Setting the second angle equal to the first gives the double and triple angle rules

0:00Six more results are about to arrive, and between them they cost one substitution. Set the second angle equal to the first in the expansions you already have. That is the whole trick, and it is worth almost nothing on its own. What it buys is what makes this worth twelve minutes. But before any of it, kill the guess that ruins this topic. Doubling the angle does not double the coordinates.

0:29Take a sixth of a half turn, thirty degrees. Its first coordinate is root three over two. Twice that is root three, which is more than one, and no coordinate on a circle of radius one is ever more than one. The doubled angle is sixty degrees, and its first coordinate is a half. Across all twenty four inputs, the doubled first coordinate equals twice the first coordinate at exactly zero of them.

1:00The doubled second coordinate equals twice the second at two, and both of those are inputs where the second coordinate is nothing. So there is nothing to double here, and everything to derive. Start with the expansion for the first coordinate of a sum. The first coordinate of A plus B is the product of the two first coordinates, less the product of the two second ones. Now set B equal to A.

1:31Both products become squares. The doubled first coordinate is the squared first coordinate, less the squared second. One substitution, and that is the first result of the six. Checked against the circle at all twenty four inputs, it holds at twenty four. That is one face of it. There are four, and knowing which to reach for is most of the skill this topic asks for. The other faces cost no new work at all.

2:03The two squared coordinates add up to one. That is the identity from the very start of all this, and it holds at all twenty four. It lets you trade either square for the other. Replace the squared second coordinate with one less the squared first, and the result becomes twice the squared first coordinate, less one. Replace the squared first instead, and it becomes one, less twice the squared second.

2:33Each of those is checked at all twenty four inputs, and each holds at all twenty four. Three faces, and the difference between them is one use of one identity. They are not three results. A student who learns them as three will not know which to reach for, and choosing is the only decision this topic actually requires. The fourth face gets rid of both coordinates. Write the first face over the identity, which is legitimate because the identity is one.

3:09Then divide the top and the bottom by the squared first coordinate. Every term becomes a quotient of the two coordinates, or a one. The doubled first coordinate is one less the squared quotient, over one plus the squared quotient. It holds at all twenty two inputs where the quotient exists. And here is the thing worth noticing about its condition. The divisor of that fraction is one plus a square, so it is never nothing, at all twenty two.

3:42The face refuses at exactly two inputs, and at both of them it is the quotient itself that refused, not anything the face added. So the condition attached to this face is precisely where the quotient fails to exist, and not one input more. The second coordinate is easier. Set the two angles equal in its expansion, and the two terms become the same term. The doubled second coordinate is twice the product of the two coordinates.

4:16Twenty four of twenty four, and it carries no condition at all, because nothing anywhere was divided by anything. Divide by the identity over the squared first coordinate, the same move as before, and you get the quotient form: twice the quotient over one plus its square. Twenty two of twenty two. Same divisor, so the same condition, and again for the same reason. A condition is not a fact about a formula.

4:48It is a receipt for a division. The doubled quotient is where this gets interesting. Set the angles equal in the quotient identity and you get twice the quotient over one less its square. The condition that comes attached to it is written on the doubled angle, and it does exactly one job. Four of the twenty four inputs make that divisor nothing, and those four are precisely the four the condition rules out.

5:20Exactly the vanishing divisor, and nothing more. Now watch what it does not do. Two of the twenty four inputs have no quotient of their own, and the condition rules out zero of them. Take a quarter turn. Doubling it gives a half turn, whose quotient is a perfectly respectable nothing. The left hand side is fine. The right hand side wants the quotient at a quarter turn, which does not exist, so it is not a number at all.

5:52Counted across all twenty four: eighteen inputs where both sides are numbers and agree at all eighteen, two where the left side lives and the right side does not, zero the other way, and four where neither does. That condition is necessary and it is not sufficient. Tripling costs the same substitution a second time. Write three A as two A plus A, and expand with the sum results you already have.

6:24Then put the doubled results into the slots where the doubled angle appears, and clean up with the identity. The tripled second coordinate comes to three times the second coordinate, less four times its cube. The tripled first comes to four times the cube of the first coordinate, less three times it. Both hold at all twenty four inputs. Now look at those two lines side by side, because they are a trap.

6:54Three and four, then four and three, with the subtraction the other way about. They look like mirror images, so people learn one and flip it. Copy the second coordinate's shape across to the first, and it is right at six of the twenty four inputs. Copy it the other way, and again six of twenty four. Right a quarter of the time, which is the most dangerous kind of wrong.

7:22The tripled quotient is three times the quotient, less its cube, over one less three times its square. Its condition is written on the tripled angle, and it rules out six of the twenty four inputs. Four of those six are where its divisor vanishes. The other two are the inputs where the quotient of A itself does not exist. So the tripled condition covers both ways the right hand side can fail, and the doubled one covers only the first.

