9709 P3 §3.3 — Compound Angle Formulae I

💡 Lead In: The Sine That Broke Arithmetic

The Scenario

A student writes down a confident claim:

$$\sin(30^{\circ}+60^{\circ})=\sin 30^{\circ}+\sin 60^{\circ}.$$

Let's check it numerically. $\sin(30^{\circ}+60^{\circ})=\sin 90^{\circ}=1$. But $\sin 30^{\circ}+\sin 60^{\circ}=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\approx 0.5+0.866=1.366$. These are not equal! The naive "sine distributes over addition" rule fails.

Review · Activating Prior Knowledge

Quick check — pick the correct statement in each question before we build today's compound-angle formulae.

Review 1 — Which set of exact values is correct?

$\sin 30^{\circ}=\dfrac{\sqrt{3}}{2}, \; \cos 30^{\circ}=\dfrac{1}{2}, \; \sin 90^{\circ}=\dfrac{1}{2}$

$\sin 30^{\circ}=\dfrac{1}{2}, \; \cos 30^{\circ}=\dfrac{1}{2}, \; \sin 60^{\circ}=1$

$\sin 30^{\circ}=\dfrac{1}{2}, \; \cos 30^{\circ}=\dfrac{\sqrt{3}}{2}, \; \sin 60^{\circ}=\dfrac{\sqrt{3}}{2}$

$\sin 60^{\circ}=\dfrac{1}{2}, \; \cos 60^{\circ}=\dfrac{\sqrt{3}}{2}, \; \cos 90^{\circ}=1$

$\sin 30^{\circ}=\dfrac{1}{2}$, $\cos 30^{\circ}=\dfrac{\sqrt{3}}{2}$ and $\sin 60^{\circ}=\dfrac{\sqrt{3}}{2}$. These exact values plug into today's formulae: for example $\sin 75^{\circ}=\sin(45^{\circ}+30^{\circ})$.

Review 2 — Which identity is true for every angle $\theta$?

$\sin^2\theta+\cos^2\theta=1$

$\sin\theta+\cos\theta=1$

$\sin^2\theta-\cos^2\theta=1$

$1+\tan^2\theta=\cos^2\theta$

The master identity $\sin^2\theta+\cos^2\theta=1$ holds for every angle and is the key tool for converting between $\sin$ and $\cos$ inside today's proofs. B is false (sines do not add that way), C has the wrong sign, and D mixes up $\tan$ — the correct form is $1+\tan^2\theta=\sec^2\theta$.

The Question

Since $\sin(A+B)\neq\sin A+\sin B$, there must be a correct expansion. Guess: which ingredients do you think appear on the right-hand side of $\sin(A+B)$? Only sines? Only cosines? A mix of both?

💡 The Bridge (click to reveal)

The correct expansions mix the sine of one angle with the cosine of the other:

$$\sin(A+B)=\sin A\cos B+\cos A\sin B, \qquad \sin(A-B)=\sin A\cos B-\cos A\sin B.$$

$$\cos(A+B)=\cos A\cos B-\sin A\sin B, \qquad \cos(A-B)=\cos A\cos B+\sin A\sin B.$$

These compound-angle formulae are today's focus.

📖 Key Concepts: Compound Angle Formulae

Compound angle formula for sine:

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

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

Compound angle formula for cosine:

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

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

Notice the pattern: the sine formulae swap $\sin \leftrightarrow \cos$ on the right-hand side, while the cosine formulae keep $\cos \cos$ and $\sin \sin$ . The sign on the left matches the sign between the terms in the sine formula; for cosine, the sign is reversed.

Activity 1 · Pair F2 Challenge Quest: Escape the Trig Chamber — Deriving cos(A+B)

Pair activity (8 min): You are trapped in the Trig Chamber! To escape, you must unlock 3 doors by solving the geometric derivation of $\cos(A+B)$ . Work with your partner — each door requires a correct answer to proceed.

Time: 08:00

🚪 Door 1: Geometric Setup

In the unit circle, two angles $A$ and $B$ are drawn from the positive x-axis. The coordinates of the point at angle $A$ are $(\cos A, \sin A)$ . What are the coordinates of the point at angle $A+B$ ?

