When the two cancellation rules hold, and the angles where the second one fails

0:00Two statements. They are written one line apart, they look like mirror images of each other, and one of them is true always while the other is true only sometimes. Here is the first. Take a number, send it in through an inverse function, and send whatever comes out back through the direct one. You get your number back. Here is the second. Take an angle, send it in through the direct function, and send whatever comes out back through the inverse one. You get an angle back, but not necessarily the one you sent.

0:38That is the whole topic. The two compositions are not the same composition, the order matters, and the second one carries a condition that the first does not. Nothing here is exotic. Two thirds of a half turn is only a sixth of a half turn past the end of the interval, and it already fails. So this cannot be learned as a formula. It has to be learned as a procedure with a check inside it, and the check is the part everybody drops.

1:12Everything asserted in this video was measured first: six families of ratios, two candidate intervals, three hundred and sixty-one angles, and every principal value hunted rather than looked up. Start with the one that needs no condition at all. Each inverse function accepts a certain set of numbers, and that set is not something to memorise. It is whatever the function can actually name an angle for. So offer the same sixty-one numbers, spread evenly from minus six to six, to all six inverse functions, and ask each one which of them it can answer.

1:57The inverse sine accepts eleven of them and the inverse cosine eleven: the ones from minus one to one. The inverse secant and the inverse cosecant accept fifty-two each, everything except the strip in the middle. The inverse tangent and the inverse cotangent accept all sixty-one. Now run the first rule on every value each one accepts. Send the number in, take the angle that comes out, and send that angle back through the direct function.

2:32Eleven tried and zero failures for the sine. Eleven and zero for the cosine. Sixty-one and zero for the tangent and for the cotangent. Fifty-two and zero for the secant and for the cosecant. Not one failure anywhere, and that is not luck. The number you handed in was, by construction, a number that inverse function accepts, so it never leaves the set the rule is stated over. That is why the first rule needs no case work. There is nowhere for the input to go wrong.

3:11Now the other order, and it behaves completely differently. Take three hundred and sixty-one angles spread evenly across three whole turns either side of zero. Send each one through the direct function and then back through the inverse. For the sine, all three hundred and sixty-one angles have a value to send. Sixty-one of them lie inside the promoted interval. And the number that come back unchanged is sixty-one. Exactly the ones inside. The number of angles from outside the interval that come back as themselves is zero.

3:52The cosine gives the same shape: three hundred and sixty-one with a value, sixty-one inside, sixty-one returned, none from outside. The tangent has three hundred and fifty-five angles with a value, because six of them sit exactly where the tangent has no value at all. Fifty-nine of those lie inside its interval, and fifty-nine come back. Every family tells the same story. The angles the second rule holds at are exactly the angles inside the promoted interval, and there is nothing else to remember.

4:33Put the two rules side by side. The number of families the first rule holds everywhere for is six. The number of families the second rule holds everywhere for is zero. Why the asymmetry? The reason is one sentence long and it is worth saying slowly. The output of an inverse function is confined to the promoted interval. That is what promotion means. Whatever you feed in, the answer comes back from inside that interval and from nowhere else.

5:09So run the first rule. The answer lands inside the interval, and then the direct function sends it to the number you started with. The confinement never gets in the way, because you never asked for a particular angle. Now run the second rule. You handed in an angle, and you want that angle back. But the answer can only come from inside the interval. If your angle was not in there, it cannot come back.

5:40It is not that the machinery breaks. It is that you asked it for something it is not allowed to produce. Which kills the next misconception straight away. When the second rule fails, the expression is not undefined. Across three turns either way, the number of angles where the rule fails for the sine is three hundred, and the number of those where the expression has no value at all is zero. Three hundred perfectly good answers, none of them the angle you sent.

6:15The cosine reads the same: three hundred failures, none of them undefined. The tangent reads two hundred and ninety-six failures, and again none undefined. There is exactly one place in this topic where an expression genuinely has no value, and it is a different problem altogether. At a quarter turn the tangent has no value, so there is nothing to send in. There is a warning that belongs on this whole method, and it is usually stated in one breath and then forgotten.

