Edexcel A-Level Pure, Year 1

Trigonometric Ratios

A-LEVEL MATHEMATICS · EDEXCEL 9MA0 · PURE YEAR 1 — CH 9

The theory, at a glance

The sine and cosine rules, area of a triangle, the ambiguous case and the shapes of the trig graphs. ✎

① The Sine Rule

Side a is always opposite angle A — that pairing is the whole rule.

Use it when you have a matching side–angle pair plus one more piece.

② The Cosine Rule

a² = b² + c² − 2bc·cos A

Rearranged to find an angle:

cos A = b² + c² − a² / 2bc

Use it for three sides, or two sides and the angle between them.

The letter on the left matches the angle: a² pairs with cos A.

A negative cos means the angle is obtuse — that's fine, not an error.

③ Area & the Ambiguous Case

Area = ½ · a · b · sin C

The angle must sit between the two sides you use.

The ambiguous case.

④ The Three Graphs

cos x is sin x shifted 90° left. Both repeat every 360°, so every trig equation has infinitely many solutions — the given range is what limits them.

★ Worked example ✎

In triangle ABC, a = 8 cm, b = 5 cm and angle C = 60°. Find c, then the area of the triangle.

I have two sides and the angle between them, so it's the cosine rule — with c² on the left to match cos C.

c² = a² + b² − 2ab·cos C

Substitute, remembering cos 60° = ½ exactly.

c² = 64 + 25 − 2(8)(5)(½)

c² = 89 − 40 = 49 → c = 7 cm

For the area, angle C sits between sides a and b — exactly what the formula needs.

Leave it exact unless told otherwise.

Area = 10√3 ≈ 17.3 cm²

Exam trap

Don't round mid-question.

A-LEVEL MATHEMATICS · EDEXCEL 9MA0 · PURE YEAR 1 — CH 10

Identities & Equations

Exact values, the CAST diagram, the two key identities and a reliable method for solving in a range. ✎

① Exact Values to Memorise

√2⁄2

√3⁄2

√3⁄3

√3

tan 90° is undefined

② The CAST Diagram

Which ratios are positive in each quadrant. Angles are measured anticlockwise from the positive x-axis:

Read it anticlockwise from the bottom-right: C, A, S, T. Use it to turn one calculator answer into all the answers in the range.

③ The Two Key Identities

sin²θ + cos²θ ≡ 1

tan θ ≡ sin θ / cos θ

Rearrange freely: sin²θ = 1 − cos²θ and cos²θ = 1 − sin²θ.

These let you get an equation into one ratio only — then it's usually a quadratic.

The ≡ sign means true for every θ, unlike an equation you solve.

④ Solving in a Given Range

Get to one ratio = number, using an identity or factorising first if needed.

Take the inverse to get the principal value from the calculator.

Use CAST to find the second angle in one turn, then add or subtract 360° to sweep the whole range.

Discard anything outside the range and list what's left in order.

2x

x + 30°

stretch the range first

Solve 2sin²x + 3cos x = 3 for 0° ≤ x ≤ 360°.

Two different ratios, so use sin²x = 1 − cos²x to get everything in cos x.

2(1 − cos²x) + 3cos x = 3

Expand and collect into a quadratic in cos x.

2 − 2cos²x + 3cos x = 3

2cos²x − 3cos x + 1 = 0

Factorise it just like an ordinary quadratic.

(2cos x − 1)(cos x − 1) = 0

cos x = ½ or cos x = 1

cos x = ½ gives 60°. Cos is also positive in the fourth quadrant, so the partner is 360° − 60°.

x = 60° and x = 300°

cos x = 1 happens at the ends of the range — both are included here.

x = 0° and x = 360°

List every solution in the range, in order.

x = 0°, 60°, 300°, 360°

Never divide by cos x to simplify — you'd lose solutions. Factorise instead. And don't forget the endpoints: 0° and 360° count when the range says ≤.

Download the PDF cheatsheet

One page, free — along with nine others.

↓ PDF

Practice, with the working

Three questions from the bank, each with its full solution.

1Easier
A triangle has two sides of length 88 cm and 1010 cm, and its area is 2424 cm². Given that the included angle is acute, find its size to 1 decimal place.
›Show solution
**Solution**
[M1] 12(8)(10)sin⁡θ=24\dfrac{1}{2}(8)(10)\sin\theta = 24.
[A1] sin⁡θ=0.6\sin\theta = 0.6.
[A1] θ=36.9°\theta = 36.9° (acute).
**Answer:** 36.9°36.9°
2Medium
In triangle PQRPQR, PQ=11PQ = 11 cm, PR=9PR = 9 cm and angle QPR=70°QPR = 70°.
›Show solution
**Part (a)**
_Find QRQR, to 3 significant figures._
[M1] QR2=112+92−2(11)(9)cos⁡70°QR^2 = 11^2 + 9^2 - 2(11)(9)\cos 70°.
[A1] =202−198cos⁡70°=134.3= 202 - 198\cos 70° = 134.3.
[A1] QR=11.6QR = 11.6 cm (3 s.f.).
**Answer:** 11.611.6 cm
**Part (b)**
_Find the area of the triangle, to 3 significant figures._
[M1] Area =12(11)(9)sin⁡70°= \dfrac{1}{2}(11)(9)\sin 70°.
[A1] =46.5= 46.5 cm² (3 s.f.).
**Answer:** 46.546.5 cm²
3Harder
A triangle has sides 44 cm, 55 cm and 88 cm.
›Show solution
**Part (a)**
_Show that the largest angle is obtuse, and find it to 1 decimal place._
[M1] Angle opposite the 8 cm side: cos⁡θ=16+25−642(4)(5)\cos\theta = \dfrac{16 + 25 - 64}{2(4)(5)}.
[A1] =−2340= -\dfrac{23}{40}.
[R1] Negative cosine ⇒\Rightarrow obtuse angle.
[A1] θ=125.1°\theta = 125.1° (1 d.p.).
**Answer:** 125.1°125.1°
**Part (b)**
_Find the area of the triangle, to 3 significant figures._
[M1] Area =12(4)(5)sin⁡125.1°= \dfrac{1}{2}(4)(5)\sin 125.1°.
[A1] =8.18= 8.18 cm² (3 s.f.).
**Answer:** 8.188.18 cm²

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