Trapping a function between two others to settle the trigonometric cases

0:00Here is a quotient that looks like every other one you have taken apart, and is not. Sine of x, over x, as x closes on nothing. Both parts head for nothing, so it is the same nothing-over-nothing you have learned to treat as a signal to factorise. Except there is nothing to factorise. A sine is not built out of powers of x, so there is no shared piece sitting on the top and the bottom waiting to be taken out.

0:31Every tool that worked on quotients of polynomials stops here. What replaces it is not more algebra - it is order. You trap the quotient between two things you can already read, drive them together, and the quotient has nowhere left to go. That is the whole idea, and the work is in earning the trapping. One thing has to be said before any of it, because it decides whether anything here is a measurement.

1:02A sine is not a fraction, and neither is a cosine, and this is a topic made entirely of comparisons between them. So nothing here is ever a decimal. Every sine, every cosine and every tangent is caught between two exact fractions - two neighbouring partial sums of its own alternating sum, which straddle the true value while the terms are still shrinking. Every comparison is then made between those brackets, not between numbers.

1:36And when two brackets overlap, the comparison refuses to answer rather than guessing. That refusal is what makes the zeros later on worth anything. Even the quarter-turn at the end of the interval is bracketed rather than written down: it is caught between two seven-place decimals, by a bracket narrower than a hundred billion billionth. Ninety-six places are swept, forty-eight either side of nothing. Of those ninety-six, the number that ARE the place is zero, and the number falling outside the interval is zero.

2:16The first of the two results you need is almost too mild to notice. Take two rules on the same set of inputs, with the first never anywhere above the second. If both settle on something at a place, then the first settling point is not above the second either. Order survives the limit. Notice what it does not say. It does not say the two settle on different values - they are perfectly free to meet.

2:46And it does not say anything about the place itself, because the ordering is required everywhere, not just there. On its own this is nearly useless. One more turn of the same idea, and it becomes the tool everything here runs on. Now three rules instead of two. The middle one lies between the outer two everywhere they are all defined. If the outer two settle on the SAME value at a place, the middle one has nowhere else to be, and settles there too.

3:21That is the trapping, and the thing worth saying out loud is what it does not require. It does not require the two outer rules to be equal. They only have to agree in the limit, at one place, and everywhere else they may be as far apart as they like. The pair used here are the cosine and the flat constant one. Measured across all ninety-six swept places, the number where those two meet is zero - they touch nowhere on the interval at all.

3:53They agree only in the limit, which is the only place the trapping asks about. So the work is to find something below the quotient and something above it. Both come out of one drawing. A circle of radius one, a radius out to the right, and a second radius lifted by an angle of x. Drop the far end of that second radius straight down onto the first, and you have a thin triangle inside the circle.

4:25Now run a tangent up from the end of the first radius until it meets the second radius extended, and you have a larger triangle containing the wedge. That corner is outside the circle, and it has to be, or nothing about the argument works. Measured, not assumed: across all forty-eight positive places, the number where that corner sits strictly outside the circle is forty-eight. The number where the point on the circle fails to be on it is zero, and the number where it would sit on a circle of a different size is also zero.

5:05The drawing is now a set of facts rather than a picture you are asked to trust. Three regions, nested one inside the next. The thin triangle, then the wedge, then the big triangle. Their areas are half the sine, half the angle, and half the tangent. The middle one is half the angle only because the angle is measured in turns of the radius rather than in degrees, and that will matter enormously in a few minutes.

5:37Now the ordering, which is the step everything downstream leans on. Across the forty-eight places, the number where the three areas come out in the wrong order is zero, and the number where the brackets are too wide to call it is also zero. That second zero is the one to look at. Ask the same forty-eight questions with only two terms of each sum and the brackets are too wide every single time - forty-eight refusals out of forty-eight.

6:10So the zeros are a finding about the shapes, not a habit of the machinery. Here is the part usually passed over in a clause. The wedge lies between the two triangles because you can see that it does. That is the argument, and it is a picture-led argument, not a proved one - the containment is read off the drawing. It happens to be true, and it is worth knowing that it is being taken on sight.

6:40Which is exactly why it is measured here instead. And the measuring is only worth something if the second half of the comparison could fail. Hand the same two comparisons a shape that is smaller than the wedge rather than larger, and the number of places that still pass is zero. The second link is doing work, and now we know it. Every one of the three areas carries the same positive factor: half, times the radius, times the radius.

7:14Strip it out of all three and what is left is bare. Sine of x, below x, below tangent of x. No areas, no circle, no picture - three quantities in a row. That is the inequality everything so far was aiming at, and it holds for every angle strictly between nothing and a quarter turn. It is worth pausing on how little is left of the drawing. The circle was scaffolding for the comparison, and it is not in the answer.

