OCR A Level Mathematics A: Reciprocal and inverse trig, identities, compound angles, R-form and proof (H240) – Revision Notes

These revision notes condense the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), first assessment 2018, sections 1.05h to 1.05q. Read the study guide for full explanations and worked examples. Section 1.05j and the basic form of 1.05o are Stage 1 (shared with AS Level); everything else here is Stage 2, A Level only.

Test yourself with the practice questions. Related pages: the OCR A Level Mathematics course page, the checklist, the pure mathematics overview and the free diagnostics.

Definitions (1.05h, 1.05i)

Function	Meaning	Undefined when	Range
sec θ	1 / cos θ	cos θ = 0	sec θ ≤ −1 or sec θ ≥ 1
cosec θ	1 / sin θ	sin θ = 0	cosec θ ≤ −1 or cosec θ ≥ 1
cot θ	1 / tan θ	sin θ = 0	all reals

Inverse	Domain	Range (principal values)
arcsin x	−1 ≤ x ≤ 1	−π/2 ≤ y ≤ π/2
arccos x	−1 ≤ x ≤ 1	0 ≤ y ≤ π
arctan x	all real x	−π/2 < y < π/2

Graph key points: arcsin passes through (−1, −π/2), (0, 0), (1, π/2). arccos passes through (−1, π), (0, π/2), (1, 0). arctan passes through the origin with asymptotes y = ±π/2. Each is a reflection in y = x of the restricted original.

Identities (1.05j, 1.05k)

Identity	Stage
tan θ = sin θ / cos θ	1
sin²θ + cos²θ = 1	1
sec²θ = 1 + tan²θ	2
cosec²θ = 1 + cot²θ	2

Compound and double angles (1.05l, 1.05m)

Formula	Result
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)
sin 2A	2 sin A cos A
cos 2A	cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A
tan 2A	2 tan A / (1 − tan²A)

Geometrical proof outline: split a triangle by an altitude into angles A and B, write the whole area as ½pq sin(A + B), then as the sum of two smaller areas. Replace B by −B for the difference formula, use cos x = sin(90° − x) for the cosine formula, and divide for tan.

Reminder. tan 15° = tan(45° − 30°) = (1 − 1/√3) / (1 + 1/√3) = (√3 − 1)/(√3 + 1) = 2 − √3.

Reminder (double angle in an equation). Solve sin 2x = sin x for 0 ≤ x ≤ 2π. Write 2 sin x cos x − sin x = 0, so sin x (2 cos x − 1) = 0. Then sin x = 0 gives x = 0, π, 2π, and cos x = 1/2 gives x = π/3, 5π/3. Five solutions in all.

Reminder (finding exact values). If sin A = 3/5 with A acute, then cos A = 4/5, so sin 2A = 2(3/5)(4/5) = 24/25 and cos 2A = 1 − 2(9/25) = 7/25. Pick the form of cos 2A that uses the value you already know.

The R-form (1.05n)

Starting form	Match	R and α
a cos θ + b sin θ = R cos(θ − α)	R cos α = a, R sin α = b	R = √(a² + b²), tan α = b/a
a cos θ + b sin θ = R sin(θ + α)	R cos α = b, R sin α = a	R = √(a² + b²), tan α = a/b
a cos θ − b sin θ = R cos(θ + α)	R cos α = a, R sin α = b	tan α = b/a

Method in steps

Choose R sin(θ ± α) or R cos(θ ± α) to suit the question. Use the form the question asks for; if it leaves the choice to you, a form whose signs match your coefficients keeps α acute.

1. Expand the target form, for example R cos θ cos α + R sin θ sin α.
2. Equate the coefficient of cos θ and the coefficient of sin θ.
3. Find R as √(a² + b²), then tan α as the ratio of the two equations.
4. Check the quadrant of α from the signs of R cos α and R sin α.
5. Use the single wave: greatest value is R, least is −R, and the wave has period 2π.

Features of the graph of y = a cos θ + b sin θ: it is a sine or cosine wave of amplitude R, period 2π, with y-intercept a. The greatest value R occurs where the bracket in R cos(θ − α) is 0, that is at θ = α. The least value −R occurs at θ = α + π.

Reminder. 8 cos θ + 6 sin θ = 10 cos(θ − α) with tan α = 6/8, α = 36.87°. To solve 8 cos θ + 6 sin θ = 7: cos(θ − 36.87°) = 0.7, so θ − 36.87° = ±45.57°, giving θ = 82.4° or 351.3° for 0 ≤ θ < 360°.

