What are the trigonometric addition formulae in Pure Mathematics 3 for Edexcel International A Levels?

In Pure Mathematics 3 for Edexcel International A Levels, the trigonometric addition formulae are used to find the sine, cosine, and tangent of the sum or difference of two angles. The key formulae are: sin(A ± B) = sinA cosB ± cosA sinB, cos(A ± B) = cosA cosB ∓ sinA sinB, and tan(A ± B) = (tanA ± tanB) / (1 ∓ tanA tanB). These formulae are essential for solving complex trigonometric problems.

How do you apply the sine addition formula in solving trigonometric problems?

To apply the sine addition formula, sin(A ± B) = sinA cosB ± cosA sinB, you need to identify the angles A and B in your problem. Substitute these angles into the formula to find the sine of their sum or difference. This technique is particularly useful in simplifying expressions and solving equations where angles are added or subtracted.

Why are trigonometric addition formulae important in Edexcel International A Levels Mathematics?

Trigonometric addition formulae are crucial in Edexcel International A Levels Mathematics because they allow students to simplify complex trigonometric expressions and solve equations involving multiple angles. These formulae are foundational for understanding more advanced topics in trigonometry and calculus, making them indispensable for students aiming to excel in Pure Mathematics 3.

Can you provide an example of using the cosine addition formula in a problem?

Certainly! Suppose you need to find the exact value of cos(75°). You can express 75° as 45° + 30° and use the cosine addition formula: cos(A + B) = cosA cosB - sinA sinB. Substituting the known values, cos(75°) = cos(45°)cos(30°) - sin(45°)sin(30°). This simplifies to (√2/2)(√3/2) - (√2/2)(1/2), resulting in (√6 - √2)/4.

How do trigonometric addition formulae relate to solving real-world problems?

Trigonometric addition formulae are used in various real-world applications, such as engineering, physics, and computer graphics. They help in calculating angles and distances in navigation, analyzing wave patterns, and designing mechanical systems. Understanding these formulae in the context of Edexcel International A Levels Mathematics equips students with the skills to tackle practical problems involving trigonometry.

Why is "Writing $\cos(A + B) = \cos A \cos B + \sin A \sin B$" a common mistake in Trigonometric addition formulae?

Why it happens: Copying the sign pattern of the sine formula. How to avoid it: Mnemonic: has a SIGN FLIP — (A + B) = A B - A B. Think 'minus for plus, plus for minus' in the cosine case. The formula booklet has it correctly — copy from there. Source: WMA13/01 examiner reports — flagged every sitting

Why is "Using $\cos 2A = \cos^{2}A - \sin^{2}A$ when integrating $\sin^{2}A$" a common mistake in Trigonometric addition formulae?

Why it happens: Default to the most-memorable form. How to avoid it: For ^2x\,dx, use ^2x = 1 - 2x2. For ^2x\,dx, use ^2x = 1 + 2x2. Each form has its purpose — pick to eliminate the unwanted term.

Why is "Quoting $R = -5$" a common mistake in Trigonometric addition formulae?

Why it happens: Both signs satisfy R^2 = 25. How to avoid it: By convention R > 0. The phase absorbs any necessary sign. Always quote the POSITIVE square root.

Why is "Quoting $\alpha$ outside $(0, \pi/2)$ for the form $R\sin(\theta + \alpha)$ with $a, b > 0$" a common mistake in Trigonometric addition formulae?

Why it happens: Just running (b/a) on the calculator without thinking. How to avoid it: When both a and b are positive, = (b/a) lies in (0, /2) — the principal-value range. If the question fixes a different range (e.g. -/2 < < /2), check the resulting form matches the original by expansion.

Why is "$\tan(A + B) = \dfrac{\tan A + \tan B}{1 + \tan A \tan B}$" a common mistake in Trigonometric addition formulae?

Why it happens: Confusing the sign pattern. How to avoid it: Mnemonic: 'Numerator follows LHS sign, denominator flips it.' (A + B) = A + B/1 - A B (plus on top, minus on bottom).

Pure Mathematics 3Edexcel International A Levels Mathematics (XMA01-YMA01)

Trigonometric addition formulae

Q: Why is "$\tan(A + B) = \dfrac{\tan A + \tan B}{1 + \tan A \tan B}$" a common mistake in Trigonometric addition formulae?

Why it happens: Confusing the sign pattern. How to avoid it: Mnemonic: 'Numerator follows LHS sign, denominator flips it.' (A + B) = A + B/1 - A B (plus on top, minus on bottom).

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Trigonometric Addition Formulae Study Notes — Edexcel IAL Mathematics P3 (WMA13/01, 2018 spec — 2026 onwards)

The compound-angle (addition) formulae for $\sin$ , $\cos$ , $\tan$ ; the double-angle identities derived from them; and the R-formula $a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha)$ that compresses a linear combination of $\sin$ and $\cos$ into a single sine wave. Mastery of these unlocks max/min analysis, exact angle values, and many P3 integrals.

At a glance

Addition formulae: $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$ .
$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$ — note the FLIPPED sign.
$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$ .
Double-angle: $\sin 2A = 2\sin A\cos A$ , $\cos 2A = 2\cos^{2}A - 1 = 1 - 2\sin^{2}A$ .
R-formula: $a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)$ , $R = \sqrt{a^{2} + b^{2}}$ , $\tan\alpha = b/a$ .
Max of $a\sin\theta + b\cos\theta$ is $R$ ; min is $-R$ .
All addition and double-angle formulae are IN the IAL formula booklet — R-formula derivation NOT.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

P3 4.1 — Use the addition formulae for $\sin(A \pm B)$ , $\cos(A \pm B)$ , $\tan(A \pm B)$ .
P3 4.2 — Derive and use the double-angle formulae.
P3 4.3 — Express $a\sin\theta + b\cos\theta$ in the forms $R\sin(\theta \pm \alpha)$ or $R\cos(\theta \pm \alpha)$ .
P3 4.4 — Use the R-formula to find maxima, minima, and to solve equations of the form $a\sin\theta + b\cos\theta = c$ .

The six addition (compound-angle) formulae

Plus the sign quirks — sin matches, cos flips, tan flips inside the denominator.

The six identities.

$\sin(A + B) = \sin A \cos B + \cos A \sin B$ $\sin(A - B) = \sin A \cos B - \cos A \sin B$ $\cos(A + B) = \cos A \cos B - \sin A \sin B$ $\cos(A - B) = \cos A \cos B + \sin A \sin B$ $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$ $\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$

Sign patterns.

$\sin$ : SIGN MATCHES the LHS sign (plus on LHS → plus inside).
$\cos$ : SIGN FLIPS (plus on LHS → minus inside the $\sin\sin$ term).
$\tan$ : sign matches in the numerator, FLIPS in the denominator.

Memorise the cos rule. The cosine formulae have a sign FLIP that catches students out every exam. Write it out: $\cos(A + B) = \cos A\cos B - \sin A\sin B$ . PLUS on LHS, MINUS inside.

Exact-value technique. Use addition formulae to compute exact values of angles not in the standard $\{0, \pi/6, \pi/4, \pi/3, \pi/2\}$ set.

Worked example. Find $\sin 75°$ exactly.

$\sin 75° = \sin(45° + 30°) = \sin 45° \cos 30° + \cos 45° \sin 30° = \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}.$

Six addition formulae — all in the booklet.
$\sin$ — sign matches. $\cos$ — sign flips.
$\tan$ : top matches LHS, bottom flips.
Use to compute exact values of awkward angles.

Double-angle formulae

Set $B = A$ in the addition formulae.

Derivation. Set $B = A$ in $\sin(A + B)$ : $\sin 2A = 2\sin A \cos A.$

Set $B = A$ in $\cos(A + B)$ : $\cos 2A = \cos^{2}A - \sin^{2}A.$

Two equivalent forms via $\sin^{2} + \cos^{2} = 1$ : $\cos 2A = 2\cos^{2}A - 1 = 1 - 2\sin^{2}A.$

For tan: $\tan 2A = \dfrac{2\tan A}{1 - \tan^{2}A}.$

Which form of $\cos 2A$ to use.

