Why none of the six ratios has an inverse on its natural domain

0:00Here is a question with a trap in it. The sine of some angle is one half. What is the angle? You know one answer straight away. Pi by six. But five pi by six also has a sine of one half. So does pi by six plus a full turn. So does five pi by six plus a full turn. Take the angles built by adding whole turns to those two, three turns each way. That is fourteen angles, all fourteen of them distinct.

0:33Ask the sine for its value at every one of them, and you get back one number. One value, fourteen inputs, and the list would go on for ever. So the question as asked has no answer. Not a hard answer — no answer, because nothing can choose. This whole topic is about what exactly goes wrong there, and about which half of a two-part test it breaks. Getting that right is the difference between the repair that follows looking like sense and looking like magic.

1:09The test comes from what an undoing function has to do. It has to take an output and hand back the input it came from. One input, definitely, and only one. That means two demands, and they are separate demands. First: no two different inputs may share an output. Otherwise the undoing has to choose between them, and it cannot. Second: every element of the declared target set must actually be reached. Otherwise the undoing is handed something it has no answer for.

1:45The first demand is called one-one. The second is called onto. A function has an inverse exactly when it passes both. So when something fails the test, the useful question is never just whether it failed. It is which half it failed. Hold on to that. It decides everything that comes after. Now the six ratios, each written out with the two sets it comes with: what it accepts, and where its values are declared to live.

2:19Sine and cosine accept every real number, and their values are declared to lie in the closed interval from minus one to one. Tangent accepts every real number except the odd multiples of half pi, and its values are declared across the whole real line. Cotangent accepts every real number except the whole multiples of pi, and its values also run across the whole line. Secant accepts the same inputs as the tangent, and cosecant the same as the cotangent.

2:56But those two have a different output set. Their values are declared on the real line with the open interval from minus one to one taken out. That is a hole punched in the middle of the line, and it is there deliberately. Six lines. Two of them accept everything, four of them do not. Three different shapes of output set. Every one of those details is about to do work.

3:26Start with the four subtractions, because they are not decoration. The tangent is the sine over the cosine. The secant is one over the cosine. Both of them stop existing wherever the cosine is nought. The cotangent is the cosine over the sine, and the cosecant is one over the sine. Both stop existing wherever the sine is nought. So the excluded inputs are not a rule to memorise. They are wherever a denominator vanishes.

4:00Take a window of angles running from minus three pi to three pi, stepping by a twelfth of pi. Seventy-three angles. Walk it and ask where the cosine is nought. Six angles, and they are the odd multiples of half pi. Ask where the sine is nought. Seven angles, and they are the whole multiples of pi. Not one angle is in both lists. And if you merge the two lists and measure the gaps, every gap is the same: half pi. The two removed sets interleave perfectly, one then the other, all the way along.

4:39That is why the sine and the cosine keep all seventy-three angles, while the tangent and the secant keep sixty-seven, and the cotangent and cosecant sixty-six. Now the output sets, which come in three shapes. The closed interval, for sine and cosine. Their values never leave it, and every point of it is reached. The whole line, for tangent and cotangent. Nothing is out of reach. And the punctured line, for secant and cosecant.

5:15That hole is worth a moment, because it is easy to assume those two behave like the tangent and take every value. They do not. Walk the whole admitted window and count how many angles the secant sends strictly inside the interval from minus one to one. Nought. Same for the cosecant. Nought. The tangent, on the same window, puts plenty of angles inside that interval — so the count really was capable of coming out otherwise.

5:49Secant and cosecant simply cannot land there, which is why their declared target has the hole in it, and why their inverses will later have a domain shaped differently from all the others. Now back to the trap from the opening, and let us make it a measurement rather than a story. Take that window of seventy-three angles again, and bucket them by the value the sine sends them to.

6:18Seventy-three angles land on just thirteen distinct values. And the biggest bucket holds seven angles — seven different inputs, one output. Do it for the cosine: thirteen values again, biggest bucket six. The tangent: sixty-seven admitted angles, eleven values, biggest bucket seven. The cotangent: sixty-six angles, eleven values, biggest bucket six. The secant: sixty-seven angles, twelve values, biggest bucket six. The cosecant: sixty-six, twelve, six. Six ratios, and every single one of them hands some output to more than one input.

7:04And this is not an artefact of a coarse grid. The window only covers three turns each way; go further and the buckets only grow. Repetition is the obstruction, and it is the same obstruction six times over. So which half of the test fails? The answer people reach for is that these functions are not onto — that there is some value the sine never manages to take. There is not. Go and look.

7:36Here is how to look properly. Do not draw the curve and squint at it. Walk the declared target set, and for each target produce an angle that lands on it, then check two separate things: that the domain admits the angle, and that the ratio really sends it there. Run that over forty-one targets spread across the closed interval, for the sine. Forty-one tried. Nought that no angle could be named for. Nought refused by the domain. Nought that landed anywhere else.

8:12The cosine: the same. The tangent and cotangent, over forty-one targets spread along the line: the same. The secant and cosecant, over forty targets outside the hole: the same again. Across all six, the number that fail to cover their declared target is nought. The number that repeat is six. And that routine is capable of failing. Hand it a producer that returns the target itself as though a number were an angle, and it reports forty of forty-one landing astray. Hand it one that always names an angle the tangent has no value at, and it reports all forty-one refused at the door.

8:59So the reading is not a rubber stamp. Each of the six is onto, exactly as declared. The half that fails, every time, is one-one. That matters, because everything that follows shrinks a domain. Nothing that follows ever touches a target set. Look at the failure from the undoing function's side for a moment, because that is where it bites. You hand it the number one half and ask for the angle.

9:30It has fourteen candidates in front of it just within three turns, and that is only where we stopped counting. Pi by six. Five pi by six. Pi by six plus a turn. And so on, in both directions, without end. Every one of them is a correct answer to the question as asked. That is exactly the problem. A function has to return one value. Not a set of values, not a favourite, and not whichever one is convenient.

10:03So there is no rule to be found here, however clever. The obstruction is not that the answer is hard to compute. The obstruction is that the question does not have one answer, and no amount of computation fixes that. Which points at the only possible repair: change the question. Give the sine a smaller set of inputs, so that only one of those candidates survives. Before the repair, it is worth being clear about what you get once a function does pass both halves.

10:40You get exactly one undoing function. Not several. Its inputs are the outputs of the original, and its outputs are the inputs of the original. The two sets swap places. It is itself one-one and onto. Undo the undoing and you are back to the function you started with. And running one after the other, in either order, leaves everything where it was — on the appropriate set each time.

11:09None of that is new here; it is what invertible means, taken from the general case. The only thing this topic adds is that none of the six ratios qualifies yet, and it says precisely why. One notation warning, and it is the most expensive misreading in the subject. A raised minus one written after a function's name means the undoing function. A raised minus one written after a number means one divided by that number.

11:44They are not the same operation, and putting them next to each other is the fastest way to see it. The inverse sine of one half is pi by six, which is about nought point five two four. One divided by the sine of pi by six is two. Nought point five two four against two. They are not close, and they are not the same kind of object: one is an angle, the other is a ratio.

12:13Across the whole closed interval those two expressions agree nowhere at all. The confusion is understandable, because the same little symbol is doing two different jobs. Read what it is attached to. So: the repair. If repetition is the problem, keep fewer inputs. Take the sine and keep only the angles from minus half pi to half pi — thirteen of them, on our step. Bucket them by output. Thirteen inputs, thirteen distinct values, biggest bucket one.

12:50One-one, at last. And nothing was reached that was not reached before: run the coverage walk again over the same forty-one targets and it is still forty-one tried, nought missed. The rule did not change either. Of the thirteen angles kept, the number sent anywhere new is nought. That is what a restriction is. Same rule, fewer inputs. The sine of pi by six is still one half. The full window held sixty more angles than the kept one, and every one of those was a duplicate answer waiting to cause trouble.

13:29Which interval to keep is the next question, and it is not obvious — it has to contain exactly one angle for each value, and there is more than one honest way to choose it. But the shape of the answer is now fixed: shrink the domain, never the target. One last thing, briefly, because it explains why the notation feels bolted on. The ratios themselves are ancient. The subject took shape in India, passed into Arabic mathematics, and reached Europe centuries later.

14:07The raised minus one is not ancient at all. It arrived in the nineteenth century, from an astronomer who wanted a compact way to write the undoing of a function. So the symbol is younger than the mathematics it sits on by well over a thousand years, which is exactly why it collides with the older habit of writing one over a quantity. And that is the topic. Six ratios, six target sets that each of them fully covers, and six failures of the same kind.

14:43Not a missing value anywhere. Just too many angles giving the same answer.