Compound Angles

A2 · P3 · 2 min

The compound-angle formulae expand $\sin(A + B)$ , $\cos(A + B)$ and $\tan(A + B)$ . They are the source of every other P3 identity: put $B = A$ and you get the double-angle formulae; write $a\sin\theta + b\cos\theta$ as one of them backwards and you get the $R$ -form.

The formulae

Key result

\sin(A \pm B) \equiv \sin A\cos B \pm \cos A\sin B

\cos(A \pm B) \equiv \cos A\cos B \mp \sin A\sin B

\tan(A \pm B) \equiv \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}

Sine keeps the sign; cosine flips it; tangent has the flipped sign in the denominator.

Exact values

Angles like $75^\circ = 45^\circ + 30^\circ$ and $15^\circ = 45^\circ - 30^\circ$ become exact.

An exact value from a sum

Find the exact value of $\sin 75^\circ$ .

Solution

\sin(45^\circ + 30^\circ) = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \tfrac{\sqrt{2}}{2} \cdot \tfrac{\sqrt{3}}{2} + \tfrac{\sqrt{2}}{2} \cdot \tfrac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}.

An exact tangent

Show that $\tan 15^\circ = 2 - \sqrt{3}$ .

Solution

\tan(45^\circ - 30^\circ) = \frac{1 - \tfrac{1}{\sqrt{3}}}{1 + \tfrac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}.

Given two ratios, find a compound one

Combining given information

$\sin A = \tfrac{3}{5}$ with $A$ acute, and $\cos B = -\tfrac{5}{13}$ with $B$ obtuse. Find the exact value of $\cos(A + B)$ and $\tan(A - B)$ .

Solution

$\cos A = \tfrac{4}{5}$ , $\tan A = \tfrac{3}{4}$ . $B$ obtuse: $\sin B = \tfrac{12}{13}$ , $\tan B = -\tfrac{12}{5}$ .

$\cos(A + B) = \tfrac{4}{5}\left(-\tfrac{5}{13}\right) - \tfrac{3}{5} \cdot \tfrac{12}{13} = -\tfrac{20}{65} - \tfrac{36}{65} = -\tfrac{56}{65}$ .

$\tan(A - B) = \dfrac{\tfrac{3}{4} + \tfrac{12}{5}}{1 - \tfrac{3}{4}\left(-\tfrac{12}{5}\right)} = \dfrac{\tfrac{63}{20}}{\tfrac{56}{20}} = \tfrac{9}{8}$ .

Solving equations

Expand the compound angle, collect terms, and reduce to a single function, usually $\tan$ .

Expanding to solve

Solve $\sin(\theta + 60^\circ) = 2\cos\theta$ for $0^\circ \leq \theta \leq 360^\circ$ .

Solution

$\sin\theta\cos 60^\circ + \cos\theta\sin 60^\circ = 2\cos\theta \Rightarrow \tfrac{1}{2}\sin\theta + \tfrac{\sqrt{3}}{2}\cos\theta = 2\cos\theta$ .

$\tfrac{1}{2}\sin\theta = \left(2 - \tfrac{\sqrt{3}}{2}\right)\cos\theta \Rightarrow \tan\theta = 4 - \sqrt{3} \approx 2.268$ .

$\theta = 66.2^\circ, 246.2^\circ$ .

Two compound angles

Solve $\cos(x - 30^\circ) = \sin(x + 30^\circ)$ for $0^\circ \leq x \leq 180^\circ$ .

Solution

$\cos x\cos 30^\circ + \sin x\sin 30^\circ = \sin x\cos 30^\circ + \cos x\sin 30^\circ$ .

$\tfrac{\sqrt{3}}{2}\cos x + \tfrac{1}{2}\sin x = \tfrac{\sqrt{3}}{2}\sin x + \tfrac{1}{2}\cos x \Rightarrow \left(\tfrac{\sqrt{3} - 1}{2}\right)\cos x = \left(\tfrac{\sqrt{3} - 1}{2}\right)\sin x \Rightarrow \tan x = 1$ .

$x = 45^\circ$ .

Proving identities

An identity with compound angles

Prove that $\sin(A + B) + \sin(A - B) \equiv 2\sin A\cos B$ .

Solution

$\sin A\cos B + \cos A\sin B + \sin A\cos B - \cos A\sin B = 2\sin A\cos B$ .

Watch out

$\sin(A + B)$ is not $\sin A + \sin B$ . Test with $A = B = 45^\circ$ : $\sin 90^\circ = 1$ but $\sin 45^\circ + \sin 45^\circ = \sqrt{2}$ .

Exam tip

When the question says "find the exact value", keep surds and rationalise the denominator if the answer is a single fraction. When it gives $\sin A$ and asks about $\cos(A + B)$ , draw right-angled triangles to get the missing ratios, and use the quadrant to fix signs.

Practice

Question

Find the exact value of $\cos 105^\circ$ .
Expand and simplify $\sin\left(x + \tfrac{\pi}{4}\right) - \sin\left(x - \tfrac{\pi}{4}\right)$ .
$\tan A = 2$ and $\tan B = \tfrac{1}{3}$ . Find $\tan(A + B)$ and hence $A + B$ if both angles are acute.
Solve $\cos(\theta + 45^\circ) = \sin\theta$ for $0^\circ \leq \theta \leq 360^\circ$ .

Answers

$\cos(60^\circ + 45^\circ) = \tfrac{\sqrt{2} - \sqrt{6}}{4}$ .
$2\cos x\sin\tfrac{\pi}{4} = \sqrt{2}\cos x$ .
$\dfrac{2 + \tfrac{1}{3}}{1 - \tfrac{2}{3}} = 7$ ; $A + B = 81.9^\circ$ .
$\tfrac{\sqrt{2}}{2}(\cos\theta - \sin\theta) = \sin\theta \Rightarrow \tan\theta = \dfrac{\sqrt{2}}{2 + \sqrt{2}} = \sqrt{2} - 1$ : $\theta = 22.5^\circ, 202.5^\circ$ .

The formulae

Exact values

Given two ratios, find a compound one

Solving equations

Proving identities

Practice

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