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

This study guide follows the OCR A Level Mathematics A specification (Version 3.1, August 2026, for first assessment in 2018). It covers sections 1.05h to 1.05q of the Pure Mathematics trigonometry content: reciprocal and inverse trigonometric functions, the Pythagorean identities, compound and double angle formulae, the R-form, trigonometric equations, proof and modelling. In the specification, 1.05j and the basic form of 1.05o are Stage 1 (shared with AS Level); everything else here is Stage 2, so A Level only. Pure Mathematics content is assessed on all three written papers.

Use it with the revision notes and the practice questions. The course page is OCR A Level Mathematics, the printable list is the checklist, and the free diagnostics show where to start. For the wider Pure Mathematics picture see the pure mathematics overview. Basic ratios, triangle rules, small angles and the basic graphs are assumed.

What this unit covers

Section	You must be able to	Stage
1.05h	Use sec, cosec, cot and arcsin, arccos, arctan, and their links to sin, cos, tan	2
1.05i	Know the graphs, domains and ranges of those functions	2
1.05j	Use sin²θ + cos²θ = 1 together with tan θ = sin θ / cos θ	1
1.05k	Use sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ	2
1.05l	Use the double angle formulae and the sin, cos, tan of A ± B	2
1.05m	Understand the geometrical proofs of those formulae	2
1.05n	Write a cos θ + b sin θ as R cos(θ ± α) or R sin(θ ± α)	2
1.05o	Solve equations with multiples of the angle, quadratics and identities	1 and 2
1.05p	Construct proofs using trigonometric identities	2
1.05q	Use trigonometric functions to model situations	2

1.05h and 1.05i: reciprocal and inverse functions

The three reciprocal functions are:

Function	Definition	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 = cos θ / sin θ	sin θ = 0	all real numbers

The graph of sec θ has vertical asymptotes wherever cos θ = 0, and it touches 1 or −1 where cos θ is 1 or −1. Between asymptotes it is a U shape (where cos is positive) or a ∩ shape (where cos is negative). cosec θ is the same picture built from sin θ, with asymptotes at 0, π, 2π and so on. cot θ falls from +∞ to −∞ between each pair of asymptotes, crosses zero where cos θ = 0, and has period π.

The inverse functions undo sin, cos and tan on a restricted domain, giving the principal values:

Function	Domain	Range
arcsin x (or sin⁻¹ x)	−1 ≤ x ≤ 1	−π/2 ≤ y ≤ π/2
arccos x (or cos⁻¹ x)	−1 ≤ x ≤ 1	0 ≤ y ≤ π
arctan x (or tan⁻¹ x)	all real x	−π/2 < y < π/2

Each graph is the reflection in the line y = x of the matching piece of the original graph. arcsin rises from (−1, −π/2) through the origin to (1, π/2). arccos falls from (−1, π) through (0, π/2) to (1, 0). arctan rises through the origin and flattens towards y = ±π/2.

Worked example 1. Given cosec θ = −13/5 and cos θ > 0, find sec θ and cot θ.

sin θ = 1 / cosec θ = −5/13
cos θ is positive, so cos θ = +√(1 − 25/169) = 12/13
sec θ = 13/12
cot θ = cos θ / sin θ = (12/13) ÷ (−5/13) = −12/5

Worked example 2. Evaluate (a) arcsin(−√3/2), (b) arccos(−1/2), (c) arcsin(sin 5π/6).

(a) The angle in −π/2 ≤ y ≤ π/2 with sine −√3/2 is −π/3. (b) The angle in 0 ≤ y ≤ π with cosine −1/2 is 2π/3. (c) sin 5π/6 = 1/2, and the principal value with sine 1/2 is π/6, not 5π/6.

In (c), arcsin always returns a value in its own range.

1.05j and 1.05k: identities

Dividing sin²θ + cos²θ = 1 by cos²θ gives sec²θ = 1 + tan²θ. Dividing by sin²θ gives cosec²θ = 1 + cot²θ. Both hold for every θ where the functions are defined. Use sin²θ + cos²θ = 1 to turn a mixed equation into one trig function, and the other two when sec, cosec, tan or cot appear.

