OCR A Level Mathematics A: Trigonometric ratios, triangle rules, small angles, graphs and basic equations (H240) – Revision Notes

These notes condense sections 1.05a to 1.05g and 1.05o of OCR A Level Mathematics A (H240), Version 3.1 (August 2026), for first assessment in 2018. Longer explanations and full worked examples are in the study guide. To test yourself afterwards, try the practice questions. Content that is A Level only (Stage 2) is labelled; everything else is Stage 1, shared with AS. Course hub: /boards/ocr/a-level/mathematics/. Checklist: /checklists/ocr/a-level/mathematics/. Unsure where to begin? The free diagnostics will show you.

Key facts at a glance

Idea	Statement
Unit circle (1.05a)	Point at angle θ is (cos θ, sin θ); tan θ = sin θ / cos θ
Sine rule (1.05b)	a / sin A = b / sin B = c / sin C
Cosine rule (1.05b)	a² = b² + c² − 2bc cos A
Area (1.05c)	½ab sin C, with C between a and b
Radians (1.05d)	180° = π; A Level only
Arc and sector (1.05d)	s = rθ and A = ½r²θ, θ in radians; A Level only
Small angles (1.05e)	sin θ ≈ θ, cos θ ≈ 1 − θ²/2, tan θ ≈ θ, θ in radians; A Level only
Periods (1.05f)	sin and cos: 360° (2π); tan: 180° (π)

Exact values

θ (degrees)	0°	30°	45°	60°	90°
θ (radians, A Level only)	0	π/6	π/4	π/3	π/2
sin θ	0	1/2	√2/2	√3/2	1
cos θ	1	√3/2	√2/2	1/2	0
tan θ	0	√3/3	1	√3	undefined

Method for any angle

1. Find the quadrant. Sine is positive in quadrants 1 and 2, cosine in 1 and 4, tangent in 1 and 3.
2. Find the reference angle: the acute angle to the x-axis.
3. Write the table value for the reference angle with the correct sign.

Reminder. tan 300°: quadrant 4, reference angle 60°, tangent negative there, so tan 300° = −√3.

Radians

degrees → radians: multiply by π/180      radians → degrees: multiply by 180/π

Reminder: 72° = 72π/180 = 2π/5, and 7π/12 = 7 × 15° = 105°.

Sector facts (A Level only): arc length s = rθ, area A = ½r²θ, perimeter = 2r + rθ.

• If a sector has area A and arc s, then A = ½rs, which helps when two of r, θ, s, A are given.
• Area of a segment = sector area − triangle area = ½r²θ − ½r² sin θ.

Reminder. r = 6 and θ = 1.5 give s = 9 and A = ½ × 36 × 1.5 = 27.

Reminder (segment). r = 10 and θ = 0.6 give segment area = ½ × 100 × (0.6 − sin 0.6) = 50 × 0.0354 = 1.77 cm².

Triangle rules

You know	Use	Watch out for
Side, opposite angle, one more piece	Sine rule	Ambiguous case when finding an angle
Two sides and the included angle	Cosine rule for the third side	Negative cos means obtuse angle
Three sides	Cosine rule rearranged	Find the largest angle first
Two sides and the included angle	Area = ½ab sin C	The angle must be between the sides

Method for a bearings problem

1. Mark a north line wherever the direction changes.
2. Mark each bearing clockwise from north.
3. Use the back bearing (add or subtract 180°) to find the interior angle at each turn.
4. Apply the sine or cosine rule.
5. Add the interior angle to, or subtract it from, a known bearing for the final answer. Write three figures, and state your answer as a distance and a bearing if both are asked for.

The ambiguous case. Given a, b and A (a opposite A), sin B = b sin A / a has solutions B and 180° − B. The second is valid only if A + (180° − B) < 180°.

Reminder. A = 50°, a = 9, b = 7. Then sin B = 7 sin 50° / 9 = 0.5958, so B = 36.6° or 143.4°. Since 50° + 143.4° is more than 180°, only B = 36.6° works. Here a is longer than b, so there is one triangle.

Reminder (area). Sides 9 and 14 and area 40: ½ × 9 × 14 × sin C = 40, so sin C = 80/126 = 0.635 and C = 39.4° or 140.6°. Give both unless told the angle is acute.

Reminder (cosine rule for an angle). Sides 5, 6, 7: cos of the angle opposite 7 is (25 + 36 − 49) / (2 × 5 × 6) = 0.2, so the angle is 78.5°.