6:52It goes like this. These results hold inside the promoted intervals, and they hold wherever the expressions are defined. Some of them fail for some inputs. Which inputs, exactly, is a question this level does not pursue. That last clause is the important one, and it is honest rather than evasive. Characterising every failure in general is genuinely harder than the material it would have to sit in. But notice what it does not say. It does not say that the check can be skipped in a particular case.

7:31It says the general theory is out of reach, and therefore that you must do the check by hand, every time, on the angle in front of you. So this is a procedure and not a formula. And a procedure with a step skipped is not a shorter procedure. It is a different one, and it gets different answers. The rest of this is that procedure: watch it fail once, then three moves, then five worked items.

8:02Here is the failure, on the smallest interesting angle. Take three fifths of a half turn, a little over a right angle, and ask for the inverse sine of its sine. Apply the second rule exactly as written. It returns the angle you sent, three fifths of a half turn. Write that down. Now check it. The promoted interval for the inverse sine runs from minus a quarter turn to a quarter turn. Three fifths of a half turn is larger than a quarter turn. It is outside.

8:39So that answer is wrong, and writing it down first was deliberate. It is the fastest way to see that the check is a real step and not decoration. Now the repair. You need a different angle with the same sine, one that does lie inside the interval. Subtract the angle from a half turn. Three fifths of a half turn becomes two fifths of a half turn, and two fifths is less than a half, so it is inside.

9:10Does it really have the same sine? Here that is a measurement rather than an identity recalled. The hunt that produced the answer came back with two fifths of a half turn, and the subtraction from a half turn landed exactly on it. That is the shape of every question of this kind. Apply, check, repair. Three moves, in order, and the order is not negotiable. First: apply the rule and write down the angle you were given. This is not the answer. It is the candidate.

9:47Second: check whether the candidate lies inside the promoted interval for that family. Two of the six intervals are open at both ends, so an angle sitting exactly on an end is outside and not inside. Third: if it is inside, you are finished. If it is outside, replace it with an angle that has the same value and does lie inside. Only the third move needs any thought, and it is the move that depends on which family you are in.

10:20One warning about the second move before we go on. The check is about the interval, not about the size of the angle. Of the three hundred failing angles for the sine, sixty of them sit within a half turn of zero. Small angles fail too, if they are on the wrong side. So: which replacement do you reach for? Four candidate moves were put through every failing angle across three turns and scored. Not chosen. Scored.

10:54Subtract the angle from a half turn. That move repairs sixty of the sine's failing angles, and zero of the cosine's. Subtract the angle from a whole turn. That repairs sixty of the cosine's, and exactly one of the sine's, which is a coincidence at a whole turn where the two moves happen to agree. Subtract a half turn from the angle. That repairs fifty-nine of the tangent's failures, and it is the tangent's move alone.

11:27And subtracting a whole turn repairs sixty-one for the sine, sixty-one for the cosine and fifty-nine for the tangent. That is the move you need when the angle is more than a turn from home. Three families, three different first moves. None of this is a matter of taste. What happens if you use the wrong one? Take two thirds of a half turn on the sine. The right move gives one third of a half turn. The tangent's move, subtracting a half turn, gives minus one third of a half turn.

12:05Same size, wrong sign, and the sign is the whole answer. That single substitution is the commonest wrong answer in this topic. One more honest number. Over three turns, the four moves between them repair one hundred and twenty of the sine's three hundred failures. They are first moves, not a complete theory, and for angles further out you reduce by whole turns first and then use one of them.

12:37Five items, run the same way every time. One. The inverse sine of the sine of two thirds of a half turn. Candidate: two thirds. Check: larger than a quarter turn, so outside. Repair: subtract from a half turn, giving one third of a half turn, which is inside. That is the answer. Two. The inverse tangent of the tangent of three quarters of a half turn. Candidate: three quarters. Check: the tangent's interval runs from minus a quarter turn to a quarter turn and is open at both ends, so three quarters is outside. Repair: subtract a half turn, giving minus one quarter of a half turn.

