Technique Masterclasses

Pure Mathematics 3

Deep Dive: Compound and Double Angle Formulae

Provenance (2026-09-29): read in full — research/veridian/WMA13-verified-facts.md (§1 for the verbatim spec 2.3 wording, §3 for the formula-booklet "must memorise" split, §4 for the mark-scheme conventions used below, §5 "2.3 Double angle, sum/difference, R sin/cos(θ±α) harmonic form" for every examiner-report and mark-scheme citation this document draws on, §7–§8 for the topic-strength and flagged-gap summary); content/veridian/curriculum/wma13/compound-and-double-angle-formulae.ts (1,007 lines, read in full, specifically to identify every derivation, worked chain, trap and MCQ it already contains and avoid duplicating any of them); content/veridian/technique-hub/wma13/codex-formula-and-method-bank.md, forge-exam-technique-guide.md and signal-frequency-and-boundaries.md (all read in full, to avoid restating their tables); content/veridian/question-bank/wma13.ts (read in full — confirmed it contains no compound- or double-angle item at all, so nothing there needed cross-referencing). No new primary-source PDF was opened for this pass; every citation below traces to one of the documents just named. Every micro-example's arithmetic was independently recomputed with Python (sympy, plus direct math evaluation) before being written in, not carried over from any file on trust. Not affiliated with or endorsed by Pearson Edexcel.

CODEX Deep-Dive: Compound and Double Angle Formulae (WMA13, spec 2.3)

0. What this document is, and what it deliberately doesn't repeat

Three documents already cover this topic well. The lesson file derives sin⁡2A\sin2A, all three equivalent forms of cos⁡2A\cos2A, and tan⁡2A\tan2A from the compound-angle formulae by substituting B=AB=A; teaches the rule for choosing which cos⁡2A\cos2A form to use; carries a full worked chain solving cos⁡2θ=sin⁡θ\cos2\theta=\sin\theta; a chain-drill deriving and stress-testing tan⁡2A\tan2A; a marked solution proving (sin⁡θ+cos⁡θ)2≡1+sin⁡2θ(\sin\theta+\cos\theta)^2\equiv1+\sin2\theta and then solving it, including the interval-widening step for a sin⁡2θ=k\sin2\theta=k equation; a beyond-spec section deriving sin⁡3θ≡3sin⁡θ−4sin⁡3θ\sin3\theta\equiv3\sin\theta-4\sin^3\theta against a real exam anchor; five named traps; and six MCQs. The CODEX formula bank gives the compact identity table and two method skeletons (an equation/identity-proof skeleton and a harmonic-form skeleton). FORGE gives the "quote the formula first" convention, the exact-answer rule for harmonic-form construction marks, and one genuinely new item not built into the lesson at all — a note that a real Jan 2022 examiner report records candidates forgetting to halve a solved 2θ2\theta back to θ\theta. SIGNAL gives the raw frequency count. None of the four states, in one place, the full real mark ladder for either of the two real exam anchors that the lesson only cites in fragments (Oct 2020 Q1 and Jan 2022 Q2); none synthesises the "which form, and how many solutions" decision procedure into a single reusable rule; and none reports, honestly, where the evidence for this topic actually runs out. That is this document's job. Where a fact or a full worked example already exists elsewhere, it is pointed at by name, not re-taught.

1. Where this sits on the paper

Spec 2.3 (paraphrased from the facts bank's §1 transcription): knowledge and use of the double-angle formulae, and of the compound-angle formulae for sin⁡(A±B)\sin(A\pm B), cos⁡(A±B)\cos(A\pm B) and tan⁡(A±B)\tan(A\pm B), including writing acos⁡θ+bsin⁡θa\cos\theta+b\sin\theta as rcos⁡(θ±α)r\cos(\theta\pm\alpha) or rsin⁡(θ±α)r\sin(\theta\pm\alpha). The guidance column names three required jobs — application to half angles, solving acos⁡θ+bsin⁡θ=ca\cos\theta+b\sin\theta=c in an interval, and proving identities such as cos⁡xcos⁡2x+sin⁡xsin⁡2x≡cos⁡x\cos x\cos2x+\sin x\sin2x\equiv\cos x — and one explicit exclusion, the t=tan⁡(12θ)t=\tan(\tfrac12\theta) substitution.

