Double Angle Formulae

The double angle formulae are special cases of the compound angle formulae with B = A. They are among the most frequently used identities at A-Level (9MA0) and appear in integration, equation solving, and proof questions.

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The Double Angle Formulae

sin(2A)

sin(2A) = 2sinA cosA

cos(2A) — Three Forms

cos(2A) = cos²A - sin²A

cos(2A) = 2cos²A - 1

cos(2A) = 1 - 2sin²A

All three are equivalent (use sin²A + cos²A = 1 to convert between them).

tan(2A)

tan(2A) = 2tanA / (1 - tan²A)

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Deriving from Compound Angle Formulae

sin(2A) = sin(A + A) = sinA cosA + cosA sinA = 2sinA cosA

cos(2A) = cos(A + A) = cosA cosA - sinA sinA = cos²A - sin²A

Using sin²A = 1 - cos²A: cos(2A) = cos²A - (1 - cos²A) = 2cos²A - 1

Using cos²A = 1 - sin²A: cos(2A) = (1 - sin²A) - sin²A = 1 - 2sin²A

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Useful Rearrangements

From cos(2A) = 2cos²A - 1:

cos²A = (1 + cos(2A))/2

From cos(2A) = 1 - 2sin²A:

sin²A = (1 - cos(2A))/2

These "half-angle" forms are essential for integration of sin²x and cos²x.

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Solving Equations

Worked Example 1

Solve cos(2x) + sin(x) = 0 for 0 <= x <= 2pi.

Use cos(2x) = 1 - 2sin²x:

1 - 2sin²x + sinx = 0

2sin²x - sinx - 1 = 0

Let s = sinx:

2s² - s - 1 = 0

(2s + 1)(s - 1) = 0

s = -1/2 or s = 1

sinx = -1/2: x = 7pi/6, 11pi/6

sinx = 1: x = pi/2

Solutions: x = pi/2, 7pi/6, 11pi/6

Worked Example 2

Solve sin(2x) = sinx for 0 <= x < 2pi.

2sinx cosx = sinx

2sinx cosx - sinx = 0

sinx(2cosx - 1) = 0

sinx = 0: x = 0, pi

cosx = 1/2: x = pi/3, 5pi/3

Solutions: x = 0, pi/3, pi, 5pi/3

Worked Example 3

Solve 3cos(2x) = 1 - cos(x) for 0 <= x <= 2pi.

Use cos(2x) = 2cos²x - 1:

3(2cos²x - 1) = 1 - cosx

6cos²x - 3 = 1 - cosx

6cos²x + cosx - 4 = 0

Using the quadratic formula with c = cosx:

c = (-1 ± sqrt(1 + 96)) / 12 = (-1 ± sqrt(97)) / 12

sqrt(97) ≈ 9.849

c = (-1 + 9.849)/12 ≈ 0.7374 or c = (-1 - 9.849)/12 ≈ -0.9041

cos(x) = 0.7374: x ≈ 0.739, 5.544 cos(x) = -0.9041: x ≈ 2.713, 3.570

Solutions: x ≈ 0.739, 2.713, 3.570, 5.544 (to 3 d.p.)

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Proving Identities

Worked Example 4

Prove that (sinA + cosA)² = 1 + sin(2A).

LHS = sin²A + 2sinA cosA + cos²A

= (sin²A + cos²A) + 2sinA cosA

= 1 + sin(2A)

= RHS

Worked Example 5

Prove that cos(4x) = 8cos^4(x) - 8cos²(x) + 1.

cos(4x) = 2cos²(2x) - 1

= 2(2cos²x - 1)² - 1

= 2(4cos^4(x) - 4cos²(x) + 1) - 1

= 8cos^4(x) - 8cos²(x) + 2 - 1

= 8cos^4(x) - 8cos²(x) + 1 = RHS

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Exam Tips

Tip	Detail
Choose the right form	Pick the form of cos(2A) that matches the other terms in the equation
Quadratic equations	Using double angle identities often creates quadratics in sinx or cosx
Factor, do not cancel	sinx(2cosx - 1) = 0 gives sinx = 0 OR cosx = 1/2; do not divide by sinx
Integration link	cos²x = (1 + cos2x)/2 and sin²x = (1 - cos2x)/2 are essential for integration
Half-angle context	If an equation involves both theta and 2theta, use double angle formulae

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Practice Problems

1. Solve sin(2x) = cos(x) for 0 <= x < 2pi. [Answer: pi/6, pi/2, 5pi/6, 3pi/2]
2. Prove that 2cos²(3x) - 1 = cos(6x).
3. Given sinA = 3/5 (A acute), find sin(2A) and cos(2A). [Answer: sin(2A) = 24/25, cos(2A) = 7/25]
4. Solve cos(2x) = cos²x for 0 <= x <= 2pi. [Answer: x = 0, pi, 2pi]
5. Show that 4sin(x)cos(x)cos(2x) = sin(4x).

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Summary

• sin(2A) = 2sinA cosA
• cos(2A) has three forms: cos²A - sin²A, 2cos²A - 1, 1 - 2sin²A
• tan(2A) = 2tanA / (1 - tan²A)
• Choose the form of cos(2A) that eliminates the trig function not present in the equation.
• Factor equations — never divide by a trig function that could be zero.

A-Level Deep Dive: Double Angle Formulae

Spec mapping

Edexcel 9MA0 specification section 5 — Trigonometry, sub-strand on double-angle formulae covers double angle formulae; use of formulae for $\sin(A \pm B)$sin(A±B), $\cos(A \pm B)$cos(A±B), $\tan(A \pm B)$tan(A±B) and the equivalent expressions for $\sin 2A$sin2A, $\cos 2A$cos2A and $\tan 2A$tan2A (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 ($\cos^2 A - \sin^2 A$cos2A−sin2A, $1 - 2\sin^2 A$1−2sin2A, $2\cos^2 A - 1$2cos2A−1), and $\tan 2A = 2\tan A/(1 - \tan^2 A)$tan2A=2tanA/(1−tan2A) — appear in the Edexcel formula booklet alongside the compound-angle formulae, so they do not need to be memorised in isolation, but fluent recall and strategic choice of which form to substitute is essential under exam time pressure. Synoptic reach: trig identities (section 5), integration of $\sin^2 x$sin2x and $\cos^2 x$cos2x via the power-reduction identities $\sin^2 x = (1 - \cos 2x)/2$sin2x=(1−cos2x)/2 and $\cos^2 x = (1 + \cos 2x)/2$cos2x=(1+cos2x)/2 (section 8), the harmonic form derivation $a\sin x + b\cos x = R\sin(x + \alpha)$asinx+bcosx=Rsin(x+α) (section 5), trigonometric equation solving over given intervals, and 9MA0-03 Mechanics — particularly simple harmonic motion, where $x(t) = A\cos(\omega t)$x(t)=Acos(ωt) leads naturally to expressions like $x^2 = A^2\cos^2(\omega t) = A^2(1 + \cos 2\omega t)/2$x2=A2cos2(ωt)=A2(1+cos2ωt)/2 when computing average kinetic energy.

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Worked example with full mark scheme

Question (8 marks): Solve the equation $2\sin 2x = \sin x$2sin2x=sinx exactly for $x \in [0, 2\pi]$x∈[0,2π].

Solution with mark scheme:

Step 1 — substitute the double-angle formula for $\sin 2x$sin2x.

$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. The whole question hinges on this single substitution. Candidates who try to keep $\sin 2x$sin2x unevaluated and "treat it as an unknown" lose this mark and every subsequent one.

Step 2 — rearrange and factorise. Do NOT divide by $\sin x$sinx.

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

M1 — taking everything to one side and factorising out $\sin x$sinx. This is the make-or-break presentation step. Candidates who divide by $\sin x$sinx to obtain $4\cos x = 1$4cosx=1 lose all the solutions where $\sin x = 0$sinx=0 — a guaranteed two-mark deduction at A1 stage.

A1 — correct factorised form $\sin x(4\cos x - 1) = 0$sinx(4cosx−1)=0.

Step 3 — solve each factor separately.

