Double Angle Formulae

Builds on: aqa-alevel-maths-pure-2 / further-trigonometry — this lesson is the A2-level deep dive.

This lesson covers the double angle formulae for $\sin$sin, $\cos$cos, and $\tan$tan — the expressions for $\sin 2A$sin2A, $\cos 2A$cos2A (in three forms), and $\tan 2A$tan2A. These are derived directly from the addition formulae and are used extensively in solving equations, proving identities, and integration.

Spec Mapping — AQA 7357 Section E Trigonometry. This lesson covers the double-angle identities $\sin 2A$sin2A, $\cos 2A$cos2A and $\tan 2A$tan2A content of Section E, at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

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Derivation from Addition Formulae

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

$\sin 2A$sin2A

$$\sin(A + A) = \sin A \cos A + \cos A \sin A = 2\sin A \cos A$$sin(A+A)=sinAcosA+cosAsinA=2sinAcosA

$\sin 2A \equiv 2\sin A \cos A$sin2A≡2sinAcosA

$\cos 2A$cos2A

$$\cos(A + A) = \cos A \cos A - \sin A \sin A = \cos^{2}A - \sin^{2}A$$cos(A+A)=cosAcosA−sinAsinA=cos2A−sin2A

$\cos 2A \equiv \cos^2 A - \sin^2 A$cos2A≡cos2A−sin2A — (Form 1)

Using $\sin^2 A + \cos^2 A \equiv 1$sin2A+cos2A≡1:

Replace $\sin^2 A$sin2A with $1 - \cos^2 A$1−cos2A:

$$\cos 2A = \cos^{2}A - (1 - \cos^{2}A) = 2\cos^{2}A - 1$$cos2A=cos2A−(1−cos2A)=2cos2A−1

$\cos 2A \equiv 2\cos^2 A - 1$cos2A≡2cos2A−1 — (Form 2)

Replace $\cos^2 A$cos2A with $1 - \sin^2 A$1−sin2A:

$$\cos 2A = (1 - \sin^{2}A) - \sin^{2}A = 1 - 2\sin^{2}A$$cos2A=(1−sin2A)−sin2A=1−2sin2A

$\cos 2A \equiv 1 - 2\sin^2 A$cos2A≡1−2sin2A — (Form 3)

All three forms of $\cos 2A$cos2A are equivalent and you should be fluent with all of them. The choice of which form to use depends on the context.

$\tan 2A$tan2A

$$\tan(A + A) = \frac{\tan A + \tan A}{1 - \tan A \tan A} = \frac{2\tan A}{1 - \tan^{2}A}$$tan(A+A)=1−tanAtanAtanA+tanA​=1−tan2A2tanA​

$\tan 2A \equiv \dfrac{2\tan A}{1 - \tan^2 A}$tan2A≡1−tan2A2tanA​

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When to Use Each Form of cos 2A

Form	Use when...
$\cos^2 A - \sin^2 A$cos2A−sin2A	Both $\sin A$sinA and $\cos A$cosA appear
$2\cos^2 A - 1$2cos2A−1	You want an expression in $\cos$cos only
$1 - 2\sin^2 A$1−2sin2A	You want an expression in $\sin$sin only

Rearranged Forms (Very Useful for Integration)

From $\cos 2A = 2\cos^2 A - 1$cos2A=2cos2A−1:

$$\cos^{2}A = \frac{1 + \cos 2A}{2}$$cos2A=21+cos2A​

From $\cos 2A = 1 - 2\sin^2 A$cos2A=1−2sin2A:

$$\sin^{2}A = (1 - \cos 2A)/2$$sin2A=(1−cos2A)/2

These rearranged forms are essential for integrating $\sin^2 x$sin2x and $\cos^2 x$cos2x, since you cannot integrate these directly.

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Finding Exact Values

Example 1: Given that $\cos\theta = \tfrac{3}{5}$cosθ=53​ and $\theta$θ is acute, find the exact values of $\sin 2\theta$sin2θ, $\cos 2\theta$cos2θ, and $\tan 2\theta$tan2θ.

First find $\sin\theta$sinθ. Since $\theta$θ is acute:

$$\begin{aligned} \sin^{2} \theta = 1 - \cos^{2} \theta = 1 - 9/25 = 16/25 \\ \sin \theta = 4/5 \end{aligned}$$sin2θ=1−cos2θ=1−9/25=16/25sinθ=4/5​

$\sin 2\theta$sin2θ:

$$\sin 2 \theta = 2\sin \theta \cos \theta = 2 \times (4/5) \times (3/5) = 24/25$$sin2θ=2sinθcosθ=2×(4/5)×(3/5)=24/25

$\cos 2\theta$cos2θ:

$$\cos 2 \theta = \cos^{2} \theta - \sin^{2} \theta = 9/25 - 16/25 = -7/25$$cos2θ=cos2θ−sin2θ=9/25−16/25=−7/25