The booklet split matters more here than almost anywhere else on this paper. The three compound-angle formulae ARE printed in the formula booklet. None of sin⁡2A\sin2A, any form of cos⁡2A\cos2A, tan⁡2A\tan2A, or a half-angle result is printed — the spec states this list "will not appear in the booklet," making it genuine memorised content that must also be rebuildable on demand under exam pressure, not a look-up fact.

Frequency. SIGNAL's own count states 2.3 (bundled with harmonic form, the sibling technique for the same spec point) was seen in 5 of the WMA13 series this course's research has reviewed so far — Oct 2020, Jan 2022, Jun 2022, Jun 2023, Jan 2024 — at an observed weight of 8–9 marks when it anchors a full question, with harmonic form specifically confirmed head-on only once (Oct 2020 Q7, 9 marks). Honest gap, stated here rather than smoothed over: of those five series, only Oct 2020 (three separate real anchors — Q1, Q5(a), Q7) and Jan 2022 (Q2) have their actual question text and mark scheme independently transcribed anywhere in the sources available to this pass. Jun 2022, Jun 2023 and Jan 2024's specific 2.3 question content is not detailed in the facts bank's own per-topic write-up — this document does not invent what those questions asked, and flags the gap again in §6.

2. The complete real theory

2.1 What's given

sin⁡(A±B)≡sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A\pm B)\equiv\sin A\cos B\pm\cos A\sin B; cos⁡(A±B)≡cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A\pm B)\equiv\cos A\cos B\mp\sin A\sin B; tan⁡(A±B)≡tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A\pm B)\equiv\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B} — all three printed on the formula sheet, none proved in this course (nothing requires it), all three used as given.

2.2 The one substitution that generates everything else

Setting B=AB=A in each of the three formulae above is the entire derivation, not three separate ones: sin⁡2A≡2sin⁡Acos⁡A\sin2A\equiv2\sin A\cos A; cos⁡2A≡cos⁡2A−sin⁡2A≡2cos⁡2A−1≡1−2sin⁡2A\cos2A\equiv\cos^2A-\sin^2A\equiv2\cos^2A-1\equiv1-2\sin^2A (one quantity, reached three times by substituting cos⁡2A+sin⁡2A≡1\cos^2A+\sin^2A\equiv1 two different ways into the first form); tan⁡2A≡2tan⁡A1−tan⁡2A\tan2A\equiv\dfrac{2\tan A}{1-\tan^2A}. The lesson file derives all three in full narrative form and is not reworked here. Independently re-checked at a fresh angle not used anywhere in the lesson, A=20°A=20°: cos⁡220°−sin⁡220°=0.76604\cos^220°-\sin^220°=0.76604, 2cos⁡220°−1=0.766042\cos^220°-1=0.76604, 1−2sin⁡220°=0.766041-2\sin^220°=0.76604, and cos⁡40°=0.76604\cos40°=0.76604 — all four agree to five decimal places, confirming the three-forms-are-one-quantity claim independently of the lesson's own A=60°A=60° check.

2.3 Choosing a form — stated as one decision rule, extended to a third context

The lesson's own rule covers two contexts: an equation already written in cos⁡θ\cos\theta (reach for 2cos⁡2θ−12\cos^2\theta-1) or already in sin⁡θ\sin\theta (reach for 1−2sin⁡2θ1-2\sin^2\theta), reserving cos⁡2θ−sin⁡2θ\cos^2\theta-\sin^2\theta for when a difference of squares is being factorised. A third, real context the lesson does not name directly: an expression already containing 1+cos⁡2θ1+\cos2\theta or 1−cos⁡2θ1-\cos2\theta as a block, most often inside a fraction rather than a bare equation. 1+cos⁡2θ≡2cos⁡2θ1+\cos2\theta\equiv2\cos^2\theta falls straight out of the 2cos⁡2θ−12\cos^2\theta-1 form rearranged; 1−cos⁡2θ≡2sin⁡2θ1-\cos2\theta\equiv2\sin^2\theta falls out of 1−2sin⁡2θ1-2\sin^2\theta rearranged the same way. This is not a new derivation — it is the identical substitution, read as "what does adding or subtracting 1 do to this form" rather than "what does this form equal." Illustrative micro-example (self-generated for this document, not sourced to any exam paper — presented as a worked identity, not an exam citation): simplify sin⁡2θ1+cos⁡2θ\dfrac{\sin2\theta}{1+\cos2\theta}. Recognise the denominator as 2cos⁡2θ2\cos^2\theta: 2sin⁡θcos⁡θ2cos⁡2θ=sin⁡θcos⁡θ=tan⁡θ\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta. Checked numerically at θ=35°\theta=35°: both sides evaluate to 0.7002080.700208 to six decimal places. This third context matters beyond identity-proving alone — the CODEX formula bank's own §2.3 skeleton for integration-by-recognition.ts notes that integrating sin⁡2x\sin^2x, tan⁡2x\tan^2x or cos⁡23x\cos^23x requires exactly this kind of double-angle conversion first, so the same choice-of-form decision this section states for 2.3 recurs, unlabelled, in a sibling lesson's own real content.

2.4 Half angles and the triple-angle extension — both fully built elsewhere, cross-referenced not reworked

Setting A=θ2A=\tfrac{\theta}{2} in the 1−2sin⁡2A1-2\sin^2A form of cos⁡2A\cos2A, rather than A=A= (a full angle), reads the same substitution backwards and produces sin⁡2(θ2)≡1−cos⁡θ2\sin^2\left(\tfrac{\theta}{2}\right)\equiv\dfrac{1-\cos\theta}{2} — the lesson derives this in full, including the matching trap for confusing "half the angle" with "half the value." The only real WMA13 exam question located anywhere in this course's research that embeds a half-angle argument is Oct 2020 Q7, and it does so inside the harmonic-form technique (θ=12t\theta=\tfrac12t), not inside a bare double-angle identity — that anchor belongs to harmonic-form.ts, per the lesson's own header note, and is not reworked here. Similarly, the triple-angle identity sin⁡3θ≡3sin⁡θ−4sin⁡3θ\sin3\theta\equiv3\sin\theta-4\sin^3\theta — real, examined content despite never being named in the spec's own guidance column (confirmed on a real 4-mark "show that" question, Oct 2020 Q5(a)) — is derived in full in the lesson's beyond-spec section, using the identical write-as-a-sum-of-known-angles method this document does not repeat. What is worth stating once, precisely, and is not stated as a rule anywhere else: the method generalises because nθ=(n−1)θ+θn\theta=(n-1)\theta+\theta for any integer n⩾2n\geqslant2, so the same compound-angle-then-double-angle-substitution chain that produces sin⁡3θ\sin3\theta from sin⁡(2θ+θ)\sin(2\theta+\theta) would, in principle, produce sin⁡4θ\sin4\theta from sin⁡(3θ+θ)\sin(3\theta+\theta) and so on. This is reported as a mathematical fact about the method's structure, not as a claim that a quadruple-angle identity has ever been examined or is likely to be — no such question has been located, and none is asserted here.

3. Every real distinct sub-case, with a short worked example each

A. Proving an identity that mixes a single and a double angle. Fully covered by the lesson's method-comparison block (Pearson's own spec-guidance identity, cos⁡xcos⁡2x+sin⁡xsin⁡2x≡cos⁡x\cos x\cos2x+\sin x\sin2x\equiv\cos x, two independent full methods) and its marked solution ((sin⁡θ+cos⁡θ)2≡1+sin⁡2θ(\sin\theta+\cos\theta)^2\equiv1+\sin2\theta). Not reworked here.

