Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 2 Trigonometry – Revision Notes

For full explanations and longer worked examples, use the study guide.

These notes cover section 2.3, Trigonometry, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). Section 2.3 is part of Pure Mathematics 2 and is examined on Paper 2 (1 hour 15 minutes, 50 marks), the AS Level Pure Mathematics route. The same outcomes are section 3.3 of Pure Mathematics 3. Paper 1 trigonometry is assumed: recap it with the Pure Mathematics 1 trigonometry revision notes. A scientific calculator is allowed, but unsupported calculator answers earn no marks.

Course links: A Level Mathematics hub, printable 9709 checklist, practice questions for this unit, and the whole-paper Pure Mathematics 2 revision notes.

2.3 at a glance

Outcome	You must be able to
Six functions	Define sec, cosec, cot; sketch and use their graphs for any angle
Pythagorean identities	Use sec²θ ≡ 1 + tan²θ and cosec²θ ≡ 1 + cot²θ
Compound angles	Expand sin(A ± B), cos(A ± B), tan(A ± B)
Double angles	Use sin 2A, cos 2A (three forms), tan 2A
R-form	Write a sin θ + b cos θ as R sin(θ ± α) or R cos(θ ± α)
Applying them	Simplify, find exact values, prove identities, solve equations

Definitions

• sec θ = 1/cos θ, undefined where cos θ = 0.
• cosec θ = 1/sin θ, undefined where sin θ = 0.
• cot θ = 1/tan θ = cos θ/sin θ, undefined where sin θ = 0.
• Each reciprocal has the same sign as its parent function in every quadrant.
• |sec θ| ≥ 1 and |cosec θ| ≥ 1 always. cot θ takes every real value.

Formulas

Identity	In MF19?
tan θ ≡ sin θ/cos θ; cos²θ + sin²θ ≡ 1	Yes
1 + tan²θ ≡ sec²θ; cot²θ + 1 ≡ cosec²θ	Yes
sin(A ± B) ≡ sin A cos B ± cos A sin B	Yes
cos(A ± B) ≡ cos A cos B ∓ sin A sin B	Yes
tan(A ± B) ≡ (tan A ± tan B)/(1 ∓ tan A tan B)	Yes
sin 2A ≡ 2 sin A cos A	Yes
cos 2A ≡ cos²A − sin²A ≡ 2cos²A − 1 ≡ 1 − 2sin²A	Yes
tan 2A ≡ 2 tan A/(1 − tan²A)	Yes
sec θ = 1/cos θ, cosec θ = 1/sin θ, cot θ = 1/tan θ	No
a sin θ + b cos θ ≡ R sin(θ + α), R = √(a² + b²)	No

Useful rearrangements (not in MF19, derived from cos 2A): cos²A ≡ (1 + cos 2A)/2 and sin²A ≡ (1 − cos 2A)/2.

Graphs

	sec x	cosec x	cot x
Period	360°	360°	180°
Asymptotes	90°, 270°	0°, 180°, 360°	0°, 180°, 360°
Turning points	(0°, 1), (180°, −1), (360°, 1)	(90°, 1), (270°, −1)	none
Symmetry	even: sec(−x) = sec x	odd	odd

Sketch the parent curve lightly first. Asymptotes sit where the parent is zero. Turning points sit where the parent is ±1.

Method in steps

Proving an identity

1. Start with the more complicated side.
2. Write sec, cosec, cot, tan in terms of sin and cos.
3. Combine fractions into one.
4. Use a Pythagorean identity to replace 1 − cos², 1 − sin², sec² − 1 and so on.
5. Finish exactly on the other side, with every line shown.

Pythagorean equation (for example tan² with sec)

1. Replace the squared function so only one function remains.
2. Rearrange to a quadratic = 0 and factorise.
3. Turn each root into cos, sin or tan.
4. Reject values outside the range (|sec| < 1, |cosec| < 1) and give a reason.
5. Find every angle in the interval.

R-form

1. Expand the required form, such as R cos(θ + α) = R cos θ cos α − R sin θ sin α.
2. Match coefficients: R cos α = …, R sin α = … .
3. R = √(a² + b²). tan α = (R sin α)/(R cos α).
4. To solve = c: divide by R, shift the interval by α, find all values, then undo the shift.
5. Greatest value R, least value −R.

Equation with 2θ and θ

1. Replace sin 2θ, cos 2θ or tan 2θ by functions of θ.
2. For cos 2θ, pick the form that leaves one function only.
3. Factorise. Never divide by a function that could be zero.

Small worked reminders

Proof. (1 − sin θ)/cos θ ≡ cos θ/(1 + sin θ): multiply top and bottom of the left side by (1 + sin θ) to get (1 − sin²θ)/(cos θ(1 + sin θ)) = cos²θ/(cos θ(1 + sin θ)) = cos θ/(1 + sin θ).

