motivated by email from Jason G.
Dr. Norman Wildberger (University of New South Wales, Australia:
personal website) has this neat idea to replace the standard notions of distance and, especially,
of trigonometry, by something which (he claims) is simpler, more intuitive and eliminates the need for the
transcendental trig functions.
(He has a book.
The first chapter in PDF format is here)
>Transenwho?
Transcendental? First we have algebraic functions. If f(x) is algebraic, if it can be computed, given the value of x, by elementary operations like addition, subtraction,
multiplication, division and extraction of integer roots (like the 3rd or 27th root; in general, by "integer roots"
I means the nth root, where n is an integer).
For example, x^{1/3} / (1+x^{2}) is an algebraic function, and so it ( 1x^{3}) and so is ...
>Okay, but what's that dental stuff.
Transcendental? Them's all the other functions, like 10^{x} or e^{x} and all the trig functions.
>You mean sine, cosine, tangent, etc.?
Yes, and ...
>But I can calculate sin(x) when x = 0 and 10^{x} when x = 3 and ...
Uh ... yes. I should have said that transcendental functions cannot be calculated with elementary operations for all values of x.
>But what about e^{x}=1 + x + x^{2}/2! + x^{3}/3! +...
Yes. I should have said transcendental can't be evaluated, for all x, with a finite number of elementary operations.
>Okay, so what's this new trig stuff?
Okay, look at Figure 1.
Do you remember Pythagorus? He says: a^{2} + b^{2} = c^{2}.
Notice that only the squares of the distances occur.
 Figure 1 
Wildberger thinks that the squares of distances are a more natural way to measure how far apart things are.
Indeed, he calls the squared distance the Quadrance.
>Quadrances? That's a strange name, isn't it?
If we're talking squares of distances between points, expressed like a^{2} + b^{2} (for example), then that's a quadratic form and ...
>Okay, quadrances it is.
Look at Figure 2 where A, B and C are the Quadrances of the three sides.
Then Pythagorus becomes: A +B = C
 Figure 2 
>So A = a^{2} and B = b^{2} and ...
Yes... and here are a few other neat formulas:
>Yeah, I like that! And the trig stuff?
In Figure 1 (where a, b and c are garden variety distances), we'd have: sin(y) = b / c
where y is the angle (in blue).
If we are to use Quadrances (instead of distances), then we should consider the ratio
b^{2} / c^{2} or B / C (using the notation of Figure 2).
That's Wildberger's measure of the angle (in blue). He calls it the Spread.
>But that's just sin^{2}, right?
Yes, but if the squares, like distance^{2} and sin^{2}, occur more often in calculations, then maybe it's time to use those
squares to provide a measure of distance and angle.
>So Quadrances and Spreads make life easier?
Look carefully at Figure 3 and, after some cerebral activity, give me the angle in blue.
>Uh ... wait! I can't find my trig text.
Okay, give me the Spread of that angle.
>It's ... let's see ... 4^{2} / 5^{2} or 16/25, right?
 Figure 3 
Right. By the way, did you ever find your trig text?
Now look at Figure 4a which gives the ordinary side lengths and angles.
Figure 4b gives the respective Quadrances and the Spreads.
So what's the relationship between the three side lengths, a, b and c, and the angles x, y and z?
>Are you kidding?
How about the three Quadrances and the three Spreads?
>I assume that p, q and r are those spreads, right?
Yes.
>I give up.
It's p / A = q / B = r / C.
>Are you kidding?
 Figure 4a
Figure 4b

>Okay, so that's pretty simple ... but what's that relationship the "old" way?
Uh ... I can't find my trig text.
Just kidding. It's the ol' Sine Law for triangles: sin(x) / a = sin(y) / b = sin(z) / c.
Okay, how about the Cosine Law? Remember that?
>Are you kidding?
The "old" Trig would say something like (refer to Figure 4a):
a^{2} = b^{2} + c^{2}  2bc cos(x)
or maybe
b^{2} = a^{2} + c^{2}  2ac cos(y)
or maybe
c^{2} = a^{2} + b^{2}  2ab cos(z).
The new Trig would say:
p = A [(A + B + C)^{2} / (4ABC)]
>That's easier?
Note that (A + B + C)^{2} / (4ABC) is so symmetrical.
It's the same number, regardless of which angle spread you're calculating.
In fact, we could put λ = (A + B + C)^{2} / (4ABC).
Then we can write:
p = λ A, q = λ B and r = λ C
>Uh ... that means that p / A = q / B = r / C = λ. Hey! I'm getting the hang of this new Trig.
Good fer you
Okay, suppose you want to find the height of a flag pole, as in Figure 5. You'd ...
>I'd stand back, say 10 metres (according to my tape measure).
Then, with my trusty transit, I'd measure the angle θ. Then H = 10 tan(θ), right?
And if θ = 40 degrees? What's H?
>I'd whip out my $100 calculator and see that tan(40) = 0.839 so H = 10(0.839) = 8.39 m.
You don't understand. In this WildbergerWorld, our transit would measure Spreads. In fact, we'd get the spread as 0.413
and our tape measure would measure in Quadrances, so we'd get 100.
>So ... uh, the height is ... uh ...
The quadrance of the height would be 100(0.413)/(10.413) = 70.4 so ...
 Figure 5 
>What's that (0.413)/(10.413) stuff?
With the new Trig, the Quadrance of the height is given by the magic formula: 100(spread)/(1spread).
>Yeah, so what's H?
I'd whip out my $9.95 calculator and calculate 70.4 = 8.39 metres.
>But that's more complicated!
But I never needed to evaluate them thar transcendental trig functions.
>But you needed a calculator!
To perform an elementary operation! A square root!
>Oh ... I see. So where do I buy those measuring tapes ... the ones that measure in Quadrances?
Soon they'll be in all the shops
 
Here are some trigtype formulas:
Notice that (in the middle one) the measure of a sum of spreads is NOT equal to the sum of the measures.
>Huh?
For two angles, p and q, the total angle is just p + q.
For two Spreads, P and Q, the total Spread isn't P+Q, but is P + Q 2PQ + 2√ ...
>Yeah, yeah! I can see the formula!
Of course, if we define P' =1 P and Q' = 1  Q, then that formula for the sum can be written as:
