Addition Formulae

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

This lesson covers the addition (compound angle) formulae for $\sin$ , $\cos$ , and $\tan$ . These formulae allow you to expand expressions like $\sin(A + B)$ and $\cos(A - B)$ , and they are fundamental to much of the trigonometry at A-Level — including double angle formulae, the $R\cos(\theta + \alpha)$ form, and proving identities.

Spec Mapping — AQA 7357 Section E Trigonometry. This lesson covers the compound-angle (addition) formulae for $\sin$ , $\cos$ and $\tan$ content of Section E, at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

The Formulae

Sine Addition Formulae

\begin{aligned} \sin(A + B) &\equiv \sin A \cos B + \cos A \sin B \\ \sin(A - B) &\equiv \sin A \cos B - \cos A \sin B \end{aligned}

Cosine Addition Formulae

\begin{aligned} \cos(A + B) &\equiv \cos A \cos B - \sin A \sin B \\ \cos(A - B) &\equiv \cos A \cos B + \sin A \sin B \end{aligned}

Note the sign pattern: in the cos formulae, the sign in the expansion is opposite to the sign in the argument.

Tangent Addition Formulae

\begin{aligned} \tan(A + B) &\equiv \frac{\tan A + \tan B}{1 - \tan A \tan B} \\ \tan(A - B) &\equiv \frac{\tan A - \tan B}{1 + \tan A \tan B} \end{aligned}

These formulae are in the AQA formula booklet, but you must be able to use them fluently.

Proof of $\cos(A - B)$ (Geometric/Unit Circle)

Consider two points $P$ and $Q$ on the unit circle, where $P$ is at angle $A$ and $Q$ is at angle $B$ from the positive $x$ -axis. Then:

P = (\cos A, \sin A) \quad \text{and} \quad Q = (\cos B, \sin B)

The distance $PQ$ can be found using the distance formula:

\begin{aligned} PQ^{2} &= (\cos A - \cos B)^{2} + (\sin A - \sin B)^{2} \\ &= \cos^{2}A - 2\cos A \cos B + \cos^{2}B + \sin^{2}A - 2\sin A \sin B + \sin^{2}B \\ &= (\cos^{2}A + \sin^{2}A) + (\cos^{2}B + \sin^{2}B) - 2(\cos A \cos B + \sin A \sin B) \\ &= 1 + 1 - 2(\cos A \cos B + \sin A \sin B) \\ &= 2 - 2(\cos A \cos B + \sin A \sin B) \end{aligned}

Alternatively, using the cosine rule in the triangle with vertices $O, P, Q$ (with $OP = OQ = 1$ and angle $POQ = A - B$ ):

PQ^{2} = 1^{2} + 1^{2} - 2(1)(1)\cos(A - B) = 2 - 2\cos(A - B)

Equating the two expressions for $PQ^2$ :

\begin{aligned} 2 - 2\cos(A - B) &= 2 - 2(\cos A \cos B + \sin A \sin B) \\ \cos(A - B) &= \cos A \cos B + \sin A \sin B \end{aligned}

The other formulae can be derived from this one by substitution.

Deriving the Other Formulae

$\cos(A + B)$

Replace $B$ with $-B$ in $\cos(A - B)$ :

\begin{aligned} \cos(A + B) &= \cos A \cos(-B) + \sin A \sin(-B) \\ &= \cos A \cos B - \sin A \sin B \end{aligned}

(using $\cos(-B) = \cos B$ and $\sin(-B) = -\sin B$ )

$\sin(A + B)$

Use the identity $\sin\theta = \cos(\tfrac{\pi}{2} - \theta)$ :

\begin{aligned} \sin(A + B) &= \cos\!\left(\frac{\pi}{2} - (A + B)\right) \\ &= \cos\!\left(\left(\frac{\pi}{2} - A\right) - B\right) \\ &= \cos\!\left(\frac{\pi}{2} - A\right)\cos B + \sin\!\left(\frac{\pi}{2} - A\right)\sin B \\ &= \sin A \cos B + \cos A \sin B \end{aligned}

$\tan(A + B)$

\begin{aligned} \tan(A + B) &= \frac{\sin(A + B)}{\cos(A + B)} \\ &= \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B} \end{aligned}

Divide numerator and denominator by $\cos A \cos B$ :

= \frac{\tan A + \tan B}{1 - \tan A \tan B}

Finding Exact Values

Example 1: Find the exact value of $\sin 75°$ .

$75° = 45° + 30°$ , so:

\begin{aligned} \sin 75^\circ &= \sin(45^\circ + 30^\circ) \\ &= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \\ &= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} \\ &= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \\ &= \frac{\sqrt{6} + \sqrt{2}}{4} \end{aligned}

Example 2: Find the exact value of $\cos 15°$ .

$15° = 45° - 30°$ , so:

\begin{aligned} \cos 15^\circ &= \cos(45^\circ - 30^\circ) \\ &= \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \\ &= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} \\ &= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \\ &= \frac{\sqrt{6} + \sqrt{2}}{4} \end{aligned}

Note: $\sin 75° = \cos 15°$ — as expected, since $75° + 15° = 90°$ .

Example 3: Find the exact value of $\tan 75°$ .

\begin{aligned} \tan 75^\circ &= \tan(45^\circ + 30^\circ) \\ &= \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \\ &= \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} \end{aligned}

Multiply numerator and denominator by $\sqrt{3}$ :

= \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

Rationalise:

\begin{aligned} &= \frac{(\sqrt{3} + 1)^{2}}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\ &= \frac{3 + 2\sqrt{3} + 1}{3 - 1} \\ &= \frac{4 + 2\sqrt{3}}{2} \\ &= 2 + \sqrt{3} \end{aligned}

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 4: Solve $\sin(x + \tfrac{\pi}{3}) = \tfrac{1}{2}$ for $0 \leq x \leq 2\pi$ .

x + \frac{\pi}{3} = \frac{\pi}{6} + 2n\pi \quad\text{or}\quad x + \frac{\pi}{3} = \pi - \frac{\pi}{6} + 2n\pi = \frac{5\pi}{6} + 2n\pi

From $x + \tfrac{\pi}{3} = \tfrac{\pi}{6}$ : $x = \tfrac{\pi}{6} - \tfrac{\pi}{3} = -\tfrac{\pi}{6}$ (not in range) From $x + \tfrac{\pi}{3} = \tfrac{5\pi}{6}$ : $x = \tfrac{5\pi}{6} - \tfrac{\pi}{3} = \tfrac{5\pi}{6} - \tfrac{2\pi}{6} = \tfrac{3\pi}{6} = \tfrac{\pi}{2}$ ✓

For $n = 1$ : From $x + \tfrac{\pi}{3} = \tfrac{\pi}{6} + 2\pi = \tfrac{13\pi}{6}$ : $x = \tfrac{13\pi}{6} - \tfrac{\pi}{3} = \tfrac{13\pi}{6} - \tfrac{2\pi}{6} = \tfrac{11\pi}{6}$ ✓ From $x + \tfrac{\pi}{3} = \tfrac{5\pi}{6} + 2\pi = \tfrac{17\pi}{6}$ : $x = \tfrac{17\pi}{6} - \tfrac{\pi}{3} = \tfrac{15\pi}{6} = \tfrac{5\pi}{2} > 2\pi$ (out of range)

Solutions: $x = \tfrac{\pi}{2}, \tfrac{11\pi}{6}$

Example 5: Given that $\sin(x + 30°) = 2\cos(x - 30°)$ , find $\tan x$ .

Expand both sides:

\begin{aligned} \sin x \cos 30^\circ + \cos x \sin 30^\circ &= 2(\cos x \cos 30^\circ + \sin x \sin 30^\circ) \\ \frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x &= 2\cdot\frac{\sqrt{3}}{2}\cos x + 2\cdot\frac{1}{2}\sin x \\ \frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x &= \sqrt{3}\cos x + \sin x \end{aligned}

Rearranging:

\begin{aligned} ( \sqrt{3}/2)\sin x - \sin x = \sqrt{3} \cos x - (1/2)\cos x \\ \sin x( \sqrt{3}/2 - 1) = \cos x( \sqrt{3} - 1/2) \\ \sin x( \sqrt{3} - 2)/2 = \cos x(2 \sqrt{3} - 1)/2 \\ \tan x = (2 \sqrt{3} - 1)/( \sqrt{3} - 2) \end{aligned}

Rationalise by multiplying by $(\sqrt{3} + 2)/(\sqrt{3} + 2)$ :

\begin{aligned} &= \frac{(2\sqrt{3} - 1)(\sqrt{3} + 2)}{(\sqrt{3} - 2)(\sqrt{3} + 2)} \\ &= \frac{2 \times 3 + 4\sqrt{3} - \sqrt{3} - 2}{3 - 4} \\ &= \frac{6 + 3\sqrt{3} - 2}{-1} \\ &= \frac{4 + 3\sqrt{3}}{-1} \\ &= -4 - 3\sqrt{3} \end{aligned}

Proving Identities Using Addition Formulae

Example 6: Prove that $\cos(A + B) + \cos(A - B) \equiv 2\cos A \cos B$ .

\begin{aligned} \text{LHS} &= (\cos A \cos B - \sin A \sin B) + (\cos A \cos B + \sin A \sin B) \\ &= 2\cos A \cos B \\ &= \text{RHS} \qquad\blacksquare \end{aligned}

Example 7: Prove that $\sin(A + B)\sin(A - B) \equiv \sin^2 A - \sin^2 B$ .

\begin{aligned} \text{LHS} &= (\sin A \cos B + \cos A \sin B)(\sin A \cos B - \cos A \sin B) \\ &= \sin^{2}A \cos^{2}B - \cos^{2}A \sin^{2}B \\ &= \sin^{2}A(1 - \sin^{2}B) - (1 - \sin^{2}A)\sin^{2}B \\ &= \sin^{2}A - \sin^{2}A \sin^{2}B - \sin^{2}B + \sin^{2}A \sin^{2}B \\ &= \sin^{2}A - \sin^{2}B \\ &= \text{RHS} \qquad\blacksquare \end{aligned}

Synoptic Links

This topic connects to:

Double-angle formulae (aqa-alevel-maths-trigonometry-depth / double-angle-formulae) — special case $A = B$ specialises the addition formulae.
Projectile motion (aqa-alevel-maths-mechanics / projectiles) — resolving velocity into perpendicular components uses $\sin(A + B)$ expansion when the launch frame is rotated.
Integration by substitution (aqa-alevel-maths-calculus-applications / integration-by-substitution) — product-to-sum identities (derived from addition formulae) convert $\sin A \cos B$ integrands into integrable form.

Summary

$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$ (note the opposite sign)
$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$
These formulae are in the AQA formula booklet.
Use them to find exact values of non-standard angles (e.g., $15°$ , $75°$ , $105°$ ).
Use them to solve equations of the form $\sin(x + \alpha) = k$ or $\cos(x - \alpha) = k$ .
Use them to prove identities.

Exam Tip: When using addition formulae, be very careful with signs — especially in the cos formula, where the sign in the expansion is opposite to the sign in the argument. Write out the formula first before substituting values. In "show that" questions, show every step clearly. The most common error is a sign mistake.

A-Level Deep Dive: Addition (Compound Angle) Formulae

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section E (Trigonometry), sub-strand E5 covers the addition formulae for sine, cosine and tangent; understand and use double-angle formulae as a special case (refer to the official specification document for exact wording). The compound-angle formulae $\sin(A \pm B)$ , $\cos(A \pm B)$ and $\tan(A \pm B)$ are printed in the AQA formula booklet, but candidates are still expected to apply them fluently in proof, equation-solving and modelling contexts. Compound-angle work is examined across Section E (trig identities and equations), Section J (integration via product-to-sum substitution), Section L (proof) and (in mechanics, 7357/2) within force-resolution problems where $R\sin(\theta + \alpha)$ harmonic form is used to model oscillations.

Worked example with full mark scheme

Question (8 marks):

(a) Without using a calculator, find the exact value of $\sin(75°)$ , expressing your answer in the form $\dfrac{\sqrt{a} + \sqrt{b}}{c}$ where $a$ , $b$ and $c$ are integers. (4)

(b) Hence, or otherwise, prove that $\sin(75°) - \sin(15°) = \dfrac{\sqrt{2}}{2}$ . (4)

Solution with mark scheme:

(a) Step 1 — express $75°$ as a sum of standard angles.

$75° = 45° + 30°$

M1 — choosing a decomposition into angles whose sine and cosine are exactly known. $75° = 60° + 15°$ does not help (since $15°$ is itself unknown), and $75° = 90° - 15°$ converts to $\cos(15°)$ which still requires compound angles. The cleanest split is $45° + 30°$ .

Step 2 — apply the addition formula for sine.

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

M1 — correct quotation of $\sin(A + B) = \sin A \cos B + \cos A \sin B$ . Sign and ordering must match exactly; writing $\sin A \sin B + \cos A \cos B$ (the formula for $\cos(A - B)$ ) loses both M1s and the subsequent A1.

Step 3 — substitute exact values.

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

$\sin(75°) = \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4}$

A1 — correct substitution and simplification of each product. Note $\dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{6}}{4}$ , not $\dfrac{\sqrt{5}}{4}$ — surd multiplication, not addition.

Step 4 — combine into requested form.

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

A1 — final form with $a = 6$ , $b = 2$ , $c = 4$ .

(b) Step 1 — find $\sin(15°)$ by analogous decomposition.

$15° = 45° - 30°$ , so:

$\sin(15°) = \sin 45° \cos 30° - \cos 45° \sin 30° = \dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6} - \sqrt{2}}{4}$

M1 — correct application of $\sin(A - B) = \sin A \cos B - \cos A \sin B$ , noting the sign change relative to part (a).

A1 — correct exact value for $\sin(15°)$ .

Step 2 — subtract.

$\sin(75°) - \sin(15°) = \dfrac{\sqrt{6} + \sqrt{2}}{4} - \dfrac{\sqrt{6} - \sqrt{2}}{4} = \dfrac{2\sqrt{2}}{4} = \dfrac{\sqrt{2}}{2}$

M1 — correct subtraction with the $\sqrt{6}$ terms cancelling.

A1 — final exact answer matching the printed result $\dfrac{\sqrt{2}}{2}$ , completing the proof.

Total: 8 marks (M4 A4).

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): Prove that, for all real $A$ and $B$ for which both sides are defined,

$\dfrac{\sin(A + B)}{\cos A \cos B} = \tan A + \tan B$

Hence find the exact value of $\tan(15°) + \tan(30°)$ in surd form, given that $\sin(45°) = \dfrac{\sqrt{2}}{2}$ and $\cos(15°)\cos(30°) = \dfrac{\sqrt{6} + \sqrt{2}}{4} \cdot \dfrac{\sqrt{3}}{2}$ . (6)

Mark scheme decomposition by AO:

M1 (AO1.1a) — quoting $\sin(A + B) = \sin A \cos B + \cos A \sin B$ .
M1 (AO1.1b) — dividing both terms of the numerator by $\cos A \cos B$ :

$\dfrac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} = \dfrac{\sin A}{\cos A} + \dfrac{\sin B}{\cos B} = \tan A + \tan B$

A1 (AO2.1) — identity established with the algebraic step explicit. AQA awards this AO2 mark for the quality of the proof — splitting the fraction term-by-term must be visible.
M1 (AO3.1a) — recognising that $A = 15°$ , $B = 30°$ gives $A + B = 45°$ , and applying the proved identity with $\sin(45°) = \dfrac{\sqrt{2}}{2}$ .
M1 (AO1.1b) — computing the denominator product $\cos(15°)\cos(30°)$ from the supplied values.
A1 (AO1.1b) — final exact value of $\tan(15°) + \tan(30°)$ in surd form.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. AQA loads compound-angle proof questions toward AO1/AO2 — the technique is procedural, but the connecting move (recognising that the printed identity is the engine of the numerical part) is the AO3 reasoning the synoptic mark rewards.

Synoptic links

Connects to:

Double-angle formulae as special case (Section E): setting $A = B$ in $\sin(A + B)$ gives $\sin(2A) = 2\sin A \cos A$ . Likewise $\cos(2A) = \cos^2 A - \sin^2 A$ , with the alternative forms $1 - 2\sin^2 A$ and $2\cos^2 A - 1$ following from the Pythagorean identity. Compound-angle formulae are the parent identities; double-angle formulae are their $A = B$ degeneration. Examiners exploit this — a question stated in compound-angle form often reduces to double-angle work.
Harmonic form $R\sin(\theta + \alpha)$ (Section E): expanding $R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$ shows that $a\sin\theta + b\cos\theta$ can always be rewritten as a single sinusoid with $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = b/a$ . This is the engine of every "find the maximum/minimum of $3\sin\theta + 4\cos\theta$ " question and underpins phasor analysis in physics.
Trigonometric equations (Section E): equations like $\sin(2x) = \cos(x)$ are unsolvable until $\sin(2x)$ is expanded as $2\sin x \cos x$ , after which the equation factorises as $\cos x (2\sin x - 1) = 0$ . Compound and double-angle expansion is the gateway technique for non-trivial trig equations.
Product-to-sum and integration (Section J): the identities $2\sin A \cos B = \sin(A + B) + \sin(A - B)$ and $2\cos A \cos B = \cos(A - B) + \cos(A + B)$ are derived directly from the addition formulae. They convert products (which cannot be integrated directly) into sums (which can), making integrals like $\int \sin(3x)\cos(x)\,dx$ accessible.

Addition Formulae

Addition Formulae

The Formulae

Sine Addition Formulae

Cosine Addition Formulae

Tangent Addition Formulae

Proof of cos⁡(A−B)\cos(A - B)cos(A−B) (Geometric/Unit Circle)

Deriving the Other Formulae

cos⁡(A+B)\cos(A + B)cos(A+B)

sin⁡(A+B)\sin(A + B)sin(A+B)

tan⁡(A+B)\tan(A + B)tan(A+B)

Finding Exact Values

Specimen Question (Modelled on the AQA Paper Format)

Solving Equations

Proving Identities Using Addition Formulae

Synoptic Links

Summary

A-Level Deep Dive: Addition (Compound Angle) Formulae

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

More in Mathematics

Proof of $\cos(A - B)$ (Geometric/Unit Circle)

$\cos(A + B)$

$\sin(A + B)$

$\tan(A + B)$