Substituting an angle for the variable to collapse a messy expression

What to assume they know

• The six promoted output intervals and the convention that fixes them
• Both cancellation rules with their conditions, from the previous topic
• Class XI: the double and half angle formulas, the compound angle formulas for sine, cosine and tangent, and the triple angle expansions
• The Pythagorean identity in all three of its forms
• Rationalising, factorising a difference of two squares, and cancelling a common factor honestly

What they should be able to do

• Carry out the basic move: name the inverse value as an angle, so that the variable becomes a trigonometric ratio of it
• Choose the substitution from the shape of the expression, and in particular from which surd appears
• Collapse the resulting expression with a Class XI identity, then cancel with the appropriate rule
• Identify, in a printed argument, the step at which the stated restriction on the variable is actually needed
• Recognise when the answer is a different inverse function from the one you started with
• Run the method on the seven simplification items and the two proof items of the chapter's second exercise
• Combine two inverse values into one, rebuilding the needed relation from a substitution and a compound angle formula
• Solve an equation whose unknown sits inside an inverse function

Where it usually goes wrong

• "Pick any substitution; they all work eventually." They do not. The surd decides. A root of one less the square under a tangent substitution produces a worse expression than the one you started with, and students conclude the method has failed when only the choice has.
• "The restriction printed beside the result is just the domain of the expression." Sometimes it is more. In Example 4 and in both proof items of Exercise 2.2 the printed range is precisely the set on which the final cancellation is legal, which is a stronger requirement than being defined.
• "Example 3 proves its own restriction is needed." It does not — it states the restriction and then never uses it, because the printed argument cancels an inverse sine against a sine without checking anything. Show the check. Then show what goes wrong past one over root two, where the expression is still perfectly well defined and the claimed answer is wrong.
• "The answer has to be in the same inverse function as the question." Example 5 begins with an inverse cotangent and ends with an inverse secant. Which function the answer wears is decided by the substitution, and several exercise items exploit that.
• "Cancelling the common factor is always safe." In Example 4 the cancelled factor is zero at one particular angle, which is why the stated interval is open at that end. A cancellation that quietly divides by zero is the standard way to lose a solution in this chapter.
• "The chapter must have a formula for adding two inverse sines." It does not. Nothing of that kind is stated anywhere in the surviving chapter, and every one of the five miscellaneous proof items has to be rebuilt from a substitution. Tell the student that plainly rather than letting them hunt for a formula that was removed.
• "Once the algebra gives a value, the equation is solved." In the last multiple-choice item the algebra gives two values and only one survives substitution. Squaring, and taking a sine of both sides, both introduce candidates that were never solutions.
• "Two roots that differ only by a sign inside them are effectively the same." The distractors in the sine-of-an-inverse-tangent item are built from that confusion. One of the two is the root of a quantity that can be negative.

Questions to check understanding

• Simplify a stated inverse expression containing a surd, naming the substitution used
• Prove a stated identity between a multiple of one inverse function and a single inverse function, and say at which step the printed restriction was needed
• Given a printed restriction, show that it is exactly the condition the final cancellation requires
• Evaluate a numerical expression combining two inverse values, by naming both as angles
• Prove that a sum of two inverse values equals a third, including the check that the sum lies in the right interval
• Solve an equation in which the unknown appears inside an inverse function, and reject any candidate that fails on substitution
• Choose the correct simplified form from four options that differ only inside a root

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.