$\tan 2\theta$tan2θ:

$$\begin{aligned} \tan \theta = \sin \theta /\cos \theta = (4/5)/(3/5) = 4/3 \\ \tan 2 \theta = 2\tan \theta /(1 - \tan^{2} \theta ) = 2(4/3)/(1 - 16/9) = (8/3)/(-7/9) = (8/3) \times (-9/7) = -24/7 \end{aligned}$$tanθ=sinθ/cosθ=(4/5)/(3/5)=4/3tan2θ=2tanθ/(1−tan2θ)=2(4/3)/(1−16/9)=(8/3)/(−7/9)=(8/3)×(−9/7)=−24/7​

Check: $\tan 2\theta = \sin 2\theta / \cos 2\theta = (24/25)/(-7/25) = -24/7$tan2θ=sin2θ/cos2θ=(24/25)/(−7/25)=−24/7 ✓

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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.

Solving Equations

Example 2: Solve $\cos 2x + 3\sin x = 2$cos2x+3sinx=2 for $0° \leq x \leq 360°$0°≤x≤360°.

Replace $\cos 2x$cos2x with $1 - 2\sin^2 x$1−2sin2x (Form 3, since we want everything in $\sin$sin):

$$\begin{aligned} 1 - 2\sin^{2}x + 3\sin x &= 2 \\ -2\sin^{2}x + 3\sin x - 1 &= 0 \\ 2\sin^{2}x - 3\sin x + 1 &= 0 \\ (2\sin x - 1)(\sin x - 1) &= 0 \end{aligned}$$1−2sin2x+3sinx−2sin2x+3sinx−12sin2x−3sinx+1(2sinx−1)(sinx−1)​=2=0=0=0​

$\sin x = \tfrac{1}{2}$sinx=21​: $x = 30°$x=30° or $x = 150°$x=150° $\sin x = 1$sinx=1: $x = 90°$x=90°

Solutions: $x = 30°, 90°, 150°$x=30°,90°,150°

Example 3: Solve $\sin 2x = \cos x$sin2x=cosx for $0 \leq x \leq 2\pi$0≤x≤2π.

$$\begin{aligned} 2\sin x \cos x = \cos x \\ 2\sin x \cos x - \cos x = 0 \\ \cos x(2\sin x - 1) = 0 \end{aligned}$$2sinxcosx=cosx2sinxcosx−cosx=0cosx(2sinx−1)=0​

$\cos x = 0$cosx=0: $x = \tfrac{\pi}{2}$x=2π​ or $x = \tfrac{3\pi}{2}$x=23π​ $\sin x = \tfrac{1}{2}$sinx=21​: $x = \tfrac{\pi}{6}$x=6π​ or $x = \tfrac{5\pi}{6}$x=65π​

Solutions: $x = \tfrac{\pi}{6}, \tfrac{\pi}{2}, \tfrac{5\pi}{6}, \tfrac{3\pi}{2}$x=6π​,2π​,65π​,23π​

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

Example 4: Solve $\cos 2x = \cos^2 x$cos2x=cos2x for $0° \leq x \leq 360°$0°≤x≤360°.

Replace $\cos 2x$cos2x with $2\cos^2 x - 1$2cos2x−1 (Form 2):

$$\begin{aligned} 2\cos^{2}x - 1 = \cos^{2}x \\ \cos^{2}x = 1 \\ \cos x = \pm 1 \end{aligned}$$2cos2x−1=cos2xcos2x=1cosx=±1​

$\cos x = 1$cosx=1: $x = 0°$x=0° or $x = 360°$x=360° $\cos x = -1$cosx=−1: $x = 180°$x=180°

Solutions: $x = 0°, 180°, 360°$x=0°,180°,360°

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Proving Identities

Example 5: Prove that $\dfrac{1 - \cos 2A}{\sin 2A} \equiv \tan A$sin2A1−cos2A​≡tanA.

$$\begin{aligned} \text{LHS} &= \frac{1 - (1 - 2\sin^{2}A)}{2\sin A \cos A} \\ &= \frac{2\sin^{2}A}{2\sin A \cos A} \\ &= \frac{\sin A}{\cos A} \\ &= \tan A \\ &= \text{RHS} \quad \blacksquare \end{aligned}$$LHS​=2sinAcosA1−(1−2sin2A)​=2sinAcosA2sin2A​=cosAsinA​=tanA=RHS■​

Example 6: Prove that $\dfrac{\sin 2A}{1 + \cos 2A} \equiv \tan A$1+cos2Asin2A​≡tanA.