Graphs and symmetry

Graph	Key points
y = sin x	passes through (0, 0), maximum 1 at 90°, minimum −1 at 270°
y = cos x	passes through (0, 1), zeros at 90° and 270°, minimum −1 at 180°
y = tan x	passes through (0, 0), asymptotes at 90° and 270°, repeats every 180°

sin(−x) = −sin x      cos(−x) = cos x      tan(−x) = −tan x
sin(180° − x) = sin x       cos(180° − x) = −cos x
tan(x + 180°) = tan x       sin(90° − x) = cos x

Reminder. If cos 20° = c, then cos 160° = −c, cos 340° = c and sin 70° = c.

Small angle approximations (A Level only)

With θ in radians and small: sin θ ≈ θ, cos θ ≈ 1 − θ²/2, tan θ ≈ θ.

1. Replace each trig function by its approximation.
2. Expand and simplify.
3. Neglect powers of θ the question tells you to ignore.

Reminder. For small θ, (1 − cos θ) / θ² ≈ (θ²/2) / θ² = 1/2, and tan 3θ − sin θ ≈ 3θ − θ = 2θ.

Solving simple equations

Steps

1. Isolate sin, cos or tan of the angle: for example sin(kθ) = c.
2. Convert the given interval for θ into the interval for kθ.
3. Take the principal value, then use the symmetry for that function: sin: α and 180° − α. cos: α and 360° − α. tan: α and α + 180°.
4. Add or subtract full periods until you leave the interval, then divide by k.
5. Quadratics: treat sin θ (or cos θ, tan θ) as a single unknown, factorise, then solve each case.

Reminder. Solve cos 3x = −0.5 for 0° ≤ x < 180°. Then 0° ≤ 3x < 540°, and cos is −0.5 at 120°, 240°, 480°. So x = 40°, 80°, 160°.

Reminder. Solve tan²θ = 3 for 0° ≤ θ < 360°: tan θ = ±√3, so θ = 60°, 120°, 240°, 300°.

Reminder (quadratic). Solve 3 tan²θ − 2 tan θ − 1 = 0 for 0° ≤ θ < 360°. Factorise: (3 tan θ + 1)(tan θ − 1) = 0. Then tan θ = 1 gives 45° and 225°, and tan θ = −1/3 gives 161.6° and 341.6°.

Reminder (radians, A Level only). 2 cos x = 1 for −π ≤ x ≤ π gives cos x = 1/2, so x = ±π/3.

Must-know distinctions

• Degrees against radians. Arc length, sector area and the small angle approximations need radians.
• Principal value against all solutions. The calculator gives one value only.
• Included against non-included angle. Included angle: cosine rule or area. Opposite pair: sine rule.
• Acute against obtuse. An angle and its supplement have the same sine, so use other information to choose.
• Period. The period of tan is half that of sin and cos.
• Exact against decimal. Exact values come from the table. Decimals come from a calculator set to the right mode, and a question that asks for “exact” wants surds or multiples of π.

Quick self-test

1. Give the exact value of tan 300°.
2. Convert 7π/12 to degrees, and 72° to radians.
3. A sector has radius 6 cm and angle 1.5 radians. Find its arc length and area.
4. In triangle ABC, a = 9, A = 40° and B = 65°. Find b to 3 s.f.
5. A triangle has sides 5, 6 and 7. Find the angle opposite the side of length 7, to 1 d.p.
6. Find the area of a triangle with sides 10 and 12 and an included angle of 30°.
7. Given cos 20° = c, write cos 160° in terms of c.
8. What is the period of y = tan x in radians?
9. For small θ, approximate tan 3θ − sin θ.
10. Solve 2 cos θ = −1 for 0° ≤ θ < 360°.
11. Solve tan x = −1 for −π < x ≤ π.
12. How many solutions does sin x = 0.7 have for 0° ≤ x < 720°?

Answers

1. −√3
2. 105°; 2π/5
3. Arc length 9 cm; area 27 cm²
4. b = 9 sin 65° / sin 40° = 12.7
5. 78.5°
6. 30
7. −c
8. π
9. 2θ
10. θ = 120°, 240°
11. x = −π/4, 3π/4
12. 4 (44.4°, 135.6°, 404.4°, 495.6°)

Where marks are usually lost

• Leaving the calculator in the wrong angle mode, especially in radian questions.
• Giving only the principal value, or stopping after one solution of two.
• Rounding too early in a multi-step triangle problem. Keep full values until the final line.
• Writing a bearing as 70° instead of 070°, or giving an angle that is not a bearing at all.
• Choosing the sine rule when the given angle is the included angle. Check which pair of side and opposite angle you have.
• Missing the second triangle in an ambiguous case, or keeping a second angle that does not fit.
• Forgetting to convert the interval when solving for a multiple angle, which loses solutions.
• Applying small angle approximations to a large angle, or keeping terms the question says to neglect.
• Giving a decimal where an exact value was asked for, or the reverse.

Official syllabus

OCR, A Level in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR.