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Notes/Mathematics/Paper 2/Trigonometry
CAIEAS Level9709§2.3

Trigonometry

sec, cosec and cot, two more Pythagorean identities, the compound and double angle formulae, and the R-form — the hardest topic on Paper 2.

30 min read 4 sub-topics
114
question parts
2021–2025 · 37 papers
11 marks
per paper
≈ 23% of the paper
2.5/3
avg difficulty
demanding
#3
most examined
of 6 topics by marks

Paper 2 does not restart trigonometry — it extends it. Everything on the Paper 1 page still applies: the graphs, the exact values, the two identities, and above all the habit of finding every angle in the stated interval rather than the one the calculator returns.

What Paper 2 adds is more ways to reduce an equation to a single trigonometric function:

  • three reciprocal functions, sec⁡\sec, cosec⁡\operatorname{cosec} and cot⁡\cot;
  • two more Pythagorean identities, both derived from sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1;
  • the compound angle expansions and the double angle formulae;
  • the R-form, which folds asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta into one wave.

This is the third-biggest topic on Paper 2 by marks and, at 2.5/32.5/3, the hardest topic on the paper. Not because any single step is difficult, but because a full question chains several of them: a double angle formula to set it up, the R-form to simplify it, interval-shifting to solve it, and often an integral at the end. The identity work in Paper 1 §04 is what makes all of that tractable, so be fluent there first.

Before you start you should be able to

  • All of Paper 1 Trigonometry — the graphs, exact values, tan⁡θ≡sin⁡θcos⁡θ\tan\theta \equiv \frac{\sin\theta}{\cos\theta}, sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1, and solving in an interval

  • Solving quadratic equations by factorising (see Quadratics)

  • Radians, since Paper 2 intervals are often given in terms of π\pi (see Circular Measure)

By the end of this page you can

  • Relate sec⁡\sec, cosec⁡\operatorname{cosec} and cot⁡\cot to cos⁡\cos, sin⁡\sin and tan⁡\tan, and sketch all three

  • Use sec⁡2θ≡1+tan⁡2θ\sec^2\theta \equiv 1 + \tan^2\theta and cosec⁡2θ≡1+cot⁡2θ\operatorname{cosec}^2\theta \equiv 1 + \cot^2\theta to reduce an equation to one function

  • Expand sin⁡(A±B)\sin(A \pm B), cos⁡(A±B)\cos(A \pm B) and tan⁡(A±B)\tan(A \pm B), with the correct signs

  • Use the double angle formulae, choosing the form of cos⁡2A\cos 2A that leaves a single function

  • Express asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta as Rsin⁡(θ±α)R\sin(\theta \pm \alpha) or Rcos⁡(θ±α)R\cos(\theta \pm \alpha), and use it to solve equations and read off maxima and minima

01

Secant, cosecant and cotangent

Syllabus requirement · §2.3

“

understand the relationship of the secant, cosecant and cotangent functions to cosine, sine and tangent, and use properties and graphs of all six trigonometric functions for angles of any magnitude.

”

Paper 2 adds three more functions, and each is simply one over one of the originals. There is nothing new to understand — but the pairing is deliberately counter-intuitive, so it is worth fixing early.

The three reciprocals
sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}

secant pairs with COSINE

cosec⁡θ=1sin⁡θ\operatorname{cosec}\theta = \frac{1}{\sin\theta}

cosecant pairs with SINE

cot⁡θ=1tan⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}

cotangent pairs with tangent

The third-letter trick

Match them by the third letter: seCant goes with cosine, coseCant goes with sine. Reading the first three letters instead ("sec ↔ sin") is the classic mix-up and it will wreck an entire question.

901802703601-1y = sec x= 1 / cos x901802703601-1y = cosec x= 1 / sin x901802703601-1y = cot x= 1 / tan xan asymptote wherever the function underneath hits zero · |sec x| and |cosec x| are never less than 1

Each reciprocal graph has an asymptote wherever the function underneath crosses zero, and touches ±1 wherever that function reaches ±1. The dashed curve in each panel is the original.

sec⁡x\sec x

cosec⁡x\operatorname{cosec} x

cot⁡x\cot x

Undefined where

cos⁡x=0\cos x = 0

sin⁡x=0\sin x = 0

sin⁡x=0\sin x = 0

Asymptotes in 0°–360°

90∘,270∘90^\circ, 270^\circ

0∘,180∘,360∘0^\circ, 180^\circ, 360^\circ

0∘,180∘,360∘0^\circ, 180^\circ, 360^\circ

Range

∣y∣⩾1|y| \geqslant 1

∣y∣⩾1|y| \geqslant 1

all real values

Period

360°

360°

180°

sec and cosec can never take a value strictly between −1 and 1 — a fact questions use to reject impossible roots.