(\cos(A+B), \sin(A+B))

(\cos(A-B), \sin(A-B))

(\cos A \cos B, \sin A \sin B)

(\cos A + \cos B, \sin A + \sin B)

💡 Hint: Unit Circle Setup

On the unit circle, every point at angle $\theta$ from the positive x-axis has coordinates $(\cos\theta, \sin\theta)$ . Identify the key points:

Point $P$ is at angle $A$ , point $Q$ is at angle $A+B$ , and the point at angle $0$ is $(1, 0)$ .

🔒 Door 2: The Rotation Trick

Consider the distance between points at angles $A+B$ and $0$ on the unit circle. This distance equals the distance between the points at angles $A$ and $-B$ (by rotation). Which equation does this give?

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

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

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

(\cos(A+B)-1)^2 + \sin^2(A+B) = (\cos A - \cos B)^2 + (\sin A + \sin B)^2

💡 Hint: The Rotation Trick

Rotate the entire diagram by $-B$ . Distances are preserved, so $|QP_0| = |Q'P'|$ . Track the coordinates of each key point before and after rotation:

Key insight: $Q$ at angle $A+B$ moves to $Q'$ at angle $A$ . $P_0$ at angle $0$ moves to $P'$ at angle $-B$ , with coordinates $(\cos B, -\sin B)$ . Since rotation preserves distance: $|QP_0| = |Q'P'|$ .

🔒 Door 3: The Final Formula

Expand both sides of the equation from Door 2 and simplify. The $\cos^2(A+B) + \sin^2(A+B) = 1$ identity cancels, and after collecting terms, what is the resulting formula for $\cos(A+B)$ ?

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

\cos(A+B) = \cos A + \cos B

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

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

🎉 Chamber Escaped!

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

Discussion: How would you derive $\sin(A+B)$ from this result? What identity connects sine and cosine?

Activity 2 · Individual B2 Pattern Detective: From cos(A+B) to sin(A+B)

Individual activity (5 min): You have already derived $\cos(A+B) = \cos A \cos B - \sin A \sin B$ . Now follow the detective trail — four clues that lead from this formula to $\sin(A+B)$ . At each clue, identify the correct expression from the suspects.

Known formula: $\cos(A+B) = \cos A \cos B - \sin A \sin B$

Clue 1: Substitute B with −B

Given:

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

.
Recall:

\cos(-B) = \cos B

(even),

\sin(-B) = -\sin B

(odd).

Replace $B$ with $-B$ everywhere. What is $\cos(A-B)$ ?

\sin A \sin B - \cos A \cos B

\cos A \cos B + \sin A \sin B

\cos A \cos B - \sin A \sin B

-\cos A \cos B + \sin A \sin B

Clue 1 / 4

Case Solved!

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

Score: 0 / 4

Discussion: Can you derive $\sin(A-B)$ using the same strategy?

Worked Example 1: Expressing as Surds Exercise 3B Q1 [adapted]

Use the compound-angle formulae to write the following as surds.

(i) $\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ})$ (ii) $\cos 135^{\circ} = \cos(90^{\circ} + 45^{\circ})$

👁 Reveal Solution

Part (i):

\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ})

Using $\sin(A+B) = \sin A \cos B + \cos A \sin B$ :

$\sin 75^{\circ} = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ}$

$= \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2}$

$= \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} = \boxed{\dfrac{\sqrt{6} + \sqrt{2}}{4}}$

Part (ii):

\cos 135^{\circ} = \cos(90^{\circ} + 45^{\circ})

Using $\cos(A+B) = \cos A \cos B - \sin A \sin B$ :

$\cos 135^{\circ} = \cos 90^{\circ} \cos 45^{\circ} - \sin 90^{\circ} \sin 45^{\circ}$

$= 0 \cdot \dfrac{\sqrt{2}}{2} - 1 \cdot \dfrac{\sqrt{2}}{2}$

$= 0 - \dfrac{\sqrt{2}}{2} = \boxed{-\dfrac{\sqrt{2}}{2}}$

Worked Example 2: Expanding Compound Expressions Exercise 3B Q2 [adapted]

Expand each of the following expressions.

(i) $\sin(\theta + 45^{\circ})$ (ii) $\cos(\theta - 30^{\circ})$ (iii) $\sin(60^{\circ} - \theta)$ (iv) $\cos(2\theta + 45^{\circ})$

👁 Reveal Solution

(i)

\sin(\theta + 45^{\circ})

$= \sin\theta \cos 45^{\circ} + \cos\theta \sin 45^{\circ} = \boxed{\dfrac{\sqrt{2}}{2}(\sin\theta + \cos\theta)}$

(ii)

\cos(\theta - 30^{\circ})

$= \cos\theta \cos 30^{\circ} + \sin\theta \sin 30^{\circ} = \boxed{\dfrac{\sqrt{3}}{2}\cos\theta + \dfrac{1}{2}\sin\theta}$

(iii)

\sin(60^{\circ} - \theta)

$= \sin 60^{\circ} \cos\theta - \cos 60^{\circ} \sin\theta = \boxed{\dfrac{\sqrt{3}}{2}\cos\theta - \dfrac{1}{2}\sin\theta}$

(iv)

\cos(2\theta + 45^{\circ})

$= \cos 2\theta \cos 45^{\circ} - \sin 2\theta \sin 45^{\circ} = \boxed{\dfrac{\sqrt{2}}{2}(\cos 2\theta - \sin 2\theta)}$

✏️ Concept Quiz

Score: 0 / 4

1. Which of the following is the expansion of

\sin(A + B)

\cos A \cos B - \sin A \sin B

\sin A \cos B - \cos A \sin B

\sin A \cos B + \cos A \sin B

\sin A + \sin B

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

. The sine formula swaps

\sin

and

\cos

between the two terms, and the sign between terms matches the sign on the left (

+

2. Which of the following is the expansion of

\cos(A - B)

\cos A \cos B + \sin A \sin B

-\cos A \cos B + \sin A \sin B

\cos A \cos B - \sin A \sin B

\cos A \sin B - \sin A \cos B

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

. The cosine formula keeps

\cos\cos

and

\sin\sin

, and the sign between terms is opposite to the sign on the left (

-

on left

\rightarrow

+

between terms).

3. Using compound angle formulae,

\sin 75^{\circ} = \sin(45^{\circ}+30^{\circ})

equals:

-\dfrac{\sqrt{2}}{2}

\dfrac{\sqrt{6} - \sqrt{2}}{4}

-\dfrac{\sqrt{6}+\sqrt{2}}{4}

\dfrac{\sqrt{6} + \sqrt{2}}{4}

\sin 75^{\circ} = \sin 45^{\circ}\cos 30^{\circ} + \cos 45^{\circ}\sin 30^{\circ} = \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2} = \dfrac{\sqrt{6}+\sqrt{2}}{4} \approx 0.966

4. The expansion of

\cos(\theta - 30^{\circ})

is:

-\dfrac{\sqrt{3}}{2}\cos\theta - \dfrac{1}{2}\sin\theta

\dfrac{\sqrt{3}}{2}\cos\theta + \dfrac{1}{2}\sin\theta

-\dfrac{\sqrt{3}}{2}\cos\theta + \dfrac{1}{2}\sin\theta

\dfrac{1}{2}\cos\theta - \dfrac{\sqrt{3}}{2}\sin\theta

\cos(\theta-30^{\circ}) = \cos\theta\cos 30^{\circ} + \sin\theta\sin 30^{\circ} = \dfrac{\sqrt{3}}{2}\cos\theta + \dfrac{1}{2}\sin\theta

. The

-

on the left gives

+

between terms for cosine.

🎉

You scored 0 / 4

📝 More Practice

Attempt each question on paper first, then click Reveal Solution to check your work.

Exercise 3B Q3 [adapted: parts (ii), (iv)] Standard

Simplify each of the following expressions.

(ii) $\cos \phi \cos 7\phi - \sin \phi \sin 7\phi$ (iv) $\cos \theta \cos \theta - \sin \theta \sin \theta$

👁 Reveal Solution

(ii)

\cos \phi \cos 7\phi - \sin \phi \sin 7\phi

This matches $\cos(A+B) = \cos A \cos B - \sin A \sin B$ with $A = \phi$ , $B = 7\phi$ .

