The Addition Formulae

Use the formulae for sin, cos and tan of (A ± B) to find exact values, derive the double-angle formulae, prove identities and solve equations.

In the exam for Edexcel Further Pure Maths 4PM1 · 25 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Why not just add?

It is tempting to think $sin (3 0^{\circ} + 6 0^{\circ}) = sin 3 0^{\circ} + sin 6 0^{\circ}$ . Test it. The left side is $sin 9 0^{\circ} = 1$ . The right side is $0.5 + 0.866 = 1.366$ . They are not equal.

So the sine of a sum needs its own formula. It mixes sines and cosines of both angles:

Formula

sin (A + B) = sin A cos B + cos A sin B

$A$ — the first angle.
$B$ — the second angle. Each term has one sin and one cos, one of A and one of B.

Formula

sin (A - B) = sin A cos B - cos A sin B

$A$ — the angle you start from.
$B$ — the angle you take away. For sin, the sign in the middle matches the sign in the bracket.

Worked example

1
Check the formula with $A = 6 0^{\circ}$ and $B = 3 0^{\circ}$ .
2
$sin 9 0^{\circ} = sin 6 0^{\circ} cos 3 0^{\circ} + cos 6 0^{\circ} sin 3 0^{\circ}$ — Put the two angles into the formula.
3
$= \frac{3}{2} \times \frac{3}{2} + \frac{1}{2} \times \frac{1}{2}$ — Exact values.
4
$= \frac{3}{4} + \frac{1}{4} = 1$ — It matches sin 90° = 1.

Rule

The 4PM1 formula sheet prints all six addition formulae: sin, cos and tan of $A + B$ and $A - B$ . You do not need to memorise them, but you must use them quickly and correctly. The double-angle formulae later in this lesson are not printed.

Key idea

Around 150 CE, Ptolemy in Alexandria used a result equivalent to these formulae to build his table of chords: one of the first trig tables, made for astronomy.

Cos and tan

Cos has its own pattern: cos with cos, sin with sin, and the sign flips.

Formula

cos (A + B) = cos A cos B - sin A sin B

$-$ — a plus in the bracket gives a minus in the middle.

Formula

cos (A - B) = cos A cos B + sin A sin B

$+$ — a minus in the bracket gives a plus in the middle.

Memory trick

Sin keeps the sign. Cos changes the sign.

Tan is sin over cos, so its formula is a fraction:

Formula

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

$A, B$ — any angles where every tan exists. The top keeps the sign; the bottom has the opposite sign.

Formula

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

$A, B$ — same pattern: minus on top, plus underneath.

Worked example

1
Find the exact value of $tan 7 5^{\circ}$ .
2
$tan 7 5^{\circ} = tan (4 5^{\circ} + 3 0^{\circ})$ — Split 75° into two angles with exact values.
3
$= \frac{1 + \frac{3}{3}}{1 - 1 \times \frac{3}{3}}$ — tan 45° = 1 and tan 30° = √3/3.
4
$= \frac{3 + 3}{3 - 3}$ — Multiply top and bottom by 3.
5
$= \frac{( 3 + 3 ) ^{2}}{9 - 3} = \frac{12 + 6 3}{6}$ — Rationalise: multiply top and bottom by 3 + √3.
6
$= 2 + 3$ — Divide both terms by 6.

Worked example

1
Use the minus formulae to find the exact values of $cos 1 5^{\circ}$ and $tan 1 5^{\circ}$ .
2
$cos (4 5^{\circ} - 3 0^{\circ}) = cos 4 5^{\circ} cos 3 0^{\circ} + sin 4 5^{\circ} sin 3 0^{\circ}$ — A minus in the bracket gives a plus in the middle.
3
$= \frac{2}{2} \times \frac{3}{2} + \frac{2}{2} \times \frac{1}{2} = \frac{6 + 2}{4}$ — Exact values, then multiply the surds.
4
$tan (4 5^{\circ} - 3 0^{\circ}) = \frac{1 - \frac{3}{3}}{1 + \frac{3}{3}} = \frac{3 - 3}{3 + 3}$ — Minus on top, plus underneath. Then multiply top and bottom by 3.
5
$= \frac{( 3 - 3 ) ^{2}}{9 - 3} = \frac{12 - 6 3}{6} = 2 - 3$ — Rationalise with 3 − √3, then divide by 6.

Common mistake

Keeping the sign for cos: $cos (A + B) = cos A cos B + sin A sin B$ is wrong. Check with $A = B = 4 5^{\circ}$ : $cos 9 0^{\circ}$ is 0, so the middle sign must be minus.

Key idea

Sin: mixed pairs, same sign. Cos: matching pairs, opposite sign. Tan: a fraction with the opposite sign underneath.

Exact values

The formulae turn angles like $1 5^{\circ}$ , $7 5^{\circ}$ and $10 5^{\circ}$ into sums or differences of $3 0^{\circ}$ , $4 5^{\circ}$ and $6 0^{\circ}$ . That gives exact answers with surds.

Worked example

1
Find the exact value of $sin 1 5^{\circ}$ .
2
$sin 1 5^{\circ} = sin (4 5^{\circ} - 3 0^{\circ})$ — 15° is 45° take away 30°.
3
$= sin 4 5^{\circ} cos 3 0^{\circ} - cos 4 5^{\circ} sin 3 0^{\circ}$ — The sin(A − B) formula.
4
$= \frac{2}{2} \times \frac{3}{2} - \frac{2}{2} \times \frac{1}{2}$ — Exact values.
5
$= \frac{6 - 2}{4}$ — Multiply: √2 × √3 = √6, and both fractions are over 4.

Another common question gives you the ratios as fractions and tells you which kind of angle each is.

Worked example

1
Angle A is acute with $sin A = \frac{3}{5}$ . Angle B is obtuse with $cos B = - \frac{5}{13}$ . Find the exact values of $sin (A + B)$ and $cos (A + B)$ .
2
$cos A = \frac{4}{5}$ — A 3, 4, 5 triangle. A is acute, so cos A is positive.
3
$sin B = \frac{12}{13}$ — A 5, 12, 13 triangle. B is obtuse, in the second quadrant, so sin B is positive.
4
$sin (A + B) = \frac{3}{5} \times (- \frac{5}{13}) + \frac{4}{5} \times \frac{12}{13}$ — sin A cos B + cos A sin B.
5
$= \frac{- 15 + 48}{65} = \frac{33}{65}$ — Multiply the fractions and add.
6
$cos (A + B) = \frac{4}{5} \times (- \frac{5}{13}) - \frac{3}{5} \times \frac{12}{13}$ — cos A cos B − sin A sin B.
7
$= \frac{- 20 - 36}{65} = - \frac{56}{65}$ — Multiply and subtract.

You can also expand with one letter and one number. This is the first step in many equations.

Worked example

1
Expand $sin (x + 3 0^{\circ})$ exactly.
2
$sin x cos 3 0^{\circ} + cos x sin 3 0^{\circ}$ — The sin(A + B) formula with A = x and B = 30°.
3
$= \frac{3}{2} sin x + \frac{1}{2} cos x$ — Put in the exact values.

Common mistake

Losing a sign from the quadrant. For an obtuse angle cos is negative but sin is positive. Decide each sign before you substitute.

Key idea

Split the angle into friendly angles, use the formula, then simplify the surds.

Double angles

Put $B = A$ in the addition formulae and you get formulae for $2 A$ . They are not on the formula sheet, and exam questions often ask you to show them from the printed formulae. So learn how to get them.

$sin (A + A) = sin A cos A + cos A sin A$ — the sin(A + B) formula with B = A.

$sin 2 A = 2 sin A cos A$ — the two terms are the same.

Formula

sin 2 A = 2 sin A cos A

$A$ — any angle. The 2 goes in front; it does not double the sine.

$cos (A + A) = cos A cos A - sin A sin A$ — the cos(A + B) formula with B = A.

$cos 2 A = cos^{2} A - sin^{2} A$ — first form.

$= cos^{2} A - (1 - cos^{2} A) = 2 cos^{2} A - 1$ — swap sin² A for 1 − cos² A.

$= (1 - sin^{2} A) - sin^{2} A = 1 - 2 sin^{2} A$ — or swap cos² A for 1 − sin² A.

Formula

cos 2 A = cos^{2} A - sin^{2} A = 2 cos^{2} A - 1 = 1 - 2 sin^{2} A

$A$ — any angle. Three forms of the same thing: pick the one that leaves the ratio you want.

Formula

tan 2 A = \frac{2 tan A}{1 - tan ^{2} A}

$A$ — any angle where the bottom is not 0. It comes from tan(A + B) with B = A.

Rule

When a question says “use formulae from the formula sheet to show that”, write the A + A line first. Mark schemes give a method mark for the addition formula with B = A, another for using $sin^{2} A + cos^{2} A = 1$ , then the accuracy mark.

Worked example

1
Given that $cos A = \frac{1}{3}$ and A is acute, find the exact values of $cos 2 A$ and $sin 2 A$ .
2
$cos 2 A = 2 (\frac{1}{3})^{2} - 1 = \frac{2}{9} - 1 = - \frac{7}{9}$ — Use the form with cos only.
3
$sin A = 1 - \frac{1}{9} = \frac{2 2}{3}$ — From sin² A + cos² A = 1. A is acute, so sin A is positive.
4
$sin 2 A = 2 \times \frac{2 2}{3} \times \frac{1}{3} = \frac{4 2}{9}$ — Two times sin A times cos A.

Common mistake

Writing $sin 2 A = 2 sin A$ . Test with $A = 9 0^{\circ}$ : $sin 18 0^{\circ} = 0$ , but $2 sin 9 0^{\circ} = 2$ .