Worked example 3 (Stage 1). Solve 2 sin²x − cos x − 1 = 0 for 0 ≤ x < 360°.

2(1 − cos²x) − cos x − 1 = 0
1 − cos x − 2cos²x = 0
2cos²x + cos x − 1 = 0
(2cos x − 1)(cos x + 1) = 0
cos x = 1/2  → x = 60°, 300°
cos x = −1   → x = 180°

Worked example 4 (radians). Solve 2 tan²θ + sec θ = 1 for 0 ≤ θ < 2π.

2(sec²θ − 1) + sec θ − 1 = 0
2sec²θ + sec θ − 3 = 0
(2sec θ + 3)(sec θ − 1) = 0
sec θ = −3/2 → cos θ = −2/3 → θ = 2.3005 or 2π − 2.3005 = 3.9827
sec θ = 1    → cos θ = 1    → θ = 0

1.05l: compound and double angle formulae

Formula	Statement
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)

Watch the sign in cos(A ± B): the sign flips. The double angle formulae are the compound ones with B = A.

Worked example 5. Find the exact value of cos 105°.

cos 105° = cos(60° + 45°)
         = cos 60° cos 45° − sin 60° sin 45°
         = (1/2)(√2/2) − (√3/2)(√2/2)
         = (√2 − √6) / 4

Worked example 6. A is acute with sin A = 4/5. B is obtuse with cos B = −5/13. Find sin(A + B).

cos A = 3/5 and sin B = 12/13. Then sin(A + B) = (4/5)(−5/13) + (3/5)(12/13) = (−20 + 36)/65 = 16/65.

Worked example 7. Solve cos 2x + 3 sin x = 2 for 0 ≤ x ≤ 360°.

Use cos 2x = 1 − 2sin²x so that only sin x remains:

1 − 2sin²x + 3 sin x − 2 = 0
2sin²x − 3 sin x + 1 = 0
(2 sin x − 1)(sin x − 1) = 0
sin x = 1/2 → x = 30°, 150°
sin x = 1   → x = 90°

Worked example 8. Solve tan(θ + 45°) = 4 for 0 ≤ θ < 180°. Expand: (tan θ + 1)/(1 − tan θ) = 4, so tan θ + 1 = 4 − 4 tan θ, giving tan θ = 3/5 and θ = 31.0°.

1.05m: geometrical proofs

You should understand where the formulae come from. Here is a proof of sin(A + B) for acute A and B.

Take a triangle XYZ with ZX = p and ZY = q. Draw the perpendicular ZW from Z to XY, with W between X and Y. Let ZW = h, ∠XZW = A and ∠WZY = B. Then ∠XZY = A + B.

Area of XYZ = ½ p q sin(A + B)
Also area = area ZXW + area ZWY
          = ½ p h sin A + ½ q h sin B
In the two right-angled triangles, h = q cos B and h = p cos A
So area = ½ p (q cos B) sin A + ½ q (p cos A) sin B
        = ½ p q (sin A cos B + cos A sin B)

Comparing the two area expressions gives sin(A + B) = sin A cos B + cos A sin B. Replacing B by −B gives sin(A − B). Then cos x = sin(90° − x) turns it into the cosine formula:

cos(A + B) = sin(90° − A − B) = sin((90° − A) − B)
           = cos A cos B − sin A sin B

Dividing sin(A + B) by cos(A + B) and then dividing top and bottom by cos A cos B gives the tan formula.

1.05n: the R-form

To write a cos θ + b sin θ as R sin(θ + α), expand R sin(θ + α) = R cos α sin θ + R sin α cos θ and match coefficients. Then R cos α = (the sin θ coefficient) and R sin α = (the cos θ coefficient), with R = √(a² + b²) and R > 0. Take α acute when both coefficients are positive; otherwise check the quadrant from the signs of R cos α and R sin α.

