Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 1 Trigonometry – Study Guide

This guide teaches 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). Pure Mathematics 1 is examined on Paper 1 (1 hour 50 minutes, 75 marks, 10 to 12 structured questions). Paper 1 is compulsory for both AS Level and A Level, and it counts for 60% of the AS Level and 30% of the A Level. A scientific calculator is allowed, but you must show your working.

Course hub: Cambridge A Level Mathematics. Printable list: 9709 checklist. To find your gaps first, take the AS 10-minute diagnostic or browse the 9709 self-check bank. For this unit there are also revision notes and a practice set.

What this unit covers

Syllabus 1.5	What you must be able to do
Graphs	Sketch and use the graphs of sin, cos and tan for angles of any size, in degrees or radians, including forms such as y = a sin bx + c and y = tan(x + k)
Exact values	Use the exact sin, cos and tan of 30°, 45°, 60° and related angles
Inverse notation	Use sin⁻¹x, cos⁻¹x and tan⁻¹x for principal values, and understand them as inverse functions
Identities	Use tan θ ≡ sin θ / cos θ and sin²θ + cos²θ ≡ 1 to prove identities, simplify and solve
Equations	Find all solutions of simple trigonometric equations in a given interval

General solutions are not required. The secant, cosecant and cotangent functions and the compound-angle formulae belong to later sections (2.3 and 3.3), not to Paper 1. Radians and the conversion 180° = π radians come from section 1.4 Circular measure; you need them here.

Angles of any size

Take a point P on a circle of radius 1, centre O. The angle θ is measured anticlockwise from the positive x-axis (clockwise for negative angles). Then P = (cos θ, sin θ) and tan θ = sin θ / cos θ.

The sign of each ratio depends on the quadrant:

Quadrant	Angles (degrees)	Positive ratios
First	0° to 90°	all three
Second	90° to 180°	sin only
Third	180° to 270°	tan only
Fourth	270° to 360°	cos only

The related acute angle is the angle between OP and the x-axis. Every ratio equals ± the same ratio of the related angle, with the sign from the table. The rules you use most:

• sin(180° − θ) = sin θ, cos(180° − θ) = −cos θ
• sin(180° + θ) = −sin θ, cos(180° + θ) = −cos θ, tan(180° + θ) = tan θ
• sin(360° − θ) = −sin θ, cos(360° − θ) = cos θ
• sin(−θ) = −sin θ, cos(−θ) = cos θ

Graphs of sin, cos and tan

	y = sin x	y = cos x	y = tan x
Period	360° (2π)	360° (2π)	180° (π)
Range	−1 ≤ y ≤ 1	−1 ≤ y ≤ 1	all real y
Zeros	0°, 180°, 360°, …	90°, 270°, …	0°, 180°, 360°, …
Asymptotes	none	none	x = 90°, 270°, …

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

• the amplitude is a, so the graph runs from c − a to c + a;
• the period is 360°/b (or 2π/b);
• c moves the graph up by c.

For y = tan(x + k), the whole graph moves k to the left, so the zeros and asymptotes move by k too. These are the transformations from section 1.2; see the functions study guide.

Worked example 1. Sketch y = 2 + 3 cos 2x for 0 ≤ x ≤ π, and find where it crosses the x-axis.

Period = 2π/2 = π, so one full cycle.
Greatest value 2 + 3 = 5 at x = 0 and x = π.
Least value 2 − 3 = −1 at x = π/2.
y = 0:  cos 2x = −2/3
        2x = cos⁻¹(−2/3) = 2.3005   or   2x = 2π − 2.3005 = 3.9827
        x = 1.15  or  x = 1.99   (3 s.f.)

The sketch starts at (0, 5), falls to (π/2, −1), and rises back to (π, 5), crossing the x-axis at x = 1.15 and x = 1.99. Label all of these points.

Worked example 2. For y = tan(x − π/3), 0 ≤ x ≤ 2π, state the intercepts and asymptotes.

y-intercept: tan(−π/3) = −√3, so (0, −√3).
Zeros: x − π/3 = 0 or π, so x = π/3 and x = 4π/3.
Asymptotes: x − π/3 = π/2 or 3π/2, so x = 5π/6 and x = 11π/6.

Exact values

θ	30° (π/6)	45° (π/4)	60° (π/3)
sin θ	1/2	1/√2 = √2/2	√3/2
cos θ	√3/2	1/√2 = √2/2	1/2
tan θ	1/√3 = √3/3	1	√3

Also: sin 0° = 0, cos 0° = 1, sin 90° = 1, cos 90° = 0, and tan 90° is undefined. You can rebuild the table from two triangles: half an equilateral triangle of side 2 (sides 1, √3, 2) and a right isosceles triangle (sides 1, 1, √2).

Worked example 3. Find the exact values of sin 300°, tan(5π/6) and cos(7π/4).

sin 300°: fourth quadrant, related angle 60°, sin negative  → −√3/2
tan(5π/6): 5π/6 = 150°, second quadrant, related 30°, tan negative  → −1/√3 = −√3/3
cos(7π/4): 7π/4 = 315°, fourth quadrant, related 45°, cos positive  → 1/√2 = √2/2

“Exact” means surds and fractions. A calculator decimal such as 0.707 scores nothing when an exact value is asked for.