Key idea

Every double-angle formula is an addition formula with B = A. If you forget one, rebuild it in one line.

Proving identities

Proofs here use the same plan as before: start with the harder side and change it, line by line, into the other side. The new move is to replace a double angle with one of its forms.

Steps

1
1. Expand any $(A + B)$ or $2 A$ with a formula.
2. Choose the cos 2A form that cancels a 1 or matches the other side.
3. Use $sin^{2} A + cos^{2} A = 1$ and $tan A = \frac{s i n A}{c o s A}$ to tidy up.
4. Stop when you reach the other side exactly.

Worked example

1
Prove that $\frac{1 - c o s 2 A}{s i n 2 A} = tan A$ .
2
$1 - cos 2 A = 1 - (1 - 2 sin^{2} A) = 2 sin^{2} A$ — Choose the form with a 1 in it, so the 1s cancel.
3
$sin 2 A = 2 sin A cos A$ — The sin double angle.
4
$LHS = \frac{2 s i n ^{2} A}{2 s i n A c o s A}$ — Put the two parts together.
5
$= \frac{s i n A}{c o s A} = tan A = RHS$ — Cancel 2 sin A from the top and the bottom.

Worked example

1
Show that $sin (A + B) + sin (A - B) = 2 sin A cos B$ .
2
$sin (A + B) = sin A cos B + cos A sin B$ — Expand the first one.
3
$sin (A - B) = sin A cos B - cos A sin B$ — Expand the second one.
4
$LHS = 2 sin A cos B = RHS$ — Add: the cos A sin B terms cancel.

Common mistake

Choosing a cos 2A form at random. In the first proof, $cos^{2} A - sin^{2} A$ would leave a messy $1 - cos^{2} A + sin^{2} A$ . Pick the form that makes the 1 disappear.

Key idea

Look at what the other side contains, then pick the form of cos 2A that moves you towards it.

Solving equations

Many equations need a formula first, so that only one ratio of one angle is left. Then solve as usual.

Worked example

1
Solve $cos (x + 6 0^{\circ}) = sin x$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
2
$cos x cos 6 0^{\circ} - sin x sin 6 0^{\circ} = sin x$ — Expand with the cos(A + B) formula.
3
$\frac{1}{2} cos x - \frac{3}{2} sin x = sin x$ — Exact values.
4
$cos x = (2 + 3) sin x$ — Multiply by 2 and collect the sin x terms on one side.
5
$tan x = \frac{1}{2 + 3} = 2 - 3$ — Divide by cos x and by (2 + √3), then rationalise.
6
$x = 1 5^{\circ}, 19 5^{\circ}$ — tan 15° = 2 − √3 (or use the calculator), then add 180°.

Worked example

1
Solve $cos 2 x = sin x$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
2
$1 - 2 sin^{2} x = sin x$ — Choose the form with sin only, to match the right side.
3
$2 sin^{2} x + sin x - 1 = 0$ — Bring everything to one side.
4
$(2 sin x - 1) (sin x + 1) = 0$ — Factorise.
5
$sin x = \frac{1}{2} \Rightarrow x = 3 0^{\circ}, 15 0^{\circ}$ — The calculator value and 180° − 30°.
6
$sin x = - 1 \Rightarrow x = 27 0^{\circ}$ — The bottom of the sine wave.

Common mistake

Leaving both $2 x$ and $x$ in one equation and trying to solve. First use a double-angle formula so every term is in x.

Common mistake

Dividing by $cos x$ or $sin x$ when it is a common factor. Take it out as a factor instead, or you lose solutions.

Key idea

Expand, make one ratio of one angle, then solve. The formula is just the first step.

Which method? Read the question

The words and the angles in the question tell you which formula to reach for.

You see	You do
“Exact value of $sin 7 5^{\circ}$ ” (or $1 5^{\circ}$ , $10 5^{\circ}$ )	Split into $3 0^{\circ}$ , $4 5^{\circ}$ , $6 0^{\circ}$ ; use the addition formula
$sin A$ and $cos B$ given, A acute, B obtuse	Triangles for the other ratios, signs from the quadrant, then the formula
$sin (x + 3 0^{\circ})$ in an equation	Expand, then collect into $tan x$ or a quadratic
“Show that $cos 2 A$ = …”	Write $cos (A + A)$ first, then $sin^{2} A + cos^{2} A = 1$
$cos 2 x$ with $sin x$ (or $cos x$ )	Pick the cos 2x form with the same ratio, then a quadratic
$sin 2 x$ with $sin x$	$2 sin x cos x$ , then take out the common factor
$sin (A + B) + sin (A - B)$	Expand both and let terms cancel

Common mistake

Using a formula from memory with the wrong sign. Copy it from the formula sheet, then substitute.

Key idea

One formula, many uses: exact values, double angles, proofs and equations all start by expanding.

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Real past-paper questions on the addition formulae

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