• Example 3, part (i) (Part I p. 28). The inverse sine of twice the variable times the root of one less its square, asserted equal to twice the inverse sine of the variable, for the variable between minus and plus one over root two. The printed argument puts the variable equal to the sine of an angle, so that the root becomes the cosine of that angle, the product becomes the sine of twice the angle, and the inverse sine of that returns twice the angle. Verified: every step holds, and the last one — cancelling the inverse sine against the sine — needs twice the angle to lie between minus half pi and half pi. That is exactly what the printed restriction on the variable delivers, since the variable is at most one over root two, so the angle is at most a quarter of a half turn. The printed argument never says this. Section 5 exists to say it.
• Example 3, part (ii) (Part I p. 28). The same left-hand side, asserted equal to twice the inverse cosine of the variable, for the variable between one over root two and one. The chapter disposes of it in one line, saying to start from the cosine substitution instead and proceed as before. Verified: with the variable equal to the cosine of an angle, the angle runs from zero to a quarter of a half turn, the root becomes the sine of the angle, the product is again the sine of twice the angle, and twice the angle now runs from zero to half pi, which is inside the interval. The two parts together are the sharpest thing on the page: one expression, two different simplifications, and which one is correct depends entirely on where the variable is..
• Example 4 (Part I p. 28). The inverse tangent of the cosine of an angle divided by one less its sine, for the angle between minus three half pi and half pi. The printed argument rewrites both parts in half-angle terms, factorises the numerator as a difference of two squares and the denominator as a perfect square, cancels the shared factor, divides through by the half-angle cosine to reach a ratio of one plus and one minus the half-angle tangent, recognises that as the tangent of a quarter of a half turn plus the half angle, and cancels. Verified, including the two things the print leaves silent: the cancelled factor vanishes exactly when the angle reaches half pi, which the open right end of the stated range excludes; and the final cancellation needs the result to lie strictly between minus and plus half pi, which the stated range delivers exactly, with nothing to spare at either end. This is the one place in the chapter where a printed restriction is precisely the condition the argument requires, and it is worth saying so out loud.
• Example 5 (Part I p. 29). The inverse cotangent of one over the root of the variable squared less one, for the variable greater than one. The chapter puts the variable equal to the secant of an angle, so the root becomes the tangent, the reciprocal becomes the cotangent, the cancellation returns the angle, and the angle is the inverse secant of the variable. Verified. Note what has happened: the item was posed with an inverse cotangent and the answer is an inverse secant. The substitution, not the question, decides which inverse function the answer is written in, and students find this genuinely destabilising.
• Which substitution to reach for (not in the book). Verified against all four choices and against every item of Exercise 2.2: a root of one less the square wants the sine or the cosine; a root of one plus the square wants the tangent; a root of the square less one wants the secant; and a ratio built from one plus and one minus the square, or from one plus and one minus the variable, also wants the tangent. Four shapes, four choices, and every item in the exercise falls into one of them.
• Exercise 2.2, the seven simplification items (Part I p. 29). Verified, all seven, by working added here: question 3 becomes half the inverse tangent of the variable, on the tangent substitution; question 4 becomes half the angle, on the half-angle identity; question 5 becomes a quarter of a half turn less the angle, after dividing top and bottom by the cosine; question 6 becomes the inverse sine of the variable over the constant, on the sine substitution; question 7 becomes three times the inverse tangent of the variable over the constant, on the tangent substitution and the triple angle expansion; question 8 evaluates to a quarter of a half turn; and question 9 evaluates to the sum of the two variables over one less their product. Question 9 is the hardest and the most instructive: it hides two substitutions inside one bracket, and the three stated conditions on the two variables are what make both of them legal.
• Exercise 2.2, the two proof items (Part I p. 29). Question 1 asserts that three times the inverse sine of the variable equals the inverse sine of the triple angle expansion, for the variable between minus and plus one half. Question 2 asserts the matching statement for the cosine, for the variable between one half and one. Verified: with the sine substitution the first reduces to cancelling an inverse sine against a sine of three times the angle, and the stated restriction is exactly what keeps three times the angle inside the promoted interval; the second works the same way against the cosine's interval from zero to pi. These two are the best items in the chapter for showing that the restriction is not decoration — better than Example 3, where the same point is available but the printed argument does not draw it.
• Exercise 2.2 question 12 (Part I p. 30). The tangent of a sum of an inverse sine and an inverse cotangent, at three fifths and three halves. Verified: name the two terms as angles; the first has tangent three quarters and the second has tangent two thirds; the compound angle formula for the tangent gives seventeen over six.
• Miscellaneous Exercise questions 3 to 7 (Part I p. 31). Five statements combining two inverse values into one — a doubled inverse sine equal to an inverse tangent; a sum of two inverse sines equal to an inverse tangent; a sum of two inverse cosines equal to an inverse cosine; an inverse cosine plus an inverse sine equal to an inverse sine; and an inverse tangent equal to a sum of an inverse sine and an inverse cosine. Verified, all five, by naming each term as an angle, computing the required ratio of the sum with a Class XI compound angle formula, and then checking that the sum still lies inside the promoted interval of the function on the other side. The arithmetic is clean in every case — the fractions are three-four-five, five-twelve-thirteen and eight-fifteen-seventeen triples, chosen so that no surd survives. See the note below: the chapter supplies no formula for any of these, and the range check is entirely the student's own.
• Miscellaneous Exercise questions 8, 9 and 10 (Part I p. 31). Three more proofs of the same kind, on substitutions rather than on numbers. Verified: question 8 opens on the tangent substitution for the root of the variable; question 9 uses the fact that one plus or minus the sine of an angle is a perfect square in the half angle, and the stated range of the angle is what fixes the sign of each root when it is taken; question 10 opens on the printed hint. Question 10 carries the only hint printed anywhere in this chapter — a bracketed instruction to set the variable equal to the cosine of twice an angle. Point that out; a lone hint is a signal about difficulty.
• Miscellaneous Exercise questions 11 and 12 (Part I p. 31). Two equations to solve. Verified: the first reduces, on the double-angle formula for the tangent, to the requirement that the cosine and the sine of the unknown agree, giving a quarter of a half turn. The second, on the tangent substitution, gives a quarter of a half turn less the angle on the left and half the angle on the right, so the angle is a sixth of a half turn and the unknown is one over root three. Both are the same method as the simplification items, run one step further.
• Miscellaneous Exercise questions 13 and 14 (Part I p. 31). Two multiple-choice items. Verified: the first asks for the sine of an inverse tangent, which the tangent substitution turns into the variable over the root of one plus its square — the fourth option, and the two wrong roots differ from it only by a sign inside the root. The second is an equation in an inverse sine whose algebra produces two candidate values, zero and one half, of which one half fails when substituted back because it makes the left side negative; the answer is zero alone, the third option. That rejection step is the whole item, and it is the clearest instance in the chapter of a range condition doing real work.

Figures to have open

• One persistent right triangle, reused throughout, in which the named angle sits and the two legs carry the variable and the surd. Every substitution in the topic can be read off it, and using one triangle rather than a fresh sketch per item is what makes the four families look like one method.
• A four-row card of surd shape against substitution for section 3, built with the repo's DataTable component.
• An interval strip for sections 5, 7 and 10, showing the variable's stated range on one line and the angle's induced range beneath it, so the reader watches one turn into the other. It must be the same strip in all three sections.
• Three worked-derivation boards for Examples 3, 4 and 5, laid out at the same width and type size. These are peers and must not be sized to their own content.
• No figure in this topic comes from the chapter. Pages 28 to 31 carry no figure at all — checked on the page image of each.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 2 "Inverse Trigonometric Functions", §2.3, Example 3 parts (i) and (ii) and Example 4, Part I p. 28
• Example 5, Part I p. 29
• Exercise 2.2, questions 1 to 9, Part I p. 29, and question 12, Part I p. 30
• Miscellaneous Exercise on Chapter 2, questions 3 to 14, Part I p. 31