Cutting the domain down until the function is one-one, and what a branch is

What to assume they know

• The graphs of all six trigonometric ratios over several turns, from Class XI
• Where each ratio attains its largest and smallest values, and where it breaks
• Radian measure and the standard angles written as fractions of pi
• Interval notation, and a set written as an interval with one point removed
• One-one, onto, and the previous topic's result that no ratio is one-one on its full domain
• The two reciprocal relations that define the cosecant and the secant

What they should be able to do

• State the two demands a surviving piece of the domain has to meet before an inverse can be built on it
• Name the interval the chapter restricts each of the six ratios to, and the two neighbouring intervals it offers alongside
• Use the chapter's word for one admissible interval, and the longer phrase for the one it singles out
• Explain why the sine and cosine intervals close at both ends, why the tangent and cotangent intervals are open, and why the cosecant and secant intervals have an interior point removed
• Locate the cut points of each family on the input axis, and say what is happening to the graph there
• Check for a proposed interval whether the ratio is one-one on it and whether it still reaches every value of the range
• Explain why the chapter's choice is a convention rather than a consequence
• Show that a mis-set interval of the same width can fail the one-one demand

Where it usually goes wrong

• "Any interval half a turn wide will do for the sine." It will not. From zero to pi is half a turn and the sine takes the value one half twice inside it, while never taking any negative value at all. Both demands fail at once. Run the counter-example; asserting the point does not stick.
• "The chapter had to pick these intervals." It did not. Each family listed is infinite, every member works, and the chapter says so before promoting one of them. What follows is a convention chosen so that everyone writes the same answer, not a theorem.
• "The endpoints are included or excluded for tidiness." They are decided by the function. Sine and cosine attain their extreme values at the ends, so dropping an end would lose a value and break the onto demand. Tangent and cotangent are undefined at the ends, so the ends cannot be kept at all.
• "The removed point in the cosecant interval is removed to make it one-one." No — it is removed because the cosecant does not exist there. One-one-ness was never in danger at that point. Students who mix the two reasons then remove points from the sine interval as well.
• "The cosecant interval loses zero, so the secant interval must lose zero too." The secant loses half pi. Each interval loses the point at which its own function breaks, and the two functions break in different places — that is the whole difference between them.
• "Restricting changes what the sine of an angle is." It does not. The rule is untouched; only the set of admissible inputs shrinks. Every Class XI identity survives intact on the surviving inputs.
• "Once restricted, the sine is onto the reals." It is onto the closed interval from minus one to one, which is all that was ever asked. Onto is always onto a stated target, and the target here is the range.
• "Because the tangent is one-one on an open interval, the sine should be too." The bracket type follows the function's behaviour at the cut, not a house style. Three different bracket shapes appear in this one section and each has its own cause.

Questions to check understanding

• Name the interval to which a stated ratio is restricted, and give one other interval that would have served
• Decide whether a proposed interval makes a stated ratio one-one, and whether it still covers the whole range, justifying each half separately
• Explain why a stated interval is closed at one end and open at the other, or why an interior point is removed
• Given a cut point, say which of the six ratios cuts there and what the graph is doing at that input
• Produce a counter-example showing that an interval of the right width need not work
• Write the arrow statement for a named inverse function, with the correct bracket shapes on both sides

