AQA A-Level Mathematics: E: Trigonometry (7357) – Revision Notes

For full explanations and worked examples, use the Trigonometry study guide. These notes condense Section E: Trigonometry (E1 to E9) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. Section E is Paper 1 content and can also be assessed on Papers 2 and 3. Test yourself afterwards with the Trigonometry practice questions. Course hub: /boards/aqa/a-level/mathematics/. Printable checklist: /checklists/aqa/a-level/mathematics/.

Formulae you must know without being given them

Appendix B of the specification says these must be used without being provided:

Result	Formula
Sine rule	a / sin A = b / sin B = c / sin C
Cosine rule	a² = b² + c² − 2bc cos A
Area of triangle	½ab sin C
Pythagorean	cos²A + sin²A ≡ 1
	sec²A ≡ 1 + tan²A
	cosec²A ≡ 1 + cot²A
Double angle	sin 2A ≡ 2 sin A cos A
	cos 2A ≡ cos²A − sin²A
	tan 2A ≡ 2 tan A / (1 − tan²A)

Also know (they follow from the above or are standard in E1 and E6):

• cos 2A ≡ 2cos²A − 1 ≡ 1 − 2sin²A
• sin(A ± B) ≡ sin A cos B ± cos A sin B
• cos(A ± B) ≡ cos A cos B ∓ sin A sin B
• tan(A ± B) ≡ (tan A ± tan B) / (1 ∓ tan A tan B)
• tan θ ≡ sin θ / cos θ
• Arc length s = rθ, sector area A = ½r²θ (θ in radians)

E1: definitions and radians

• On the unit circle, the point at angle θ is (cos θ, sin θ). This defines sin and cos for every angle, including negative and reflex.
• Signs by quadrant (0 to 2π): first, all positive; second, sin positive; third, tan positive; fourth, cos positive.
• π rad = 180°, so 1 rad ≈ 57.3°. Multiply degrees by π/180 to convert.
• Segment area = sector − triangle = ½r²θ − ½r² sin θ.

E2: small angles (θ in radians)

Function	Approximation
sin θ	θ
cos θ	1 − θ²/2
tan θ	θ

Substitute the whole argument: sin 4θ ≈ 4θ, cos 3θ ≈ 1 − 9θ²/2.

E3: graphs and exact values

Graph	Period	Range	Key features
y = sin x	2π	[−1, 1]	odd: sin(−x) = −sin x
y = cos x	2π	[−1, 1]	even: cos(−x) = cos x
y = tan x	π	all reals	asymptotes at x = π/2 + nπ

y = sin kx and y = cos kx have period 2π/k; y = tan kx has period π/k.

Exact values to learn:

θ	0	π/6	π/4	π/3	π/2
sin	0	½	√2/2	√3/2	1
cos	1	√3/2	√2/2	½	0
tan	0	√3/3	1	√3	—

Use the related acute angle and the quadrant sign for multiples, e.g. sin(4π/3) = −√3/2.

E4: reciprocal and inverse functions

Function	Domain	Range
sec x = 1/cos x	x ≠ π/2 + nπ	≤ −1 or ≥ 1
cosec x = 1/sin x	x ≠ nπ	≤ −1 or ≥ 1
cot x = cos x / sin x	x ≠ nπ	all reals
arcsin x	[−1, 1]	[−π/2, π/2]
arccos x	[−1, 1]	[0, π]
arctan x	all reals	(−π/2, π/2)

• sec x has a local minimum of 1 where cos x = 1 and a local maximum of −1 where cos x = −1.
• cot x has asymptotes where sin x = 0 and zeros where cos x = 0.
• Inverse graphs: reflect the restricted original in y = x.

E6: R-form in steps

To write a cos θ + b sin θ in a given form:

1. Expand the target, e.g. R cos(θ − α) = R cos θ cos α + R sin θ sin α.
2. Match coefficients: R cos α = a, R sin α = b.
3. R = √(a² + b²); tan α = b/a, checking the signs put α in the stated interval.
4. Max value R, min value −R. Set the bracket equal to 0 or π (for cos) or ±π/2 (for sin) to find where they occur.

Geometric proofs: you must understand the diagram proof of sin(A + B) and cos(A + B) (two right-angled triangles stacked at the origin), not just quote the results.