Solving equations (1.05o)

Method in steps

1. Rewrite using an identity so that one trig function remains.
2. Treat as a quadratic (factorise or use the formula) and solve for that function.
3. Reject any value outside the range of the function, such as sec θ = 1/3.
4. For a multiple or shifted angle, widen the interval first.
5. Find the principal value, list every solution in the widened interval, then undo the multiple.
6. Never divide by a trig function that could be zero; factorise instead.

Reminder. Solve cosec²θ = 3 cot θ + 1 for 0 < θ < 2π.

1 + cot²θ = 3 cot θ + 1
cot θ (cot θ − 3) = 0
cot θ = 0 → θ = π/2, 3π/2
cot θ = 3 → tan θ = 1/3 → θ = 0.3217, 3.4633

Reminder (tan with a multiple). Solve tan 2x = −1 for 0 ≤ x < 180° by the same widening: 2x runs from 0 to 360°, tan is −1 at 135° and 315°, so x = 67.5° and 157.5°. Note that tan has period 180°, so the next value of 2x after 135° is 315°, not 225°.

Proof (1.05p)

Method in steps

1. Pick the more complicated side and start there.
2. Replace every double angle, tan or reciprocal using a known identity.
3. Simplify step by step; do not work on both sides at once.
4. Finish with the exact form of the other side and write “as required”.
5. For an inequality, use the R-form or a completed square and state why the bound holds.

Reminder. Prove that tan x + cot x ≡ 2 cosec 2x. Start with sin x / cos x + cos x / sin x = (sin²x + cos²x) / (sin x cos x) = 1 / (sin x cos x) = 2 / (2 sin x cos x) = 2 / sin 2x = 2 cosec 2x.

Modelling (1.05q)

A model a + R sin(kt + α) has mean level a, amplitude R and period 2π/k. Its greatest value is a + R and its least is a − R. To find when the quantity reaches a given level, solve the sine equation and list every solution in the stated time interval. A model written as p sin kt + q cos kt is converted to the R-form first, which makes the greatest and least values and the times they occur easy to read off. When resolving a force at angle θ above the horizontal, the components are P cos θ and P sin θ.

Must-know distinctions

• arcsin x is an angle; cosec x is 1/sin x. The notation sin⁻¹ x means arcsin x, not 1/sin x.
• arcsin(sin x) = x only when x is in the principal range.
• cos 2A has three forms; choose the one that leaves the function you want.
• cos(A + B) has a minus between the products; sin(A + B) has a plus.
• sin²θ + cos²θ = 1 is Stage 1; the sec and cosec identities are Stage 2.

Quick self-test

1. State the domain and range of arccos x.
2. Find the exact value of cot(3π/4).
3. Given tan A = 1/3, find tan 2A.
4. Simplify (1 − cos 2x) / sin 2x.
5. Find the exact value of sin 75°.
6. Write 5 cos θ − 12 sin θ in the form R cos(θ + α), with α in radians to 3 decimal places.
7. Find the greatest value of 1 / (20 + 5 cos θ + 12 sin θ).
8. Solve sin 3x = 0.5 for 0 ≤ x ≤ 180°.
9. How many solutions does cos 3x = 0.2 have for 0 ≤ x < 360°?
10. Evaluate arctan(−√3), giving the answer in terms of π.

Answers

1. Domain −1 ≤ x ≤ 1, range 0 ≤ y ≤ π.
2. cos(3π/4) = −√2/2 and sin(3π/4) = √2/2, so −1.
3. (2/3) / (1 − 1/9) = (2/3) × (9/8) = 3/4.
4. 1 − cos 2x = 2sin²x and sin 2x = 2 sin x cos x, so the result is tan x.
5. sin(45° + 30°) = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4.
6. R cos α = 5, R sin α = 12, so R = 13 and tan α = 12/5: 13 cos(θ + 1.176).
7. The denominator is 20 + 13 cos(θ − α), least value 7, so the greatest value is 1/7.
8. 3x = 30°, 150°, 390°, 510°, so x = 10°, 50°, 130°, 170°.
9. 3x runs over 0 to 1080°, which holds three full turns with two solutions each, so 6.
10. −π/3.

Where marks are usually lost

• Giving 7π/6 for arcsin(sin 7π/6) instead of the principal value.
• Writing cos(A + B) = cos A cos B + sin A sin B.
• Cancelling a factor of cos x or sin x and so losing solutions.
• Keeping an impossible root such as sec θ = 1/3 or sin θ = 1.5 without comment.
• Not widening the interval for a multiple angle, so solutions are missing.
• Quoting α in degrees when the question asks for radians, or the reverse.
• Starting a proof from both sides at once.
• Rounding R or α early and then losing accuracy in the final answers.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Sections 1.05h to 1.05q.