The six principal value branches, and the table that fixes them

0:00Six inverse trigonometric functions. Each one needs two things written down before it can be used: the numbers it will accept, and the angles it is allowed to answer with. Six rows, two columns, twelve cells. Laid out like that it looks like twelve separate facts to be memorised, and almost everybody memorises them that way. They are not twelve facts. Across all six rows the answer column uses only two intervals. Two.

0:33One runs from minus a quarter turn to a quarter turn. The other runs from zero to a half turn. Every single row answers inside one of those two, with at most a couple of points taken out. And the points that get taken out are not decided by anyone. They are exactly the places where the ratio in that row has no value at all. So the whole block comes down to this: pick one of two intervals, then let the ratio itself punch its own holes in it.

1:07That is the claim, and in this video it is not asserted. It is derived. Two intervals are named at the start and nothing else is; every row you will see comes out of a measurement. Start with a single row, because a row says less than people think it does. It says two things. What the inverse function accepts, and what it returns. Nothing else. It does not tell you how to compute anything.

1:38Take the inverse sine. It accepts every number from minus one to one, ends included, and it returns an angle between minus a quarter turn and a quarter turn, ends included. Read that as a promise about answers. Hand this function a number in the accepted set, and the angle it hands back will be in the returned set. Always. That is the entire contract. The second half is where all the trouble lives, so it is worth being blunt about it. The returned set is not a suggestion, and it is not a range of typical answers. It is a hard boundary.

2:21If an angle you have written down sits outside that set, it is not the answer to that expression, however correct the arithmetic that produced it. Everything else in this video is about how the six returned sets get chosen, and about what goes wrong when a student remembers the accepted set and forgets the returned one. Take the accepted sets first, because they are the easy column and they have a shape.

2:52Ask each ratio what values it actually takes. Not what it should take — sweep it across several turns and watch. The sine and the cosine never leave the stretch from minus one to one. Nothing they do goes outside it, and everything inside it happens. The cosecant and the secant do the exact opposite. Neither of them ever produces a number smaller than one in size. They live outside the unit and never enter it.

3:24The tangent and the cotangent take every number there is. So there are three shapes, and each is shared by exactly two functions. That is the whole input column: three facts, not six. It is also the first place students lose marks, and the mistake is always the same one. The inverse secant is handed a half, or a third, and asked for an angle. There is no such angle. No angle has a secant of a half, because the secant never comes inside the unit. The expression is not hard — it is empty.

4:04Two rows share a set of inputs, and the two rows sharing yours are not always the two you expect. Now the answer column, which is where the real structure is. Only two intervals will be used, so put them both on one axis and look at them together. The first runs from minus a quarter turn to a quarter turn. It is symmetric about zero, and half of it lies below zero.

4:34The second runs from zero to a half turn. It sits entirely at or above zero. Not one angle in it is negative. That difference is not a detail. It is the single most consequential fact in this whole topic, and we will come back to it twice. Both intervals are a half turn wide. Both are, in a sense, one full sweep of the circle folded in half — long enough to reach every value once, short enough not to repeat.

5:07Three of the six rows will end up on the first. Three will end up on the second. And which functions land where is not a choice anybody made. That is the part worth proving rather than announcing, so proving it is what comes next. Before the choosing, one more measurement, and it explains every bracket in the block. Walk each anchor and ask the ratio, at each angle, whether it has a value there at all.

5:38The sine and the cosine always do. Nothing is removed from their intervals, and both ends stay. The cosecant has no value at one point inside the first anchor — the middle, where the sine it is built on vanishes. One point out of the middle, and the two ends survive. The secant has no value at one point inside the second anchor, again the middle. Same shape, different interval.

6:07The tangent has no value at either end of the first anchor. Both ends come off, and nothing is taken from the middle. The cotangent has no value at either end of the second anchor. Both ends come off there too. Three shapes: nothing removed, one point out of the middle, both ends off. Each shape used by exactly two rows. So the brackets were never a convention. A square bracket means the ratio has a value at that end. A round one means it does not, and there is nothing to keep.

