Trigonometry

This lesson covers Trigonometry as required by the A-Level Mathematics Pure 1 specification. Trigonometry extends beyond right-angled triangles to include trigonometric functions, identities, equations, radians, and the sine and cosine rules. This topic is fundamental and appears in many areas of A-Level Mathematics, including calculus and mechanics.

Spec Mapping — AQA 7357 Section E Trigonometry. This lesson covers exact values, radians, the sine and cosine rules, basic identities and solution of trigonometric equations in Section E. Refer to the official AQA specification document for exact wording.

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Exact Trigonometric Values

You must know the exact values of sine, cosine, and tangent for key angles:

Angle	sin θ	cos θ	tan θ
0°	0	1	0
30°	1/2	√3/2	1/√3 = √3/3
45°	√2/2	√2/2	1
60°	√3/2	1/2	√3
90°	1	0	undefined

Exam Tip: These exact values frequently appear in exam questions. Memorise them and be prepared to use them without a calculator.

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Radians

Radians are an alternative unit for measuring angles. One complete revolution = 2π radians.

Degrees	Radians
30°	π/6
45°	π/4
60°	π/3
90°	π/2
180°	π
360°	2π

Conversion: Radians = Degrees × π/180, and Degrees = Radians × 180/π.

Arc Length and Sector Area

For a sector with radius r and angle θ (in radians):

Formula	Description
s = rθ	Arc length
A = ½r²θ	Sector area

Example: A sector has radius 6 cm and angle π/3 radians.

$$\begin{aligned} \text{Arc length} &= 6 \times \dfrac{\pi}{3} = 2\pi\,\text{cm} \\ \text{Area} &= \tfrac{1}{2} \times 36 \times \dfrac{\pi}{3} = 6\pi\,\text{cm}^{2} \end{aligned}$$Arc lengthArea​=6×3π​=2πcm=21​×36×3π​=6πcm2​

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The Sine and Cosine Rules

Sine Rule

a/sin A = b/sin B = c/sin C

Use when you know two angles and one side (AAS/ASA), or two sides and a non-included angle (SSA — beware the ambiguous case).

Cosine Rule

a² = b² + c² − 2bc cos A (finding a side)

cos A = (b² + c² − a²)/(2bc) (finding an angle)

Use when you know two sides and the included angle (SAS), or three sides (SSS).

Area of a Triangle

Area = ½ab sin C

Example: Find the area of a triangle with sides 8 cm and 11 cm, included angle 30°.

$\text{Area} = \tfrac{1}{2} \times 8 \times 11 \times \sin 30^{\circ} = \tfrac{1}{2} \times 8 \times 11 \times \tfrac{1}{2} = 22\,\text{cm}^{2}$Area=21​×8×11×sin30∘=21​×8×11×21​=22cm2

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Trigonometric Graphs

Function	Period	Range	Symmetry
y = sin x	360° (2π)	−1 ≤ y ≤ 1	Odd function: sin(−x) = −sin x
y = cos x	360° (2π)	−1 ≤ y ≤ 1	Even function: cos(−x) = cos x
y = tan x	180° (π)	All real numbers	Odd function: tan(−x) = −tan x

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Trigonometric Identities

Fundamental Identities

Identity	Form
Pythagorean identity	sin²θ + cos²θ ≡ 1
Tangent identity	tan θ ≡ sin θ / cos θ

Example: Prove that (1 − cos²θ)/cos²θ ≡ tan²θ.

LHS = (1 − cos²θ)/cos²θ
    = sin²θ/cos²θ          (using sin²θ + cos²θ = 1)
    = tan²θ = RHS ∎

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Solving Trigonometric Equations

The CAST Diagram

The CAST diagram shows which trigonometric functions are positive in each quadrant:

        90°
    S   |   A       A = All positive (0° to 90°)
        |           S = Sin positive (90° to 180°)
180° ———+——— 0°     T = Tan positive (180° to 270°)
        |           C = Cos positive (270° to 360°)
    T   |   C
       270°

Example: Solve sin x = 1/2 for 0° ≤ x ≤ 360°.

Principal value: x = 30°. Sin is positive in Q1 and Q2: x = 30° or x = 150°.

Example: Solve 2cos²x − cos x − 1 = 0 for 0° ≤ x ≤ 360°.