Difference of squares. cos⁴θ − sin⁴θ = (cos²θ − sin²θ)(cos²θ + sin²θ) = cos 2θ × 1 = cos 2θ.

Simplifying. cos(x + 60°) + cos(x − 60°) = 2 cos x cos 60° = cos x, because the sin x sin 60° terms cancel.

Reciprocal equation. Solve cosec θ = 2.5 cot θ for 0° < θ < 360°. Write 1/sin θ = 2.5 cos θ/sin θ. sin θ cannot be 0 here (cosec θ would be undefined), so multiplying by sin θ is safe: cos θ = 0.4, giving θ = 66.4°, 293.6°.

Exact value from cos 2A. sin²22.5° = (1 − cos 45°)/2 = (1 − √2/2)/2 = (2 − √2)/4.

R-form. 5 sin θ + 2 cos θ = R sin(θ + α): R cos α = 5, R sin α = 2, so R = √29 and α = 21.80°. The greatest value √29 occurs at θ = 68.2°, the least −√29 at θ = 248.2°.

Must-know distinctions

• sec θ vs cos⁻¹θ. sec θ is 1/cos θ. cos⁻¹θ is the inverse function (an angle).
• cosec pairs with sin. The “co” functions do not pair with the “co” parents: cosec goes with sin, sec with cos.
• cos(A + B) vs cos(A − B). The sign in the middle is the opposite of the sign in the bracket.
• sin 2A vs 2 sin A. sin 2A = 2 sin A cos A, never 2 sin A.
• Three forms of cos 2A. Use 1 − 2sin²A with sin terms, 2cos²A − 1 with cos terms.
• R sin(θ + α) vs R cos(θ − α). Different forms give different α. Expand the one asked for.
• Identity vs equation. An identity (≡) is true for every θ and is proved. An equation (=) is true for some θ and is solved.
• No solution vs one solution. sec θ = 0.5 has no solution; sec θ = 1 does (cos θ = 1).

Quick self-test

1. Find the exact value of sec 300°.
2. Find the exact value of cot(−π/6).
3. Given cosec θ = 3 and θ is obtuse, find the exact value of cot θ.
4. Find the exact value of cos 75°.
5. Simplify 2 sin 3x cos 3x.
6. Show that (1 − cos 2θ)/sin 2θ simplifies to a single function of θ.
7. Write cos²θ in terms of cos 2θ.
8. Express sin θ + √3 cos θ as R sin(θ + α), with R > 0 and 0° < α < 90°.
9. Hence find the greatest and least values of 1/(3 − sin θ − √3 cos θ).
10. Solve sec x = −2 for 0 ≤ x ≤ 2π.
11. Solve sin 2θ = sin θ for 0° ≤ θ ≤ 180°.
12. Given tan A = 3 and tan B = 1/2, find tan(A − B).

Answers

1. cos 300° = 1/2, so sec 300° = 2.
2. tan(−π/6) = −1/√3, so cot(−π/6) = −√3.
3. cot²θ = cosec²θ − 1 = 8. θ is obtuse, so cot θ < 0: cot θ = −2√2.
4. cos(45° + 30°) = (√2/2)(√3/2) − (√2/2)(1/2) = (√6 − √2)/4.
5. sin 6x (sin 2A with A = 3x).
6. (2sin²θ)/(2 sin θ cos θ) = tan θ.
7. cos²θ = (1 + cos 2θ)/2.
8. R cos α = 1, R sin α = √3: R = 2, α = 60°.
9. The denominator is 3 − 2 sin(θ + 60°), which runs from 1 to 5. Greatest value 1, least value 1/5.
10. cos x = −1/2: x = 2π/3, 4π/3.
11. sin θ(2 cos θ − 1) = 0: θ = 0°, 60°, 180°.
12. (3 − 1/2)/(1 + 3/2) = 1.

Where marks are usually lost

• Pressing cos⁻¹ on a sec value such as 1.5, instead of taking cos θ = 2/3 first.
• Keeping a root such as cosec θ = 1/3 or sec θ = −0.4, or rejecting it with no reason.
• Writing sin(A + B) as sin A + sin B, or using a plus sign in cos(A + B).
• Picking the wrong form of cos 2θ, so the equation still has both sin θ and cos θ.
• Cancelling a common factor such as sin θ or tan θ and losing solutions.
• Giving α to 1 decimal place and then using the rounded value, so the final angle is out by 0.1°.
• Not shifting the interval for θ − α or 2θ, so a solution is missed or one outside the range is kept.
• In a “prove” question, working on both sides at once or skipping the Pythagorean step.
• Giving a decimal where the question asks for an exact value.

Next steps

Work through the practice questions, then mixed Paper 2 work in the Pure Mathematics 2 practice set. The A Level route covers the same content in the Pure Mathematics 3 revision notes. Test yourself with the free AS diagnostic or the 9709 self-check bank.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 2, Pure Mathematics 2 (for Paper 2): section 2.3 Trigonometry.