Two more identities from the same equation, and the angles they hold for

0:00Here is the one relation that every right triangle satisfies. The square on one leg, plus the square on the other, equals the square on the longest side. Last time we divided that through by the square of the longest side, and the first identity fell out. But look at what the move actually was. We picked one of the three sides, and divided every term by its square. There are three sides. So there are three divisions.

0:31And there are three identities, which is the whole of this video. I measured all of what follows rather than assuming it. Eight hundred and nineteen corners, taken off a few hundred figures, and most of those corners are not right angles at all. At every one of them I took the two sides meeting there and the side facing it, and divided the relation through by each of the three in turn.

0:59All three balance at the fifty two corners where the angle turned out to be a right angle, measured rather than assumed. And all three fail at every one of the six hundred and sixty three corners where it is not. So none of the three is a fact about triangles. All three are the right angle, written down three different ways. Same triangle, same relation. This time divide every term by the square of the side that touches the angle.

1:33The first term is that side over itself, squared. One. The second is the side facing the angle, over the side touching it, squared. That is the tangent of A, squared. The third is the longest side over the side touching the angle, squared. That is the secant of A, squared. So one plus the squared tangent equals the squared secant. Nothing new went in. The same equation, a different quantity underneath.

2:05And notice what decided the names. Divide by the longest side and you get a sine and a cosine, because the longest side is what those two stand over. Divide by the side touching the angle and you get a tangent and a secant, because that side is what those two stand over. The division does not only produce a statement. It chooses which two ratios the statement is about.

2:33Which means you never have to remember which identity pairs which ratios. Put a side underneath, and the two ratios standing on it are the two you get. Now the part that gets read as fine print, and is not. That identity is claimed from zero degrees up to ninety, with ninety left out. Why leave one out? Open the angle towards a right angle and watch the triangle. The side touching the angle shrinks. At ninety degrees it has gone.

3:08And both ratios in the identity were standing on that side. The tangent is the facing side over it. The secant is the longest side over it. Both become a division by nothing, so neither has a value at all. Here is the sentence to be careful with. At ninety degrees this identity is not false. It is not being asserted. There is nothing there to be true or wrong about, because the two quantities it talks about do not exist.

3:41The routine that checked all of this can give three answers rather than two. It holds, it fails, or there is nothing there to evaluate. At ninety degrees, asked about this identity, it gives the third answer. One divisor left. The side facing the angle. Divide every term by the square of that. The first term is the side touching the angle, over the side facing it, squared. That is the cotangent of A, squared.

4:15The second is the facing side over itself. One. The third is the longest side over the facing side, squared. The squared cosecant. So the squared cotangent plus one equals the squared cosecant. Three divisions. Three identities. All of them out of the same single line. And the pattern in the names held again: the two ratios that came out are exactly the two that stand on the side we divided by.

4:48This one has a missing endpoint too, and it is at the other end. Close the angle towards nothing instead of opening it. Now it is the side facing the angle that shrinks. At zero degrees it has gone. The cotangent stands on it. The cosecant stands on it. A division by nothing again, twice. So this identity is claimed from just above zero up to ninety, with ninety included this time and zero left out.

5:19Which is the exact mirror image of the last one. That is a thing to prove rather than a thing to notice, and it is what the next two minutes are for. Put the three ranges on one axis, running from zero to ninety. The first identity, the one over the longest side, is closed at both ends. The second, over the side touching the angle, is open at ninety.

5:47The third, over the side facing the angle, is open at zero. Here is the claim that holds the three together. The angle an identity is not claimed at is exactly the angle where the side it divided by has collapsed. I checked that as a table rather than as a remark. Three identities, two ends each. Six cells. In two of them the divisor had gone and the identity had nothing to say. In the other four the divisor was still there and the identity held.

6:22The two cells that would break the claim are both empty. There is no case where the divisor vanished and the identity survived, and no case where the divisor was fine and the identity went missing. So a range is not a warning stapled to a formula. It is the receipt for the division that produced it. And the first identity keeps both of its ends for exactly one reason. It divided by the longest side, and the longest side is the one that never collapses.

6:58One more thing about the second identity, because it gets written two ways. One plus the squared tangent equals the squared secant. And: the squared secant minus the squared tangent equals one. Those are the same statement. Move one term across the equals sign. It is worth saying out loud, because meeting the two of them a week apart is how one identity turns into two formulas in a notebook.

