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A Level Maths 9709 · Paper 3 Pure Mathematics 3 · 3.3 (also Paper 2) · ~8 min read

Further trigonometry

sec, cosec and cot and their graphs, the identities 1 + tan²θ ≡ sec²θ and 1 + cot²θ ≡ cosec²θ, the addition and double-angle formulae, and expressing a sin θ + b cos θ as R sin(θ ± α) to solve equations and find maxima.

sec, cosec, cotPythagorean identitiesAddition formulaeDouble-angle formulaeExact valuesProving identitiesR sin(θ + α)Solving equations
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • sec θ = 1/cos θ, cosec θ = 1/sin θ, cot θ = 1/tan θ. Each has asymptotes where its parent function is zero.
  • Divide sin²θ + cos²θ ≡ 1 by cos²θ to get 1 + tan²θ ≡ sec²θ; by sin²θ to get 1 + cot²θ ≡ cosec²θ.
  • Addition and double-angle formulae are in the formula list; using them quickly is the skill. cos 2A has three forms: pick the one that leaves a single trig function.
  • a sin θ + b cos θ = R sin(θ + α) with R = √(a² + b²) and tan α = b/a. It turns a two-function equation into a one-function equation, and gives the maximum R and minimum −R.

By the end you can

  • Use sec, cosec and cot and sketch their graphs.
  • Use the Pythagorean, addition and double-angle identities to simplify, prove and evaluate exactly.
  • Express a sin θ + b cos θ in the forms R sin(θ ± α) or R cos(θ ± α).
  • Solve trig equations in degrees or radians using these identities.
  • Find maximum and minimum values of expressions using the R form.

1Reciprocal functions

sec, cosec and cot

FunctionDefinitionUndefined whereRange
sec θ1/cos θcos θ = 0 (θ = 90°, 270°, …)sec θ ≤ −1 or sec θ ≥ 1
cosec θ1/sin θsin θ = 0 (θ = 0°, 180°, …)cosec θ ≤ −1 or cosec θ ≥ 1
cot θ1/tan θ = cos θ/sin θsin θ = 0all real values
  • Graphs: sec θ is ∪-shaped where cos θ > 0 and ∩-shaped where cos θ < 0, touching ±1 where cos θ = ±1. cosec θ is the same shape shifted by 90°.
1 + tan²θ ≡ sec²θ ; 1 + cot²θ ≡ cosec²θ FORMULA LIST (MF19)
divide sin²θ + cos²θ ≡ 1 by cos²θ or by sin²θ
Worked example

Solve 3 sec²θ − 5 tan θ − 5 = 0 for 0° ≤ θ ≤ 360°.

  1. 3(1 + tan²θ) − 5 tan θ − 5 = 0 ⇒ 3tan²θ − 5 tan θ − 2 = 0 M1
  2. (3 tan θ + 1)(tan θ − 2) = 0 M1
  3. tan θ = 2 ⇒ θ = 63.4°, 243.4°; tan θ = −⅓ ⇒ θ = 161.6°, 341.6° A1 A1
Worked example

Solve tan θ + cot θ = 4 for 0° < θ < 180°.

  1. t + 1/t = 4 ⇒ t² − 4t + 1 = 0 ⇒ t = 2 ± √3 M1 A1
  2. θ = 75° or θ = 15° A1 (tan 75° = 2 + √3 exactly)

2Addition and double-angle formulae

The formulas

sin(A ± B) ≡ sin A cos B ± cos A sin B FORMULA LIST (MF19)
cos(A ± B) ≡ cos A cos B ∓ sin A sin B FORMULA LIST (MF19)
note the sign swaps
tan(A ± B) ≡ (tan A ± tan B)/(1 ∓ tan A tan B) FORMULA LIST (MF19)
sin 2A ≡ 2 sin A cos A ; cos 2A ≡ cos²A − sin²A ≡ 2cos²A − 1 ≡ 1 − 2sin²A ; tan 2A ≡ 2 tan A/(1 − tan²A) FORMULA LIST (MF19)
put B = A in the addition formulae
  • Choosing the cos 2A form: if the equation has sin x, use 1 − 2sin²x; if it has cos x, use 2cos²x − 1.
  • For integration: cos²x ≡ ½(1 + cos 2x) and sin²x ≡ ½(1 − cos 2x). See Further integration.
Worked example

Find the exact value of cos 75°.

  1. cos(45° + 30°) = cos 45° cos 30° − sin 45° sin 30° M1
  2. = (√2/2)(√3/2) − (√2/2)(½) M1
  3. = (√6 − √2)/4 A1
Worked example

Solve sin(x + 30°) = 2 cos x for 0° ≤ x ≤ 360°.