If the rest of the expression has $\cos^{2}$ : use $\cos 2A = 2\cos^{2}A - 1$ .
If $\sin^{2}$ : use $1 - 2\sin^{2}A$ .
For 'show that' or algebra: use whatever gives the cleanest collapse.

Worked example. Prove $\dfrac{1 - \cos 2A}{\sin 2A} \equiv \tan A$ .

Numerator: $1 - \cos 2A = 1 - (1 - 2\sin^{2}A) = 2\sin^{2}A$ .

Denominator: $\sin 2A = 2\sin A \cos A$ .

LHS = $\dfrac{2\sin^{2}A}{2\sin A\cos A} = \dfrac{\sin A}{\cos A} = \tan A$ . ✓

Half-angle relations. Rearranging the double-angle gives: $\sin^{2}A = \dfrac{1 - \cos 2A}{2}, \qquad \cos^{2}A = \dfrac{1 + \cos 2A}{2}.$

These are essential for integrating $\sin^{2}$ and $\cos^{2}$ (see P3 Integration 3).

$\sin 2A = 2\sin A\cos A$ .
$\cos 2A$ has THREE forms — pick to suit the problem.
$\tan 2A = \dfrac{2\tan A}{1 - \tan^{2}A}$ .
$\sin^{2}A = \dfrac{1 - \cos 2A}{2}$ — essential for integration.

The R-formula: combining $a\sin\theta + b\cos\theta$

One sine wave with amplitude $R$ and phase shift $\alpha$ .

Claim. For any constants $a, b$ (not both zero), there exist $R > 0$ and $\alpha$ such that $a\sin\theta + b\cos\theta = R\sin(\theta + \alpha).$

Derivation. Expand the RHS: $R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha = (R\cos\alpha)\sin\theta + (R\sin\alpha)\cos\theta.$

Match coefficients: $R\cos\alpha = a, \qquad R\sin\alpha = b.$

Square and add: $R^{2}(\cos^{2}\alpha + \sin^{2}\alpha) = a^{2} + b^{2} \Rightarrow R = \sqrt{a^{2} + b^{2}}$ (take positive root).

Divide: $\tan\alpha = \dfrac{b}{a}$ .

Worked example. Express $3\sin\theta + 4\cos\theta$ in the form $R\sin(\theta + \alpha)$ .

$R = \sqrt{9 + 16} = 5$ .
$\tan\alpha = 4/3 \Rightarrow \alpha = \arctan(4/3) \approx 0.927$ rad.

So $3\sin\theta + 4\cos\theta = 5\sin(\theta + 0.927)$ .

Other equivalent forms.

Form	Setup	Useful when...
$R\sin(\theta + \alpha)$	$R\cos\alpha = a, R\sin\alpha = b$	$a, b > 0$
$R\sin(\theta - \alpha)$	$R\cos\alpha = a, R\sin\alpha = -b$	$b < 0$
$R\cos(\theta - \alpha)$	$R\cos\alpha = b, R\sin\alpha = a$	the cosine version is cleaner

Max and min.

Max of $a\sin\theta + b\cos\theta = R$ (when $\sin(\theta + \alpha) = 1$ ).
Min $= -R$ .
Max reached at $\theta = \pi/2 - \alpha$ .
Min reached at $\theta = -\pi/2 - \alpha$ .

Solving equations. $a\sin\theta + b\cos\theta = c$ becomes $R\sin(\theta + \alpha) = c$ , so $\sin(\theta + \alpha) = c/R$ — a single-variable trig equation. Solvable provided $|c| \leq R$ .

$a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)$ .
$R = \sqrt{a^{2} + b^{2}}$ , $\tan\alpha = b/a$ .
Max value $= R$ , min value $= -R$ .
Use to solve $a\sin\theta + b\cos\theta = c$ .

R-formula applications

Max/min in modelling; solving equations; finding ranges.

Application 1: Maximum / minimum in modelling.

A pendulum's horizontal displacement is $x(t) = 3\sin t + 4\cos t$ . Find the maximum displacement.

Rewrite: $x(t) = 5\sin(t + 0.927)$ .
Max = $5$ (amplitude).
Reached when $t + 0.927 = \pi/2 \Rightarrow t = 0.644$ .

Application 2: Solving $a\sin\theta + b\cos\theta = c$ .