7:28The range travels with it, by the way. Rearranging changes nothing about which angles you may claim it at. I put both forms through the same comparison on every column of values, including the ends where neither form has anything to say. They give the same verdict every time, gaps included. And to be sure that comparison is capable of disagreeing, I ran the rearrangement done with the wrong sign through it as well.

8:00That one disagrees, in exactly the places it should. Now what these are for. Suppose you are told one ratio, and nothing else at all. Say the tangent is one over root three. Squared, that is a third. The cotangent is its reciprocal, so the squared cotangent is three. One plus a third is four thirds, and by the second identity that is the squared secant. The squared cosine is the reciprocal of that. Three quarters.

8:34The squared sine is one minus three quarters, by the first identity. A quarter. And the squared cosecant is the reciprocal of that. Four. Six values out of one, using nothing but the identities and the reciprocal pairings. That is the thirty degree column, complete. And this is not a trick that happens to work at thirty degrees. I ran the same chain from the tangent, from the cotangent, and from the secant, at every one of the hundred and four acute readings in the population.

9:11All three reproduced the six measured values every time. Started from the wrong one of the six, they do not - except on the eighteen readings where the two legs are equal, so the tangent and the cotangent are the same number and a crossed wire cannot be caught by arithmetic. So far the identities have been things to use. Now they are things to prove with. And a proof of an identity is a different shape from solving an equation.

9:44You do not write down the thing you want and push it around until it looks true. You start on one side, and work it down to the other. The discipline matters, because assuming what you are proving is the commonest way one of these quietly goes wrong. Three collapses are coming. Each begins with an expression that looks like nothing in particular, and ends at a single number. And the interesting question about them is not whether they collapse.

10:16It is which of them actually needed an identity. That question has an answer you can measure, and the measurement is the last thing in this video. First one. A secant, times one minus the sine, times a secant plus a tangent. Three brackets with nothing obviously in common, and no reason to expect them to go anywhere. Convert everything to sines and cosines. That is the default move, and it is right more often than not.

10:50The secant is one over the cosine. The tangent is the sine over the cosine. So the third bracket becomes one plus the sine, all over the cosine. Collect what is left. One minus the sine, times one plus the sine, over the squared cosine. The top is a sum times a difference, so it is one minus the squared sine. And there is the identity, used once. One minus the squared sine is the squared cosine.

11:23Squared cosine over squared cosine. One. I ran that product on sixteen figures whose three sides are whole numbers, and it came to one on all sixteen of them. Second one, and this is the one worth slowing down for. A cotangent minus a cosine, over a cotangent plus a cosine. Show that it equals a cosecant minus one, over a cosecant plus one. The instinct here is to reach for an identity. Resist it, and look at the expression first.

12:00Write the cotangent as the cosine over the sine. The top is now the cosine over the sine, minus the cosine. Take the cosine out in front, and what is left is one over the sine, minus one. The bottom is the same thing with a plus. So the cosine is a common factor above and below, and it cancels. And one over the sine is the cosecant. That is the entire proof. No identity was used anywhere in it.

12:33Here is how I know that, rather than believe it. I evaluated all three collapses twice over. Once on sixteen genuine pairs, and once on six pairs of numbers that are not a sine and a cosine of anything. The first collapse holds on the genuine pairs and fails on all six of the others. The third one does the same. This one holds on all twenty two. It never needed the relation between the sine and the cosine, because it was never doing anything but factoring.

13:10Third one, and it teaches the move that matters most. A sine minus a cosine plus one, over a sine plus a cosine minus one. Show that it equals one over a secant minus a tangent. Look at the target before you touch the expression. The target is written in secants and tangents. So convert into secants and tangents, and do that before simplifying anything at all. Divide the top and the bottom by the cosine.

13:42The top becomes a tangent, minus one, plus a secant. The bottom becomes a tangent, plus one, minus a secant. Now multiply above and below by a tangent minus a secant. The bottom collapses to a squared tangent minus a squared secant, and the second identity turns that into minus one. The rest is tidying up. But the lesson is the first move and not the algebra. Look at which ratios the answer is written in, and convert into that vocabulary before you start.

14:21So which of the three do you reach for? It is not luck and it is not memory. It is vocabulary. Look at which ratios the expression already contains. A sine and a cosine, and it is the first identity. A tangent and a secant, and it is the second. A cotangent and a cosecant, and it is the third. And if a difference of one and a squared ratio is sitting anywhere on the page, that is an identity waiting to be replaced.

14:54One equation. Three sides to divide it by. Three identities. And three ranges that are nothing more than the receipts for the divisions that made them. Divide by a side, and you may claim the result only for as long as that side is there.