6:45The block of six rows is two anchors and three ways of losing points, and every one of those losses was forced by a ratio dividing by zero. Now the choosing. Six ratios, two anchors: twelve possible pairings. Six of them are the block, and the other six are wrong. The question is what separates them. One test. Walk the anchor, collect the values the ratio takes, and count how many angles land on the most popular value. If that count is one, the ratio never repeats itself there.

7:26Run it on all twelve. Six pass. Exactly six, one per ratio, and they are precisely the six rows. The other six fail, and two of the failures are worth watching because they look like they should work. The tangent on the interval from zero to a half turn. Its middle is missing, so it does not blow up twice, and it rises steadily on each piece. It looks fine.

7:54It is not fine. The tangent of zero is zero, and the tangent of a half turn is also zero. Both ends are kept, both ends give the same answer, and an undoing map handed zero would have to choose between them. The cotangent fails the same way on the other anchor. Its cotangent at minus a quarter turn is zero, and at a quarter turn it is zero again. Two angles, one value.

8:26One test, run twelve times, and the six rows fall out. Nothing was assumed. There is a second demand on a branch, and it is a different demand, so it is worth measuring separately rather than folding it in. The inverse function has to have an answer for every number its row accepts. Not most of them. Every one. Check it by hunting. Take each accepted value in turn, search the branch for an angle the ratio sends there, and then check two further things separately: that the angle really is inside the branch, and that the ratio really does send it to that value.

9:09All six rows pass. Every accepted value has an angle, that angle is inside, and it lands. And that hunt is capable of failing, which is what makes the result mean something. Hand it a lazy searcher that offers the middle of the branch no matter what it was asked for, and of the forty-one values it is asked to reach, forty come back with an angle that does not land there.

9:38Now the interesting part. Run the reaching test on all twelve pairings, and eight of them pass, not six. So reaching everything is the weaker demand. It clears the two failures we just watched, because the tangent from zero to a half turn does reach every number, and so does the cotangent on the other anchor. They were thrown out for repeating, not for missing anything. Not one pairing in the twelve is one-one but short of inputs. Every rejection in the whole block is a repetition.

10:16So the block is settled. Now the convention that makes it binding, because without it the block is only advice. The convention is this. When an inverse trigonometric symbol is written with no branch named, the promoted branch is meant. Not any branch. That one. That single sentence is what allows the question to have one answer, and it is the sentence students skip. Here is why it is needed. Ask for an angle whose sine is one over root two, and count how many angles do it. Over ten turns either side of zero, there are twenty.

10:57Twenty correct angles. Any of them satisfies the equation. Not one of them is more true than another. Of those twenty, the promoted branch keeps exactly one, and that one is the principal value. So a principal value is not simply the answer. It is the answer subject to a constraint, and the constraint is the returned set of its own row. Take the constraint away and there is no single answer to take. Which means that if you cannot say which interval your answer had to lie in, you have not finished the question.

11:38The method is three moves, and it never changes. Move one. Give the answer a name. Say the thing you want is an angle, and call it theta. Move two. Turn the inverse statement into a direct one. If theta is the inverse sine of one over root two, then the sine of theta is one over root two. The inverse symbol is now gone, and what is left is an ordinary trigonometric equation.

12:09Move three. Among all the angles that satisfy it, take the one inside the promoted interval. Run it. The sine of theta is one over root two. The sine of an eighth of a turn is one over root two — that is a standard angle you already know. Is an eighth of a turn inside the interval from minus a quarter turn to a quarter turn? It is, comfortably. So that is the principal value, and it is pi by four.

12:41Notice what move three did and did not do. It did not check whether the angle was reasonable, or small, or positive. It checked membership of one specific set, and that is all it ever checks. On a positive argument this feels like a formality. That is exactly why the minus sign comes next. Same three moves, and now a minus sign, which is where every mark in this topic is actually won or lost.

13:14Find the inverse cotangent of minus one over root three. Move one, name it theta. Move two, the cotangent of theta is minus one over root three. Move three needs a standard angle, so start from the positive one. The cotangent of a sixth of a turn — pi by three — is one over root three. Now the sign. Subtract an angle from a half turn and the cotangent flips sign, so the cotangent of a half turn minus pi by three is minus one over root three. That angle is two pi by three.