Solve $3\sin\theta + 4\cos\theta = 2$ for $0 \leq \theta \leq 2\pi$ .

$5\sin(\theta + 0.927) = 2 \Rightarrow \sin(\theta + 0.927) = 0.4$ .
Let $\phi = \theta + 0.927$ . $\sin\phi = 0.4$ .
Principal value: $\phi_{1} = \arcsin(0.4) \approx 0.412$ . Second solution in $[0, 2\pi]$ : $\phi_{2} = \pi - 0.412 \approx 2.730$ .
Account for the shift: $\phi$ ranges over $[0.927, 2\pi + 0.927]$ . Need to add $2\pi$ to $\phi_{1}$ to put it in range: $\phi_{1}' \approx 6.696$ .
Convert back: $\theta = \phi - 0.927$ .

Solutions: $\theta \approx 1.804, \;\; 5.768$ .

Application 3: Range of an expression.

The expression $5 - 3\sin\theta - 4\cos\theta$ has range $[5 - R, 5 + R] = [0, 10]$ .

Why? $3\sin\theta + 4\cos\theta$ ranges over $[-5, 5]$ . Subtract from $5$ : $[5 - 5, 5 + 5] = [0, 10]$ .

Application 4: Inequalities. $a\sin\theta + b\cos\theta \geq c$ becomes $R\sin(\theta + \alpha) \geq c$ , an inequality in a single sine. Solve and convert back.

Use R-formula to find max/min of $a\sin + b\cos$ .
Solve equations by reducing to one trig function.
Range of $k - a\sin\theta - b\cos\theta$ is $[k - R, k + R]$ .
Inequalities work the same way.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$ (sign matches).
$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$ (sign flips).
$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$ .
$\sin 2A = 2\sin A\cos A$ .
$\cos 2A = \cos^{2}A - \sin^{2}A = 2\cos^{2}A - 1 = 1 - 2\sin^{2}A$ .
$\tan 2A = \dfrac{2\tan A}{1 - \tan^{2}A}$ .
$a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)$ , $R = \sqrt{a^{2} + b^{2}}$ , $\tan\alpha = b/a$ .
Max value $= R$ , min value $= -R$ .

How it’s examined

Trigonometric addition formulae appear on EVERY WMA13/01 paper, often as a multi-part question worth $10$ – $15$ marks. Typical setup: part (a) 'show that $a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)$ ' ( $4$ marks for $R$ and $\alpha$ ); part (b) 'find the maximum value' ( $2$ marks); part (c) 'solve $a\sin\theta + b\cos\theta = k$ ' ( $5$ - $6$ marks); part (d) 'sketch / state range / sketch and shade'. Marks lost: students forget to state both solutions in the $[0, 2\pi]$ interval or use degrees vs radians inconsistently. Always quote the form requested.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA13/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Trigonometric addition formulae

Step-by-step solutions to past-paper-style questions on trigonometric addition formulae, written exactly the way a tutor would explain them at the board.