$= \cos(\phi + 7\phi) = \boxed{\cos 8\phi}$

(iv)

\cos \theta \cos \theta - \sin \theta \sin \theta

This matches $\cos(A+B) = \cos A \cos B - \sin A \sin B$ with $A = \theta$ , $B = \theta$ .

$= \cos(\theta + \theta) = \boxed{\cos 2\theta}$

Exercise 3B Q4 (i) Standard

Solve the following equation for values of $\theta$ in the range $0^{\circ} \leqslant \theta \leqslant 180^{\circ}$ .

(i) $\cos(60^{\circ} + \theta) = \sin \theta$

💡 Hint

Expand $\cos(60^{\circ}+\theta)$ using the compound angle formula, then collect $\sin\theta$ and $\cos\theta$ terms. Divide through by $\cos\theta$ to get an equation in $\tan\theta$ .

👁 Reveal Solution

Step 1: Expand the left side

$\cos(60^{\circ}+\theta) = \cos 60^{\circ}\cos\theta - \sin 60^{\circ}\sin\theta = \dfrac{1}{2}\cos\theta - \dfrac{\sqrt{3}}{2}\sin\theta$

Step 2: Set equal to

\sin\theta

and rearrange

$\dfrac{1}{2}\cos\theta - \dfrac{\sqrt{3}}{2}\sin\theta = \sin\theta$

$\dfrac{1}{2}\cos\theta = \sin\theta\left(1 + \dfrac{\sqrt{3}}{2}\right) = \sin\theta \cdot \dfrac{2+\sqrt{3}}{2}$

$\cos\theta = \sin\theta(2+\sqrt{3})$

Step 3: Solve for

\tan\theta

$\tan\theta = \dfrac{1}{2+\sqrt{3}} = \dfrac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})} = \dfrac{2-\sqrt{3}}{1} = 2-\sqrt{3}$

Since $\tan 15^{\circ} = 2 - \sqrt{3}$ , we get $\boxed{\theta = 15^{\circ}}$ .

Exercise 3B Q4 (v) Medium

Solve the following equation for values of $\theta$ in the range $0^{\circ} \leqslant \theta \leqslant 180^{\circ}$ .

(v) $\sin \theta = \cos(\theta + 120^{\circ})$

👁 Reveal Solution

Step 1: Expand

\cos(\theta + 120^{\circ})

$\cos(\theta+120^{\circ}) = \cos\theta\cos 120^{\circ} - \sin\theta\sin 120^{\circ} = -\dfrac{1}{2}\cos\theta - \dfrac{\sqrt{3}}{2}\sin\theta$

Step 2: Set equal to

\sin\theta

and rearrange

$\sin\theta = -\dfrac{1}{2}\cos\theta - \dfrac{\sqrt{3}}{2}\sin\theta$

$\sin\theta + \dfrac{\sqrt{3}}{2}\sin\theta = -\dfrac{1}{2}\cos\theta$

$\sin\theta \cdot \dfrac{2+\sqrt{3}}{2} = -\dfrac{1}{2}\cos\theta$

$\tan\theta = \dfrac{-1}{2+\sqrt{3}} = -(2-\sqrt{3}) = \sqrt{3}-2$

Step 3: Solve in the given range

Since $\tan\theta = \sqrt{3}-2 \approx -0.268 < 0$ and $0^{\circ} \leqslant \theta \leqslant 180^{\circ}$ , $\theta$ is in quadrant II.

$\theta = 180^{\circ} - 15^{\circ} = \boxed{165^{\circ}}$

Exercise 3B Q5 (i) Medium

Solve the following equation for values of $\theta$ in the range $0 \leqslant \theta \leqslant \pi$ . (When the range is given in radians, the solutions should be in radians, using multiples of $\pi$ where appropriate.)

