Compound Angle Formulae

The compound angle (or addition) formulae express sin(A ± B), cos(A ± B) and tan(A ± B) in terms of sin and cos of A and B individually. These are given in the Edexcel formula booklet and are essential for A-Level (9MA0).

The Six Compound Angle Formulae

Sine

sin(A + B) = sinA cosB + cosA sinB

sin(A - B) = sinA cosB - cosA sinB

Cosine

cos(A + B) = cosA cosB - sinA sinB

cos(A - B) = cosA cosB + sinA sinB

Tangent

tan(A + B) = (tanA + tanB) / (1 - tanA tanB)

tan(A - B) = (tanA - tanB) / (1 + tanA tanB)

Using the Formulae to Find Exact Values

Worked Example 1

Find the exact value of sin(75 degrees) = sin(pi/12 x 5) = sin(45 + 30).

sin(75) = sin(45 + 30) = sin45 cos30 + cos45 sin30

= (sqrt(2)/2)(sqrt(3)/2) + (sqrt(2)/2)(1/2)

= sqrt(6)/4 + sqrt(2)/4

= (sqrt(6) + sqrt(2))/4

Worked Example 2

Find the exact value of cos(15 degrees) = cos(45 - 30).

cos(15) = cos45 cos30 + sin45 sin30

= (sqrt(2)/2)(sqrt(3)/2) + (sqrt(2)/2)(1/2)

= sqrt(6)/4 + sqrt(2)/4

= (sqrt(6) + sqrt(2))/4

Note: sin(75) = cos(15), as expected since 75 + 15 = 90.

Worked Example 3

Find the exact value of tan(105 degrees) = tan(60 + 45).

tan(105) = (tan60 + tan45) / (1 - tan60 x tan45)

= (sqrt(3) + 1) / (1 - sqrt(3))

Rationalise: multiply top and bottom by (1 + sqrt(3)):

= (sqrt(3) + 1)(1 + sqrt(3)) / (1 - sqrt(3))(1 + sqrt(3))

= (sqrt(3) + 3 + 1 + sqrt(3)) / (1 - 3)

= (4 + 2sqrt(3)) / (-2)

= -2 - sqrt(3)

Proving Identities Using Compound Angles

Worked Example 4

Prove that sin(x + pi/6) + sin(x - pi/6) = sin(x).

LHS = [sinx cos(pi/6) + cosx sin(pi/6)] + [sinx cos(pi/6) - cosx sin(pi/6)]

= 2sinx cos(pi/6)

= 2sinx x (sqrt(3)/2)

= sqrt(3) sinx

Hmm, this gives sqrt(3) sinx, not sinx. Let me reconsider the identity. The correct identity is:

sin(x + pi/6) + sin(x - pi/6) = sqrt(3) sin(x)

LHS = 2sin(x)cos(pi/6) = 2sin(x) x sqrt(3)/2 = sqrt(3) sin(x) = RHS

Worked Example 5

Show that cos(A + B) + cos(A - B) = 2cosA cosB.

cos(A + B) = cosA cosB - sinA sinB cos(A - B) = cosA cosB + sinA sinB

Adding: cos(A + B) + cos(A - B) = 2cosA cosB

Solving Equations

Worked Example 6

Solve sin(x + pi/4) = cos(x) for 0 <= x < 2pi.

Expand the left side:

sinx cos(pi/4) + cosx sin(pi/4) = cosx

(sqrt(2)/2)sinx + (sqrt(2)/2)cosx = cosx

(sqrt(2)/2)sinx = cosx - (sqrt(2)/2)cosx

(sqrt(2)/2)sinx = cosx(1 - sqrt(2)/2)

(sqrt(2)/2)sinx = cosx(2 - sqrt(2))/2

sinx/cosx = (2 - sqrt(2))/sqrt(2)

tanx = (2 - sqrt(2))/sqrt(2) = 2/sqrt(2) - 1 = sqrt(2) - 1

x = arctan(sqrt(2) - 1) = pi/8

And x = pi + pi/8 = 9pi/8

Worked Example 7

Given that sin(A) = 3/5 where A is acute, and cos(B) = 12/13 where B is acute, find sin(A + B).