$$\begin{aligned} \text{LHS} &= \frac{2\sin A \cos A}{1 + 2\cos^{2}A - 1} \\ &= \frac{2\sin A \cos A}{2\cos^{2}A} \\ &= \frac{\sin A}{\cos A} \\ &= \tan A \\ &= \text{RHS} \qquad\blacksquare \end{aligned}$$LHS​=1+2cos2A−12sinAcosA​=2cos2A2sinAcosA​=cosAsinA​=tanA=RHS■​

Example 7: Show that $\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$cos4θ=8cos4θ−8cos2θ+1.

$$\begin{aligned} \cos 4\theta &= \cos 2(2\theta) = 2\cos^{2}(2\theta) - 1 \\ \cos 2\theta &= 2\cos^{2}\theta - 1, \ \text{so} \ \cos^{2}(2\theta) = (2\cos^{2}\theta - 1)^{2} \\[4pt] \cos 4\theta &= 2(2\cos^{2}\theta - 1)^{2} - 1 \\ &= 2(4\cos^{4}\theta - 4\cos^{2}\theta + 1) - 1 \\ &= 8\cos^{4}\theta - 8\cos^{2}\theta + 2 - 1 \\ &= 8\cos^{4}\theta - 8\cos^{2}\theta + 1 \qquad\blacksquare \end{aligned}$$cos4θcos2θcos4θ​=cos2(2θ)=2cos2(2θ)−1=2cos2θ−1, so cos2(2θ)=(2cos2θ−1)2=2(2cos2θ−1)2−1=2(4cos4θ−4cos2θ+1)−1=8cos4θ−8cos2θ+2−1=8cos4θ−8cos2θ+1■​

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Half-Angle Substitution (Additional Technique)

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

$$\begin{aligned} \sin A &= \frac{2t}{1 + t^{2}} \\ \cos A &= \frac{1 - t^{2}}{1 + t^{2}} \\ \tan A &= \frac{2t}{1 - t^{2}} \end{aligned}$$sinAcosAtanA​=1+t22t​=1+t21−t2​=1−t22t​​

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

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Synoptic Links

This topic connects to:

• Addition formulae (aqa-alevel-maths-trigonometry-depth / addition-formulae) — these are the source from which all double-angle results derive.
• Integration of $\sin^2 x$sin2x and $\cos^2 x$cos2x (aqa-alevel-maths-calculus-applications / integration-trigonometric) — $\sin^2 x = (1 - \cos 2x)/2$sin2x=(1−cos2x)/2 turns a non-elementary integral into a standard one.
• $R\cos(\theta - \alpha)$Rcos(θ−α) form (aqa-alevel-maths-trigonometry-depth / rcos-alpha-form) — the harmonic form sometimes requires double-angle identities to extract amplitude or phase.

Summary

• $\sin 2A = 2\sin A \cos A$sin2A=2sinAcosA
• $\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$cos2A=cos2A−sin2A=2cos2A−1=1−2sin2A
• $\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$tan2A=1−tan2A2tanA​
• The three forms of $\cos 2A$cos2A are all equivalent — choose the form that suits the problem.
• $\cos^2 A = (1 + \cos 2A)/2$cos2A=(1+cos2A)/2 and $\sin^2 A = (1 - \cos 2A)/2$sin2A=(1−cos2A)/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$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 $\cos 2A$cos2A. Ask yourself: "Do I want everything in terms of $\sin$sin, $\cos$cos, or both?" Also remember: when solving equations, factorise rather than dividing — dividing by $\sin x$sinx or $\cos x$cosx 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 $\sin(A \pm B)$sin(A±B), $\cos(A \pm B)$cos(A±B) and $\tan(A \pm B)$tan(A±B); understand geometrical proofs of these formulae (refer to the official specification document for exact wording). The double angle results — $\sin 2A = 2 \sin A \cos A$sin2A=2sinAcosA, the three forms of $\cos 2A$cos2A, and $\tan 2A = \dfrac{2 \tan A}{1 - \tan^2 A}$tan2A=1−tan2A2tanA​ — 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.

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Worked example with full mark scheme

Question (8 marks): Solve $2 \sin 2x = \sin x$2sin2x=sinx for $x \in [0, 2\pi]$x∈[0,2π], 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$2sin2x=sinx⟹2(2sinxcosx)=sinx⟹4sinxcosx=sinx

M1 — correct substitution $\sin 2x = 2 \sin x \cos x$sin2x=2sinxcosx from the formula booklet. The two factors of 2 combine to give $4 \sin x \cos x$4sinxcosx.

Step 2 — rearrange to a product form.

$4 \sin x \cos x - \sin x = 0 \implies \sin x (4 \cos x - 1) = 0$4sinxcosx−sinx=0⟹sinx(4cosx−1)=0

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

A1 — correct factorised form.

Step 3 — solve each factor.

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

From $4 \cos x - 1 = 0$4cosx−1=0: $\cos x = \tfrac{1}{4}$cosx=41​, so $x = \arccos(\tfrac{1}{4}) \approx 1.318$x=arccos(41​)≈1.318 or $x = 2\pi - 1.318 \approx 4.965$x=2π−1.318≈4.965.

