Compound Angle Identities

Compound angle identities expand expressions like $cos (α - β)$ and $sin (α + β)$ . The derivation of $cos (α - β)$ is taught for understanding but is not examined in the IEB exam, and the expansions are used for exact values, proofs and equations throughout Paper 2.

Deriving cos(α - β)

Place $P (cos α; sin α)$ and $Q (cos β; sin β)$ on the unit circle, so that angle $P O Q = α - β$ .

Distance formula:

$P Q^{2} = (cos α - cos β)^{2} + (sin α - sin β)^{2} = 2 - 2 (cos α cos β + sin α sin β)$

using $sin^{2} + cos^{2} = 1$ twice.

Cosine rule in triangle $P O Q$ with $O P = O Q = 1$ :

$P Q^{2} = 1 + 1 - 2 cos (α - β)$

Equating the two expressions:

$cos (α - β) = cos α cos β + sin α sin β$

Work through this derivation so you understand where the formulae come from. The IEB does not examine the derivation of the compound and double angle formulae.

Expansions of sin and cos of sums and differences

From $cos (α - β)$ , replacing $β$ with $- β$ and using co-functions gives the full set:

$cos (α - β) = cos α cos β + sin α sin β$
$cos (α + β) = cos α cos β - sin α sin β$
$sin (α + β) = sin α cos β + cos α sin β$
$sin (α - β) = sin α cos β - cos α sin β$

Exact values. Write the angle as a sum or difference of special angles.

$cos 75° = cos (45° + 30°) = \frac{1}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{1}{2} = \frac{6 - 2}{4}$

Recognising expansions. $sin 80° cos 20° - cos 80° sin 20° = sin (80° - 20°) = sin 60° = \frac{3}{2}$ . Spotting this pattern is often the whole question.

Key terms

Compound angle: An angle written as a sum or difference, such as $α + β$ .
Unit circle: A circle with radius 1 centred at the origin; a point on it at angle $θ$ is $(cos θ; sin θ)$ .
Exact value: A value written with surds and fractions, not a rounded decimal.

Common mistakes

Writing $cos (α + β) = cos α + cos β$ .
Getting the sign wrong in the cosine expansions: $cos (α - β)$ has a plus sign.
Swapping the order of terms in $sin (α - β)$ , which changes the sign.
Not finding the second ratio (for example $cos A$ from $sin A$ ) with the correct sign for the quadrant.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

Without using a calculator, determine the exact value of $cos 75°$ .

Worked solution

Hidden - try the question first

Solution

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

Question 2

5 marks total

routine

Prove that $sin (x + 30°) + cos (x + 60°) = cos x$ .

Worked solution

Hidden - try the question first

Solution

LHS $= sin x cos 30° + cos x sin 30° + cos x cos 60° - sin x sin 60°$
$= \frac{3}{2} sin x + \frac{1}{2} cos x + \frac{1}{2} cos x - \frac{3}{2} sin x$
$= cos x =$ RHS

Question 3

6 marks total

complex

(a) Prove that $cos (45° + x) cos (45° - x) = \frac{1}{2} cos 2 x$ . (4)

(b) Hence, without using a calculator, determine the value of $cos 75° cos 15°$ . (2)

Worked solution

Hidden - try the question first

Solution

(a) LHS $= sin x cos 30° + cos x sin 30° + cos x cos 60° - sin x sin 60°$
$= (\frac{1}{2} cos x - \frac{1}{2} sin x) (\frac{1}{2} cos x + \frac{1}{2} sin x)$
$= \frac{1}{2} cos^{2} x - \frac{1}{2} sin^{2} x$
$= \frac{1}{2} cos 2 x =$ RHS
(b) $x = 30°$ : $cos 75° cos 15° = \frac{1}{2} cos 60°$
$= \frac{1}{4}$

Question 4

8 marks total

problem-solving

$sin A = \frac{3}{5}$ with $A$ obtuse, and $cos B = \frac{12}{13}$ with $B$ acute. Without using a calculator, determine:

(a) $sin (A + B)$ (5)

(b) $cos (A - B)$ (3)

Worked solution

Hidden - try the question first

Solution

(a) $= sin x cos 30° + cos x sin 30° + cos x cos 60° - sin x sin 60°$ in quadrant 2: $cos A = - \frac{4}{5}$
$B$ in quadrant 1: $sin B = \frac{5}{13}$
$sin (A + B) = sin A cos B + cos A sin B$
$= \frac{3}{5} \cdot \frac{12}{13} + (- \frac{4}{5}) \cdot \frac{5}{13}$
$= \frac{36 - 20}{65} = \frac{16}{65}$
(b) $cos (A - B) = cos A cos B + sin A sin B$
$= - \frac{4}{5} \cdot \frac{12}{13} + \frac{3}{5} \cdot \frac{5}{13}$
$= \frac{- 48 + 15}{65} = - \frac{33}{65}$

FAQs

Is the compound angle proof examinable in IEB Grade 12?

No. The derivation of the compound and double angle formulae is not examined in the IEB exam. You need to apply the expansions fluently.

Are the compound angle formulae on the IEB formula sheet?

Yes, the expansions are on the formula sheet. You still need to apply them fluently.

How do I find an exact value like sin 15°?

Write it as a difference of special angles, such as $45° - 30°$ , then expand and simplify.

Deriving cos(α - β)

Expansions of sin and cos of sums and differences

Key terms

Common mistakes

Practice questions

FAQs

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