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.

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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.

Arc length = 6 × π/3 = 2π cm
Area = ½ × 36 × π/3 = 6π cm²

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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°.

Area = ½ × 8 × 11 × sin 30° = ½ × 8 × 11 × 1/2 = 22 cm²

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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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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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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.

5. Section J — Numerical methods: Newton-Raphson on $f(x) = \cos x - x$f(x)=cosx−x converges to the Dottie number $x \approx 0.739$x≈0.739 rad, the unique fixed point of cosine. The iteration only makes sense in radians.

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Mark-scheme literacy

AQA trigonometry questions distribute AO marks roughly as follows: