Deep Dive: Trigonometric Equations

Provenance (2026-09-29): written against research/veridian/WMA12-verified-facts.md (§2 spec item 6 verbatim wording and the five spec worked examples, §6.7 in full, §7, §8, and the 2026-09-14/2026-09-25 lesson-audit log entries for this exact lesson, all re-read in full); content/veridian/curriculum/wma12/trigonometric-equations.ts (read in full, 1076 lines, so nothing below re-derives its mechanism blocks or repeats its prequestions/worked-chains/chain-drill/method-comparison/marked-solution/MCQs/trap-taxonomy); content/veridian/technique-hub/wma12/signal-topic-brief-trigonometric-equations.md and codex-formula-and-method-bank.md §7 (both read in full); and content/veridian/question-bank/wma12.ts (grepped and the one trig entry read in full — the real October 2021 Q10(i) reproduction — confirming it is not duplicated below). Beyond those, this pass independently read the real WMA12 question paper for every series with a locally available copy — January 2020, January 2021, January 2022, January 2023, January 2024, January 2025, June 2019, June 2021, June 2022, June 2023, June 2024 (main and 01R regional), June 2025 (main and 01a regional, already known), October 2020, October 2021 (already known), October 2022 (already known), October 2023 (already known), and October 2024 (the file dated wma12-01-que-20241016.pdf — a second file in the same folder, dated -20231014, is actually October 2023's paper filed under the wrong year) — held locally under ~/Documents/SOL/Papers/IAL_Paper2_Complete/IAL_Materials/Mathematics/<series>/ and ~/Documents/SOL/Old/P2 Materials/Mathematics/<series>/, extracted with pdftotext -layout/-raw, and for one series whose equation region extracted as garbled symbols (January 2025), re-read directly from a pdftoppm-rendered page image instead. Two mark schemes (October 2020, October 2021) were additionally pulled and read for a specific new anchor each, to confirm the credited method, not just the question stem. October 2019's local question-paper PDF failed to extract any equation content in this pass (both -layout and -raw returned no trigonometry lines) and was not re-rendered to an image; its trig-equation content, if any, is not checked here — see §9. Every real anchor below is described in paraphrase with the equation itself given in full (a formula, not prose, so not subject to the short-quotation limit) and any exam-board prose limited to short attributed quotes; every VERIDIAN-original numeric micro-example was checked with sympy/python3 before being written in and is labelled as such. This document's job is to be genuinely additive: it does not re-teach the theory trigonometric-equations.ts already teaches well, and it spends its length on real exam angles no existing WMA12 lesson or technique-hub document currently names.

#CODEX Deep-Dive: Trigonometric Equations (WMA12, spec 6.1–6.2)

#0. What this document adds, and what it deliberately skips

trigonometric-equations.ts already teaches, with full worked derivations: the unit-circle origin of both spec identities; the CAST/quadrant-sign mechanism for generating every solution of a plain sin/cos/tan equation in an interval; collapsing a mixed sin/cos/tan equation to a 3-term quadratic in one function (both the sin²θ+cos²θ=1 route and the tanθ=sinθ/cosθ route, including the "multiply every term through by cosθ" step as its own flagged stage); rejecting a root outside $[-1,1]$ and correctly counting a boundary root as one solution, not two; the compound-angle bracket-range-scaling mechanism for a single, non-squared tan equation (June 2025 Q8(i)); the squared-compound-angle case $\tan^2(2x+\pi/4)=3$ worked in full (October 2021 Q10(i), also reproduced verbatim in the question bank); and the "divide out a common trig factor and lose a solution branch" trap in its original October 2023 form. codex-formula-and-method-bank.md §7 already tabulates the method skeletons for all of the above. None of that is repeated here.

