Double Angle Formulae

This lesson covers the double angle formulae for sin, cos, and tan — the expressions for sin 2A, cos 2A (in three forms), and tan 2A. These are derived directly from the addition formulae and are used extensively in solving equations, proving identities, and integration.

Derivation from Addition Formulae

The double angle formulae are simply the addition formulae with B = A.

sin 2A

sin(A + A) = sin A cos A + cos A sin A = 2sin A cos A

sin 2A ≡ 2sin A cos A

cos 2A

cos(A + A) = cos A cos A − sin A sin A = cos²A − sin²A

cos 2A ≡ cos²A − sin²A — (Form 1)

Using sin²A + cos²A ≡ 1:

Replace sin²A with 1 − cos²A:

cos 2A = cos²A − (1 − cos²A) = 2cos²A − 1

cos 2A ≡ 2cos²A − 1 — (Form 2)

Replace cos²A with 1 − sin²A:

cos 2A = (1 − sin²A) − sin²A = 1 − 2sin²A

cos 2A ≡ 1 − 2sin²A — (Form 3)

All three forms of cos 2A are equivalent and you should be fluent with all of them. The choice of which form to use depends on the context.

tan 2A

tan(A + A) = (tan A + tan A) / (1 − tan A tan A) = 2tan A / (1 − tan²A)

tan 2A ≡ 2tan A / (1 − tan²A)

When to Use Each Form of cos 2A

Form	Use when...
cos²A − sin²A	Both sin A and cos A appear
2cos²A − 1	You want an expression in cos only
1 − 2sin²A	You want an expression in sin only

Rearranged Forms (Very Useful for Integration)

From cos 2A = 2cos²A − 1:

cos²A = (1 + cos 2A)/2

From cos 2A = 1 − 2sin²A:

sin²A = (1 − cos 2A)/2

These rearranged forms are essential for integrating sin²x and cos²x, since you cannot integrate these directly.

Finding Exact Values

Example 1: Given that cos θ = 3/5 and θ is acute, find the exact values of sin 2θ, cos 2θ, and tan 2θ.

First find sin θ. Since θ is acute:

sin²θ = 1 − cos²θ = 1 − 9/25 = 16/25
sin θ = 4/5

sin 2θ:

sin 2θ = 2sin θ cos θ = 2 × (4/5) × (3/5) = 24/25

cos 2θ:

cos 2θ = cos²θ − sin²θ = 9/25 − 16/25 = −7/25

tan 2θ:

tan θ = sin θ/cos θ = (4/5)/(3/5) = 4/3
tan 2θ = 2tan θ/(1 − tan²θ) = 2(4/3)/(1 − 16/9) = (8/3)/(−7/9) = (8/3) × (−9/7) = −24/7

Check: tan 2θ = sin 2θ/cos 2θ = (24/25)/(−7/25) = −24/7 ✓

Solving Equations

Example 2: Solve cos 2x + 3sin x = 2 for 0° ≤ x ≤ 360°.

Replace cos 2x with 1 − 2sin²x (Form 3, since we want everything in sin):

1 − 2sin²x + 3sin x = 2
−2sin²x + 3sin x − 1 = 0
2sin²x − 3sin x + 1 = 0
(2sin x − 1)(sin x − 1) = 0

sin x = 1/2: x = 30° or x = 150° sin x = 1: x = 90°

Solutions: x = 30°, 90°, 150°

Example 3: Solve sin 2x = cos x for 0 ≤ x ≤ 2π.

2sin x cos x = cos x
2sin x cos x − cos x = 0
cos x(2sin x − 1) = 0

cos x = 0: x = π/2 or x = 3π/2 sin x = 1/2: x = π/6 or x = 5π/6

Solutions: x = π/6, π/2, 5π/6, 3π/2

Important: Do not divide both sides by cos x — you would lose the solutions where cos x = 0. Always factorise instead.

Example 4: Solve cos 2x = cos²x for 0° ≤ x ≤ 360°.

Replace cos 2x with 2cos²x − 1 (Form 2):

2cos²x − 1 = cos²x
cos²x = 1
cos x = ±1

cos x = 1: x = 0° or x = 360° cos x = −1: x = 180°

Solutions: x = 0°, 180°, 360°

Proving Identities

Example 5: Prove that (1 − cos 2A) / sin 2A ≡ tan A.

LHS = (1 − (1 − 2sin²A)) / (2sin A cos A)
    = 2sin²A / (2sin A cos A)
    = sin A / cos A
    = tan A
    = RHS ∎

Example 6: Prove that sin 2A / (1 + cos 2A) ≡ tan A.