B. Solving an equation that reduces to a 3-term quadratic in one trig function, real anchor Oct 2020 Q1 — no full mark ladder built anywhere yet. The lesson's trap-taxonomy and beyond-spec sections both quote fragments of this question's examiner report and mark scheme (the bracket-ambiguity trap specifically), but neither lays out the complete five-mark ladder in one place, and the worked-chain block instead uses a different, original equation (cos⁡2θ=sin⁡θ\cos2\theta=\sinθ). The real question: "Solve, for 0°⩽x<360°0°\leqslant x<360°, the equation 2cos⁡2x=7cos⁡x2\cos2x=7\cos x." Substituting cos⁡2x≡2cos⁡2x−1\cos2x\equiv2\cos^2x-1 (matching the fact the rest of the equation is already in cos⁡x\cos x) gives 2(2cos⁡2x−1)=7cos⁡x2(2\cos^2x-1)=7\cos x, a genuine 3-term quadratic once rearranged: 4cos⁡2x−7cos⁡x−2=04\cos^2x-7\cos x-2=0. Factorising, (4cos⁡x+1)(cos⁡x−2)=0(4\cos x+1)(\cos x-2)=0 [check: 4cos⁡2x−8cos⁡x+cos⁡x−2=4cos⁡2x−7cos⁡x−24\cos^2x-8\cos x+\cos x-2=4\cos^2x-7\cos x-2 ✓], giving cos⁡x=−14\cos x=-\tfrac14 or cos⁡x=2\cos x=2 (rejected outright — cosine is bounded by ±1\pm1). Solving cos⁡x=−14\cos x=-\tfrac14: x=arccos⁡(−14)=104.5°x=\arccos(-\tfrac14)=104.5° or x=360°−104.5°=255.5°x=360°-104.5°=255.5°, both independently re-verified by direct computation to match the real mark scheme's own stated answer. Real mark ladder, paraphrased rather than quoted at length: a method mark for the correct substitution with the bracket intact around (2cos⁡2x−1)(2\cos^2x-1) — the mark scheme explicitly scores a version missing that bracket as worth nothing unless the identity was stated first — a further method mark for solving the resulting quadratic, an accuracy mark for reaching cos⁡x=−14\cos x=-\tfrac14 specifically (not left as a value of a substituted variable), a dependent method mark for taking arccos⁡\arccos of at least one value, and a final accuracy mark for both angles with nothing extra in range. The real examiner report records this as generally well answered, with roughly two-thirds of candidates scoring full marks — a minority error, not the modal outcome, worth stating precisely rather than implying this is a heavily-failed question.

C. A reciprocal-identity first step collapsing into a sin⁡2θ=k\sin2\theta=k equation, with two of its three real error findings never built into any WMA13 lesson at all — real anchor Jan 2022 Q2. This anchor sits just outside spec 2.3's own boundary in its first half (the reciprocal identity itself is spec 2.1, and belongs properly to reciprocal-and-inverse-trigonometric-functions.ts) but crosses fully into 2.3 territory the moment the equation is rewritten in terms of 2θ2\theta — the facts bank's own scope note says exactly this. The real question, part (b): having shown 8cos⁡θ=3csc⁡θ8\cos\theta=3\csc\theta reduces to sin⁡2θ=34\sin2\theta=\tfrac34, find the smallest positive solution in degrees to one decimal place. Independently recomputed: θ=12arcsin⁡(0.75)=24.295...°\theta=\tfrac12\arcsin(0.75)=24.295...°, rounding to 24.3°24.3° — matching the real mark scheme's stated answer exactly. Two real, examiner-report-confirmed findings anchor this sub-case, neither built into the lesson's own trap-taxonomy under its own name: (1) a candidate who correctly solves for 2θ2\theta but reports the doubled value instead of dividing by 2 reaches 48.6°48.6°, not 24.3°24.3° — independently confirmed by the same calculation, doubling the answer above gives 48.59...°48.59...°, matching what the real examiner report records candidates actually writing. FORGE's own guide names this finding once, in passing, but it is not built into the lesson as a named trap the way the topic's other five traps are. (2) The question asks for the smallest positive solution specifically — singular — so a candidate who correctly finds the second, larger solution too but fails to reject it loses the final mark despite doing strictly more correct mathematics than the question asked for. This is the mirror image of the far more commonly discussed "missing a second solution" trap that dominates this course's other trig content: here, the failure is not discarding an extra genuine solution once the question has explicitly restricted which one it wants.

D. A fraction that collapses via a 1±cos⁡2θ1\pm\cos2\theta block — see §2.3 for the fully worked micro-example (sin⁡2θ/(1+cos⁡2θ)≡tan⁡θ\sin2\theta/(1+\cos2\theta)\equiv\tan\theta), presented there as the natural extension of the form-choice rule rather than repeated here.

