Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 1 Trigonometry – Revision Notes

These notes condense section 1.5 Trigonometry of Pure Mathematics 1 in the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). The content is examined on Paper 1 (1 hour 50 minutes, 75 marks), which every AS and A Level candidate takes. A scientific calculator is allowed, but exact values must be found by hand. For full explanations and worked examples, use the study guide.

Course hub: Cambridge A Level Mathematics. Printable list: 9709 checklist. Quick check: AS 10-minute diagnostic and the 9709 self-check bank. Exam-style questions: trigonometry practice set.

The five things 1.5 asks for

1. Sketch and use graphs of sin, cos, tan, any angle, degrees or radians.
2. Exact values for 30°, 45°, 60° and related angles.
3. sin⁻¹x, cos⁻¹x, tan⁻¹x as principal values of inverse functions.
4. The identities tan θ ≡ sin θ / cos θ and sin²θ + cos²θ ≡ 1.
5. All solutions of simple equations in a given interval (no general solutions).

sec, cosec, cot and compound angles are not in 1.5. They come in Pure Mathematics 2 (2.3) and Pure Mathematics 3 (3.3).

Key facts table

Fact	Value	In MF19?
tan θ	sin θ / cos θ	Yes
Pythagorean identity	sin²θ + cos²θ ≡ 1	Yes
Principal range of sin⁻¹x	−π/2 ≤ sin⁻¹x ≤ π/2	Yes
Principal range of cos⁻¹x	0 ≤ cos⁻¹x ≤ π	Yes
Principal range of tan⁻¹x	−π/2 < tan⁻¹x < π/2	Yes
Exact values of 30°, 45°, 60°	see below	No, learn them
Radians	180° = π	No, learn it

Exact values

	0	π/6 (30°)	π/4 (45°)	π/3 (60°)	π/2 (90°)
sin	0	1/2	√2/2	√3/2	1
cos	1	√3/2	√2/2	1/2	0
tan	0	√3/3	1	√3	undefined

Memory aid: the sin row is √0/2, √1/2, √2/2, √3/2, √4/2. The cos row is the same list backwards.

Signs and related angles

Positive ratios by quadrant, going anticlockwise from 0°: all, then sin, then tan, then cos.

Method: exact value of any angle

1. Reduce the angle to 0° to 360° by adding or subtracting 360°.
2. Find the quadrant.
3. Find the related acute angle (the angle to the x-axis).
4. Take the ratio of the related angle; attach the sign for that quadrant.

Reminder. cos 200° = −cos 20° (third quadrant), sin 290° = −sin 70° (fourth), tan 140° = −tan 40° (second).

Graph facts

	sin x	cos x	tan x
Period	2π (360°)	2π (360°)	π (180°)
Range	[−1, 1]	[−1, 1]	all reals
Symmetry	odd: sin(−x) = −sin x	even: cos(−x) = cos x	odd
Asymptotes	none	none	x = π/2 + nπ

For y = a sin bx + c or y = a cos bx + c (a, b > 0):

Feature	Value
Greatest value	c + a
Least value	c − a
Period	2π/b (or 360°/b)
Cycles in 0 to 2π	b

y = tan(x + k): shift left by k. Zeros where x + k = 0, π, 2π, …; asymptotes where x + k = π/2, 3π/2, …

Reminder. y = 1 + 4 sin 3x for 0° ≤ x ≤ 120°: period 360° ÷ 3 = 120°, so one cycle. It starts at (0, 1), reaches its greatest value 5 at (30°, 5), its least value −3 at (90°, −3), and ends at (120°, 1).

y = −a cos x starts at a minimum, not a maximum. A sketch must show the start and end points, the maximum and minimum points and the axis intercepts, with coordinates.

Principal values

The calculator’s inverse key gives the principal value only. It is the answer to sin⁻¹x, cos⁻¹x or tan⁻¹x, but usually only one answer to an equation.

• sin⁻¹(−1/2) = −π/6 (negative, because the range includes negatives)
• cos⁻¹(−1/2) = 2π/3 (never negative)
• cos⁻¹(cos θ) = θ only when 0 ≤ θ ≤ π

The graph of y = cos⁻¹x is the reflection of y = cos x, for 0 ≤ x ≤ π, in y = x.

Identities: method box

Proving an identity:

1. Start with the more complicated side.
2. Combine fractions, expand brackets, or write tan as sin/cos.
3. Replace 1 − cos²θ by sin²θ (or 1 − sin²θ by cos²θ) when it appears.
4. Finish exactly at the other side. Never cross-multiply both sides.

Finding one ratio from another:

1. Use sin²θ + cos²θ = 1 to find the size.
2. Use the quadrant to fix the sign.
3. Use tan = sin/cos for the third ratio.

Equations: method box

Equation type	First step
sin x = k	x₁ = sin⁻¹k, x₂ = 180° − x₁ (or π − x₁)
cos x = k	x₁ = cos⁻¹k, x₂ = 360° − x₁ (or 2π − x₁)
tan x = k	x₁ = tan⁻¹k, then add 180° (or π)
a sin x = b cos x	divide by cos x → tan x = b/a
mix of sin² and cos	replace sin² by 1 − cos² → quadratic in cos
mix of cos² and sin	replace cos² by 1 − sin² → quadratic in sin
common factor sin x or cos x	factorise, set each factor to zero
sin(bx + k) = c	change interval for bx + k first

Then add or subtract 360° (or 2π) to find every value in the interval, and undo any multiple or shift.

Reminder (common factor). Solve 7 sin x cos x = 2 sin x for 0° ≤ x ≤ 360°.

sin x (7 cos x − 2) = 0
sin x = 0:     x = 0°, 180°, 360°
cos x = 2/7:   x = 73.4°, 360° − 73.4° = 286.6°

Five solutions. Cancelling sin x would lose three.

Reminder (±). Solve tan²x = 3 for 0 ≤ x ≤ π.

tan x = √3 or tan x = −√3
x = π/3 or x = π − π/3 = 2π/3

A square root always gives two signs.

Must-know distinctions

• sin⁻¹x vs (sin x)⁻¹. sin⁻¹x is the inverse function; (sin x)⁻¹ = 1/sin x.
• sin²x vs sin x². sin²x = (sin x)²; sin x² means the sine of x².
• Identity vs equation. An identity is true for all θ; you prove it. An equation is true for some θ; you solve it.
• Principal value vs all solutions. tan⁻¹k is one number; tan x = k has a solution every 180°.
• Degrees vs radians. The interval tells you which to use. 0 ≤ x ≤ 2π means radians.
• Period of sin bx vs tan bx. 360°/b for sin and cos; 180°/b for tan.
• “Exact” vs “3 s.f.” Exact means surds, fractions and multiples of π.
• Dividing vs factorising. Divide by cos x only when cos x = 0 cannot be a solution; otherwise factorise.

Quick self-test

1. Find the exact value of cos 330°.
2. Find the exact value of tan(2π/3).
3. Find the exact value of sin(−5π/4).
4. State the period of y = 4 tan 3x, in degrees.
5. State the range of y = 5 − 2 sin x.
6. Find the exact value of cos⁻¹(−√3/2).
7. Find tan⁻¹(1/√3) in degrees.
8. Given sin θ = 0.6 and 90° < θ < 180°, find tan θ.
9. Simplify (1 − sin²θ)/cos θ.
10. Solve tan x = −2 for 0° ≤ x ≤ 360°.
11. Solve 2 sin(x + 30°) = 1 for 0° ≤ x ≤ 360°.
12. State the number of solutions of cos 4x = 0.3 for 0° ≤ x ≤ 180°.

Answers

1. Fourth quadrant, related 30°, cos positive: √3/2.
2. Second quadrant, related π/3, tan negative: −√3.
3. −5π/4 is the same as 3π/4 (second quadrant), sin positive: √2/2.
4. 180° ÷ 3 = 60°.
5. sin x runs from −1 to 1, so 3 ≤ y ≤ 7.
6. 5π/6 (must lie in 0 to π).
7. 30°.
8. cos θ = −0.8 (second quadrant), so tan θ = 0.6 ÷ (−0.8) = −0.75.
9. (1 − sin²θ)/cos θ = cos²θ/cos θ = cos θ.
10. tan⁻¹(−2) = −63.4°; add 180° and 360°: x = 116.6°, 296.6°.
11. sin(x + 30°) = 1/2 with 30° ≤ x + 30° ≤ 390°: x + 30° = 30°, 150°, 390°, so x = 0°, 120°, 360°.
12. 0° ≤ 4x ≤ 720° is two full cycles, each with two solutions: 4.

Where marks are usually lost

• Giving only the calculator’s principal value when the interval contains two or more solutions.
• Not extending the interval for 2x or 3x, then missing solutions beyond 360°.
• Leaving answers as values of 2x instead of dividing back to x.
• Answers in degrees when the interval is in radians, or the reverse.
• Cancelling a common factor of sin x or cos x and losing the zero solutions.
• Keeping an impossible root such as cos x = −3 without rejecting it.
• Writing cos θ = +√(1 − sin²θ) without checking the quadrant sign.
• In a proof, working on both sides at once or starting from the answer.
• A sketch with no coordinates at the end points, turning points or intercepts.
• Rounding a principal value to 1 d.p. before adding 180° or dividing by 2.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027, Version 4, Cambridge Assessment International Education (part of Cambridge University Press & Assessment). Section 1.5 Trigonometry, in 1 Pure Mathematics 1 (for Paper 1).