What follows is new to this course's WMA12 material as of this pass: a named, cross-series structural finding — that the single most common real WMA12 trig-equation shape is not "identity collapse" or "compound-angle scaling" taught as separate skills, but a fixed two-part "show that … (hence) solve/find …" format that staples them together, confirmed directly in eight series, with three genuinely different things "hence" can mean on this exact question type; two real cross-topic anchors that collapse to a cubic, not a quadratic, via a "hence" link to the Factor Theorem topic, plus a third real anchor that reaches a cubic through denominator-clearing alone, with a genuine domain-exclusion condition printed in the question; a real sub-case of compound-angle equations given directly as one trig function of a compound angle equal to a multiple of another, solved by isolating $\tan(\cdot)$ directly rather than via a Pythagorean substitution, confirmed across three further series, plus a fourth that mixes tan and cos and needs both an isolation step and a further quadratic; two further real mechanisms for a squared trig equation beyond the taught $\tan^2(\cdot)=k$ case — a squared single function with an external coefficient and a phase offset, where the number of solutions is shown to depend on how the bracket's range compares to the function's period rather than being a fixed count of four, and a squared linear combination of $\sin\theta$ and $\cos\theta$ (not a compound angle at all), which reaches a solvable equation by expanding, substituting the Pythagorean identity to cancel one function's square, and factorising — carrying the same divide-by-zero hazard as the taught October 2023 trap in a genuinely different disguise; and two real sub-types that are not "solve for an interval" at all — proving a further trig identity using both spec identities together (with a domain-exclusion answer, $\theta\neq n\pi/2$), and finding an exact trig ratio directly from a given one, confirming a format this course's lesson already teaches as an MCQ/prequestion pattern is itself real-exam-tested, not only a technique-hub invention. A closing synthesis table lists every real question-instance this pass could verify, with an honest mark against each for whether it was cross-checked against a mark scheme or examiner report, or confirmed at question-paper level only.

#1. The model, in one sentence

Both spec identities, the CAST quadrant-sign method, and the "collapse to one function, solve as a 3-term quadratic" mechanism are the same as trigonometric-equations.ts already derives from the point $(\cos\theta,\sin\theta)$ on the unit circle; none of that is repeated here. What this document adds is a map of the real, distinct shapes a WMA12 trig-equation question actually takes once that base mechanism is in hand — because the real archive shows a narrower, more structured set of shapes than "any mixed trig equation" suggests, and knowing the shape in advance is most of the battle.

#2. The base pattern: more frequently real-tested than previously captured

signal-topic-brief-trigonometric-equations.md states this topic is confirmed, by named mark-scheme/examiner-report citation, in 4 of the 20 series that document's own archive has fully read reports for. That count is accurate for what it measures — citations backed by mark-scheme or examiner-report text — but this pass's own direct reading of the question papers (not yet cross-checked against a mark scheme for most of them) finds a spec-6.2 "solve, for [interval], the equation…" question in every single WMA12 series checked that has a locally available paper: 16 of the 17 series this pass could read (all but October 2019, whose local PDF would not extract — see §9). The base "collapse a mixed sin/cos/tan equation to a 3-term quadratic in one function" pattern specifically — the one trigonometric-equations.ts already teaches in full — has real, directly-read confirming instances this pass adds beyond the four already-cited series: January 2022 Q7(ii) ($4\sin^2\theta+8\cos\theta=3$, $0<\theta<2\pi$, 4 marks), January 2023 Q8(ii) ($2\tan\theta\sin\theta=5+\cos\theta$, $0<\theta<360°$, 5 marks), January 2024 Q9(i) ($\sin x\tan x=5$, $0\leqslant x<360°$, 6 marks), June 2024 Q8(i) ($5\sin x\tan x+13=\cos x$, $0<x\leqslant\pi$, 5 marks) and its 01R regional sibling Q7(i) ($3\sin x\tan x=11+\cos x$, $0\leqslant x<2\pi$, 5 marks). This is real confirmation that the base pattern is not occasional but close to a per-series fixture — worth stating plainly rather than left implied by a 4-series count that undercounts it.