Inverse notation: sin⁻¹x, cos⁻¹x, tan⁻¹x

sin, cos and tan are many-one, so they have no inverse on their full domains. Restricting the domain makes each one-one. The inverse then gives one answer, the principal value:

Notation	Domain	Principal value range
sin⁻¹x	−1 ≤ x ≤ 1	−π/2 ≤ sin⁻¹x ≤ π/2
cos⁻¹x	−1 ≤ x ≤ 1	0 ≤ cos⁻¹x ≤ π
tan⁻¹x	all real x	−π/2 < tan⁻¹x < π/2

These ranges are in the MF19 list of formulae. Your calculator’s sin⁻¹, cos⁻¹ and tan⁻¹ keys return these values. The graph of y = sin⁻¹x is the reflection of y = sin x, for −π/2 ≤ x ≤ π/2, in the line y = x, as for any inverse function.

Note that sin⁻¹x is the inverse function, not 1/sin x.

Worked example 4. Find the exact values of sin⁻¹(−√3/2), cos⁻¹(−1/√2), tan⁻¹(−1) and sin⁻¹(sin(5π/6)).

sin⁻¹(−√3/2) = −π/3         (must lie in −π/2 to π/2)
cos⁻¹(−1/√2) = 3π/4         (must lie in 0 to π)
tan⁻¹(−1)    = −π/4
sin⁻¹(sin(5π/6)) = sin⁻¹(1/2) = π/6, not 5π/6, because 5π/6 is outside the principal range.

The two identities

tan θ ≡ sin θ / cos θ and sin²θ + cos²θ ≡ 1

Both are in MF19. Rearranged forms you will use: sin²θ ≡ 1 − cos²θ and cos²θ ≡ 1 − sin²θ. The sign ≡ means true for every θ where both sides are defined.

Worked example 5 (proof). Prove that 1/cos θ − cos θ ≡ sin θ tan θ.

LHS = 1/cos θ − cos θ
    = (1 − cos²θ)/cos θ          (common denominator)
    = sin²θ / cos θ              (sin²θ + cos²θ ≡ 1)
    = sin θ × (sin θ / cos θ)
    = sin θ tan θ = RHS

Start from one side and work to the other. Do not treat the identity as an equation and do the same thing to both sides.

Worked example 6 (using an identity). θ is obtuse and sin θ = 5/13. Find the exact values of cos θ and tan θ.

cos²θ = 1 − 25/169 = 144/169, so cos θ = ±12/13.
θ is obtuse (second quadrant), so cos θ < 0:  cos θ = −12/13.
tan θ = (5/13) ÷ (−12/13) = −5/12.

Solving trigonometric equations

Method:

1. Rearrange to sin(…) = k, cos(…) = k or tan(…) = k.
2. If the angle is a multiple or shift, such as 3x or x − 40°, change the interval to match.
3. Find the principal value with sin⁻¹, cos⁻¹ or tan⁻¹.
4. Use the graph or quadrant rules for the other values in the interval.
5. Undo the multiple or shift, and check each answer lies in the original interval.

Worked example 7 (multiple angle). Solve 2 cos 3x = −1 for 0° ≤ x ≤ 180°.

cos 3x = −1/2, and 0° ≤ 3x ≤ 540°.
Principal value: cos⁻¹(−1/2) = 120°.
Other values: 360° − 120° = 240°, then 120° + 360° = 480°. (600° is too big.)
3x = 120°, 240°, 480°
x = 40°, 80°, 160°

Worked example 8 (quadratic in sin θ). Solve 2cos²θ + 3 sin θ = 3 for 0° ≤ θ ≤ 360°.

2(1 − sin²θ) + 3 sin θ − 3 = 0
−2sin²θ + 3 sin θ − 1 = 0
2sin²θ − 3 sin θ + 1 = 0
(2 sin θ − 1)(sin θ − 1) = 0
sin θ = 1/2:  θ = 30°, 150°
sin θ = 1:    θ = 90°
θ = 30°, 90°, 150°

Replace cos²θ with 1 − sin²θ so the equation has one function only. The factorising is the same as in quadratics.

Worked example 9 (using tan). Solve 3 sin x = 2 cos x for 0 ≤ x ≤ 2π.

Divide by cos x (cos x = 0 is not a solution, since then sin x = ±1 and 3 ≠ 0):
tan x = 2/3
x = tan⁻¹(2/3) = 0.588   or   x = 0.588 + π = 3.73   (3 s.f.)

Dividing is safe here only because cos x = 0 does not satisfy the original equation. If cos x or sin x is a common factor, factorise instead. For example, sin θ tan θ = 3 sin θ gives sin θ(tan θ − 3) = 0, so sin θ = 0 (θ = 0°, 180°, 360°) or tan θ = 3 (θ = 71.6°, 251.6°). Dividing by sin θ loses the first three.

Calculator and accuracy

• Put your calculator in the right mode. If the interval is in radians, give answers in radians.
• Give non-exact answers to 3 significant figures, or angles in degrees to 1 decimal place, unless the question says otherwise.
• Keep full calculator values until the end. Rounding a principal value early can change the last figure of later answers.
• The syllabus says no marks are given for unsupported answers from a calculator, so write down the equation you solved and the principal value.

Common errors

• Stopping at the calculator’s principal value and missing the second (or third) solution.
• Forgetting to change the interval for 2x or 3x, so solutions are missed.
• Changing the interval but not dividing back at the end.
• Mixing degrees and radians in one answer, for example x = π/6 and x = 150°.
• Writing sin⁻¹x as 1/sin x.
• Taking cos θ = +12/13 without checking the quadrant.
• Cancelling sin θ or cos θ and losing the solutions where it equals zero.
• Keeping a value such as sin θ = 1.5 (or cos θ = −3), which has no solutions.

Next steps

Use the revision notes for tables, method boxes and a quick self-test, then work through the trigonometry practice set. Trigonometry also appears in the Pure Mathematics 1 mixed practice. If you go on to Pure Mathematics 2 or Pure Mathematics 3, these skills are the base for sec, cosec, cot and the compound-angle formulae.

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).