From $\sin x = 0$sinx=0 on $[0, 2\pi]$[0,2π]: $x = 0, \pi, 2\pi$x=0,π,2π.

M1 — using $\sin x = 0$sinx=0 to obtain principal and additional solutions in the closed interval. Note both endpoints are included because the interval is closed.

A1 — three correct solutions $\{0, \pi, 2\pi\}${0,π,2π} from the first factor.

From $4\cos x - 1 = 0$4cosx−1=0, so $\cos x = \tfrac{1}{4}$cosx=41​. The principal value is $x = \arccos(1/4)$x=arccos(1/4). Since $\cos x = \tfrac{1}{4} > 0$cosx=41​>0, solutions lie in the first and fourth quadrants:

$x = \arccos(1/4) \quad \text{and} \quad x = 2\pi - \arccos(1/4)$x=arccos(1/4)andx=2π−arccos(1/4)

M1 — correct use of the symmetry $\cos(2\pi - x) = \cos x$cos(2π−x)=cosx to find the second solution within $[0, 2\pi]$[0,2π].

A1 — both solutions $\arccos(1/4)$arccos(1/4) and $2\pi - \arccos(1/4)$2π−arccos(1/4) stated exactly (the question asks for exact answers, so leaving $\arccos(1/4)$arccos(1/4) unevaluated is correct — converting to a decimal would lose this mark).

Step 4 — collect all solutions.

$x \in \left\{0, \arccos\tfrac{1}{4}, \pi, 2\pi - \arccos\tfrac{1}{4}, 2\pi\right\}$x∈{0,arccos41​,π,2π−arccos41​,2π}

A1 — all five solutions stated, none missing, none extraneous.

Total: 8 marks (M4 A4). The 5 marks for the $\sin x = 0$sinx=0 branch (M1 substitute, M1 factorise, A1 factorised form, M1 solve, A1 list) plus 3 marks for the $\cos x = 1/4$cosx=1/4 branch (M1 symmetry, A1 exact form, A1 final collation).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): (a) Show that the equation $\cos 2\theta + 3\sin\theta = 2$cos2θ+3sinθ=2 can be written as a quadratic in $\sin\theta$sinθ. (2)

(b) Hence solve $\cos 2\theta + 3\sin\theta = 2$cos2θ+3sinθ=2 exactly for $\theta \in [0, 2\pi]$θ∈[0,2π]. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — selecting the form $\cos 2\theta = 1 - 2\sin^2\theta$cos2θ=1−2sin2θ because it eliminates $\cos\theta$cosθ and matches the $\sin\theta$sinθ already present.
• A1 (AO1.1b) — substituting and rearranging to $2\sin^2\theta - 3\sin\theta + 1 = 0$2sin2θ−3sinθ+1=0 in the form requested.

(b)

• M1 (AO1.1b) — factorising the quadratic: $(2\sin\theta - 1)(\sin\theta - 1) = 0$(2sinθ−1)(sinθ−1)=0.
• A1 (AO1.1b) — identifying $\sin\theta = \tfrac{1}{2}$sinθ=21​ or $\sin\theta = 1$sinθ=1.
• M1 (AO2.5) — using sine symmetry to find all values of $\theta$θ in $[0, 2\pi]$[0,2π] for each root: $\sin\theta = \tfrac{1}{2}$sinθ=21​ gives $\theta = \pi/6, 5\pi/6$θ=π/6,5π/6; $\sin\theta = 1$sinθ=1 gives $\theta = \pi/2$θ=π/2 only.
• A1 (AO1.1b) — final list $\theta \in \{\pi/6, \pi/2, 5\pi/6\}$θ∈{π/6,π/2,5π/6}, all in exact form.

Total: 6 marks split AO1 = 5, AO2 = 1. AO2 emerges in the symmetry/principal-value reasoning when only one of the two roots produces a single value (the boundary case $\sin\theta = 1$sinθ=1 has just one solution, $\theta = \pi/2$θ=π/2, in $[0, 2\pi]$[0,2π] — recognising this without inventing a phantom second solution earns the AO2 mark).