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

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

M1 — using $\cos x = \tfrac{1}{4}$cosx=41​, recognising the symmetry $\cos x = \cos(2\pi - x)$cosx=cos(2π−x) to find both solutions in the interval.

A1 — $x \approx 1.32$x≈1.32 to 3 s.f.

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

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.

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Specimen question modelled on the AQA 7357/2 format

Question (6 marks): Given that $\cos 2\theta = \dfrac{7}{25}$cos2θ=257​ and $\theta$θ is acute:

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

(b) Hence find the exact value of $\tan 2\theta$tan2θ. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — selecting the form $\cos 2\theta = 1 - 2 \sin^2 \theta$cos2θ=1−2sin2θ from the three available. This form is chosen because it isolates $\sin \theta$sinθ directly.
• M1 (AO1.1b) — solving $\tfrac{7}{25} = 1 - 2 \sin^2 \theta$257​=1−2sin2θ giving $\sin^2 \theta = \tfrac{9}{25}$sin2θ=259​.
• A1 (AO1.1b) — $\sin \theta = \tfrac{3}{5}$sinθ=53​ (positive root because $\theta$θ is acute).

(b)

• M1 (AO2.1) — using $\sin 2\theta = 2 \sin \theta \cos \theta$sin2θ=2sinθcosθ with the Pythagorean identity to find $\cos \theta = \tfrac{4}{5}$cosθ=54​, so $\sin 2\theta = 2 \cdot \tfrac{3}{5} \cdot \tfrac{4}{5} = \tfrac{24}{25}$sin2θ=2⋅53​⋅54​=2524​.
• M1 (AO1.1b) — forming $\tan 2\theta = \dfrac{\sin 2\theta}{\cos 2\theta}$tan2θ=cos2θsin2θ​.
• A1 (AO1.1b) — $\tan 2\theta = \dfrac{24/25}{7/25} = \dfrac{24}{7}$tan2θ=7/2524/25​=724​.

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.

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Synoptic links

Connects to:

1. Compound angle formulae (Section E): the double angle results are special cases of the compound angle formulae with $A = B$A=B. Setting $B = A$B=A in $\sin(A + B) = \sin A \cos B + \cos A \sin B$sin(A+B)=sinAcosB+cosAsinB instantly yields $\sin 2A = 2 \sin A \cos A$sin2A=2sinAcosA. This derivation is examinable as a "show that" question, and seeing the connection unlocks proofs of higher multiple-angle formulae like $\sin 3x$sin3x.

2. Integration via power reduction (Section H): integrals such as $\int \sin^2 x \, dx$∫sin2xdx are intractable in their direct form, but the rearrangement $\sin^2 x = \dfrac{1 - \cos 2x}{2}$sin2x=21−cos2x​ — derived directly from $\cos 2x = 1 - 2 \sin^2 x$cos2x=1−2sin2x — converts them into routine trigonometric integrals: $\int \sin^2 x \, dx = \tfrac{x}{2} - \tfrac{\sin 2x}{4} + C$∫sin2xdx=2x​−4sin2x​+C. Without comfortable manipulation of double-angle identities, every $\sin^2$sin2, $\cos^2$cos2, or $\sin x \cos x$sinxcosx integrand stalls.

3. Harmonic form $R \sin(x + \alpha)$Rsin(x+α) (Section E): when expressing $a \sin x + b \cos x$asinx+bcosx as a single sinusoid, candidates often encounter expressions of the form $\sin x \cos x$sinxcosx inside a transformation, which collapses via $\sin 2x = 2 \sin x \cos x$sin2x=2sinxcosx. Recognising this hidden double-angle structure halves the algebra.

4. Simple harmonic motion (7357/3 Mechanics, Section P): SHM displacement $x = a \cos(\omega t)$x=acos(ωt) gives velocity $\dot x = -a\omega \sin(\omega t)$x˙=−aωsin(ωt), and energy expressions $\tfrac{1}{2} m \dot x^2 \propto \sin^2(\omega t)$21​mx˙2∝sin2(ωt) are simplified via the power-reduction identity. The total energy emerging as a constant — independent of $t$t — depends on $\sin^2 + \cos^2 = 1$sin2+cos2=1 combined with the double angle expansion.

5. Differentiation of trigonometric composites (Section G): derivatives like $\dfrac{d}{dx}(\sin^2 x) = 2 \sin x \cos x = \sin 2x$dxd​(sin2x)=2sinxcosx=sin2x make the chain rule visibly equivalent to a double-angle restatement. This connection explains why the derivative of $\sin^2 x$sin2x has period $\pi$π rather than $2\pi$2π: the doubling of frequency is built into the algebra.

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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 $\cos 2A$cos2A 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)