Worked example 9. Write 7 sin x − 24 cos x as R sin(x − α), with R > 0 and 0 < α < π/2. Hence solve 7 sin x − 24 cos x = 10 for 0 ≤ x < 360°, and state the greatest value.

R sin(x − α) = R cos α sin x − R sin α cos x. So R cos α = 7 and R sin α = 24.

R = √(49 + 576) = 25
tan α = 24/7 → α = 1.287 rad (73.74°)
25 sin(x − 73.74°) = 10 → sin(x − 73.74°) = 0.4
x − 73.74° = 23.58° or 156.42°
x = 97.3° or 230.2°

The greatest value of 7 sin x − 24 cos x is 25, at x = α + 90° = 163.7°; the least value is −25. The graph is a sine wave of amplitude 25, shifted right by α, crossing the y-axis at −24.

For R cos(θ ± α), match R cos θ cos α ∓ R sin θ sin α. For example √3 cos θ − sin θ = 2 cos(θ + π/6).

1.05o: equations with multiples of the angle

For sin(kθ + c), cos(kθ + c) or tan(kθ + c), first widen the interval: if 0 ≤ x < 360°, then 2x + 20° runs from 20° to 740°. Solve for the whole expression, then undo the shift and the multiple.

Worked example 10. Solve cos(2x + 20°) = 0.4 for 0 ≤ x < 360°.

2x + 20° runs from 20° to 740°
cos⁻¹ 0.4 = 66.42°, so 2x + 20° = 66.42°, 293.58°, 426.42°, 653.58°
2x = 46.42°, 273.58°, 406.42°, 633.58°
x = 23.2°, 136.8°, 203.2°, 316.8°

A doubled angle gives twice as many solutions.

1.05p: proof

Proofs of identities start from one side and reach the other, using the identities above. Do not work on both sides at once.

Worked example 11. Prove that cosec 2x + cot 2x ≡ cot x.

LHS = 1/sin 2x + cos 2x / sin 2x
    = (1 + cos 2x) / sin 2x
    = 2cos²x / (2 sin x cos x)
    = cos x / sin x = cot x = RHS

For an inequality, use the R-form: since sin x + cos x = √2 sin(x + π/4) and sin never exceeds 1, sin x + cos x ≤ √2 for all x.

1.05q: modelling

Periodic situations such as tides, sound waves and alternating quantities are modelled by a + R sin(kt + α). The constant a is the mean level, R is the amplitude and 2π/k is the period.

Worked example 12. The depth of water, d metres, in a harbour t hours after midnight is d = 7 + 2.4 sin(πt/6) + 1.8 cos(πt/6). Find the greatest and least depths, and when the depth is 8.5 m for 0 ≤ t < 12.

R = √(2.4² + 1.8²) = 3, tan α = 1.8/2.4 → α = 0.6435
d = 7 + 3 sin(πt/6 + 0.6435)
Greatest 7 + 3 = 10 m, least 7 − 3 = 4 m
Greatest depth when πt/6 + 0.6435 = π/2 → t = 1.771 (01:46)
d = 8.5 → sin(πt/6 + 0.6435) = 0.5
πt/6 + 0.6435 = 5π/6 or 13π/6
t = 3.771 or 11.771, that is 03:46 and 11:46

In mechanics, a force P at angle θ above the horizontal has components P cos θ and P sin θ. Resolving first, then using the R-form, finds the best angle: cos θ + μ sin θ has greatest value √(1 + μ²).

Common errors

• Using arcsin(sin x) = x for any x; it only holds in the principal range.
• Writing cos(A + B) = cos A cos B + sin A sin B.
• Dividing both sides by cos x, which loses the solutions where cos x = 0.
• Finding R sin(x + α) with α negative or obtuse without checking signs.
• Solving cos(2x + 20°) = 0.4 and giving only two answers.

When you are ready, try the practice questions.

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.