Understanding Section 13 of the 1799 F.T.A.

Lemma: Let m denote any positive whole number.  Then the function
sin φ - sin mφ + sin(m-1)φ is divisible by xx - 2cosφ rx + rr.

Proof: For m=1, the function is =0 and therefore divisible by any factor.  For m=2, the quotient is sin φ, and for any greater value the quotient will be
sin φ + sin 2φ + sin 3φ + etc. + sin(m-1)φ It is easily confirmed that the product of this function multiplied by xx - 2 cos φ rx + rr is equal to the given function.

What's a lemma?

Lemma means, "you'll see what I mean..."  In sections 13 and 14, Gauss is making a shocking transition into his complex domain, but since he "considers it worth the trouble to show how" the fundamental theorem "can easily be elicited without their [imaginary numbers] help," he has to construct a shadow of the complex domain to allow his real proof to be expressible in understandable terms.  Imagine telling someone how to make a hammer before building the Grand Coulee Dam.

What does he mean by "divisible by?"

((Put in the Kästner on bases for numbers and long division.))

What does "+ etc. +" mean?

Throughout the paper, Gauss writes "..." or "etc." in equations.  This means that there is a pattern in what he is writing, and that the number of things could vary and they could not all be written out.  1, 2, 3, 4, ... 99, 100 for example.  When in doubt, look at the first and last terms:
    1, 2, 3, 4, ... n-1, n
is, when n is 3, just:
    1, 2, 3,
even though 4 was written originally (to make sure that you got the pattern).

More on division (to be included in or be a substitute for the earlier section).

In division, you have the following:
   Dividend / Divisor = Quotient.
So in 10 / 5 = 2, 10 is the divident, 5 is the divisor, and 2 is the quotient.

In multiplication, you have
  Factor x Factor = Product.
  
To see whether a division is correct, you can multiply the quotient by the divisor to see if you indeed get the dividend, just as you could see that 2 x 5 is indeed 10.
  Quotient x Divisor = Dividend.

Gauss's proof in §13

m=1

Go ahead and put in 1 for m in our given function:
  sin φ - sin mφ + sin (m-1)φ
putting in 1 for m gives us:
  sin φ - sin 1φ + sin (1-1)φ
which is:
  sin φ x - sin φ x + sin 0φ r.
is 1 (think about Bernoulli's spiral: what do you get if you do 0 rotation?) and the sine of 0φ is sin 0, which is 0 (think of the circle), giving:
  sin φ x - sin φ 1 x + 0
  sin φ x - sin φ x
  0.
Zero is divisible by anything, since anything goes into it zero times, so our function is divisible by our divisor.

m=2

First, let's put in 2 for m in our dividend:
  sin φ - sin 2φ + sin (2-1)φ
which simplifies to:
  sin φ - sin 2φ x + sin 1 φ
which is:
  sin φ - sin 2φ rx + sin φ
  
  Now, Gauss says that the quotient is sin φ, so let's multiply our divisor (xx - 2 cosφ rx + rr) by our quotient to see if we get our given function:
  sin φ (xx - 2 cos φ rx + rr).
When multiplying in this fashion, the factor "distributes" itself to all the terms in the other factor.  Think of the example of reducing  a photograph to half its size.  Every object in the photograph will become half as large.  The same thing happens here.  Here we are applying a sin φ "reduction" to our xx - 2 cos φ rx + rr "photograph."
  sin φ (xx - 2 cos φ rx + rr) = sin φ xx - sin φ   x   2 cos φ rx + sin φ rr.

To see if Gauss is right about this, let's compare to see if the quotient x divisor really equals the dividend.
  sin φ - sin 2φ rx + sin φ =?=  sin φ xx - sin φ   x   2 cos φ rx + sin φ rr.
The first and last terms are no problem, both being sin φ (or xx) and sin φ (or rr), respectively.  But look at the middle terms: they do not seem to be equal:
  sin 2φ rx  =?=  sin φ   x   2 cos φ rx.
Now looking at the right hand side (RHS) of this, we can rearrange the multiplications to have 2 sin φ cos φ rx instead of sin φ   x   2 cos φ rx, so:
  sin 2φ rx  =?=  2 sin φ cos φ rx,
and not worrying about the rx part, we have to figure out:
  sin 2φ  =?= 2 sin φ cos φ,
which you can actually verify geometrically:

Greater values of m

Let's first look at what we get doing the multiplications for some greater values of m:

m=3

Dividend:
  sin φ - sin 3φ + sin (3-1)φ which is:
  sin φ - sin 3φ + sin 2φ
  
Quotient x Divisor:
  We'll have to determine what the quotient is.  Look at the first two and last two terms as it is written in Gauss's paper:
  sin φ + sin 2φ + etc. + sin 3φ + sin (3-1)φ
  sin φ x + sin 2φ r + etc. + sin 3φ + sin 2φ r.
As you can see, the second-to-last term seems very strange, and the last term is the same as the second term: so the quotient has only two terms: (sin φ x + sin 2φ r).  Multiplying the quotient by the divisor, we have:
  (sin φ x + sin 2φ r)   x   (xx - 2 cos φ rx + rr).
If the multiplication is giving you trouble, think of having three-and-a-half of 10.  That means 3x10 and x 10 added together.  So we will do:
  (sin φ x + sin 2φ r)   x   (xx - 2 cos φ rx + rr).
  sin φ x   x   (xx - 2 cos φ rx + rr) + sin 2φ r   x   (xx - 2 cos φ rx + rr), expanding to:
  sin φ xxx - 2 sin φ cos φ rxx + sin φ rrx   +   sin 2φ rxx - 2 sin 2φ cosφ rrx + sin 2φ rrr.
Now, since 2 sin φ cos φ = sin 2φ:
  sin φ xxx - sin 2φ rxx + sin φ rrx  +  sin 2φ rxx - 2 sin 2φ cos φ rrx + sin 2φ rrr.
The underlined terms cancel each other, leaving:
  sin φ + sin φ x - 2 sin 2φ cos φ x + sin 2φ .
  
Comparing dividend and quotient x divisor:
  sin φ - sin 3φ + sin 2φ   =?=  sin φ + sin φ - 2 sin 2φ cos φ + sin 2φ
The first and last terms are the same, leaving us with just:
  - sin 3φ   =?=  sin φ - 2 sin 2φ cos φ
Note that all the terms include which we don't have to concern ourselves with.  Five's being three plus two does not depend on what it is that we have five, three, and two of.  Ignoring (or dividing everything by it,) leaves us with:
  - sin 3φ  =?=  sin φ - 2 sin 2φ cos φ,
which, adding sin 3φ to both sides, becomes:
  0  =?=  sin 3φ - 2 sin 2φ cos φ + sin φ.
This looks more difficult to figure out!

An unanswered question

Let's not try do determine if this is true just yet.  Instead, keep trying higher values of m to see what sorts of patterns we come up with.  Remember, Gauss is saying that it is "easily confirmed" that the quotient times the divisor give us our given function (dividend).  Press on!

m=4

We get for the dividend:
  sin φ - sin 4φ + sin 3φ
and for the quotient:
  sin φ + sin 2φ rx + sin 3φ
(Figure out on your own why there are three terms in the quotient.)  Our quotient times our divisor is:
  (sin φ + sin 2φ rx + sin 3φ ) x (xx - 2 cos φ rx + rr), which is
  sin φ x  (xx - 2 cos φrx + rr) + sin 2φ rx  x  (xx - 2 cos φ rx + rr) + sin 3φ x  (xx - 2 cos φ rx + rr).
(Think of a three-by-three FOIL-type grid if this multiplication doesn't make sense from what's been said.)  This gives us:
      sin φ - 2 sin φ cos φ + sin φ
  + sin 2φ - 2 sin 2φ cos φ + sin 2φ
  + sin 3φ - 2 sin 3φ cos φ + sin 3φ

Now look at our product in a grid:
  
And you'll note that terms that have the same powers of r and x variables appear as "/"-shaped diagonals on the grid.  Comparing the dividend with the product of the quotient x divisor, we see that the first and last terms of the dividend and the product (sin φ and sin 3φ ) are the same and cancel.
Look at the first diagonal next to sin φ
  sin 2φ - 2 sin φ cos φ
Based on our trigonometry work from m=2, we know that these terms subtract each other to give us zero.  The next diagonal is (leaving out the common to all):
  sin 3φ - 2 sin 2φ cos φ + sin φ.
This is the return of our unanswered question from m=3!  For the m=3 case to be correct, the diagonal must also be zero.  That is:
  sin 3φ - 2 sin 2φ cos φ + sin φ must be =0.

This leaves as the only thing that doesn't disappear in the product:
  sin 2φ - 2 sin 3φ cos φ
and the only thing that doesn't disappear in the dividend:
  - sin 4φ
So for m=4, it is necessary that
  sin 2φ - 2 sin 3φ cos φ = - sin 4φ
or, forgetting about and adding sin 4φ to both sides:
  sin 4φ - 2 sin 3φ cos φ + sin 2φ = 0
must be true, if Gauss is correct in this proof.

The general case

Compare the unresolved questions for m=3, m=4, and m=5 (which we didn't go over here, but you can do on your own, along with trying out m=6).  Each higher power of m depends on all the earlier unresolved questions and introduces a new one as well.
  sin 3φ - 2 sin 2φ cos φ + sin φ  =?=  0
  sin 4φ - 2 sin 3φ cos φ + sin 2φ  =?=  0
  sin 5φ - 2 sin 4φ cos φ + sin 3φ  =?=  0
  sin 6φ - 2 sin 5φ cos φ + sin 4φ  =?=  0
A clear pattern!  Can you figure out a way of writing the general case?  Do it before reading on!




  sin Aφ - 2 sin (A-1)φ cos φ + sin (A-2)φ  =?=  0.
Reorganize this to get:
  sin Aφ = 2 sin (A-1)φ cos φ - sin (A-2)φ.
Splitting up 2 sin (A-1)φ cos φ into sin (A-1)φ cos φ + sin (A-1)φ cos φ, and subtracting one from both sides, we get:
  sin Aφ = sin (A-1)φ cos φ + sin (A-1)φ cos φ - sin (A-2)φ
  sin Aφ - sin (A-1)φ cos φ = sin (A-1)φ cos φ - sin (A-2)φ
or
  sin Aφ  -  cos φ sin (A-1)φ = cos φ sin (A-1)φ  -  sin (A-2)φ
This we will prove geometrically:

Mark off angles Aφ, (A-1)φ, and (A-2)φ on the circle.  We can get sin Aφ and sin (A-2)φ easily.  To get the
  cos φ sin (A-1)φ
piece, connect the points on the circumference at Aφ and (A-2)φ, cutting the line from the origin to the (A-1)φ spot on the circumference.  The triangle formed from the origin, the Aφ, and the cut spot on (A-1)φ is a right triangle (figure out why!), and has angle φ.  That makes the base of the triangle have length cos φ.  Now revisit the (A-1)φ angle.  The line dropped directly down is sin (A-1)φ.  Now that we have cut the radius of the circle going out to (A-1)φ to have a length of just cos φ, the line dropped from this cut will have a length of cos φ sin (A-1)φ.  (See Tennenbaum's "Cardan's Paradox to the Complex Domain" to help you with this if it seems unclear.)
  Now we have sin Aφ, sin (A-2)φ, and cos φ sin (A-1)φ.  Drawing horizontals from (A-2)φ to cos φ sin (A-1)φ and from cos φ sin (A-1)φ to Aφ, we create the pink segments which are [sin Aφ - cos φ sin (A-1)φ] and [cos φ sin (A-1)φ - sin (A-2)φ].  Ah!  Now we have a geometric understanding of what the general case for section 13 is.  We needed to know that sin Aφ  -  cos φ sin (A-1)φ = cos φ sin (A-1)φ  -  sin (A-2)φ, so it comes to determining whether the pink segments are of equal size.
  Reflection (literally) gives us to realize that the hypotenuses of the two triangles formed with pink sides are both sin φ.  Since the triangles are at the same angle (being formed by the same line as hypotenuse), the pink (left) sides must be of equal length, meaning that we've demonstrated section 13!
  To review, the equality of the pink sides let us know that sin Aφ  -  cos φ sin (A-1)φ = cos φ sin (A-1)φ  -  sin (A-2)φ, which was the general form for the required trigonometric identity required by the higher values of m.  We proved m=1 and m=2, so we can now consider Gauss's section "easily demonstrated," although not so easily.