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Synoptic links

Connects to:

1. Compound-angle formulae (section 5): the double-angle results are the special case of compound-angle with $A = B$A=B. $\sin(A + B) = \sin A\cos B + \cos A\sin B$sin(A+B)=sinAcosB+cosAsinB collapses to $\sin 2A = 2\sin A\cos A$sin2A=2sinAcosA when $B = A$B=A. Similarly $\cos(A + B) = \cos A\cos B - \sin A\sin B$cos(A+B)=cosAcosB−sinAsinB becomes $\cos 2A = \cos^2 A - \sin^2 A$cos2A=cos2A−sin2A. The Pythagorean identity $\sin^2 A + \cos^2 A = 1$sin2A+cos2A=1 then generates the other two forms of $\cos 2A$cos2A.

2. Integration of $\sin^2 x$sin2x and $\cos^2 x$cos2x (section 8): the power-reduction identities $\sin^2 x = (1 - \cos 2x)/2$sin2x=(1−cos2x)/2 and $\cos^2 x = (1 + \cos 2x)/2$cos2x=(1+cos2x)/2 are pure rearrangements of $\cos 2x = 1 - 2\sin^2 x$cos2x=1−2sin2x and $\cos 2x = 2\cos^2 x - 1$cos2x=2cos2x−1. Without these, integrating $\sin^2 x$sin2x is impossible — $\int\sin^2 x\,dx = \int\tfrac{1 - \cos 2x}{2}\,dx = \tfrac{x}{2} - \tfrac{\sin 2x}{4} + C$∫sin2xdx=∫21−cos2x​dx=2x​−4sin2x​+C.

3. Trigonometric equations with multiple solutions (section 5): double-angle substitutions multiply the count of solutions because they convert an equation in $2x$2x to one in $x$x, doubling the period density. Always check the interval for $2x$2x against the interval for $x$x — if $x \in [0, 2\pi]$x∈[0,2π], then $2x \in [0, 4\pi]$2x∈[0,4π], and you must find every $2x$2x-solution in that doubled interval before halving.

4. Harmonic form derivation (section 5): $a\sin x + b\cos x = R\sin(x + \alpha)$asinx+bcosx=Rsin(x+α) is itself a compound-angle expansion, and squaring such expressions produces $(a\sin x + b\cos x)^2 = a^2\sin^2 x + 2ab\sin x\cos x + b^2\cos^2 x$(asinx+bcosx)2=a2sin2x+2absinxcosx+b2cos2x — three terms that all reduce via double-angle identities ($2\sin x\cos x = \sin 2x$2sinxcosx=sin2x, $\sin^2 x$sin2x and $\cos^2 x$cos2x via power reduction).

5. Mechanics: simple harmonic motion (9MA0-03): an SHM displacement $x(t) = A\cos(\omega t)$x(t)=Acos(ωt) has kinetic energy proportional to $\sin^2(\omega t)$sin2(ωt). Average KE over a period uses $\overline{\sin^2(\omega t)} = \tfrac{1}{2}$sin2(ωt)​=21​, derived from $\sin^2(\omega t) = (1 - \cos 2\omega t)/2$sin2(ωt)=(1−cos2ωt)/2 and the fact that $\overline{\cos 2\omega t} = 0$cos2ωt=0 over a full period. The double-angle identity is doing the physics work.

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Mark-scheme literacy

Double-angle questions on 9MA0-02 typically split AO marks as:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Correct substitution of a double-angle identity, factorising, solving the resulting quadratic, listing solutions over the requested interval
AO2 (reasoning / interpretation)	25–35%	Choosing the strategic form of $\cos 2x$cos2x, justifying why factorisation is preferred over division, recognising boundary cases ($\sin\theta = 1$sinθ=1 produces a single solution, not two)
AO3 (problem-solving)	0–10%	Multi-step problems combining double-angle with compound-angle, integration setup, or "show that …" proofs requiring a specific path

Examiner-rewarded phrasing: "using $\cos 2x = 1 - 2\sin^2 x$cos2x=1−2sin2x to convert the equation into a quadratic in $\sin x$sinx"; "factorising rather than dividing by $\sin x$sinx to preserve all solutions"; "since the interval for $x$x is $[0, 2\pi]$[0,2π], the interval for $2x$2x is $[0, 4\pi]$[0,4π], so we must find solutions over the doubled range before halving". Phrases that lose marks: "dividing through by $\sin x$sinx" (silent solution loss); "$\cos 2x = 2\cos x$cos2x=2cosx" (simply wrong — the candidate has confused the double-angle identity with linearity); "$x = \arccos(1/4) = 1.318$x=arccos(1/4)=1.318 rad" when the question demands exact form.

