek mathematics, as all this has been known by ancient greeks and other early high cultures.

As a side note, ask your parents what they learnt in school or university. As time is limited for the school education a lot of that geometric elementary treatment (like cone cuts) has been replaced by more algebraic or computer science oriented stuff. Beginners in linear algebra today are more likely to get algebraic stuff there than analytic geometry, as more math students rather take computer science than physics as side lecture.

Regards, Marc

This was not meant to be novel, posted 9 Jan 2001 at 03:16 UTC by Crutcher » (Journeyer)

It was meant to be useful. I am not delving into deeper mathematics with this, I am presenting an introduction to Trig in terms of geometric construction, precisely because most people agree with /you/ about algebraic introductions, and I dont.

I, and some others, have difficulty memorizing formulae "just because that's how it is", but I rarely have any trouble with derivable functions, especially geometrically derivable ones, even in much higher dimensions. Hence, I often forgot the tan/sec/cot/csc parts of the trig equalities, because I couldnt seem the relationship internally. I looked for a good geometric treatment of this, and could not find one, so I wrote one. It is not new math, it is an introduction to trig (its in the damn title.)


Fashion/trend in Math education: Geometry gets replaced by Algebra, posted 9 Jan 2001 at 10:55 UTC by mvw » (Journeyer)

I was just wondering and wanted to note that the perpective in maths education seems to have shifted from geometry to algebra.

As physicist I prefer the geometric approach of course :-)

I found out that knew less than the generations before, when I learned about motion in gravitational fields, some of the algebraic curvers that show up here were teached in school in the 70ies but were missing in my school curriculum.

We also typicall learn not much tricks in solving differential equations analytically, probably because we get teached how to use a computer.

Or read the Feynmans biography, the part were they described how they calculated by hand, rulers, whatever and in teams. But it is clear that if new stuff is added to the curriculum, old stuff must vanish. That's life.

I also like to know where certain concepts or notations come from. So I have a couple of books on that subject.

Regards, Marc

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