7:55The counts say it plainly: eighteen inputs where both sides live and agree at all eighteen, zero where the left lives and the right does not, and six where neither does. Zero, against the doubled quotient's two. That is not a principle you can lean on. It is an accident of which numbers happen to fall where, and knowing it is an accident is worth more than a rule you would have to trust.

8:26Put the two on the same line of inputs and the difference is visible. The doubled quotient's condition strikes out four inputs. The tripled quotient's strikes out six. The two sets have zero inputs in common, which is worth a second's pause, because they are conditions on the same formula family. And the two inputs where the quotient itself fails sit inside the tripled set and outside the doubled one.

8:56One line of inputs holds the whole of the difference. Both conditions are true. Only one of them is enough. Back to the four faces, and the decision they exist to make. If an expression has a squared first coordinate in it and you want something linear, use the face with twice the squared first coordinate. Rearranged, that says the squared first coordinate is one plus the doubled first coordinate, all halved.

9:28It holds at all twenty four inputs. If it is a squared second coordinate you want to flatten, use the other one: the squared second coordinate is one less the doubled first, halved. Also twenty four of twenty four. That is the move that turns a sum of three squared cosines into three linear terms in one step. And if you want the plain difference of two squares, you already had it in the first face.

9:58Four faces, one decision, and the decision is: what shape do I want the answer to be in? Now run one of those rearrangements the other way. The squared second coordinate is one less the doubled first coordinate, halved. Read left to right, that computes a doubled value from a single one. Read right to left, it computes a HALVED value from a whole one, and it is exactly the same line.

10:29Half angles are not a new topic. They are this result, read backwards. But there is a catch, and it is the catch that costs more marks than anything else here. What comes out is a SQUARE. To get the value itself you have to take a root, and a root has two signs. The doubled angle cannot tell you which. Doubling reaches only twelve of the twenty four inputs, and at eleven of those twelve, the two inputs that double onto it disagree about the sign of the second coordinate.

11:04The information is genuinely not there. So where does the sign come from? From the halved angle's own quadrant, and from nothing else. This is the step people skip, and skipping it is the error. Suppose an angle lies between a half turn and three quarters of a turn. That is the third quadrant, where both coordinates are negative. Halve the range. Between a quarter turn and three eighths of a turn.

11:35That is the SECOND quadrant, where the second coordinate is positive and the first is negative. Halving a third-quadrant angle does not give you a third-quadrant angle. It gives you a second-quadrant one, with one sign flipped from what you started with. Locate the half before you take any root. Not after. Here it is on real numbers. An angle whose quotient is three quarters, lying between a half turn and three quarters of a turn.

12:10First, the whole angle: it is in the third quadrant, so its first coordinate is negative, and from the quotient it comes to minus four fifths. Second, and before anything is halved, locate the half: second quadrant. Now the two rearranged faces. The squared second coordinate of the half is one less minus four fifths, halved, which is nine tenths. The squared first coordinate of the half is one plus minus four fifths, halved, which is one tenth.

12:45Their quotient is nine, so the halved quotient is three or minus three, and the second quadrant says minus three. Two more of the same shape. An angle in the second quadrant whose first coordinate is minus three fifths halves into the first quadrant, giving four fifths, one fifth, and a quotient of two. An angle in the third quadrant whose first coordinate is minus one third halves into the second, giving two thirds, one third, and a quotient of minus root two.

13:21Every one of them: quadrant of the angle, quadrant of the half, then the roots. In that order. One more, and it runs the doubled quotient backwards instead. What is the quotient at an eighth of a half turn? That is not one of the twenty four inputs, so nothing can be looked up. But doubling it gives a quarter of a half turn, whose quotient is one. So call the unknown quotient t, and the doubling rule says twice t over one less t squared equals one.

13:56Multiply out and you have a quadratic. Its two roots are root two less one, and minus one less root two. Feed each of them back through the doubling rule and both give exactly one, so the arithmetic alone cannot choose between them. The angle is in the first quadrant, where the quotient is positive. Root two less one is positive. Minus one less root two is not. The quadrant chooses, again, and the answer is root two less one.

14:30The machine does not stop at three. For four times the angle, double the doubled quotient. It agrees with the circle at all eighteen inputs where both sides are numbers, and at six more the left side is a number while the route through the doubling is not, because it passed through a step that can refuse. Every stage you route through is a stage that can turn you away.

14:57For four times the angle without quotients, take the face with the doubled second coordinate in it and feed it the doubled angle. Twenty four of twenty four. And for six times, run the tripling on the doubled angle and substitute. Out comes a polynomial in the first coordinate alone, with thirty two, minus forty eight, eighteen and minus one on the sixth, fourth, second and constant terms. Twenty four of twenty four.

15:30So: one substitution, six results, four faces and a family of multiples that never ends. And every condition any of them carries is a receipt for a division somebody did. Read the receipt and you never have to memorise the condition.