A specific Edexcel pattern: when a question contains a single trig term (e.g. only $\sin\theta$sinθ or only $\cos\theta$cosθ on one side after the double-angle is in play), the correct form of $\cos 2\theta$cos2θ is the one that keeps that single term. Faced with $\cos 2\theta + 3\sin\theta = 2$cos2θ+3sinθ=2, choose $\cos 2\theta = 1 - 2\sin^2\theta$cos2θ=1−2sin2θ (matches $\sin\theta$sinθ). Faced with $\cos 2\theta = 5\cos\theta - 3$cos2θ=5cosθ−3, choose $\cos 2\theta = 2\cos^2\theta - 1$cos2θ=2cos2θ−1 (matches $\cos\theta$cosθ). The "wrong" choice generates an equation in both $\sin\theta$sinθ and $\cos\theta$cosθ, forcing a Pythagorean conversion that wastes time and risks errors.

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Grade-band model answers

3-mark question

Question: Given that $\sin x = \tfrac{3}{5}$sinx=53​ and $x$x is acute, find the exact value of $\sin 2x$sin2x.

Grade C response (~210 words):

If $\sin x = \tfrac{3}{5}$sinx=53​ and $x$x is acute, then I can use Pythagoras to find $\cos x$cosx. Drawing a right-angled triangle with opposite $3$3 and hypotenuse $5$5 gives adjacent $= 4$=4 (using the $3$3-$4$4-$5$5 triple), so $\cos x = \tfrac{4}{5}$cosx=54​. Since $x$x is acute, $\cos x$cosx is positive, so the sign is correct.

Now using the double-angle formula:

$\sin 2x = 2\sin x\cos x = 2 \cdot \tfrac{3}{5} \cdot \tfrac{4}{5} = \tfrac{24}{25}$sin2x=2sinxcosx=2⋅53​⋅54​=2524​.

So $\sin 2x = \tfrac{24}{25}$sin2x=2524​.

Examiner commentary: Full marks (3/3). The candidate correctly identifies $\cos x = \tfrac{4}{5}$cosx=54​ from the Pythagorean triple, justifies the positive sign with the "acute" condition, and applies the double-angle identity cleanly. The arithmetic is correct and the answer is in exact form. This is the standard Grade C answer for a procedural question — efficient and correct. Some candidates lose a mark by computing $\cos x$cosx as $\sqrt{1 - 9/25} = \sqrt{16/25}$1−9/25​=16/25​ but failing to specify the positive root explicitly; the "since $x$x is acute" justification protects against that.

Grade A response (~250 words):*

Given $\sin x = \tfrac{3}{5}$sinx=53​ with $x$x acute, I need $\cos x$cosx to apply $\sin 2x = 2\sin x\cos x$sin2x=2sinxcosx.

By the Pythagorean identity $\sin^2 x + \cos^2 x = 1$sin2x+cos2x=1:

$\cos^2 x = 1 - \sin^2 x = 1 - \dfrac{9}{25} = \dfrac{16}{25}$cos2x=1−sin2x=1−259​=2516​

So $\cos x = \pm\tfrac{4}{5}$cosx=±54​. Since $x$x is acute, $x \in (0, \pi/2)$x∈(0,π/2), where cosine is positive, so $\cos x = \tfrac{4}{5}$cosx=54​.

Applying the double-angle formula $\sin 2x = 2\sin x\cos x$sin2x=2sinxcosx:

$\sin 2x = 2 \cdot \dfrac{3}{5} \cdot \dfrac{4}{5} = \dfrac{24}{25}$sin2x=2⋅53​⋅54​=2524​