Substituting an angle for the variable to collapse a messy expression

0:00Here are five expressions. They are meant to look like five separate problems, and for most people they succeed. One has a root of one less the square of the variable sitting inside an inverse sine. One has a root of one plus the square. One has a root of the square less one. One has no root at all, just a cosine over one less a sine. One has a constant standing beside the variable.

0:30Five shapes, five different-looking arguments, and the usual response is to hunt for five different tricks. They are one problem. Every single one of them is solved by the same move, made in the same place, for the same reason. The move is this. Somewhere in the expression there is an inverse value. Stop treating it as a number and give it a name as an angle. The variable then becomes a ratio of that angle, the awkward part becomes a second ratio of the same angle, and something you already know collapses the two into one.

1:11What makes this worth doing slowly is not the move. It is the thing that comes after the move, which is the step almost everybody skips, and which is the entire reason these expressions come with ranges attached. Watch the move once, slowly, on the first of them. The variable is sitting under a root, inside an inverse sine, multiplied by itself in a way that is going nowhere. So name the angle. Let the angle be whatever the inverse sine of the variable is. That single sentence does two things at once.

1:52First, it says the sine of the angle is the variable. The variable is now a ratio. Second, and this is the part that pays, one less the square of the variable is now one less the square of a sine, and the root of that is a cosine. The root is gone. Not simplified, not approximated. Gone, because a root of a difference of two squares that add to one is exactly what a right triangle is for.

2:23So the argument has become twice a sine times a cosine of the same angle. That is a thing with a name. It is the sine of twice the angle. The whole expression is now the inverse sine of the sine of twice the angle, and every trace of the root has disappeared. Which raises the only question left. Can you cancel an inverse sine against a sine? Hold that. It is the whole video.

2:54Before the cancelling, the choice. Which substitution do you make? There is a widespread belief that this is where the cleverness lives, that you look at the problem and see it. You do not have to see anything. The shape of the awkward part tells you, and it tells you in four cases and no more. So do not take my word for it. Take four shapes, hand each of them to four substitutions, and record what actually comes out.

3:25A root of one less the square. A root of one plus the square. A root of the square less one. And a ratio built from one less over one plus the square. The four substitutions on offer: let the variable be a sine, a cosine, a tangent, or a secant. That is sixteen pairings. Of the sixteen, exactly five open the shape into a single ratio with no root left in it. The other eleven leave the root standing, and no amount of algebra afterwards rescues them.

4:02The five: a root of one less the square opens under a sine, into a cosine, and under a cosine, into a sine. A root of one plus the square opens under a tangent, into a secant. A root of the square less one opens under a secant, into a tangent. And the ratio shape opens under a tangent, into the cosine of twice the angle. Three of the four shapes have exactly one substitution that works on them. There is nothing to choose. You read the shape and the shape answers.

4:41Back to the first one, standing at the inverse sine of the sine of twice the angle. The temptation is enormous. Inverse sine, sine, they cancel, the answer is twice the angle, which is twice the inverse sine of the variable. Done. That answer is correct. It is also, as written, unjustified -- and the restriction attached to the problem is the thing that would justify it. Because an inverse sine composed with a sine gives back the angle you put in only when that angle already lies between minus a quarter turn and a quarter turn. Outside that, it gives back a different angle with the same sine.

5:28Here the angle being cancelled is twice the auxiliary angle. So the condition is not about the variable at all. It is about twice the angle. Now measure it. Let the variable run across the stated range, from minus one over root two up to one over root two, and watch where twice the angle goes. It runs from minus a quarter turn to a quarter turn. Not roughly. Exactly. The room left over at the bottom end is zero, and the room at the top end is zero.

6:05The restriction is not a safety margin someone added. It is the cancellation's condition, translated word for word into a statement about the variable. And it earns its keep. Step past one over root two and walk two hundred and forty-two values up to one: every one of them disagrees, and at the far end the two sides are a full half turn apart. Which leads somewhere uncomfortable. If that formula fails above one over root two, the expression itself does not. It is perfectly well defined all the way to one.

6:47So what is it up there? Make the other substitution. Let the variable be a cosine instead of a sine. The root of one less the square is now a sine, the argument is twice a cosine times a sine, and it is still the sine of twice the angle. Same argument, same collapse. But the angle is now the inverse cosine of the variable, and as the variable runs from one over root two up to one, twice that angle runs from zero to a quarter turn.