From sin(A) = 3/5: cos(A) = 4/5 (using Pythagoras) From cos(B) = 12/13: sin(B) = 5/13 (using Pythagoras)

sin(A + B) = sinA cosB + cosA sinB

= (3/5)(12/13) + (4/5)(5/13)

= 36/65 + 20/65

= 56/65

Exam Tips

Tip	Detail
Formula booklet	These formulae are given — but you must be fluent in using them
Sign patterns	cos has the opposite sign in the middle: cos(A + B) has minus, cos(A - B) has plus
Exact values	Combine with exact values for pi/6, pi/4, pi/3 to find non-standard angles
Right triangle	If given sin or cos of an angle, draw a right triangle to find the other ratio
Watch the question	"Show that" and "prove that" require you to demonstrate every step

Practice Problems

Find the exact value of cos(75 degrees). [Answer: (sqrt(6) - sqrt(2))/4]
Find the exact value of sin(pi/12). [Answer: (sqrt(6) - sqrt(2))/4]
Given sinA = 5/13 (A acute) and cosB = 3/5 (B acute), find cos(A - B). [Answer: 56/65]
Solve cos(x + pi/3) = sin(x) for 0 <= x < 2pi. [Answer: x = pi/12, 13pi/12]
Prove that sin(A + B) + sin(A - B) = 2sinA cosB.

Summary

Compound angle formulae express trig of sums/differences in terms of individual angles.
sin(A ± B) = sinA cosB ± cosA sinB
cos(A ± B) = cosA cosB -/+ sinA sinB (opposite sign to what you might expect)
tan(A ± B) = (tanA ± tanB) / (1 -/+ tanA tanB)
These are in the formula booklet but must be applied confidently.

A-Level Deep Dive: Compound Angle Formulae

Spec mapping

Edexcel 9MA0 Pure Mathematics specification, section 5 — Trigonometry, sub-strand 5.5 (Compound angle formulae) covers the addition formulae for sine, cosine and tangent (refer to the official specification document for exact wording). The formulae $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$ , $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$ and $\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$ ARE listed in the Edexcel formula booklet under "Trigonometric identities" — but candidates are still expected to apply them fluently, recognise when each is needed, and substitute exact values without prompting. The synoptic web is dense: this section underpins 5.6 (double-angle formulae, the special case $A = B$ ), 5.7 (harmonic form $R\sin(\theta + \alpha)$ uses $\sin(A + B)$ in reverse), 5.8 (trigonometric equations that require expansion or contraction of compound expressions), section 8 (Differentiation of products like $\sin x \cos x$ via reformulation), section 9 (Integration by reverse-product-to-sum identities) and — for Further Maths candidates — the Euler-formula derivation $e^{i(A + B)} = e^{iA} \cdot e^{iB}$ that generates every compound-angle identity in a single line of complex-number algebra.

Worked example with full mark scheme

Question (8 marks):

(a) Show that $\sin 75°$ can be written in the form $\dfrac{\sqrt{6} + \sqrt{2}}{4}$ . (4)

(b) Hence prove the identity $\sin(x + 30°) + \cos(x + 60°) \equiv \cos x$ for all real $x$ , and use it to simplify $\sin 75° + \cos 105°$ exactly. (4)

Solution with mark scheme:

(a) Step 1 — decompose 75° into a sum of standard angles.

Write $75° = 45° + 30°$ (or $75° = 30° + 45°$ — either decomposition is acceptable). Apply the compound-angle formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$ :

$\sin 75° = \sin(45° + 30°) = \sin 45° \cos 30° + \cos 45° \sin 30°$

M1 — correct decomposition into a sum of two standard angles whose sine and cosine values are exactly known, and explicit use of the compound-angle formula. The phrase "using the compound-angle formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$ " is examiner-rewarded for clarity. Common loss point: candidates write $\sin 75° = \sin 45° + \sin 30°$ , which is the textbook misconception (sine is not linear) and earns zero.

Step 2 — substitute exact values.

From the standard exact-value table: $\sin 45° = \dfrac{\sqrt{2}}{2}$ , $\cos 30° = \dfrac{\sqrt{3}}{2}$ , $\cos 45° = \dfrac{\sqrt{2}}{2}$ , $\sin 30° = \dfrac{1}{2}$ .

$\sin 75° = \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2}$

M1 — correct substitution of exact values into the formula. A frequent slip: writing $\cos 30° = \dfrac{1}{2}$ (confusing it with $\sin 30°$ ). Always sketch the special-angle table at the top of the answer to avoid this.