  1. sin x cos 30° + cos x sin 30° = 2 cos x M1
  2. (√3/2) sin x = (3/2) cos x ⇒ tan x = 3/√3 = √3 M1 A1
  3. x = 60°, 240° A1
Worked example

Solve cos 2x = 3 sin x − 1 for 0° ≤ x ≤ 360°.

  1. 1 − 2sin²x = 3 sin x − 1 ⇒ 2sin²x + 3 sin x − 2 = 0 M1
  2. (2 sin x − 1)(sin x + 2) = 0; sin x = −2 is impossible M1
  3. sin x = ½ ⇒ x = 30°, 150° A1 A1
Worked example

Prove that sin 2θ/(1 + cos 2θ) ≡ tan θ.

  1. LHS = 2 sin θ cos θ/(1 + 2cos²θ − 1) M1
  2. = 2 sin θ cos θ/(2cos²θ) = sin θ/cos θ = tan θ = RHS A1
Common mistakeWriting cos 2θ as 2 cos θ, or sin 3α as 3 sin α. Only the formulae above are valid. The 2024 Paper 3 report also notes candidates who expanded sin(A + B) correctly but forgot to put tan 45° = 1 straight in.

3The R formula

a sin θ + b cos θ = R sin(θ + α)

  • Expand and compare: R sin(θ + α) = R sin θ cos α + R cos θ sin α. Match coefficients: R cos α = a and R sin α = b.
  • So: R² = a² + b² (square and add), and tan α = b/a (divide). Take α acute when a, b > 0.
a sin θ + b cos θ ≡ R sin(θ + α), R = √(a² + b²), tan α = b/a
similar forms R sin(θ − α), R cos(θ ± α): expand and compare the same way
θy90°180°270°360°5−5max 5 at θ = 36.9°3 sin θ + 4 cos θ = 5 sin(θ + 53.1°)Dashed: 3 sin θ and 4 cos θ. Their sum is a single sine wave of amplitude R = √(3² + 4²) = 5.
3 sin θ + 4 cos θ is a single sine wave with amplitude 5, shifted left by 53.1°.
Worked example

(a) Express 3 sin θ + 4 cos θ in the form R sin(θ + α). (b) Solve 3 sin θ + 4 cos θ = 2 for 0° < θ < 360°. (c) Find the greatest value of 1/(7 − 3 sin θ − 4 cos θ).

  1. (a) R = √(9 + 16) = 5; tan α = 4/3 ⇒ α = 53.13° M1 A1 A1
  2. (b) sin(θ + 53.13°) = 0.4; θ + 53.13° lies between 53.13° and 413.13° M1
  3. θ + 53.13° = 156.42° or 383.58° ⇒ θ = 103.3°, 330.4° A1 A1
  4. (c) 7 − 5 sin(θ + α) is least when the sine is 1: 7 − 5 = 2, so the greatest value is ½ B1
Common mistakeMissing a solution because the interval for θ + α wasn't adjusted. Write the new interval first, as in (b).
Linking questionPaper 1's basics are in Trigonometry; derivatives of trig functions are in Further differentiation; and trig substitutions and identities are used in Further integration.

Test yourself — cover the answers first

Quick check

  1. Define cosec θ.
  2. State the identity linking tan²θ and sec²θ.
  3. Write cos 2A in terms of sin A only.
  4. Find R if 5 sin θ − 12 cos θ = R sin(θ − α).
  5. What is the minimum value of 3 sin θ + 4 cos θ?

Exam-style

  1. (a) Express 2 cos x − 5 sin x in the form R cos(x + α), with 0° < α < 90°. (b) Hence solve 2 cos x − 5 sin x = 3 for 0° < x < 360°.
  2. Prove that cosec 2θ + cot 2θ ≡ cot θ.
  3. Solve 2 cos 2θ + 5 cos θ = 1 for 0 ≤ θ ≤ 2π.
Show answers
  1. 1/sin θ
  2. 1 + tan²θ ≡ sec²θ
  3. 1 − 2sin²A
  4. 13
  5. −5
  6. (a) √29 cos(x + 68.20°)
    (b) cos(x + 68.20°) = 0.5571 ⇒ x + 68.20° = 303.85° or 416.15° ⇒ x = 235.7°, 347.9°
  7. (1 + cos 2θ)/sin 2θ = 2cos²θ/(2 sin θ cos θ) = cos θ/sin θ = cot θ
  8. 4cos²θ + 5 cos θ − 3 = 0 ⇒ cos θ = (−5 + √73)/8 = 0.443 ⇒ θ = 1.11, 5.17
Practise this chapter
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.