13:55And here is the move that matters. Two pi by three is not the only angle with that cotangent. Minus pi by three has it too, and over three turns either side of zero there are six angles that do. Which one survives is decided by the row, and the inverse cotangent answers from zero to a half turn. Of those six, exactly one is in there, and it is two pi by three. So that is the principal value.

14:29The arithmetic was never the hard part. The hard part was that a minus sign in the question does not mean a minus sign in the answer, and for this function it never can. That last remark generalises, and it generalises into the most useful rule in this topic. It is also a rule the block never states out loud. Three of the six rows answer inside the interval from minus a quarter turn to a quarter turn: the inverse sine, the inverse cosecant, the inverse tangent.

15:08Three answer inside the interval from zero to a half turn: the inverse cosine, the inverse secant, the inverse cotangent. Now feed every one of them a negative number — not one example each, but a whole sweep of negative arguments across everything each of them accepts. The first three answer with a negative angle. Every time, without exception, across the whole sweep. The second three never answer below a quarter turn. Every single answer is more than a quarter turn, and almost every one is obtuse — strictly between a quarter turn and a half turn.

15:52There is exactly one place where obtuse is the wrong word. At the argument minus one, the inverse cosine answers a half turn exactly, and so does the inverse secant. Straight, not obtuse. Two answers out of the whole sweep. So the misconception has a name and a cure. The inverse cosine of a negative number is not negative. It cannot be, because there is no negative angle anywhere in the set it answers from.

16:25And the mirror of the rule holds too: hand those same three a positive number and not one of them ever answers above a quarter turn. Two anchors, one split, and on half of a typical exercise that split is the entire difference between right and wrong. Expressions in this topic are often sums, and sums invite a shortcut that does not work. The shortcut is to combine the arguments first and take one inverse at the end. It is wrong, and it is wrong badly enough to be worth measuring.

17:04Take sixty-four pairs of inputs, add each pair inside the inverse sine, then add the two inverse sines separately, and compare. Eight pairs agree, and all eight are the trivial ones where the two inputs cancel to zero. Among the pairs that do not cancel, none agree. Not one. The largest disagreement is about a tenth of a radian, which is easily large enough to move an answer from one option to another.

17:36So the method is: settle each term inside its own row first, then add. Take the inverse tangent of one, plus the inverse cosine of minus one half, plus the inverse sine of minus one half. Term one is pi by four. Term two takes a negative argument into a row anchored at zero, so it is obtuse: two pi by three. Term three takes a negative argument into a row anchored below zero, so it is negative: minus pi by six.

18:12Three different rows, three different rules, in one expression. Add them and you get three pi by four. One more. The inverse cosine of one half, plus twice the inverse sine of one half. That is pi by three, plus two lots of pi by six, which is two pi by three. Settle, then add. Never the other way round. Two last questions, both of which test the block and nothing else.

18:43First: which set does the inverse sine answer from? Four candidates. Zero to a half turn with the ends kept. Minus a quarter turn to a quarter turn with the ends kept. Then the same two intervals again with their ends dropped. Put all four through the two tests. Only one survives, and it is the second. Watch how the other three die, because the two ways of dying are the two traps.

19:13The right numbers with the wrong brackets: minus a quarter turn to a quarter turn, ends dropped. It never repeats itself, so it passes the first test. But dropping those two ends loses the only angles that answer minus one and one, so two accepted inputs are left with no answer at all. The wrong numbers, with either bracket: zero to a half turn. Now the sine repeats. Ask it for nine tenths and two different angles in that interval answer, sitting about nine tenths of a radian apart. An undoing map cannot pick.

19:54So two options differing only in bracket shape are not the same option. Read the brackets aloud. Last question. The inverse tangent of root three, minus the inverse secant of minus two. The first term is pi by three. The second has a negative argument going into a row anchored at zero, so it is obtuse: the cosine of the answer is minus one half, and the answer is two pi by three.

20:27Pi by three minus two pi by three is minus pi by three. Two intervals, three ways of trimming them, one convention that makes the block binding, and one rule about negative arguments. That is the whole of it, and it fits in a sentence you can rebuild from scratch.