LHS = 2sin A cos A / (1 + 2cos²A − 1)
    = 2sin A cos A / (2cos²A)
    = sin A / cos A
    = tan A
    = RHS ∎

Example 7: Show that cos 4θ = 8cos⁴θ − 8cos²θ + 1.

cos 4θ = cos 2(2θ) = 2cos²(2θ) − 1
cos 2θ = 2cos²θ − 1, so cos²(2θ) = (2cos²θ − 1)²

cos 4θ = 2(2cos²θ − 1)² − 1
       = 2(4cos⁴θ − 4cos²θ + 1) − 1
       = 8cos⁴θ − 8cos²θ + 2 − 1
       = 8cos⁴θ − 8cos²θ + 1 ∎

Half-Angle Substitution (Additional Technique)

Sometimes it is useful to express sin, cos, and tan in terms of t = tan(A/2):

sin A = 2t/(1 + t²)
cos A = (1 − t²)/(1 + t²)
tan A = 2t/(1 − t²)

These follow from the double angle formulae with A replaced by A/2.

Summary

sin 2A = 2sin A cos A
cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A
tan 2A = 2tan A / (1 − tan²A)
The three forms of cos 2A are all equivalent — choose the form that suits the problem.
cos²A = (1 + cos 2A)/2 and sin²A = (1 − cos 2A)/2 — essential for integration.
When solving equations, do not divide by a trig function — always factorise.
Use double angle formulae to prove identities by replacing 2A terms with single-angle expressions.

Exam Tip: The double angle formulae are in the AQA formula booklet. The most common mistake is choosing the wrong form of cos 2A. Ask yourself: "Do I want everything in terms of sin, cos, or both?" Also remember: when solving equations, factorise rather than dividing — dividing by sin x or cos x will lose solutions where that function is zero.

A-Level Deep Dive: Double Angle Formulae

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section E (Trigonometry), Year 2 content covers proofs involving trigonometric functions and identities; understand and use double angle formulae; use of formulae for $\sin(A \pm B)$ , $\cos(A \pm B)$ and $\tan(A \pm B)$ ; understand geometrical proofs of these formulae (refer to the official specification document for exact wording). The double angle results — $\sin 2A = 2 \sin A \cos A$ , the three forms of $\cos 2A$ , and $\tan 2A = \dfrac{2 \tan A}{1 - \tan^2 A}$ — sit in the AQA formula booklet, so candidates are not expected to memorise them, but they are expected to choose the appropriate form rapidly under exam pressure. Double-angle work is examined directly in 7357/2 and underpins integration questions in 7357/2 (power reduction), harmonic-form questions, and SHM modelling in 7357/3 Mechanics.

Worked example with full mark scheme

Question (8 marks): Solve $2 \sin 2x = \sin x$ for $x \in [0, 2\pi]$ , giving exact answers where possible and decimal answers to 3 s.f. otherwise.

Solution with mark scheme:

Step 1 — apply the double angle formula.

$2 \sin 2x = \sin x \implies 2(2 \sin x \cos x) = \sin x \implies 4 \sin x \cos x = \sin x$

M1 — correct substitution $\sin 2x = 2 \sin x \cos x$ from the formula booklet. The two factors of 2 combine to give $4 \sin x \cos x$ .

Step 2 — rearrange to a product form.

$4 \sin x \cos x - \sin x = 0 \implies \sin x (4 \cos x - 1) = 0$

M1 — rearranging to one side and factorising out $\sin x$ . This is the critical move. Many candidates divide both sides by $\sin x$ , which loses every $x$ for which $\sin x = 0$ . Factorising preserves all roots.

A1 — correct factorised form.

Step 3 — solve each factor.

From $\sin x = 0$ : $x = 0$ , $\pi$ , $2\pi$ in the given interval.

From $4 \cos x - 1 = 0$ : $\cos x = \tfrac{1}{4}$ , so $x = \arccos(\tfrac{1}{4}) \approx 1.318$ or $x = 2\pi - 1.318 \approx 4.965$ .

M1 — solving $\sin x = 0$ giving all three boundary/interior values in $[0, 2\pi]$ .

A1 — the three exact values $0$ , $\pi$ , $2\pi$ .

M1 — using $\cos x = \tfrac{1}{4}$ , recognising the symmetry $\cos x = \cos(2\pi - x)$ to find both solutions in the interval.

A1 — $x \approx 1.32$ to 3 s.f.

A1 — $x \approx 4.97$ to 3 s.f., with all five solutions listed: $x = 0,\, 1.32,\, \pi,\, 4.97,\, 2\pi$ .

Total: 8 marks (M4 A4). Note the mark distribution: half the marks are for method (correct identity, factorisation, branch handling), the other half for accuracy (each of the five roots earns or loses an A mark). The question rewards completeness — missing one root costs an A1.

Specimen question modelled on the AQA 7357/2 format

Question (6 marks): Given that $\cos 2\theta = \dfrac{7}{25}$ and $\theta$ is acute:

(a) Find the exact value of $\sin \theta$ . (3)

(b) Hence find the exact value of $\tan 2\theta$ . (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — selecting the form $\cos 2\theta = 1 - 2 \sin^2 \theta$ from the three available. This form is chosen because it isolates $\sin \theta$ directly.
M1 (AO1.1b) — solving $\tfrac{7}{25} = 1 - 2 \sin^2 \theta$ giving $\sin^2 \theta = \tfrac{9}{25}$ .
A1 (AO1.1b) — $\sin \theta = \tfrac{3}{5}$ (positive root because $\theta$ is acute).

