The addition (compound angle) formulae are sin(A ± B) = sin A cos B ± cos A sin B, cos(A ± B) = cos A cos B ∓ sin A sin B and tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B). They give exact values for angles such as 75° and 15°, let you prove identities, and turn equations containing sin(x + 30°) or cos(x − 45°) into ones you can solve.
The addition formulae
- 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) = (tan A + tan B) / (1 − tan A tan B)
- tan(A − B) = (tan A − tan B) / (1 + tan A tan B)
In the sine formula the sign matches the one in the bracket. In the cosine formula it flips: cos(A + B) has a minus in the middle, cos(A − B) has a plus.
What are the addition formulae used for?
They expand the sine, cosine or tangent of a sum or difference of two angles into single-angle terms. That does three jobs at A-Level: finding exact values of angles that are not on the standard list, proving trig identities, and solving equations where the angle has been shifted, such as sin(x + 60°) = sin x.
The formulae appear in the exam formula booklet, but the sign change in the cosine formula still costs marks every year, because students write it from memory under pressure instead of checking.
How do you find exact values like sin 75° and cos 15°?
Split the angle into two angles whose sines and cosines you know exactly: 75° = 45° + 30°, 15° = 45° − 30°, 105° = 60° + 45°. Expand with the right formula, substitute the exact values, and simplify the surds. Answers usually come out in a form like (√6 + √2) / 4, and the question will normally tell you the form it wants.
Finding sin(A + B) from two given ratios
A standard question gives one ratio for A and one for B and asks for something like sin(A + B) or cos(A − B). For each angle, sketch a right-angled triangle and use Pythagoras to find the missing side, then fix each sign from the information about the angle. Only then substitute.
That sign step is where marks go. If A is obtuse, cos A is negative, and that single word changes the whole answer. With sin A = 3/5, the triangle gives cos A = ±4/5, and the quadrant decides which.
Proving identities with the addition formulae
Two moves cover almost every proof. Going forwards, expand each compound bracket and watch terms cancel. Going backwards, recognise a whole expression as the expanded side of a formula and compress it into a single bracket.
The backwards move is the one students miss. An expression like cos x cos 2x + sin x sin 2x has exactly the pattern cos A cos B + sin A sin B, so it compresses to cos(A − B). Adding or subtracting two expansions also makes terms cancel in pairs, which is why identities such as sin(A + B) + sin(A − B) = 2 sin A cos B collapse so neatly.
Solving compound-angle equations
You cannot solve an equation containing cos(x + 60°) directly. Expand it, substitute the exact values, and collect the sin x and cos x terms. Then divide every term by cos x to turn it into a single equation in tan x.
Divide by cos x only, and only when it cannot be zero. Dividing by sin x or by a bracket risks throwing solutions away.
Harder questions hide the formula. If an equation is full of terms like sin x cos y, dividing every term by sin x sin y produces cotangents, and dividing by cos x cos y produces tangents. Decide which by looking at what the question asks you to find.
Worked examples
Example 1. Find the exact value of sin 75°, giving your answer in the form (√a + √b) / 4.
- Split the angle: 75° = 45° + 30°.
- Sine formula (the sign matches): sin 75° = sin 45° cos 30° + cos 45° sin 30°.
- Substitute exact values: (√2/2)(√3/2) + (√2/2)(1/2).
- Simplify: √6/4 + √2/4 = (√6 + √2) / 4, so a = 6 and b = 2.
Answer: sin 75° = (√6 + √2) / 4.
Marks: M1 for the formula, M1 for exact values, A1 for the simplified answer.
Example 2. Given that sin A = 3/5 and sin B = 5/13, where A and B are both acute, find the exact value of sin(A + B).
- Triangle for A: sides 3, 4, 5, so cos A = 4/5 (positive, as A is acute).
- Triangle for B: sides 5, 12, 13, so cos B = 12/13 (positive, as B is acute).
- Use sin(A + B) = sin A cos B + cos A sin B.
- Substitute: (3/5)(12/13) + (4/5)(5/13) = 36/65 + 20/65 = 56/65.
Answer: sin(A + B) = 56/65.
Marks: M1 for both triangles, M1 for the formula, A1 for the value.
Example 3. Prove that cos x cos 2x + sin x sin 2x ≡ cos x.
- Recognise the pattern cos A cos B + sin A sin B, with A = x and B = 2x.
- That pattern is cos(A − B), so the left-hand side is cos(x − 2x) = cos(−x).
- The cosine curve is symmetric about the y-axis, so cos(−x) = cos x, which is the right-hand side.
Marks: M1 for recognising the pattern, A1 for compressing it, A1 for using cos(−x) = cos x.
Example 4. Solve cos(x + 60°) + sin x = 0 for 0° ≤ x < 360°.
- Expand (the cosine sign flips): cos x cos 60° − sin x sin 60° + sin x = 0.
- Exact values: ½ cos x − (√3/2) sin x + sin x = 0.
- Collect: ½ cos x + (1 − √3/2) sin x = 0.
- Divide every term by cos x: ½ + (1 − √3/2) tan x = 0, so tan x = −(2 + √3).
- 2 + √3 = tan 75°, so the base angle is 75°. Tangent is negative, so x = 180° − 75° = 105° and x = 105° + 180° = 285°.
Answer: x = 105°, 285°.
Marks: M1 for expanding, M1 for collecting and dividing, A1 for both solutions.
Common mistakes
- Getting the cosine sign the wrong way round. cos(A + B) = cos A cos B − sin A sin B. The sign flips in cosine, but not in sine.
- Writing sin(A + B) = sin A + sin B. Sine does not distribute over a bracket. For example, sin 90° = 1, but sin 60° + sin 30° is about 1.37.
- Ignoring the quadrant. When an angle is obtuse its cosine is negative. Fix every sign before you substitute.
- Dividing by sin x or by a bracket. Divide only by cos x, and only when cos x cannot be zero, otherwise solutions are lost.
- Leaving surds unsimplified. Exact-value answers usually have a required form, such as (√a + √b) / 4. Match it exactly to earn the final mark.
How the addition formulae are examined
This is Pure 3 content in Edexcel International A-Level (paper WMA13), in the chapter on trigonometric addition formulae, and Year 2 Pure content in Edexcel A-Level (9MA0). Questions usually ask for one of these:
- an exact value of an angle such as 75°, 15° or 105°
- sin(A ± B) or cos(A ± B) from given ratios, with obtuse angles to catch out signs
- a proof that compresses an expression into a single bracket
- a compound-angle equation solved by expanding and reducing to tan x = k
Try the Pure 3 addition formulae exam questions with full mark schemes, work through the trigonometric formulae worksheet, or get unlimited practice in the A-Level trigonometry revision engine.
Next steps: the double angle formulae come straight from these with B = A, and the R sin(θ + α) form uses them to combine two waves into one.
Frequently asked questions
Are the addition formulae in the formula booklet?
Yes. The formula booklet gives the addition formulae, but learn the sign pattern anyway: the cosine version flips the sign, and that is the mistake examiners see most often.
What is the difference between the addition formulae and the double angle formulae?
The double angle formulae are the special case of the addition formulae where both angles are equal. Put B = A in sin(A + B) and you get sin 2A = 2 sin A cos A.
How do you remember the cos(A + B) formula?
Cosine pairs cos with cos and sin with sin, then flips the sign: cos(A + B) = cos A cos B − sin A sin B. Sine mixes them and keeps the sign: sin(A + B) = sin A cos B + cos A sin B.
Why does dividing by cos x work in compound-angle equations?
Because sin x / cos x = tan x. Dividing every term by cos x turns an equation in both sin x and cos x into a single equation in tan x, which you can solve directly.