Let c = cos x: 2c² − c − 1 = 0 → (2c + 1)(c − 1) = 0.

c = −1/2 → x = 120° or x = 240°. c = 1 → x = 0° or x = 360°.

Solutions: x = 0°, 120°, 240°, 360°.

Example: Solve tan 2x = √3 for 0 ≤ x ≤ 2π.

2x = π/3, π/3 + π, π/3 + 2π, π/3 + 3π
2x = π/3, 4π/3, 7π/3, 10π/3
x = π/6, 2π/3, 7π/6, 5π/3

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Synoptic Links

This topic connects to:

• Addition and double-angle formulae (aqa-alevel-maths-trigonometry-depth / addition-formulae) — direct extension of the identities introduced here.
• Projectile motion (aqa-alevel-maths-mechanics / projectiles) — resolving velocities and ranges rely on sin/cos of the launch angle.
• Trigonometric differentiation and integration (aqa-alevel-maths-pure-2 / further-trigonometry) — calculus of sin and cos depends on radian measure introduced here.

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Summary

• Know the exact values of sin, cos, and tan for 0°, 30°, 45°, 60°, 90°.
• Convert between degrees and radians using the factor π/180.
• Use s = rθ and A = ½r²θ for arc length and sector area (θ in radians).
• Use the sine rule (AAS, SSA) and cosine rule (SAS, SSS) for non-right-angled triangles.
• The key identities are sin²θ + cos²θ ≡ 1 and tan θ ≡ sin θ/cos θ.
• Solve trig equations using the CAST diagram or graph symmetry.

Exam Tip: When solving trigonometric equations, always state the range of θ and give all solutions within that range. Show the CAST diagram or reference to graph symmetry. A common mistake is to find only one solution when multiple solutions exist in the given interval. For quadratic trig equations, substitute (e.g. let c = cos x) to make the factoring clearer.

A-Level Deep Dive: Trigonometry

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section E (Trigonometry). This section requires fluency with radian measure (and arc length $s = r\theta$s=rθ, sector area $A = \tfrac{1}{2}r^2\theta$A=21​r2θ), exact values at $0, \pi/6, \pi/4, \pi/3, \pi/2$0,π/6,π/4,π/3,π/2, the standard identities $\sin^2\theta + \cos^2\theta \equiv 1$sin2θ+cos2θ≡1 and $\tan\theta \equiv \sin\theta/\cos\theta$tanθ≡sinθ/cosθ, the double-angle and compound-angle formulae (given in the AQA formula booklet), and the inverse functions $\arcsin$arcsin, $\arccos$arccos, $\arctan$arctan with their restricted ranges. Synoptic load is heavy: differentiating and integrating trigonometric functions in section H assumes radians (the result $\frac{d}{dx}\sin x = \cos x$dxd​sinx=cosx holds only for $x$x in radians). Pure 2 builds on this with the harmonic form $a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha)$asinθ+bcosθ≡Rsin(θ+α), which appears in modelling questions on Paper 2 and underpins simple harmonic motion in Paper 3 Mechanics. The AQA formula booklet lists the compound-angle and double-angle formulae but not the Pythagorean identity or the exact values — these must be memorised.

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Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked example with full mark scheme

Question (8 marks): Solve the equation $\sin 2x + \sin x = 0$sin2x+sinx=0 for $x$x in the interval $0 \leq x < 2\pi$0≤x<2π, giving exact answers in radians.

Solution with mark scheme:

Step 1 — apply the double-angle formula.

Using $\sin 2x \equiv 2\sin x \cos x$sin2x≡2sinxcosx from the formula booklet:

$2\sin x \cos x + \sin x = 0$2sinxcosx+sinx=0

M1 — correct substitution of the double-angle identity. A common error is to write $\sin 2x = 2\sin x$sin2x=2sinx, dropping the cosine factor; this loses M1 immediately.

Step 2 — factorise.

$\sin x \,(2\cos x + 1) = 0$sinx(2cosx+1)=0

M1 — factorising out the common $\sin x$sinx. Candidates who divide both sides by $\sin x$sinx lose all subsequent solutions where $\sin x = 0$sinx=0 — a critical mark-loss pattern.

A1 — correct factorised form.

Step 3 — solve each factor.

From $\sin x = 0$sinx=0 in $[0, 2\pi)$[0,2π): $x = 0, \pi$x=0,π.

A1 — both solutions from the first factor.