Step 3 — simplify.

$\sin 75° = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$

A1 — correct intermediate $\dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4}$ (using $\sqrt{2} \cdot \sqrt{3} = \sqrt{6}$ ).

A1 — final answer combined over a common denominator in the requested form $\dfrac{\sqrt{6} + \sqrt{2}}{4}$ .

(b) Step 1 — expand each compound expression.

Apply $\sin(x + 30°) = \sin x \cos 30° + \cos x \sin 30°$ and $\cos(x + 60°) = \cos x \cos 60° - \sin x \sin 60°$ :

$\sin(x + 30°) = \dfrac{\sqrt{3}}{2}\sin x + \dfrac{1}{2}\cos x$

$\cos(x + 60°) = \dfrac{1}{2}\cos x - \dfrac{\sqrt{3}}{2}\sin x$

M1 — both expansions correct, with exact values substituted.

Step 2 — add the two expressions.

$\sin(x + 30°) + \cos(x + 60°) = \left(\dfrac{\sqrt{3}}{2}\sin x - \dfrac{\sqrt{3}}{2}\sin x\right) + \left(\dfrac{1}{2}\cos x + \dfrac{1}{2}\cos x\right) = \cos x$

A1 — correct cancellation of the $\sin x$ terms and combination of the $\cos x$ terms, producing $\cos x$ as required. Note the use of " $\equiv$ " in the question — this signals an identity for all $x$ , so the proof must hold without assuming any particular value.

Step 3 — apply to $\sin 75° + \cos 105°$ .

Set $x = 45°$ in the identity: $\sin(45° + 30°) + \cos(45° + 60°) = \cos 45°$ , i.e. $\sin 75° + \cos 105° = \cos 45° = \dfrac{\sqrt{2}}{2}$ .

M1 — correct identification of $x = 45°$ as the value that produces the requested expression.

A1 — exact answer $\dfrac{\sqrt{2}}{2}$ (or equivalent $\dfrac{1}{\sqrt{2}}$ ).

Total: 8 marks (M4 A4, split as shown).

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): Given that $\sin A = \dfrac{3}{5}$ and $\cos B = \dfrac{5}{13}$ , where $A$ is acute and $B$ is acute:

(a) Find the exact value of $\cos A$ and $\sin B$ . (2)

(b) Hence find the exact value of $\sin(A + B)$ . (2)

Mark scheme decomposition by AO:

(a)

B1 (AO1.1a) — $\cos A = \dfrac{4}{5}$ using $\sin^2 A + \cos^2 A = 1$ and the acute-angle constraint to take the positive root.
B1 (AO1.1a) — $\sin B = \dfrac{12}{13}$ similarly.

(b)

M1 (AO1.1b) — applying $\sin(A + B) = \sin A \cos B + \cos A \sin B = \dfrac{3}{5} \cdot \dfrac{5}{13} + \dfrac{4}{5} \cdot \dfrac{12}{13}$ .
A1 (AO1.1b) — exact value $\dfrac{15 + 48}{65} = \dfrac{63}{65}$ .

(c)

M1 (AO2.1) — applying $\cos(A - B) = \cos A \cos B + \sin A \sin B$ (sign flips from $-$ to $+$ inside the formula). The reasoning required is that all four constituent values are now known, so direct substitution gives the answer with no further work.
A1 (AO2.4) — $\dfrac{4}{5} \cdot \dfrac{5}{13} + \dfrac{3}{5} \cdot \dfrac{12}{13} = \dfrac{20 + 36}{65} = \dfrac{56}{65}$ , with a brief written justification ("using the compound-angle formula for $\cos(A - B)$ and the values found in part (a)").

Total: 6 marks split AO1 = 4, AO2 = 2. This is the canonical Paper 2 compound-angle structure: a Pythagorean-identity warm-up, a procedural $\sin(A + B)$ , then an AO2-flavoured "without further calculation" prompt that rewards candidates who recognise the same four numbers can be reused.