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 sine, restricted (§2.2, Part I p. 19). The chapter takes the closed interval from minus half pi to half pi, and reports that on it the sine becomes one-one while still reaching everything from minus one to one. Verified: across that interval the sine climbs steadily from minus one to one and never repeats a value, and both extreme values are attained at the two endpoints, so both demands hold and neither end can be dropped without losing a value.
• The sine's family (§2.2, Part I p. 19). Three intervals are offered alongside, each a half turn wide, running end to end: the one below, the chapter's own, and the one above. Verified: the sine is one-one on each and covers the whole of minus one to one on each, because consecutive odd multiples of half pi bracket exactly one rise or one fall of the wave. The first interval of this first list is misprinted — see the note below and use the corrected form.
• The naming (§2.2, Part I p. 19, both terms set in italic). Each admissible interval yields one branch; the branch whose outputs run from minus half pi to half pi is the principal value branch. The chapter then writes the inverse sine as a map from the closed interval minus one to one, into that branch. Note the direction of the arrow: the interval of angles is now the output side.
• The cosine, restricted (§2.2, Part I p. 21). The interval kept is from zero to pi, closed at both ends, with the neighbours running from minus pi to zero and from pi to two pi. Verified: on zero to pi the cosine falls steadily from one to minus one without repeating, and both extremes are attained at the ends. From here on the chapter says bijective where Part I p. 19 said one-one and onto; the two say the same thing.
• The cosecant, restricted (§2.2, Part I pp. 21–22). The chapter first records the domain of the cosecant as the real line with the whole multiples of pi removed, and its range as the real line with the open interval from minus one to one removed. It then keeps the closed interval from minus half pi to half pi with the point zero taken out, and offers two neighbours built the same way, each losing the multiple of pi that sits inside it. Verified: zero has to go because the cosecant is undefined there, and the two endpoints stay because the cosecant equals minus one and one at them, which are values the inverse must be able to return.
• The secant, restricted (§2.2, Part I pp. 22–23). Domain: the real line with the odd multiples of half pi removed. Range: the real line with the open interval from minus one to one removed. The interval kept runs from zero to pi with half pi taken out; the neighbours run from minus pi to zero without minus half pi, and from pi to two pi without three half pi. Verified: the removed point is exactly where the secant breaks, and the endpoints survive because the secant is one there and minus one at pi.
• The tangent, restricted (§2.2, Part I pp. 23–24). Domain: the real line with the odd multiples of half pi removed; range: the whole real line. The interval kept is from minus half pi to half pi with neither end included, and the neighbours are the two adjacent open intervals of the same width. Verified: the ends must go because the tangent is undefined at them, and nothing is lost by dropping them, since the tangent already runs through every real number strictly inside.
• The cotangent, restricted (§2.2, Part I pp. 24–25). Domain: the real line with the whole multiples of pi removed; range: the whole real line. The interval kept is from zero to pi with neither end included, neighbours likewise. Verified: same reasoning as the tangent, shifted a quarter turn.
• Where the cuts fall (not in the book). Verified against all six: for the sine and the cosecant the cuts sit at the odd multiples of half pi; for the cosine and the secant they sit at the whole multiples of pi; for the tangent they sit at the odd multiples of half pi and for the cotangent at the whole multiples of pi. In every case the cut is placed where the graph either turns around or breaks — the two places a function can stop being one-one. That single sentence organises the whole of §2.2 and the chapter never writes it.
• A mis-set interval, to be checked (not in the book). Take the sine on the closed interval from zero to pi, which is a half turn wide, like the chapter's. Verified: the sine of pi by six and the sine of five pi by six are both one half, and both inputs lie inside, so the first demand fails; and the sine never goes below zero there, so the second demand fails too. Width is not the test. Use this immediately after the family in section 4, because students routinely conclude that any half turn will do.
• The chapter's own inverse declarations (§2.2, Part I pp. 19, 21, 22, 23, 24 and 25). Six arrow statements, one per ratio, each naming the domain of the inverse on the left and the promoted branch on the right. They are collected into a single table on Part I p. 25, which is the subject of the next topic; here they should appear one at a time as each restriction is settled, so that the table arrives as a summary rather than as a list to memorise.

Figures to have open

• One long input axis, reused throughout, carrying the sine wave with the cut points marked at the odd multiples of half pi. Sections 3, 4 and 11 all work on this same drawing, and it must not be redrawn between them, or the family stops looking like one family.
• The same axis with the cosine wave and its cuts at the whole multiples of pi, drawn to the same scale and placed directly beneath the sine, so the quarter turn offset is visible rather than asserted.
• Four short strips for section 7 and section 8 — cosecant, secant, tangent, cotangent — each showing one admissible piece with its endpoints drawn as filled or hollow dots and its interior hole, if any, drawn as a hollow dot. The dot convention has to be the same in all four.
• A single legend card for section 9 showing a filled endpoint, a hollow endpoint and an interior hole, with the cause written beside each.
• No figure here is copied from the chapter. The chapter's own figures on Part I pp. 20–25 are graphs of the inverse functions and belong to the fourth topic of this module.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 2 "Inverse Trigonometric Functions", §2.2 Basic Concepts: the sine restricted and the two terms introduced, Part I p. 19
• The cosine restricted and the cosecant begun, Part I p. 21; the cosecant completed, Part I p. 22
• The secant restricted, Part I pp. 22–23; the tangent, Part I pp. 23–24; the cotangent, Part I pp. 24–25
• The six arrow statements, Part I pp. 19, 21, 22, 23, 24 and 25