From $2\cos x + 1 = 0$2cosx+1=0, so $\cos x = -\tfrac{1}{2}$cosx=−21​.

M1 — rearranging to $\cos x = -1/2$cosx=−1/2 correctly.

Step 4 — find all solutions in the interval.

$\cos x = -\tfrac{1}{2}$cosx=−21​ has principal value $x = 2\pi/3$x=2π/3 (since $\cos(\pi/3) = 1/2$cos(π/3)=1/2 and cosine is negative in quadrants II and III). The second solution in $[0, 2\pi)$[0,2π) is $x = 2\pi - 2\pi/3 = 4\pi/3$x=2π−2π/3=4π/3.

A1 — $x = 2\pi/3$x=2π/3.

A1 — $x = 4\pi/3$x=4π/3.

Step 5 — present full solution set.

$x = 0, \quad \tfrac{2\pi}{3}, \quad \pi, \quad \tfrac{4\pi}{3}$x=0,32π​,π,34π​

A1 — complete solution set with no extraneous values and no missing values.

Total: 8 marks (M3 A5). A* candidates also state the working interval explicitly and verify by substitution at least once.

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks):

(a) Show that the equation $2\cos^2\theta - 3\sin\theta = 0$2cos2θ−3sinθ=0 can be written as $2\sin^2\theta + 3\sin\theta - 2 = 0$2sin2θ+3sinθ−2=0. (2)

(b) Hence solve $2\cos^2\theta - 3\sin\theta = 0$2cos2θ−3sinθ=0 for $0 \leq \theta \leq \pi$0≤θ≤π, giving exact answers. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — using $\cos^2\theta = 1 - \sin^2\theta$cos2θ=1−sin2θ to replace $\cos^2\theta$cos2θ.
• A1 (AO2.1) — rearranging $2(1 - \sin^2\theta) - 3\sin\theta = 0$2(1−sin2θ)−3sinθ=0 to $2\sin^2\theta + 3\sin\theta - 2 = 0$2sin2θ+3sinθ−2=0 with sign manipulation shown.

(b)

• M1 (AO1.1b) — factorising $(2\sin\theta - 1)(\sin\theta + 2) = 0$(2sinθ−1)(sinθ+2)=0.
• A1 (AO1.1b) — $\sin\theta = 1/2$sinθ=1/2 accepted; $\sin\theta = -2$sinθ=−2 rejected with reason "since $|\sin\theta| \leq 1$∣sinθ∣≤1".
• A1 (AO1.1b) — $\theta = \pi/6$θ=π/6.
• A1 (AO2.5) — $\theta = 5\pi/6$θ=5π/6 (second solution in $[0, \pi]$[0,π] from $\theta = \pi - \pi/6$θ=π−π/6).

Total: 6 marks split AO1 = 4, AO2 = 2. Notice the AO2 marks reward (i) algebraic manipulation that produces the printed form, and (ii) recognising the second branch of the sine curve in the given interval.

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Synoptic links

1. Section H — Differentiation: $\tfrac{d}{dx}\sin x = \cos x$dxd​sinx=cosx and $\tfrac{d}{dx}\cos x = -\sin x$dxd​cosx=−sinx hold only when $x$x is in radians. Every chain-rule problem involving trigonometric inner functions assumes radian measure silently — answer in degrees and the calculus is wrong by a factor of $\pi/180$π/180.

2. Section I — Integration: integrals such as $\int \sin^2 x\,dx$∫sin2xdx require the double-angle identity $\sin^2 x \equiv \tfrac{1}{2}(1 - \cos 2x)$sin2x≡21​(1−cos2x) to convert into integrable form. Without identity fluency, the integration is impossible.

3. Pure 2 — Harmonic form ($R\sin(\theta + \alpha)$Rsin(θ+α)): writing $3\sin\theta + 4\cos\theta$3sinθ+4cosθ as $5\sin(\theta + \alpha)$5sin(θ+α) uses $R = \sqrt{a^2 + b^2}$R=a2+b2​ and $\tan\alpha = b/a$tanα=b/a — a direct application of compound-angle identities in reverse.

4. Paper 3 Mechanics — Simple harmonic motion: $x(t) = A\cos(\omega t + \phi)$x(t)=Acos(ωt+ϕ) models oscillating systems. The angular frequency $\omega$ω has units rad/s, period $T = 2\pi/\omega$T=2π/ω — radian measure is non-negotiable.