E. An equation whose quadratic factor lands on an extreme value of cosine, not sine — a genuinely fresh check that the lesson's own single-solution-at-a-boundary rule holds symmetrically. The lesson's marked solution establishes this for sin⁡θ=−1\sin\theta=-1 (one solution, θ=270°\theta=270°, because −1-1 is touched once per revolution, unlike an ordinary value which is touched twice). No equivalent worked check exists anywhere in the sources reviewed for the boundary case on cosine specifically. Fresh micro-example (self-generated, not sourced to a real paper): solve 2cos⁡2θ+cos⁡θ=12\cos2\theta+\cos\theta=1 for 0°⩽θ<360°0°\leqslant\theta<360°. Substituting cos⁡2θ≡2cos⁡2θ−1\cos2\theta\equiv2\cos^2\theta-1 (matching the rest of the equation, already in cos⁡θ\cos\theta): 2(2cos⁡2θ−1)+cos⁡θ=1⇒4cos⁡2θ+cos⁡θ−3=02(2\cos^2\theta-1)+\cos\theta=1\Rightarrow4\cos^2\theta+\cos\theta-3=0. Factorising, (4cos⁡θ−3)(cos⁡θ+1)=0(4\cos\theta-3)(\cos\theta+1)=0 [check: 4cos⁡2θ+4cos⁡θ−3cos⁡θ−3=4cos⁡2θ+cos⁡θ−34\cos^2\theta+4\cos\theta-3\cos\theta-3=4\cos^2\theta+\cos\theta-3 ✓], giving cos⁡θ=34\cos\theta=\tfrac34 or cos⁡θ=−1\cos\theta=-1. cos⁡θ=34\cos\theta=\tfrac34 gives the ordinary reflected pair, θ=41.4°\theta=41.4° and θ=318.6°\theta=318.6° (independently recomputed); cos⁡θ=−1\cos\theta=-1 gives exactly one solution in the interval, θ=180°\theta=180° — the boundary value of cosine, touched once per revolution at the far side of the unit circle, exactly the same structural reason sin⁡θ=−1\sin\theta=-1 contributes only once, now confirmed for the other function.

4. The solution-count check, synthesised across every real anchor above

No single document currently states, as one explicit rule, roughly how many final solutions a 2.3-style equation should produce before the algebra is finished — a fast, checkable way to notice a missing or extra solution before it costs the final mark. Every worked example above and in the lesson is consistent with the same pattern: once a double-angle substitution collapses an equation to a 3-term quadratic in one ratio, each root of that quadratic that satisfies ∣k∣<1|k|<1 contributes an ordinary reflected pair over one full period of the substituted variable, each root at exactly ∣k∣=1|k|=1 contributes exactly one solution, and any root with ∣k∣>1|k|>1 is rejected outright. Where the substituted variable is 2θ2\theta rather than θ\theta, the interval must be doubled before this count is applied (the lesson's own marked solution does this explicitly), and every resulting value must be halved and re-checked against the original interval before being accepted. Checked against all real anchors gathered in this document: Oct 2020 Q1 (§3.B) — one root rejected, one root gives an ordinary pair, 2 solutions, confirmed; the fresh §3.E equation — one root gives an ordinary pair, one root lands on the boundary and gives a single solution, 3 solutions total, confirmed; Jan 2022 Q2 (§3.C) — the equation itself only ever asks for the smaller of an ordinary pair, so the rule's role there is to flag that a full, unrestricted solve would produce 2 values before the question's own "smallest positive" instruction discards one.

5. Ledger of every real exam anchor gathered for this specific topic

SeriesQuestionMarksWhat it actually testsSolutions (§4 rule)
Oct 2020Q15Equation-solving via double-angle substitution to a 3TQ in cos⁡x\cos x2 — confirmed
Oct 2020Q5(a)4Triple-angle derivation, sin⁡3x≡3sin⁡x−4sin⁡3x\sin3x\equiv3\sin x-4\sin^3x (fully worked in the lesson, not reworked here)n/a — identity proof, no equation
Oct 2020Q79Harmonic form Rcos⁡(x−α)R\cos(x-\alpha), with a half-angle argument in the modelling context — belongs to harmonic-form.tsn/a — out of this document's scope
Jan 2022Q25 total (2.1+2.3 combined)Reciprocal identity collapsing to sin⁡2θ=k\sin2\theta=k; "smallest positive solution" instruction; divide-by-2 recovery step1 asked for, out of 2 that exist

Jun 2022, Jun 2023 and Jan 2024 are named by SIGNAL as containing a 2.3 anchor but are not included in this ledger — their specific question content is not available in any source reviewed for this pass (§1, §6).

6. What remains open — reported, not smoothed over

Sources