Synoptic links

Connects to:

Section 5.6 — Double-angle formulae: setting $A = B$ in the compound-angle formulae produces $\sin 2A = 2\sin A \cos A$ , $\cos 2A = \cos^2 A - \sin^2 A$ and $\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$ instantly. The double-angle formulae are not separate facts to memorise — they are corollaries of the compound-angle formulae, and exam questions sometimes test whether candidates can derive them on the spot.
Section 5.7 — Harmonic form $R\sin(\theta + \alpha)$ : the formula $R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$ is the compound-angle formula run forwards with $R\cos\alpha$ and $R\sin\alpha$ as the unknowns to be determined from $a\sin\theta + b\cos\theta$ . Confidence reading this expansion in both directions — expand and contract — is the key skill of section 5.7.
Section 5.8 — Trigonometric equations: equations of the form $\sin(\theta + 30°) = \dfrac{1}{2}$ are solved by isolating $\theta + 30°$ as a single argument, but problems involving $\sin\theta + \cos\theta = 1$ require contraction into harmonic form via the compound-angle formula in reverse. Recognising which direction to apply the formula is the AO2 skill examiners reward.
Section 9 — Integration via product-to-sum identities: integrals like $\int \sin 3x \cos 5x \, dx$ are evaluated by rewriting $\sin A \cos B = \dfrac{1}{2}[\sin(A + B) + \sin(A - B)]$ , an identity derived directly by adding the $\sin(A + B)$ and $\sin(A - B)$ formulae. Without the compound-angle formulae the product cannot be integrated in closed form by elementary means.
Further Maths — Euler's formula and complex numbers: $e^{i(A + B)} = e^{iA} \cdot e^{iB}$ expanded via $e^{i\theta} = \cos\theta + i\sin\theta$ gives $\cos(A + B) + i\sin(A + B) = (\cos A + i\sin A)(\cos B + i\sin B)$ . Equating real and imaginary parts produces both $\cos(A + B)$ and $\sin(A + B)$ formulae simultaneously — the cleanest derivation in mathematics.

Mark-scheme literacy

Compound-angle questions on 9MA0 split AO marks roughly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Quoting and applying the correct compound-angle formula, substituting exact values, simplifying surd expressions arising from products like $\sqrt{2} \cdot \sqrt{3}$
AO2 (reasoning / interpretation)	25–35%	Choosing which decomposition ( $75° = 45° + 30°$ vs $75° = 60° + 15°$ ), recognising when to apply the formula in reverse to contract an expression, justifying sign choices via the acute / obtuse / quadrant constraint
AO3 (problem-solving)	5–15%	Multi-step problems that require the compound-angle formula plus a Pythagorean identity, plus simplification — i.e. extended-response questions where the strategy itself is part of the assessment

Examiner-rewarded phrasing: "using the compound-angle formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$ "; "since $A$ is acute, $\cos A > 0$ , so we take the positive root"; "rewriting $75°$ as the sum $45° + 30°$ to use known exact values". Phrases that lose marks: writing " $\sin(A + B) = \sin A + \sin B$ " (linearity error, instant zero); leaving an answer in unsimplified form like $\dfrac{\sqrt{2}\sqrt{3}}{4} + \dfrac{\sqrt{2}}{4}$ rather than $\dfrac{\sqrt{6} + \sqrt{2}}{4}$ ; using degrees in one line and radians in the next without a unit shift.

A specific Edexcel pattern: questions framed "show that …" require every step to be visible, including the substitution of exact values. Skipping from " $\sin(45° + 30°)$ " to the final answer in one jump risks losing the M-marks even if the answer is correct.

Grade-band model answers

3-mark question

Question: Find the exact value of $\cos 15°$ .

Grade C response (~210 words):

Write $15° = 45° - 30°$ and use the compound-angle formula:

$\cos(A - B) = \cos A \cos B + \sin A \sin B$ .

So $\cos 15° = \cos 45° \cos 30° + \sin 45° \sin 30°$ .

Substituting: $\cos 45° = \dfrac{\sqrt{2}}{2}$ , $\cos 30° = \dfrac{\sqrt{3}}{2}$ , $\sin 45° = \dfrac{\sqrt{2}}{2}$ , $\sin 30° = \dfrac{1}{2}$ .

Compound Angle Formulae

Compound Angle Formulae

The Six Compound Angle Formulae

Sine

Cosine

Tangent

Using the Formulae to Find Exact Values

Worked Example 1

Worked Example 2

Worked Example 3

Proving Identities Using Compound Angles

Worked Example 4

Worked Example 5

Solving Equations

Worked Example 6

Worked Example 7

Exam Tips

Practice Problems

Summary

A-Level Deep Dive: Compound Angle Formulae

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

More in Mathematics