Two more Pythagorean identities

Divide sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1 by cos⁡2θ\cos^2\theta and you get one; divide it by sin⁡2θ\sin^2\theta and you get the other. They are not new facts, just the old one rearranged — which is worth knowing, because it means you can rebuild them if memory fails.

sec⁡2θ≡1+tan⁡2θ\sec^2\theta \equiv 1 + \tan^2\theta

divide the Pythagorean identity by cos²θ

cosec⁡2θ≡1+cot⁡2θ\operatorname{cosec}^2\theta \equiv 1 + \cot^2\theta

divide it by sin²θ instead

What they are for

Exactly what sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1 was for in Paper 1: removing one function so the equation has only one left. An equation containing both sec⁡2θ\sec^2\theta and tan⁡θ\tan\theta becomes a quadratic in tan⁡θ\tan\theta the moment you replace sec⁡2θ\sec^2\theta with 1+tan⁡2θ1 + \tan^2\theta.

Worked example9709/21 M/J 2022 Q26 marks

(a) Express the equation 7tan⁡θ+4cot⁡θ−13sec⁡θ=07\tan\theta + 4\cot\theta - 13\sec\theta = 0 in terms of sin⁡θ\sin\theta only. [3]
(b) Hence solve the equation for 0∘<θ<360∘0^\circ < \theta < 360^\circ. [3]

Show full working
  1. 1

    (a) Write every term in sin⁡\sin and cos⁡\cos — the reflex move from §04: 7sin⁡θcos⁡θ+4cos⁡θsin⁡θ−13cos⁡θ=0\frac{7\sin\theta}{\cos\theta} + \frac{4\cos\theta}{\sin\theta} - \frac{13}{\cos\theta} = 0

  2. 2

    Multiply through by sin⁡θcos⁡θ\sin\theta\cos\theta to clear every denominator: 7sin⁡2θ+4cos⁡2θ−13sin⁡θ=07\sin^2\theta + 4\cos^2\theta - 13\sin\theta = 0

    Note the third term: 13/cos θ × sin θ cos θ = 13 sin θ, so the cosine disappears from it entirely.

  3. 3

    Only the cos⁡2θ\cos^2\theta is now in the way, so replace it with 1−sin⁡2θ1 - \sin^2\theta: 7sin⁡2θ+4−4sin⁡2θ−13sin⁡θ=0  ⇒  3sin⁡2θ−13sin⁡θ+4=07\sin^2\theta + 4 - 4\sin^2\theta - 13\sin\theta = 0 \;\Rightarrow\; 3\sin^2\theta - 13\sin\theta + 4 = 0

  4. 4

    (b) Factorise the quadratic in sin⁡θ\sin\theta: (3sin⁡θ−1)(sin⁡θ−4)=0(3\sin\theta - 1)(\sin\theta - 4) = 0.

  5. 5

    Reject sin⁡θ=4\sin\theta = 4 — sine never exceeds 1. That leaves sin⁡θ=13\sin\theta = \tfrac13.

    Stating the rejection is creditworthy, and it is the check that stops you hunting for angles that do not exist.

  6. 6

    sin⁡−1(13)=19.5∘\sin^{-1}\left(\tfrac13\right) = 19.5^\circ, and sine's partner is 180∘−19.5∘=160.5∘180^\circ - 19.5^\circ = 160.5^\circ. Both lie in the interval.

Answer

(a) 3sin⁡2θ−13sin⁡θ+4=03\sin^2\theta - 13\sin\theta + 4 = 0 (b) θ=19.5∘\theta = 19.5^\circ and 160.5∘160.5^\circ

Part (a) is worth three of the six marks and is pure §04 technique — convert to sin⁡\sin and cos⁡\cos, clear fractions, use the Pythagorean identity. The new functions only changed what you start from.

Practise sec, cosec and cotReal past-paper questions · Secant, cosecant and cotangent functions and their graphs

The rest of this note

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Can you do all of these?

  • Write down sec⁡\sec, cosec⁡\operatorname{cosec} and cot⁡\cot in terms of cos⁡\cos, sin⁡\sin and tan⁡\tan without hesitating

  • Derive sec⁡2θ≡1+tan⁡2θ\sec^2\theta \equiv 1 + \tan^2\theta from sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1

  • Solve 2sec⁡2θ−tan⁡θ=52\sec^2\theta - \tan\theta = 5 over a given interval

  • Expand cos⁡(A−B)\cos(A - B) with the correct sign, and check it with A=B=45∘A = B = 45^\circ

  • Choose the right form of cos⁡2A\cos 2A for an equation written in sin⁡\sin

  • Express 3sin⁡θ+4cos⁡θ3\sin\theta + 4\cos\theta as Rsin⁡(θ+α)R\sin(\theta + \alpha) and state its maximum

  • Solve an R-form equation, remembering to shift the interval and shift back

Now do the questions
114 real Paper 2 parts from 2021–2025, sorted by difficulty, with mark schemes
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