13:25Look at that answer. A positive angle went in and a negative angle came out. That is not a mistake, and it is the item that makes people distrust the whole method. The output is confined to the promoted interval and nothing else constrains it. If the interval holds negative angles and the answer belongs there, the answer is negative. Three. The inverse cosine of the cosine of seven sixths of a half turn, offered with four options. Candidate: seven sixths. Check: the cosine's interval runs from zero to a half turn, so seven sixths is outside. Repair: subtract from a whole turn, giving five sixths of a half turn.

14:16The first option offered is seven sixths, the unrepaired candidate, sitting there for anyone who stops after move one. The gap between the trap and the answer is exactly one third of a half turn. Four. The sine of a third of a half turn, minus the inverse sine of minus one half. The inner term is an inverse value with a negative argument, and it comes to minus one sixth of a half turn.

14:47Subtracting a negative adds. A third plus a sixth is a half, and the sine of a quarter turn is one. No repair anywhere in that one. It needs the interval only to get the inner sign right. Five. The inverse tangent of root three, minus the inverse cotangent of minus root three. The first term is one third of a half turn. The second is the one that catches people. The inverse cotangent's interval runs from zero to a half turn and holds no negative angle at all, so even with a negative argument the answer is positive: five sixths of a half turn.

15:32One third minus five sixths is minus one half of a half turn. Minus a quarter turn. Now angles more than a whole turn from home, where the first move is arithmetic. The inverse cosine of the cosine of thirteen sixths of a half turn. Thirteen sixths is a whole turn plus one sixth. A whole turn changes nothing, so the answer is one sixth of a half turn. And the inverse tangent of the tangent of seven sixths of a half turn. Seven sixths is a half turn plus one sixth. The tangent is unchanged by a half turn, so the answer is again one sixth of a half turn.

16:21Two different reductions, a whole turn in one and a half turn in the other, landing on the same angle. That is an accident of the numbers, and it is worth saying out loud, because anyone who notices it and is not told will go looking for a pattern that is not there. The scoring confirms it. The move that repairs the first is subtracting a whole turn. The move that repairs the second is subtracting a half turn. Different moves, same destination, and no rule connecting them.

16:57The general recipe for a distant angle is: bring it within a turn first, then check, then repair if you still need to. Two arithmetic steps and one decision. There is a short form these two rules usually get quoted in, and it is where the check goes to die. The short form says: the sine of the inverse sine of a number is the number, and the inverse sine of the sine of an angle is the angle. Two clean lines, one under the other.

17:32Both conditions have been dropped. Write them back in. The first line holds for every number from minus one to one, which is every number the inverse sine accepts, so that condition is the whole of its domain and costs you nothing. The second holds for every angle from minus a quarter turn to a quarter turn, which is a slice of the sine's angles and not all of them.

17:59That is the asymmetry in one glance. One condition is a whole domain. The other is a slice. Anyone revising from the short form alone will get the first kind of question right and the second kind wrong, and will not know why. So when you write these two lines down for yourself, write the two intervals underneath them. They are not decoration. One of them is the entire content of the second rule.

18:31What has this actually settled? Two compositions that look like mirror images are not mirror images. One holds on the whole domain of the inverse function. The other holds on a slice of the direct function's angles, and that slice is the promoted interval. When the second one fails, the expression still has a value. Three hundred failing angles for the sine across three turns, and zero of them undefined.

19:03The check is per angle and it is not optional, because the general question of which inputs survive is left open at this level. What you are given is a procedure, and the check is a step in it. The repair is three moves. Apply, check, replace. And the replacement depends on the family: subtract from a half turn for the sine, subtract from a whole turn for the cosine, and shift by a half turn for the tangent.

19:35Two of the six intervals are open at their ends, so an angle sitting exactly on an end is outside them. Hold those, and every question of this shape is the same question. Drop the check, and you will get wrong every one of them that lies outside the interval, which across three turns is three hundred angles out of three hundred and sixty-one.