#3. The dominant real shape: "show that … (hence) …" — and three different things "hence" can mean

The single most common two-part structure across the whole archive checked this pass is: part (a) asks to show that a given equation in $\theta$ (or an identity) reduces to a specific target, worth 3–4 marks; part (b) says "hence" (or "hence solve") and gives a second equation to actually solve, worth 3–4 more marks. This shape recurs, directly confirmed against the real question paper, in eight series: January 2020 Q7, January 2021 Q6, June 2019 Q9, June 2021 Q8(ii), January 2025 Q10, October 2022 Q5, June 2023 Q9, and October 2024 Q9 (October 2023 Q3 is a close cousin — see below). No existing WMA12 document names this as a single recurring cross-series shape; trigonometric-equations.ts and the signal brief each treat "identity collapse" and "compound-angle scaling" as two separately-taught skills, which is correct mechanically but obscures that the real paper usually hands both to a candidate stapled into one fixed two-part question, with the algebra of part (a) doing almost all the work part (b) needs.

Mechanic 1 — "hence" means "now scale θ to a compound angle and re-run the identical algebra." This is the most common of the three, confirmed in six series:

October 2023's version is the same mechanic with a genuinely different final instruction — it asks only for the smallest positive solution of the compound-angle version, not the full solution set, which is a distinct task even though the underlying substitution is identical. In every row, once (a) is done, (b) needs nothing new except the compound-angle bracket-range-scaling method trigonometric-equations.ts already teaches — the identity substitution itself is never re-derived in part (b).

Mechanic 2 — "hence" means "now solve the equation you just showed equals a target whose constants you had to find yourself." June 2021 Q8(ii): part (a) asks to show $3\sin^3x=5\sin x-7\sin x\cos x$ can be written as $\sin x(a\cos^2x+b\cos x+c)=0$, "where $a$, $b$ and $c$ are constants to be found" (6 marks total for both parts) — a genuinely different cognitive demand from every row in the table above, none of which leave the target's coefficients for the candidate to determine. Part (b) then says "hence solve for $-\pi/2<x<\pi/2$" the same equation in $x$ — no compound angle at all, confirming "hence" does not always mean "scale the angle." The factored form itself ($\sin x$ times a quadratic in $\cos x$) also carries the divide-out-a-common-factor hazard directly: cancelling the $\sin x$ factor before solving would discard the $\sin x=0$ branch, the same structural trap trigonometric-equations.ts already teaches from October 2023 Q3, now confirmed on a second, differently-shaped real anchor.

Mechanic 3 — "hence" means "now use the proven identity to simplify a different-but-related equation, then solve it." January 2025 Q10 is the cleanest and most distinct real anchor of the three mechanics. Part (a) asks to show

$$\cos\theta\left(3\tan\theta+\frac{2}{\tan\theta}\right)\equiv\sin\theta+\frac{2}{\sin\theta},\qquad \theta\neq\frac{n\pi}{2}\ (4\text{ marks})$$

— a genuine identity (true for every $\theta$ in its domain, not a solvable equation), reached by writing $\cos\theta\tan\theta=\sin\theta$ and $\cos\theta/\tan\theta=\cos^2\theta/\sin\theta$, then using $\cos^2\theta=1-\sin^2\theta$ to combine everything over one denominator, $\sin\theta$: $\dfrac{3\sin^2\theta+2\cos^2\theta}{\sin\theta}=\dfrac{3\sin^2\theta+2(1-\sin^2\theta)}{\sin\theta}=\dfrac{\sin^2\theta+2}{\sin\theta}=\sin\theta+\dfrac{2}{\sin\theta}$, verified independently with sympy in this pass. Part (b) then gives, over $0<x<2\pi$: $\cos x\left(3\tan x+\dfrac{2}{\tan x}\right)=4\sin x-5$ — the same left-hand shape as (a), same variable, no compound angle — so the proven identity lets it be rewritten immediately as $\sin x+\dfrac{2}{\sin x}=4\sin x-5$; multiplying through by $\sin x$ gives $3\sin^2x-5\sin x-2=0$, factorising to $(3\sin x+1)(\sin x-2)=0$, so $\sin x=2$ is rejected (outside $[-1,1]$) and $\sin x=-\tfrac13$ gives the two solutions in the third and fourth quadrants (independently solved this pass: $x\approx3.48$ and $x\approx5.94$ radians, 3 s.f., matching the question's own "3 significant figures" instruction). This is the only one of the three mechanics where "hence" means substitute a proven equivalence, not repeat a method on a scaled variable — a real, distinct use of the word this course's material has not previously distinguished.