(i) $\sin\left(\theta + \dfrac{\pi}{4}\right) = \cos \theta$

👁 Reveal Solution

Step 1: Expand

\sin\left(\theta + \dfrac{\pi}{4}\right)

$= \sin\theta\cos\dfrac{\pi}{4} + \cos\theta\sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}\sin\theta + \dfrac{\sqrt{2}}{2}\cos\theta$

Step 2: Set equal to

\cos\theta

and rearrange

$\dfrac{\sqrt{2}}{2}\sin\theta + \dfrac{\sqrt{2}}{2}\cos\theta = \cos\theta$

$\dfrac{\sqrt{2}}{2}\sin\theta = \cos\theta\left(1 - \dfrac{\sqrt{2}}{2}\right) = \cos\theta \cdot \dfrac{2-\sqrt{2}}{2}$

$\sin\theta = \cos\theta \cdot \dfrac{2-\sqrt{2}}{\sqrt{2}} = \cos\theta(\sqrt{2}-1)$

$\tan\theta = \sqrt{2}-1$

Step 3: Solve in the given range

Since $\tan\left(\dfrac{\pi}{8}\right) = \sqrt{2}-1$ (this is the half-angle of $\dfrac{\pi}{4}$ ),

$\boxed{\theta = \dfrac{\pi}{8}}$

Exercise 3B Q6 (i)-(iii) Medium

The diagram shows three points $A(-4, 1)$ , $B(0, 4)$ and $C(6, -2)$ joined to form a triangle. The angles $\alpha$ and $\beta$ and the point $P$ are also shown in the diagram.

(i) Show that $\sin \alpha = \dfrac{4}{5}$ and write down the value of $\cos \alpha$ .

(ii) Find the values of $\sin \beta$ and $\cos \beta$ .

(iii) Show that $\sin(\alpha + \beta) = \dfrac{7}{5\sqrt{2}}$

💡 Hint

$\alpha$ is the angle that $BA$ makes with the vertical through $B$ . Compute $|BA|$ from the coordinates. Then $\sin\alpha = \dfrac{\text{horizontal component}}{|BA|}$ and $\cos\alpha = \dfrac{\text{vertical component}}{|BA|}$ .

Similarly, $\beta$ is the angle that $BC$ makes with the vertical through $B$ . Compute $|BC|$ .

For part (iii), use $\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$ .

👁 Reveal Solution

Part (i): Show

\sin\alpha = \dfrac{4}{5}

, find

\cos\alpha

$\overrightarrow{BA} = A - B = (-4-0,\ 1-4) = (-4, -3)$ , so $|BA| = \sqrt{16+9} = 5$ .

The angle $\alpha$ is between $BA$ and the vertical through $B$ . The horizontal component is $4$ and the vertical component is $3$ .

$\sin\alpha = \dfrac{4}{5}$ (as required). Since $\alpha$ is acute, $\cos\alpha = \dfrac{3}{5}$ .

Part (ii): Find

\sin\beta

and

\cos\beta

$\overrightarrow{BC} = C - B = (6-0,\ -2-4) = (6, -6)$ , so $|BC| = \sqrt{36+36} = 6\sqrt{2}$ .

The angle $\beta$ is between $BC$ and the vertical through $B$ . Horizontal component $= 6$ , vertical component $= 6$ .

$\sin\beta = \dfrac{6}{6\sqrt{2}} = \dfrac{1}{\sqrt{2}} = \boxed{\dfrac{\sqrt{2}}{2}}$ $\cos\beta = \dfrac{6}{6\sqrt{2}} = \boxed{\dfrac{\sqrt{2}}{2}}$

Part (iii): Show

\sin(\alpha+\beta) = \dfrac{7}{5\sqrt{2}}

$\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$

$= \dfrac{4}{5}\cdot\dfrac{\sqrt{2}}{2} + \dfrac{3}{5}\cdot\dfrac{\sqrt{2}}{2} = \dfrac{4\sqrt{2}+3\sqrt{2}}{10} = \dfrac{7\sqrt{2}}{10}$

Rationalising: $\dfrac{7\sqrt{2}}{10} = \dfrac{7\sqrt{2}}{10} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{7\cdot 2}{10\sqrt{2}} = \dfrac{14}{10\sqrt{2}} = \dfrac{7}{5\sqrt{2}}$ $\checkmark$

Exit Ticket

Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.

I can state the compound angle formulae for

\sin(A \pm B)

and

\cos(A \pm B)

. I can derive

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

using a geometric argument. I can express

\sin 75^{\circ}

and

\cos 135^{\circ}

as surds using compound angle formulae. I can expand

\sin(\theta \pm C)

and

\cos(\theta \pm C)

for a given angle

C