1Exact value of $\sin 75°$ using the addition formula (3 marks, P3)
Core• Adapted from WMA13/01 January 2024 Q2• addition formula, exact value
Question
Find the exact value of $\sin 75°$ , giving your answer in simplified surd form.
Step-by-step solution
1. Step 1
  $75° = 45° + 30°$ . Use $\sin(A + B) = \sin A \cos B + \cos A \sin B$ .
2. Step 2
  $\sin 75° = \sin 45° \cos 30° + \cos 45° \sin 30°$ .
  $= \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2}$
3. Step 3
  $= \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ .
Answer
$\sin 75° = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ .
Examiner tip
M1 for splitting as $45° + 30°$ (or $60° + 15°$ etc.). A1 for both correct exact values. A1 for the fully-simplified surd form. Common errors: writing $\dfrac{\sqrt{2} + \sqrt{6}}{2}$ (forgetting the second halving) or leaving in unsimplified form.
2Derive $\cos 2A$ in three forms and prove an identity (5 marks, P3)
Extended• Adapted from WMA13/01 June 2024 Q6• double angle, derivation, identity
Question
(a) Use $\cos(A + B) = \cos A \cos B - \sin A \sin B$ to derive $\cos 2A$ in three equivalent forms. (b) Hence prove $\dfrac{1 - \cos 2A}{\sin 2A} \equiv \tan A$ .
Step-by-step solution
1. Step 1
  (a) Set $B = A$ : $\cos 2A = \cos^{2}A - \sin^{2}A$ .
2. Step 2
  Using $\sin^{2}A = 1 - \cos^{2}A$ : $\cos 2A = \cos^{2}A - (1 - \cos^{2}A) = 2\cos^{2}A - 1$ . Using $\cos^{2}A = 1 - \sin^{2}A$ : $\cos 2A = 1 - 2\sin^{2}A$ .
3. Step 3
  (b) Numerator: $1 - \cos 2A = 1 - (1 - 2\sin^{2}A) = 2\sin^{2}A$ . Denominator: $\sin 2A = 2\sin A\cos A$ .
4. Step 4
  Combine:
  $\dfrac{2\sin^{2}A}{2\sin A \cos A} = \dfrac{\sin A}{\cos A} = \tan A$
5. Step 5
  Identity proved. ∎
Answer
(a) $\cos 2A = \cos^{2}A - \sin^{2}A = 2\cos^{2}A - 1 = 1 - 2\sin^{2}A$ . (b) $\dfrac{1 - \cos 2A}{\sin 2A} = \dfrac{2\sin^{2}A}{2\sin A \cos A} = \tan A$ .
Examiner tip
(a) M1 for setting $B = A$ in $\cos(A + B)$ . A1 each for the two alternative forms (B1 B1). (b) M1 substitution into numerator and denominator. A1 simplification. A1 final identity statement. Pick the right form of $\cos 2A$ for the problem: here using $1 - 2\sin^{2}A$ kills the $1$ in the numerator cleanly.
3Express $3\sin\theta + 4\cos\theta$ in the form $R\sin(\theta + \alpha)$ (4 marks, P3)
Extended• Adapted from WMA13/01 January 2023 Q8(a)• R-formula, asin+bcos
Question
Express $3\sin\theta + 4\cos\theta$ in the form $R\sin(\theta + \alpha)$ , where $R > 0$ and $0 < \alpha < \tfrac{\pi}{2}$ . Give $\alpha$ to 3 d.p.
Step-by-step solution
1. Step 1
  Expand $R\sin(\theta + \alpha) = R\cos\alpha \cdot \sin\theta + R\sin\alpha \cdot \cos\theta$ . Match coefficients:
  $R\cos\alpha = 3, \quad R\sin\alpha = 4$
2. Step 2
  Find $R$ by squaring and adding:
  $R^{2} = 3^{2} + 4^{2} = 25 \Rightarrow R = 5$
3. Step 3
  Find $\alpha$ by dividing:
  $\tan\alpha = \dfrac{4}{3} \Rightarrow \alpha = \arctan(4/3) \approx 0.927 \text{ rad}$
4. Step 4
  So $3\sin\theta + 4\cos\theta \approx 5\sin(\theta + 0.927)$ .
Answer
$R = 5$ , $\alpha \approx 0.927$ . So $3\sin\theta + 4\cos\theta \approx 5\sin(\theta + 0.927)$ .
Examiner tip
M1 expand $R\sin(\theta + \alpha)$ and equate coefficients. A1 for $R = 5$ . M1 for $\tan\alpha = 4/3$ . A1 for $\alpha \approx 0.927$ . Always keep $R > 0$ (so $R = +5$ , not $-5$ ). The form determines the sign convention: $R\sin(\theta + \alpha)$ gives $+$ inside; $R\sin(\theta - \alpha)$ would give $a\sin - b\cos$ .
4Use the R-formula to solve and find a maximum (7 marks, P3)
Extended• Adapted from WMA13/01 June 2024 Q9• R-formula, solve, max-min
Question
(a) Using $3\sin\theta + 4\cos\theta \equiv 5\sin(\theta + 0.927)$ , find the maximum value of $3\sin\theta + 4\cos\theta$ and the corresponding value of $\theta$ in the interval $0 \leq \theta \leq 2\pi$ . (b) Solve $3\sin\theta + 4\cos\theta = 2$ for $0 \leq \theta \leq 2\pi$ .
Step-by-step solution
1. Step 1
  (a) $5\sin(\theta + 0.927)$ has max $5$ , reached when $\sin(\theta + 0.927) = 1$ , i.e. $\theta + 0.927 = \tfrac{\pi}{2}$ .
  $\theta = \tfrac{\pi}{2} - 0.927 \approx 0.644$
2. Step 2
  (b) $5\sin(\theta + 0.927) = 2 \Rightarrow \sin(\theta + 0.927) = 0.4$ .
3. Step 3
  Let $\phi = \theta + 0.927$ . Then $\sin\phi = 0.4$ gives principal value $\phi_{1} = \arcsin(0.4) \approx 0.412$ . Second solution: $\phi_{2} = \pi - 0.412 \approx 2.730$ . Within $0 \leq \theta \leq 2\pi$ , we need $0.927 \leq \phi \leq 2\pi + 0.927 \approx 7.210$ .
4. Step 4
  $\phi_{1} \approx 0.412$ is below $0.927$ — add $2\pi$ : $\phi_{1}' \approx 6.696$ , in range. $\phi_{2} \approx 2.730$ , in range. Convert back: $\theta = \phi - 0.927$ .
  $\theta \approx 2.730 - 0.927 = 1.804, \quad \theta \approx 6.696 - 0.927 = 5.768$
5. Step 5
  Two solutions in $[0, 2\pi]$ .
Answer
(a) Max $= 5$ at $\theta \approx 0.644$ . (b) $\theta \approx 1.804$ or $\theta \approx 5.768$ .
Examiner tip
(a) M1 max value = $R = 5$ . A1 value of $\theta$ . (b) M1 setting $5\sin(\theta + \alpha) = 2$ . A1 $\sin(\theta + \alpha) = 0.4$ . M1 finding both principal-value solutions. A1 each $\theta$ value. Always handle the angle shift carefully — adjust by $2\pi$ if your $\phi$ falls outside the corresponding $\theta$ range.
5Use $\tan(A + B)$ to find an exact angle (4 marks, P3)
Extended• Adapted from WMA13/01 January 2024 Q3• tan addition, exact value
Question
Given $\tan A = 2$ and $\tan B = \tfrac{1}{3}$ , find the exact value of $\tan(A + B)$ .
Step-by-step solution
1. Step 1
  Use $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$ .
2. Step 2
  Substitute:
  $\tan(A + B) = \dfrac{2 + 1/3}{1 - 2 \cdot 1/3} = \dfrac{7/3}{1/3} = 7$
3. Step 3
  $A + B = \arctan 7$ (if needed). Exact value asked is $\tan(A + B) = 7$ .
Answer
$\tan(A + B) = 7$ .
Examiner tip
M1 quote the formula. A1 substitution. A1 simplification. A1 final value. The formula IS in the IAL booklet but writing it down still earns the method mark cleanly. Watch the sign of the denominator: it is $1 - \tan A \tan B$ (minus), not plus.

Key Formulae — Trigonometric addition formulae

The formulae you need to memorise for trigonometric addition formulae on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Sine addition formulae
$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
$A, B$
any angles (radians or degrees)
When to use
Splitting an angle into a sum/difference of known angles; deriving double-angle formulae; expanding $R\sin(\theta + \alpha)$ . IN the IAL formula booklet.
Example
$\sin 75° = \sin(45° + 30°) = \sin 45° \cos 30° + \cos 45° \sin 30° = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ .
Cosine addition formulae
$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$
$A, B$
any angles
When to use
Note the SIGN FLIP in the $\sin\sin$ term — opposite of the LHS sign. IN the IAL formula booklet.
Example
$\cos(60° - 30°) = \cos 60° \cos 30° + \sin 60° \sin 30° = \tfrac{1}{2} \cdot \tfrac{\sqrt{3}}{2} + \tfrac{\sqrt{3}}{2} \cdot \tfrac{1}{2} = \tfrac{\sqrt{3}}{2} = \cos 30°$ . ✓
Tangent addition formulae
$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$
$A, B$
angles, with $A \pm B \neq \tfrac{\pi}{2} + k\pi$
When to use
Whenever you know $\tan A$ and $\tan B$ separately. IN the IAL formula booklet.
Example
$\tan A = 2$ , $\tan B = 1/3 \Rightarrow \tan(A + B) = (2 + 1/3)/(1 - 2/3) = 7$ .
Double-angle formulae
$\sin 2A = 2\sin A \cos A, \quad \cos 2A = \cos^{2}A - \sin^{2}A = 2\cos^{2}A - 1 = 1 - 2\sin^{2}A, \quad \tan 2A = \dfrac{2\tan A}{1 - \tan^{2}A}$
$A$
any angle (with appropriate domain for $\tan$ )
When to use
Derived from the addition formulae by setting $B = A$ . The three forms of $\cos 2A$ each have a use — pick the one that matches the form of the rest of the equation. NOT in the formula booklet (derive from addition formulae if needed).
Example
Integrating $\sin^{2}x$ : use $\sin^{2}x = \tfrac{1 - \cos 2x}{2}$ , so $\int \sin^{2}x\,dx = \tfrac{x}{2} - \tfrac{\sin 2x}{4} + C$ .
R-formula ( $a\sin\theta + b\cos\theta$ )
$a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha), \quad R = \sqrt{a^{2} + b^{2}}, \quad \tan\alpha = \dfrac{b}{a}$
$R$
amplitude, $R > 0$
$\alpha$
phase shift, $0 < \alpha < \tfrac{\pi}{2}$ if $a, b > 0$
When to use
Whenever the question gives $a\sin\theta + b\cos\theta$ — combines into a single sine with amplitude $R$ and phase $\alpha$ . Max of the expression is $R$ , min is $-R$ . NOT in the formula booklet — memorise and derive.
Example
$3\sin\theta + 4\cos\theta = 5\sin(\theta + 0.927)$ ; max $= 5$ , min $= -5$ .

Key Definitions and Keywords — Trigonometric addition formulae

Definitions to memorise and the exact keywords mark schemes credit for trigonometric addition formulae answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Addition (compound-angle) formula
Examiner keyword
An identity that expresses $\sin(A \pm B)$ , $\cos(A \pm B)$ or $\tan(A \pm B)$ in terms of trig functions of $A$ and $B$ separately. All six identities are listed in the IAL formula booklet.
Double-angle formula
Examiner keyword
A special case of the addition formula with $A = B$ . Common forms: $\sin 2A = 2\sin A\cos A$ ; $\cos 2A$ has three equivalent expressions; $\tan 2A = \dfrac{2\tan A}{1 - \tan^{2}A}$ . Derive from the addition formula in proofs.
R-formula / harmonic form
Examiner keyword
The rewrite $a\sin\theta + b\cos\theta \equiv R\sin(\theta + \alpha)$ (or equivalently $R\cos(\theta - \beta)$ ) where $R = \sqrt{a^{2} + b^{2}} > 0$ is the amplitude. Useful for finding max/min and for solving equations with a sine and cosine mixed.
Amplitude $R$ and phase shift $\alpha$
Examiner keyword
In the R-form, $R$ is the maximum value of the expression and $\alpha$ is the horizontal shift (delay) of the equivalent single sine wave.

Common Mistakes and Misconceptions — Trigonometric addition formulae

The traps other students keep falling into on trigonometric addition formulae questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Writing $\cos(A + B) = \cos A \cos B + \sin A \sin B$
WMA13/01 examiner reports — flagged every sitting
Why it happens
Copying the sign pattern of the sine formula.
How to avoid it
Mnemonic: $\cos$ has a SIGN FLIP — $\cos(A + B) = \cos A \cos B - \sin A \sin B$ . Think 'minus for plus, plus for minus' in the cosine case. The formula booklet has it correctly — copy from there.
✕Using $\cos 2A = \cos^{2}A - \sin^{2}A$ when integrating $\sin^{2}A$
Why it happens
Default to the most-memorable form.
How to avoid it
For $\int \sin^{2}x\,dx$ , use $\sin^{2}x = \tfrac{1 - \cos 2x}{2}$ . For $\int \cos^{2}x\,dx$ , use $\cos^{2}x = \tfrac{1 + \cos 2x}{2}$ . Each form has its purpose — pick to eliminate the unwanted term.
✕Quoting $R = -5$
Why it happens
Both signs satisfy $R^{2} = 25$ .
How to avoid it
By convention $R > 0$ . The phase $\alpha$ absorbs any necessary sign. Always quote the POSITIVE square root.
✕Quoting $\alpha$ outside $(0, \pi/2)$ for the form $R\sin(\theta + \alpha)$ with $a, b > 0$
Why it happens
Just running $\arctan(b/a)$ on the calculator without thinking.
How to avoid it
When both $a$ and $b$ are positive, $\alpha = \arctan(b/a)$ lies in $(0, \pi/2)$ — the principal-value range. If the question fixes a different range (e.g. $-\pi/2 < \alpha < \pi/2$ ), check the resulting form matches the original by expansion.
✕ $\tan(A + B) = \dfrac{\tan A + \tan B}{1 + \tan A \tan B}$
Why it happens
Confusing the sign pattern.
How to avoid it
Mnemonic: 'Numerator follows LHS sign, denominator flips it.' $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$ (plus on top, minus on bottom).

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Trigonometric addition formulae — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Trigonometric addition formulae

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Exact value of sin⁡75°\sin 75°sin75° using the addition formula (3 marks, P3)

2Derive cos⁡2A\cos 2Acos2A in three forms and prove an identity (5 marks, P3)

3Express 3sin⁡θ+4cos⁡θ3\sin\theta + 4\cos\theta3sinθ+4cosθ in the form Rsin⁡(θ+α)R\sin(\theta + \alpha)Rsin(θ+α) (4 marks, P3)

4Use the R-formula to solve and find a maximum (7 marks, P3)

5Use tan⁡(A+B)\tan(A + B)tan(A+B) to find an exact angle (4 marks, P3)

Sine addition formulae

Cosine addition formulae

Tangent addition formulae

Double-angle formulae

R-formula (asin⁡θ+bcos⁡θa\sin\theta + b\cos\thetaasinθ+bcosθ)

Addition (compound-angle) formula

Double-angle formula

R-formula / harmonic form

Amplitude RRR and phase shift α\alphaα

✕Writing cos⁡(A+B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A + B) = \cos A \cos B + \sin A \sin Bcos(A+B)=cosAcosB+sinAsinB

✕Using cos⁡2A=cos⁡2A−sin⁡2A\cos 2A = \cos^{2}A - \sin^{2}Acos2A=cos2A−sin2A when integrating sin⁡2A\sin^{2}Asin2A

✕Quoting R=−5R = -5R=−5

✕Quoting α\alphaα outside (0,π/2)(0, \pi/2)(0,π/2) for the form Rsin⁡(θ+α)R\sin(\theta + \alpha)Rsin(θ+α) with a,b>0a, b > 0a,b>0

✕tan⁡(A+B)=tan⁡A+tan⁡B1+tan⁡Atan⁡B\tan(A + B) = \dfrac{\tan A + \tan B}{1 + \tan A \tan B}tan(A+B)=1+tanAtanBtanA+tanB​

Practice questions

Past paper style quiz

Video lesson

Trigonometric addition formulae — frequently asked questions

1Exact value of $\sin 75°$ using the addition formula (3 marks, P3)

2Derive $\cos 2A$ in three forms and prove an identity (5 marks, P3)

3Express $3\sin\theta + 4\cos\theta$ in the form $R\sin(\theta + \alpha)$ (4 marks, P3)

5Use $\tan(A + B)$ to find an exact angle (4 marks, P3)

R-formula ( $a\sin\theta + b\cos\theta$ )

Amplitude $R$ and phase shift $\alpha$

✕Writing $\cos(A + B) = \cos A \cos B + \sin A \sin B$

✕Using $\cos 2A = \cos^{2}A - \sin^{2}A$ when integrating $\sin^{2}A$

✕Quoting $R = -5$

✕Quoting $\alpha$ outside $(0, \pi/2)$ for the form $R\sin(\theta + \alpha)$ with $a, b > 0$

✕ $\tan(A + B) = \dfrac{\tan A + \tan B}{1 + \tan A \tan B}$