7:46One more thing before it can be used: the quotient we care about has the angle underneath, and this chain does not. Two steps get you there, and each one is a place where people quietly go wrong. First, divide the whole chain by the sine. The sine is positive on this range, so the order is untouched, and the leading entry becomes exactly one. Measured: the number of places where that first entry comes out as anything but one is zero - and divide by one instead of by the sine and it is forty-eight.

8:25Second, take reciprocals. That REVERSES the chain, and the number of places where the division breaks the order is zero, while the number where the reversed chain fails to climb is also zero. Invert without turning the chain round and the number that still climb is zero, which is the mistake the reversal exists to prevent. And the condition is not quite the one usually given. Of two ordered chains that straddle nothing, the number that survive inverting is zero - but of two that are entirely negative, both survive.

9:05What the step needs is one sign throughout, and all positive is simply the case at hand. Out comes the statement everything has been building to. The cosine, below the quotient, below one. And the condition attached to it is not about x - it is about the size of x, bars and all. That is not decoration. Every step of the areas argument only ever saw positive angles: there is no wedge on the other side of the drawing.

9:39The negative half arrives by a different route entirely. Turn the sign of the input round and measure what moves: the number of places where the quotient changes is zero, the number where the cosine changes is zero, and the number where the sine changes is forty-eight. The top and the bottom both flip, so the quotient does not, and the whole statement carries across unaltered. Drop those bars and you have proved exactly half of it.

10:13Now the trapping, applied. The quotient is caught between the cosine and one across every swept place - the number that fail the trapping is zero. Both outer rules head for one. So the quotient does too, and it never had to be evaluated where it has no value. Watch that as a measurement rather than a phrase. Six bounds are offered, each ten times tighter than the last, and what is recorded is how far into a halving run you must walk before the whole remaining tail is inside.

10:49One step, then three, then four, six, eight and nine. The number of bounds it never gets inside is zero. Put the same six to something trapped between minus one and one that flips at every step, and the number it gets inside is zero - and to a reading whose lower end is always inside and whose upper end never is, also zero. Now the assumption hiding underneath all of it.

11:19The wedge had area half the angle because the angle was measured in turns of the radius. Measure the same angle in degrees and that step is simply false, and the answer changes. Run the identical machinery in degrees and the quotient settles between nought point nought one seven four and nought point nought one seven five. That is a hundred and eightieth of the constant that sets the quarter turn, and it is nowhere near one.

11:51Put the same six bounds around one to the degree version and the number it ever gets inside is zero. So the famous answer is not a fact about sines. It is a fact about sines measured in the one unit that makes the wedge come out as half the angle. There is a second standard result, and the good news is that it needs no new picture. One minus the cosine, over x, as x closes on nothing.

12:22One identity turns the top into twice a squared half-angle sine. Split that into the first quotient at half the angle, multiplied by one more sine at half the angle - the first heads for one, the second for nothing, and the product for nothing. Both halves of that are checked rather than asserted. The identity holds at all forty-eight places and the number that fail is zero, while the same check with a third-angle instead of a half fails at all forty-eight.

12:56And the splitting is exact algebra in free letters, not a bracket at all: across a hundred and twenty-one pairs the number failing is zero, and the same line written with a thirding fails on all hundred and twenty-one. Scored against the six bounds it walks in three, six, nine, thirteen, sixteen and nineteen steps. One result, one identity, and no second squeeze. That pattern - rewrite it into copies of something already settled - is the whole of what follows.

13:33Sine of four x over sine of two x: make each part into the standard shape with its own angle underneath, and a two falls out. The splitting is exact algebra again, failing on none of the hundred and twenty-one pairs and failing on all of them when the inner multiple is written wrongly. Scored, it walks in three, five, six, eight, ten and eleven steps. Tangent over x is the first quotient times one over the cosine: it walks in one, three, five, six, eight and ten steps, and the number of bounds around TWO that it clears is zero.

14:15Six of these ratios were built and read, each with its own multiple on top and underneath. Each settles on its own ratio, and the number that settle on one instead is zero. The same quotient with a square underneath settles on a half. And a limit asked at a half-turn rather than at nothing moves there by naming the gap, which is a change of letter and nothing more.

14:44So what has actually been bought. A quotient that no factorisation could reach has been settled by ordering alone. The recipe: trap it, drive the outer two together, and read off where the middle one had to go. Now the honest part, which is about the checking rather than the mathematics. Every value here is a bracket, and a bracket can prove that one thing is below another when the two do not overlap.

15:14It can never prove two things exactly equal - only that their difference holds nothing, as far as the precision carried. So every equality above is either exact algebra in free letters, where no bracket appears at all, or is stated beside a deliberately wrong version that the same check throws out. That is a limit of the apparatus, not a fact about sines, and it is said out loud rather than hidden.

15:45The picture was scaffolding, the trapping was the argument, and the measuring is what turned a drawing into something you can check.