E7: solving equations in steps

1. Use an identity to get one function: replace cos²x by 1 − sin²x, tan²x by sec²x − 1, and so on.
2. Factorise. Never divide by a trig expression that could be zero.
3. Reject impossible roots: |sin x| ≤ 1, |cos x| ≤ 1, |sec x| ≥ 1, |cosec x| ≥ 1.
4. For f(kx + c), change the interval first: if 0 ≤ x < 2π then c ≤ kx + c < 2πk + c.
5. Use symmetry: sin x = s gives x and π − x; cos x = c gives x and 2π − x; tan x = t gives x and x + π. Then add or subtract the period.
6. Give answers in the units of the interval, to 3 s.f. or exactly as asked.

Worked reminder (factorise, don’t divide). Solve 2 sin x cos x = cos x for 0 ≤ x ≤ 2π.

2 sin x cos x − cos x = 0
cos x (2 sin x − 1) = 0
cos x = 0: x = π/2, 3π/2
sin x = ½: x = π/6, 5π/6

Dividing by cos x at the start would lose π/2 and 3π/2.

Worked reminder (E5, choose the sign). θ is obtuse and cos θ = −3/5. Then sin²θ = 1 − 9/25 = 16/25, and sin θ > 0 in the second quadrant, so sin θ = 4/5, tan θ = −4/3 and cosec θ = 5/4.

E8: proof in steps

• Start from one side (usually the messier one) and finish with “= RHS”.
• Common moves: write everything in sin and cos; combine fractions; use 1 − sin²θ = cos²θ; factorise.
• Don’t cross-multiply an identity or work on both sides at once.

E9: context reminders

• Model h = p + q cos(kt): mean level p, amplitude |q|, period 2π/k.
• Resolve a force F at angle θ to a direction: F cos θ along it, F sin θ perpendicular to it.
• State the units of t and h in your answer.

Must-know distinctions

• sin⁻¹x vs (sin x)⁻¹: arcsin x is the inverse function; cosec x = 1/sin x.
• Degrees vs radians: arc length, sector area, small angles and calculus all need radians.
• Identity (≡) vs equation (=): an identity is true for all values; an equation is solved for particular ones.
• Principal value vs all solutions: arcsin, arccos and arctan return one value; the question usually wants all in the interval.
• R cos(θ − α) vs R sin(θ + α): the same R, but α comes from different coefficient pairs.

Quick self-test

1. Convert 135° to radians.
2. A sector has radius 9 cm and angle 0.8 rad. Find its arc length and area.
3. Find tan(4π/3) exactly.
4. Find cosec(−π/6).
5. Find arctan(−1).
6. Use a small angle approximation to simplify 1 − cos 4θ.
7. State the period of y = cos 3x.
8. Expand cos(x + π/3).
9. Given sin A = 1/3, find cos 2A.
10. Write cos x − √3 sin x in the form R cos(x + α), with R > 0 and 0 < α < π/2.
11. State the range of arccos x.
12. How many solutions does sin 3x = ½ have for 0 ≤ x < 2π?

Answers

1. 3π/4
2. Arc = 9 × 0.8 = 7.2 cm; area = ½ × 81 × 0.8 = 32.4 cm²
3. √3 (third quadrant, tan positive)
4. −2
5. −π/4
6. 1 − (1 − 16θ²/2) = 8θ²
7. 2π/3
8. ½cos x − (√3/2) sin x
9. 1 − 2(1/9) = 7/9
10. R = √(1 + 3) = 2; R cos α = 1 and R sin α = √3, so α = π/3: 2 cos(x + π/3)
11. 0 ≤ arccos x ≤ π
12. Six (3x runs over [0, 6π): x = π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18)

Where marks are usually lost

• Leaving the calculator in degrees for a radian interval, or giving a mixture of degrees and radians.
• Giving only the principal value from arcsin or arccos when the interval holds more solutions.
• Dividing an equation such as sin x cos x = sin x by sin x and losing the sin x = 0 solutions.
• Not extending the interval for a multiple angle, so solutions beyond 2π are missed.
• Keeping impossible roots such as cos x = 2 or sec x = 0.5 without comment.
• Choosing the wrong sign for a reciprocal value: check the quadrant before taking ±√ in sec²x = 1 + tan²x.
• Writing approximations like sin 3θ ≈ θ instead of 3θ.
• In “prove” questions, working on both sides or skipping the step that uses an identity.
• In R cos(θ − α) form, finding α from tan α = a/b instead of b/a, or giving α in degrees when the interval is in radians.
• Using the sine rule for an angle without checking whether the obtuse alternative is possible.

For the link between small angles and differentiating sin x from first principles, see the Differentiation revision notes. For examiner-style setting out of a “show that”, see the exam preparation guide. When you’re ready, the free 10-minute diagnostics help you find other gaps.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018, for A-level exams June 2018 onwards: Section 3.6 E: Trigonometry (E1 to E9), and Appendix B: mathematical formulae and identities.