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

What to assume they know

• The six promoted output intervals, from the third topic of the previous module
• The convention that an unqualified inverse symbol means the promoted branch
• Class XI identities relating an angle to its supplement, its negative, and itself shifted by a half or a whole turn
• The standard angles and their ratios
• Composition of two functions, and what it means for one to undo the other

What they should be able to do

• State both cancellation rules with the exact set of inputs each is valid on
• Explain why the first rule needs no case work and the second does
• Decide, for a given angle, whether the second rule applies directly
• Repair a failing expression by replacing the angle with one that shares the same trigonometric ratio and lies inside the promoted interval
• Choose the right replacement identity for the sine family, the cosine family and the tangent family
• Evaluate an expression whose angle is more than a full turn outside the promoted interval
• Say what the chapter's own caveat paragraph is warning about, and what it declines to specify
• Recognise that the chapter summary states the rules without their conditions, and restore the conditions when revising from it

Where it usually goes wrong

• "Both rules are the same rule written two ways." They are two different compositions and only one of them is unconditional. The order matters, and the whole topic exists because it does.
• "The second rule fails only for large angles." It fails for every angle outside the promoted interval, including small ones on the wrong side. Two pi by three is nothing like large.
• "When the second rule fails, the expression is undefined." It is perfectly well defined and has a value; the value is simply not the angle you fed in. That distinction has to be made explicitly or students will start writing that the expression does not exist.
• "Repair by subtracting pi, always." Subtracting a half turn is the tangent's repair. The sine wants the supplement, and the cosine wants subtraction from a whole turn. Three families, three moves, and using the tangent's move on a sine question gives the wrong sign.
• "The answer must have the same sign as the angle fed in." Question 11 sends in a positive angle and returns a negative one. The output is confined to the promoted interval and nothing else constrains it.
• "The chapter's summary states the rules without conditions, so the conditions are optional." The conditions are printed in the section itself, on Part I p. 27. The summary is a reminder, not a restatement, and a student who revises only from it will get question 10 wrong.
• "Since the chapter says it will not go into which inputs work, the question does not matter." The chapter declines to characterise the failures in general. It does not decline to require the check in a specific case, and every item in this block turns on that check.
• "Once the angle is inside the promoted interval, check nothing else." For the tangent and the cotangent the interval is open, so an angle sitting exactly at half pi is not inside it, and the expression that would produce it is not defined at all.

Questions to check understanding

• State both cancellation rules with the exact input set for each
• Evaluate an inverse-of-direct expression whose angle lies inside the promoted interval, and one whose angle lies outside it
• Evaluate such an expression where the angle exceeds a whole turn
• Choose the correct value from four options, where one option is the unrepaired angle — the form of Exercise 2.2 question 13
• Combine two inverse values, at least one with a negative argument, and simplify
• Given an expression and a proposed answer, decide whether the answer lies in the promoted interval and correct it if not
• Explain in one sentence why the two rules have different conditions

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• The two displayed rules (§2.3, Part I p. 27). The chapter first recalls that naming the inverse sine of a number is the same act as saying that the sine of the answer is that number, and then displays the pair: the sine of the inverse sine of a number returns the number, for every number from minus one to one; and the inverse sine of the sine of an angle returns the angle, for every angle from minus half pi to half pi. Both conditions are printed here, and they are different in kind: the first covers the whole of the inverse sine's domain, and the second covers only a slice of the sine's.
• The same pair, stated once on Part I p. 19. The identical two statements appear earlier, at the end of the sine discussion, with the same two conditions. The explanation may cite either; Part I p. 27 is where the section that uses them begins.
• The extension sentence (§2.3, Part I p. 27). Rather than write the pair out for the other five functions, the chapter says once that matching results hold for them on appropriate sets of inputs. Verified by construction: the pattern is that the outer-function-first rule holds across the whole domain of the inverse, and the inverse-first rule holds only on that function's own promoted interval — the closed interval from zero to pi for the cosine, the open one from minus half pi to half pi for the tangent, and so on down the table. The chapter leaves this to the reader.
• The caveat paragraph (§2.3, Part I p. 27). Before anything is displayed, the chapter warns that what follows holds inside the promoted branches and wherever the expressions are defined; that some results will fail for some inputs; and that it will not pursue which inputs, on the grounds that the question sits outside the book. Read that last clause aloud in the explanation. It is the chapter telling the student that the conditions matter and that it is not going to supply them, and it is the reason the second rule has to be handled by a recipe rather than by memory.
• Miscellaneous Example 6 (Part I p. 30). The inverse sine of the sine of three pi by five. The chapter's own solution has an unusual and very useful shape: it applies the second rule first and gets three pi by five, then observes that three pi by five is not in the promoted interval, then replaces the angle by its supplement — two pi by five — which has the same sine and does lie in the interval, and concludes. Verified: the sine of three pi by five and the sine of two pi by five are equal because the two angles add to a half turn, and two pi by five is less than half pi, so the answer is two pi by five. Build section 6 on this, and keep the chapter's order: wrong step, catch, repair.
• Exercise 2.2 question 10 (Part I p. 30). The inverse sine of the sine of two pi by three. Verified: two pi by three exceeds half pi, so replace it by its supplement, pi by three, which has the same sine and lies inside. The answer is pi by three.
• Exercise 2.2 question 11 (Part I p. 30). The inverse tangent of the tangent of three pi by four. Verified: three pi by four is outside the tangent's promoted interval, and the tangent is unchanged by a shift of a half turn, so subtract pi to reach minus pi by four, which is inside. The answer is minus pi by four — and note that the answer is negative while the angle fed in was positive. This is the item that makes students distrust the whole method, so spend time on it.
• Exercise 2.2 question 13 (Part I p. 30). The inverse cosine of the cosine of seven pi by six, offered as a multiple-choice item with four options. Verified: seven pi by six exceeds pi, so it is outside the cosine's promoted interval; subtracting it from a whole turn gives five pi by six, which has the same cosine and does lie between zero and pi. The answer is five pi by six, which is the second option. The first option is the unrepaired angle, offered precisely to catch the student who stops at the first move.
• Exercise 2.2 question 14 (Part I p. 30). The sine of pi by three less the inverse sine of minus one half. Verified: the inner term is minus pi by six, so the expression is the sine of pi by three plus pi by six, that is the sine of half pi, which is one. The fourth option. This one needs the branch table and no repair at all, and is worth showing beside question 13 for the contrast.
• Exercise 2.2 question 15 (Part I p. 30). The inverse tangent of root three less the inverse cotangent of minus root three. Verified: the first term is pi by three. For the second, the cotangent of the answer is minus root three, and the promoted interval for the inverse cotangent runs from zero to pi and contains no negative angle, so the answer is five pi by six. The difference is pi by three minus five pi by six, which is minus half pi — the second option.
• Miscellaneous Exercise questions 1 and 2 (Part I p. 31). The inverse cosine of the cosine of thirteen pi by six, and the inverse tangent of the tangent of seven pi by six. Verified: thirteen pi by six is a full turn plus pi by six, and a full turn changes nothing, so the first answer is pi by six. Seven pi by six is a half turn plus pi by six, and a half turn leaves the tangent unchanged, so the second answer is also pi by six. Two different reductions, the same final angle — an accident of the numbers that is worth pointing out, because a student who notices it will otherwise suspect a pattern that is not there.
• The Summary bullets (Part I p. 32). Four bullets, all about the sine: the two directions of the defining equivalence, and the two cancellation rules. Read off the printed page, the two rules are printed here with no condition attached, under a single introductory line about suitable inputs. That is a trap for a student revising from the Summary alone.

Figures to have open

• One angle dial or one angle axis, reused for every repair in the topic, with the promoted interval shaded. Sections 3, 6, 8, 9 and 10 must all work on the same drawing, or the repairs will look like six unrelated tricks instead of one move applied six times.
• A pair-of-boxes diagram for section 1 showing the two orders of composition, with the set each order starts from labelled. Not in the book; the chapter displays the two lines and draws nothing.
• Three small angle pictures for section 8 — an angle and its supplement, an angle and its reflection through a whole turn, an angle and its shift by a half turn — drawn at one size and one radius, as siblings.
• A reproduction-free treatment of the Summary bullets for section 11: set the four statements as added cards and add the conditions in a second colour. Do not photograph the page.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 2 "Inverse Trigonometric Functions", §2.2, the first statement of the two rules, Part I p. 19
• §2.3 Properties of Inverse Trigonometric Functions, the caveat paragraph, the displayed pair and the extension sentence, Part I p. 27
• Miscellaneous Example 6, Part I p. 30
• Exercise 2.2, questions 10, 11, 13, 14 and 15, Part I p. 30
• Miscellaneous Exercise on Chapter 2, questions 1 and 2, Part I p. 31
• Summary, the four sine bullets, Part I p. 32