7:23Which is inside the cancellation's interval again, with zero to spare at both ends again. So the same expression equals twice the inverse sine of the variable on one stretch, and twice the inverse cosine on the other. Test it at four fifths. Twice the inverse sine gives one point eight five. Twice the inverse cosine gives one point two nine. The expression itself gives one point two nine. One expression. Two simplifications. Neither is more correct than the other, and which one applies is decided entirely by where the variable is sitting.

8:06If you ever wondered why these problems arrive with ranges bolted on, that is why. The range is not decoration on the answer. The range chooses the answer. The fourth of the five had no root in it at all. A cosine over one less a sine, sitting inside an inverse tangent. There is nothing to substitute here, because the variable is already an angle. The move is the same in spirit: rewrite everything in terms of half of it.

8:41The cosine of the angle is the square of the cosine of half of it minus the square of the sine of half. One less the sine of the angle is the square of the cosine of half minus twice the product plus the square of the sine of half. The top is a difference of two squares, so it factors into the sum times the difference. The bottom is a perfect square, the difference times itself.

9:10One factor is common. Cancel it. What is left is the sum over the difference, and dividing top and bottom by the cosine of half turns that into one plus the tangent of half over one less the tangent of half. Which is the tangent of an eighth of a turn plus half the angle. So the whole expression is an eighth of a turn plus half the angle. Clean. And it hides two questions, and both of them are about the factor we cancelled.

9:45That cancelled factor was the cosine of half the angle minus the sine of half the angle. You may divide by it only where it is not zero. So hunt its zeros rather than assume it has none. Sweep the whole neighbourhood of the stated range and find every angle where it vanishes. There are two of them. One at a quarter turn. One at three quarters of a turn on the negative side.

10:14And those two angles are precisely the two ends of the range the problem was given. The range is not an interval someone chose. It is the stretch between two consecutive places where the cancellation is illegal. It gets better. Walk the range with both ends included, two hundred and forty-two points, and ask the original expression for a value at each. It answers at two hundred and forty of them. At the two ends it has no value at all, because the bottom of the fraction is zero there.

10:52The factor that had to be cancelled and the point where the expression breaks are the same two points. That is not a coincidence, it is the same fact seen twice. And the answer itself, an eighth of a turn plus half the angle, runs from minus a quarter turn to a quarter turn across that range, with zero room at each end of the interval the last cancellation needs.

11:20Same story as before. The restriction is exactly the condition, with nothing to spare. The third shape was a root of the square less one, and it came wrapped in an inverse cotangent. The shape says secant, so let the variable be a secant. The root of the square less one becomes a tangent, and one over that is a cotangent. So the expression is the inverse cotangent of the cotangent of the angle, which for a variable above one is just the angle.

11:56And the angle was defined by the variable being its secant. So the answer is the inverse secant of the variable. Read that again. The problem was posed with an inverse cotangent. The answer is written with an inverse secant. Not a rearrangement of the same function, a different function entirely. Check it. Two hundred and forty-two values above one, and the expression agrees with the inverse secant at every one of them, largest disagreement zero. Compare it against the inverse cotangent it was posed in, and it disagrees.

12:38This is worth sitting with, because it is the clearest evidence that the method is not symbol-shuffling. Naming the angle changes which function the answer naturally wears. Two more, and these are the ones where the restriction does its work in plain sight. First: three times the inverse sine of the variable equals the inverse sine of three times the variable less four times its cube, for the variable between minus a half and a half.

13:12Name the angle, and three times the variable less four times its cube is the sine of three times the angle. So the right side is the inverse sine of the sine of three times the angle, and the cancellation needs three times the angle inside the usual interval. Let the variable run from minus a half to a half. Three times the angle runs from minus a quarter turn to a quarter turn. Zero room at the bottom, zero room at the top.

13:46Second: three times the inverse cosine, with four times the cube less three times the variable, for the variable from a half up to one. Same structure, cosine instead of sine, and again three times the angle lands exactly on the interval the cosine's cancellation needs, with nothing to spare at either end. And both of them fail outside. Two hundred and forty-two values past each stated boundary, and every single one disagrees, by as much as a whole turn.

14:23The restriction was never a hint about where the problem is interesting. It is the licence for the final step, and without it the final step is just a hope. At this point the method should feel routine, so here is a run of it. A root of one plus the square, less one, over the variable. Tangent substitution, half-angle identity, and it collapses to half the inverse tangent of the variable.

14:55A root of one less a cosine over one plus a cosine. Half-angle again, and it collapses to half the angle. A cosine less a sine over a cosine plus a sine. Divide top and bottom by the cosine, and it becomes an eighth of a turn less the angle. The variable over a root of a constant squared less the variable squared. The shape says sine, but scaled: let the variable be the constant times a sine. It collapses to the inverse sine of the variable over the constant.