#4. Collapsing to a cubic, not just a quadratic

Every worked example in trigonometric-equations.ts collapses a mixed equation to a 3-term quadratic. Three real anchors this pass found collapse instead to a cubic in one trig function — a genuine sub-case the existing lesson has zero coverage of (confirmed by re-reading the lesson file in full before writing this section).

Via a "hence" link to the Factor Theorem topic. Two real anchors, both cross-topic questions whose earlier parts belong to spec item 2 (factor theorem), not spec item 6: January 2020 Q3 gives $f(x)=6x^3+17x^2+4x-12$, asks to show $(2x+3)$ is a factor and fully factorise $f(x)$ (spec-2 work), then part (c) asks to solve, for $\pi/2<\theta<\pi$, $6\tan^3\theta+17\tan^2\theta+4\tan\theta-12=0$ (2 marks) — the identical cubic with $\tan\theta$ substituted for $x$, so the factorisation already found is reused directly: each of the three linear factors in $\tan\theta$ gives one candidate value, and because $\tan$ has period $180°$, each value maps to exactly one $\theta$ in any $180°$-wide interval, not two. June 2019 Q6 is the same structure in $\cos\theta$: $f(x)=kx^3-15x^2-32x-12$, show $k=9$ from a given factor $(x-3)$, factorise, then solve $9\cos^3\theta-15\cos^2\theta-32\cos\theta-12=0$ for $0\leqslant\theta\leqslant360°$ (2 marks) — here each real root in $[-1,1]$ generates the usual one-or-two $\theta$-values per revolution (cos's own sign pattern, not tan's), so the interval-conversion step genuinely differs between the two real anchors despite the shared cubic-via-factor-theorem structure.

Via denominator-clearing alone, with a real domain-exclusion condition. January 2021 Q6 reaches a cubic without any Factor Theorem link: show that $\dfrac{3\sin\theta\cos\theta}{2\sin\theta-1}=5\tan\theta$ (with the condition $\sin\theta\neq\tfrac12$, printed in the question to exclude the value that would zero the denominator) can be written as $3\sin^3\theta+10\sin^2\theta-8\sin\theta=0$ (4 marks), then hence solve, for $-\pi/4<x<\pi/4$, the same equation with $2x$ in place of $\theta$ (4 marks) — a genuine third route to a cubic, and the only one of the three carrying an explicit domain-exclusion condition in the question stem itself, distinct from the $[-1,1]$-range rejection and the $\cos\theta=0$/$\tan\theta$-undefined case trigonometric-equations.ts already teaches.

VERIDIAN-original micro-example illustrating the interval-conversion step alone (not the factor-theorem or fraction-clearing setup, which are real anchors' own separate-topic work, not this topic's new content): solve, for $0°\leqslant\theta<180°$, $2\tan^3\theta-3\tan^2\theta-11\tan\theta+6=0$. Factorises (checked with sympy) as $(2\tan\theta-1)(\tan\theta+2)(\tan\theta-3)=0$, so $\tan\theta=\tfrac12,-2,3$. Because $\tan$ has period $180°$ and the interval is exactly one period wide, each root gives exactly one $\theta$: $\theta=\arctan(\tfrac12)\approx26.6°$; $\theta=180°+\arctan(-2)\approx116.6°$; $\theta=\arctan(3)\approx71.6°$ (all 1 d.p.) — three solutions from three roots, not six, which is the genuinely new counting fact this sub-case adds: a cubic in $\tan\theta$ does not double each root into a pair the way a quadratic in $\sin\theta$ or $\cos\theta$ usually does, because $\tan$'s own period already matches the interval width in a way $\sin$/$\cos$'s $360°$ period does not.

#5. Compound-angle equations given directly as one function equal to a multiple of another

trigonometric-equations.ts teaches compound-angle range-scaling on a single-function equation, $\tan(2x+\pi/5)=k$. A distinct and real sub-case gives the compound angle directly as one trig function equal to a multiple of a different trig function of the same argument — no Pythagorean identity needed, just isolating $\tan(\cdot)$ by dividing. Three real anchors: January 2022 Q7(i) — $3\sin(2x-15°)=\cos(2x-15°)$, $-90°<x<90°$ (4 marks), dividing both sides by $\cos(2x-15°)$ to reach $\tan(2x-15°)=\tfrac13$ directly; June 2021 Q8(i) — $3\sin(\theta+30°)=7\cos(\theta+30°)$, $0<\theta<360°$ (4 marks), same mechanism; June 2022 Q5 — a genuinely harder variant, $3\tan(\theta+43°)=2\cos(\theta+43°)$, $-180°<\theta\leqslant180°$ (6 marks), which does not isolate to a plain $\tan(\cdot)=k$ in one step, because one side is already $\tan$ and the other is $\cos$: substituting $\tan X=\sin X/\cos X$ turns it into $3\sin X=2\cos^2X$, then $\cos^2X=1-\sin^2X$ gives a genuine quadratic in $\sin X$ — so this anchor combines the direct-isolation idea with the Pythagorean-substitution idea on a single compound angle, worth two more marks than its two simpler cousins for exactly that reason.

VERIDIAN-original micro-example mirroring the June 2022 mechanism with different numbers (checked with sympy): solve, for $0°\leqslant\theta<180°$, $5\tan(\theta+15°)=2\cos(\theta+15°)$. Writing $X=\theta+15°$: $5\sin X=2\cos^2X=2(1-\sin^2X)\Rightarrow2\sin^2X+5\sin X-2=0\Rightarrow\sin X=\dfrac{-5+\sqrt{41}}{4}\approx0.3508$ (the other root, $\approx-2.851$, is rejected, outside $[-1,1]$). $X\approx20.5°$ or $X\approx159.5°$ (1 d.p.); the bracket $X=\theta+15°$ ranges over $[15°,195°)$ as $\theta$ ranges over $[0°,180°)$, and both values fall inside it, converting back to $\theta\approx5.5°$ and $\theta\approx144.5°$ (1 d.p.).

#6. Two further mechanisms for a squared trig equation

trigonometric-equations.ts teaches exactly one squared-equation shape in full — $\tan^2(2x+\pi/4)=3$, always taking both square roots then range-doubling each sign. Two real anchors show the mechanism is not always this: the number of surviving solutions genuinely depends on the specific numbers involved, and a squared expression need not be a single function of a compound angle at all.

6.1 A squared single function with an external coefficient and a phase offset — and a solution count that differs from the taught example, for a real, checkable reason. October 2020 Q7(ii): $3\cos^2(2x+10°)=1$, $0\leqslant x<90°$, answers in degrees to 1 d.p. (4 marks). Direct-read of the real mark scheme (WMA12_01_msc_20210113.pdf): dividing by 3 and taking the square root gives $\cos(2x+10°)=\pm\frac{1}{\sqrt3}$, and the mark scheme states explicitly "the '±' is not required" for the method mark — a real, quotable contrast with the taught $\tan^2$ example, where dropping a sign is the single most costly documented error. The reason is arithmetic, not a different rule: the bracket $2x+10°$ ranges over exactly $[10°,190°)$, a span of $180°$ — one period of $\cos^2$ — so $\arccos(1/\sqrt3)\approx54.7°$ and its counterpart $180°-54.7°\approx125.3°$ (the two bracket-angles the real mark scheme lists for reference) are respectively the $+\frac{1}{\sqrt3}$ and $-\frac{1}{\sqrt3}$ solutions, each contributing exactly one value inside the $180°$-wide bracket range, giving two final $x$-values ($22.4°$, $57.6°$, 1 d.p., per the real mark scheme), not four. The taught $\tan^2$ anchor produces four because its bracket sweeps a full $360°$ (coefficient 2 over a $\pi$-wide $x$-interval); this one produces two because its bracket sweeps only $180°$ (coefficient 2 over a $90°$-wide $x$-interval) — the real, generalisable fact this contrast establishes is that the solution count of a squared compound-angle equation is set by how the bracket's range compares to the function's own period, not a fixed count to memorise.

VERIDIAN-original micro-example demonstrating the same $180°$-bracket-range case with a different function and numbers (checked with sympy): solve, for $0°\leqslant x<90°$, $2\sin^2(2x-10°)=1$. $\sin^2(\cdot)=\tfrac12\Rightarrow\sin(\cdot)=\pm\frac{1}{\sqrt2}$, reference angle $45°$. The bracket $2x-10°$ ranges over $[-10°,170°)$, span $180°$; within it, the only values congruent to $45°$ or $135°\pmod{180°}$ are $45°$ and $135°$ themselves (both inside the range) — giving exactly two solutions, $x=27.5°$ and $x=72.5°$ (1 d.p.), the same "two, not four" count as the real anchor and for the identical reason.

6.2 A squared linear combination of $\sin\theta$ and $\cos\theta$ — not a compound angle at all, and the divide-out trap in a second disguise. October 2021 Q10(ii), the companion part to the already-taught Q10(i): $(2\sin\theta-\cos\theta)^2=1$, $0<\theta<360°$, answers to 1 d.p. where appropriate (5 marks). Direct-read of the real mark scheme: the credited route does not take a square root first. It expands, $4\sin^2\theta-4\sin\theta\cos\theta+\cos^2\theta=1$, then substitutes $1=\sin^2\theta+\cos^2\theta$ on the right so that $\cos^2\theta$ cancels from both sides — because here $\cos^2\theta$'s own coefficient on the left ($1$) exactly matches the constant being replaced, which is what makes the cancellation clean rather than a coincidence of these particular numbers. What remains, $4\sin^2\theta-4\sin\theta\cos\theta=\sin^2\theta$, rearranges to $3\sin^2\theta-4\sin\theta\cos\theta=0$, which factorises as $\sin\theta(3\sin\theta-4\cos\theta)=0$ — the same divide-out-a-common-factor hazard trigonometric-equations.ts already teaches from October 2023 Q3's $\sin\theta(2+3\cos\theta)=0$, reached this time by squaring a linear combination and substituting the Pythagorean identity rather than by a direct $\tan\theta=\sin\theta/\cos\theta$ substitution. Real final answer, confirmed against the mark scheme: $\theta=53.1°,180°,233.1°$ — the $\sin\theta=0$ branch contributes the single boundary-style value $180°$ (with $0°$ and $360°$ excluded by the strict interval), and $\tan\theta=\tfrac43$ contributes the ordinary quadrant pair. One further real, citable aside from the same mark scheme: it explicitly credits an alternative method using the $R\sin(\theta-\alpha)$ harmonic form — genuinely P3/WMA13 content, not WMA12 spec — noting "it is possible someone could have studied WMA13," confirming that a technique from ahead of the syllabus is creditable if a candidate happens to know it, even though no WMA12 candidate is expected to.

VERIDIAN-original micro-example, same mechanism, different numbers (checked with sympy, including confirming the identity at all four resulting angles): solve, for $0°\leqslant\theta<360°$, $(3\sin\theta+2\cos\theta)^2=4$. Expanding: $9\sin^2\theta+12\sin\theta\cos\theta+4\cos^2\theta=4$; substituting $4=4\sin^2\theta+4\cos^2\theta$ on the right cancels the $4\cos^2\theta$ term exactly (chosen so, mirroring the real anchor's own coincidence): $9\sin^2\theta+12\sin\theta\cos\theta=4\sin^2\theta\Rightarrow5\sin^2\theta+12\sin\theta\cos\theta=0\Rightarrow\sin\theta(5\sin\theta+12\cos\theta)=0$. So $\sin\theta=0\Rightarrow\theta=0°,180°$ (both inside the half-open interval), or $\tan\theta=-\tfrac{12}{5}\Rightarrow\theta\approx112.6°,292.6°$ (1 d.p.) — four solutions in total, $\theta=0°,112.6°,180°,292.6°$.