(b)

M1 (AO2.1) — using $\sin 2\theta = 2 \sin \theta \cos \theta$ with the Pythagorean identity to find $\cos \theta = \tfrac{4}{5}$ , so $\sin 2\theta = 2 \cdot \tfrac{3}{5} \cdot \tfrac{4}{5} = \tfrac{24}{25}$ .
M1 (AO1.1b) — forming $\tan 2\theta = \dfrac{\sin 2\theta}{\cos 2\theta}$ .
A1 (AO1.1b) — $\tan 2\theta = \dfrac{24/25}{7/25} = \dfrac{24}{7}$ .

Total: 6 marks split AO1 = 5, AO2 = 1. The AO2 mark is awarded for the synoptic step linking double-angle and Pythagorean identities — exactly the kind of "joined-up" reasoning AQA rewards in Year 2 trigonometry.

Synoptic links

Connects to:

Compound angle formulae (Section E): the double angle results are special cases of the compound angle formulae with $A = B$ . Setting $B = A$ in $\sin(A + B) = \sin A \cos B + \cos A \sin B$ instantly yields $\sin 2A = 2 \sin A \cos A$ . This derivation is examinable as a "show that" question, and seeing the connection unlocks proofs of higher multiple-angle formulae like $\sin 3x$ .
Integration via power reduction (Section H): integrals such as $\int \sin^2 x \, dx$ are intractable in their direct form, but the rearrangement $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ — derived directly from $\cos 2x = 1 - 2 \sin^2 x$ — converts them into routine trigonometric integrals: $\int \sin^2 x \, dx = \tfrac{x}{2} - \tfrac{\sin 2x}{4} + C$ . Without comfortable manipulation of double-angle identities, every $\sin^2$ , $\cos^2$ , or $\sin x \cos x$ integrand stalls.
Harmonic form $R \sin(x + \alpha)$ (Section E): when expressing $a \sin x + b \cos x$ as a single sinusoid, candidates often encounter expressions of the form $\sin x \cos x$ inside a transformation, which collapses via $\sin 2x = 2 \sin x \cos x$ . Recognising this hidden double-angle structure halves the algebra.
Simple harmonic motion (7357/3 Mechanics, Section P): SHM displacement $x = a \cos(\omega t)$ gives velocity $\dot x = -a\omega \sin(\omega t)$ , and energy expressions $\tfrac{1}{2} m \dot x^2 \propto \sin^2(\omega t)$ are simplified via the power-reduction identity. The total energy emerging as a constant — independent of $t$ — depends on $\sin^2 + \cos^2 = 1$ combined with the double angle expansion.
Differentiation of trigonometric composites (Section G): derivatives like $\dfrac{d}{dx}(\sin^2 x) = 2 \sin x \cos x = \sin 2x$ make the chain rule visibly equivalent to a double-angle restatement. This connection explains why the derivative of $\sin^2 x$ has period $\pi$ rather than $2\pi$ : the doubling of frequency is built into the algebra.

Mark-scheme literacy

Double-angle questions on 7357/2 split AO marks across all three objectives more evenly than algebraic topics:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Selecting the correct double-angle form, solving the resulting equation, listing all roots in the stated interval
AO2 (reasoning / interpretation)	25–35%	Choosing which form of $\cos 2A$ minimises algebra; combining double-angle with Pythagorean identities; recognising hidden double-angle structures
AO3 (problem-solving)	10–20%	Multi-step questions linking trigonometry to integration, mechanics, or proof; "Hence solve" sequences where part (a) sets up the identity used in part (b)

Examiner-rewarded phrasing: "using the form $\cos 2\theta = 1 - 2\sin^2 \theta$ because the equation involves $\sin\theta$ only"; "since $2x \in [0, 4\pi]$ , we obtain four solutions for $2x$ , halved to give the original $x$ values"; "rejecting $\cos\theta = -1/4$ because $\theta$ is acute". Phrases that lose marks: silently dividing by $\sin x$ (loses entire branch); writing " $\cos 2x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$ " without specifying which form is being used at each stage; stating "two solutions" when the doubled interval contains four.

A specific AQA pattern to watch: when the equation involves $\sin 2x$ for $x \in [0, 2\pi]$ , the substituted variable $2x$ ranges over $[0, 4\pi]$ — a double period. Candidates routinely miss the second-period roots and lose 2 A marks. Always sketch the new interval before solving.

Grade-band model answers

3-mark question

Question: Given $\sin\theta = \tfrac{5}{13}$ where $\theta$ is acute, find the exact value of $\sin 2\theta$ .

Grade C response (~150 words):

$\sin 2\theta = 2 \sin\theta \cos\theta$ . Need $\cos\theta$ .

Double Angle Formulae

Double Angle Formulae

Derivation from Addition Formulae

sin 2A

cos 2A

tan 2A

When to Use Each Form of cos 2A

Rearranged Forms (Very Useful for Integration)

Finding Exact Values

Solving Equations

Proving Identities

Half-Angle Substitution (Additional Technique)

Summary

A-Level Deep Dive: Double Angle Formulae

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357/2 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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