15:33And a cubic over a cubic, which looks the worst of the lot and is the tamest: let the variable be the constant times a tangent, and the whole thing is the tangent of three times the angle. It collapses to three times the inverse tangent of the variable over the constant. Every one of those checks out at two hundred and forty-two points with zero disagreement. And the scaling matters. Feed a hundred and eighty-one angles through the correct scaled substitution on the fourth one and every one lands. Feed the same angles through the wrong substitution of the four and a hundred and eighty of them fail.

16:19A hundred and eighty out of a hundred and eighty-one. The one survivor is the angle zero, where both sides are zero anyway. Which is a small lesson in itself: checking a method at one convenient value proves nothing. Now a harder kind of problem, where two inverse values have to be added together. The tangent of the inverse sine of three fifths plus the inverse cotangent of three halves. There is no formula available for that. So build it.

16:55Name both. The first angle has sine three fifths, so a three-four-five triangle gives its tangent as three quarters. The second has cotangent three halves, so its tangent is two thirds. The tangent of the sum is the sum over one less the product, and that is seventeen over six. The same construction handles a whole family. A doubled inverse sine of three fifths comes out as an inverse tangent of twenty-four over seven. An inverse sine of eight seventeenths plus an inverse sine of three fifths gives seventy-seven over thirty-six.

17:39Two inverse cosines, four fifths and twelve thirteenths, give thirty-three over sixty-five. An inverse cosine of twelve thirteenths plus an inverse sine of three fifths gives fifty-six over sixty-five. And an inverse sine of five thirteenths plus an inverse cosine of three fifths gives sixty-three over sixteen. Three triangles do all of it: three-four-five, five-twelve-thirteen, and eight-fifteen-seventeen. But there is a step in every one of those that nobody wrote down, and it is the same step we have been chasing all video.

18:22Computing the ratio of the sum tells you the ratio. It does not tell you the angle. To go back from the ratio to an inverse value, the sum has to still be inside that function's own interval. For all five of those, it is. Check performed, check passed, nothing to see. So build a sixth from the same two triangles, with both cosines turned negative. Its cosine is thirty-three over sixty-five, the same fraction as one of the five above.

18:57But the sum is five point two four, and the inverse cosine of thirty-three over sixty-five is one point zero four. Those are not the same angle. They are a full turn apart. The ratio was right. The angle was wrong. And nothing in the calculation of the ratio could possibly have told you, because the ratio genuinely is thirty-three over sixty-five. That is the check. Compute the ratio, then ask separately whether the sum is somewhere the inverse function can name.

19:36Two last problems, where the point is not to simplify but to solve. Twice the inverse tangent of the cosine of an unknown equals the inverse tangent of twice the cosecant of it. Take the tangent of both sides using the doubling formula, and the requirement reduces to the cosine and the sine of the unknown being equal. The unknown is an eighth of a turn, and it is the only solution on the whole stretch.

20:08The other: the inverse tangent of one less the variable over one plus the variable equals half the inverse tangent of the variable. Substitute, and the left side is an eighth of a turn less the angle while the right is half the angle. So the angle is a twelfth of a turn and the variable is one over root three. And the last one, which is the whole video in miniature.

20:37The inverse sine of one less the variable, minus twice the inverse sine of the variable, equals a quarter turn. Do the algebra and two candidates fall out: zero, and a half. Put them back in. At zero, the equation holds exactly. At a half, the left side comes out at minus two point zero nine, which is not a quarter turn and is not close. One of the two candidates survives. The other was produced by a step that was legal as algebra and illegal as an equation, and the only way to find that out is to substitute back.

21:20So what have we actually got. One move. When an inverse value is in the way, name it as an angle. The variable becomes a ratio, the awkward part becomes another ratio of the same angle, and an identity you already know closes the gap. One rule for choosing. Read the shape of the awkward part. A root of one less the square wants a sine or a cosine. A root of one plus the square wants a tangent. A root of the square less one wants a secant. A ratio built from one less over one plus wants a tangent. Five of sixteen pairings work, and three of the four shapes leave you no choice at all.

22:07And one check, which is the part that separates people who can do these from people who can only start them. After the collapse, the last step is a cancellation, and a cancellation has a condition. Ask what interval it needs, then ask where the stated range actually puts you. Every time we measured it in this video, the answer was the same: the range lands exactly on the interval, with zero room at both ends.

22:40That is not a coincidence and it is not generosity. The range was reverse-engineered from the cancellation. Which means the restriction was never extra information. It was the argument, written in the margin, waiting for someone to read it.