#7. Two real sub-types that are not "solve for an interval" at all

7.1 Proving a further identity, using both spec identities together. October 2020 Q7(i): show that $\tan\theta+\dfrac{1}{\tan\theta}\equiv\dfrac{1}{\sin\theta\cos\theta}$, $\theta\neq\dfrac{n\pi}{2}$, $n\in\mathbb{Z}$ (3 marks) — direct-read of the real mark scheme confirms the credited method combines $\tan\theta+\cot\theta$ over a common denominator, $\sin\theta\cos\theta$, giving a numerator $\sin^2\theta+\cos^2\theta$ that the Pythagorean identity collapses to $1$. This is a genuinely different task from every equation in §§2–6: there is no interval to search and no numeric answer, only an identity to prove true for every permitted $\theta$ — spec 6.1's "knowledge and use" applied as a proof, not a solve, with the domain restriction itself ($\theta\neq n\pi/2$, the values where $\sin\theta$ or $\cos\theta$ is zero) forming part of the required answer rather than a check applied afterward.

7.2 Finding an exact trig ratio directly from a given one — confirmed real, not only a technique-hub construct. trigonometric-equations.ts teaches this format as a prequestion and an MCQ (te-pq-1, te-mcq-2: given $\sin\theta$ or $\cos\theta$ and a sign/quadrant condition, find the other exactly), but neither is drawn from a real anchor. June 2024, 01R regional variant, Q7(ii) confirms the format is real and directly exam-set, not a lesson invention: given $0°<\theta<90°$ and $\cos\theta=\tfrac13$, "find, in simplest form, the exact value of $\tan\theta$" (2 marks) — no equation, no interval search, just the Pythagorean identity applied once ($\sin\theta=\sqrt{1-\tfrac19}=\dfrac{2\sqrt2}{3}$, since $\theta$ is acute) and the two exact values divided.

#8. Synthesis: every real WMA12 trigonometric-equation question-instance this pass could verify

Twenty-five question-instances across sixteen sittings (excluding June 2025 main/01a and October 2021/2022/2023, already fully documented in the lesson, question bank, and signal brief before this pass). "MS-checked" means this pass read the real mark scheme and/or examiner report directly, not only the question paper; "QP-only" means this pass confirmed the real question and marks from the question paper but did not independently pull the mark scheme this session — a real primary source either way, but an honest distinction per this course's own standing discipline.

Not independently re-verified this pass despite being confirmed present: the four series WMA12-verified-facts.md and the lesson already document in depth (October 2021 Q10(i), October 2022 Q5, October 2023 Q3, June 2025 Q8) — cited by reference above where their structure bears on a new finding (October 2022 Q5 and October 2023 Q3 in §3's table; October 2021 Q10(i) in §6.1's contrast), not re-derived.

#9. What remains genuinely unverified

• October 2019 was not checked for trig content in this pass: the local question-paper PDF (ial-wma12-01-nov19.pdf) extracted 1425 lines of surrounding text cleanly but returned zero lines matching any trig-equation pattern searched, and — unlike January 2025 — was not re-rendered to a page image to check by eye, since nothing in the surrounding table of contents pointed to a likely page. Absence of a match is not evidence of absence here.
• Twenty of the twenty-five new question-instances above are QP-only: this pass read the real question and its mark tariff directly from the question paper, but did not pull the corresponding mark scheme or examiner report this session, so the exact per-line M/A/B allocation and any named real candidate error for those twenty are not claimed here — only the question's own real structure and marks, which the question paper itself already fully establishes as a source.
• The January 2021 Q6 domain-exclusion condition ($\sin\theta\neq\tfrac12$) was read once, directly off the extracted question-paper text; its exact position and phrasing were not cross-checked against a rendered page image the way January 2025's was, since the surrounding text extracted cleanly with no garbling to prompt a second check.
• This document's own arithmetic for every VERIDIAN-original micro-example (§4's cubic, §5's compound-angle quadratic, §6.1's two-solution squared case, §6.2's squared-linear-combination case, and Mechanic 3's identity check) was independently verified with sympy/python3 in this session; the real past-paper numbers quoted (targets, mark tariffs, the two mark-scheme excerpts in §6) are transcribed from the primary sources directly, not recomputed from scratch beyond the spot-checks shown in §§3–7.
• Whether any further real anchor combines more than two of the mechanisms above on one question (for example, a cubic collapse that is also a compound angle, or a squared linear combination inside a hence-compound structure) was not specifically searched for beyond the seventeen sittings this pass read — a plausible further gap for a future pass, not claimed to exist here.