AS & A Level · Practice questions · Compound angle formulae

Expand the bracket.

Six original Cambridge-style questions on the compound angle formulae: exact values, identities, equations and a 'show that, hence solve' question, at Paper 2 and Paper 3 standard.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Expand with exact values, then collect.

Exact values: split the angle into two from \(30^\circ, 45^\circ, 60^\circ\) (or \(\dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}\)).
Signs: sine keeps the sign, cosine flips it; tangent keeps it on top and flips it underneath.
Given ratios: find the missing ratios from a right-angled triangle, then fix signs by quadrant.
Equations: expand, collect \(\sin x\) and \(\cos x\), divide to get \(\tan x = k\).
Tan equations: clear the fraction and solve the quadratic in \(\tan x\).

Recall

[2 marks]

Find the exact value of \(\cos 15^\circ\).

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

Calculation

[3 marks]

Find the exact value of \(\tan 75^\circ\), giving your answer in the form \(a + \sqrt{b}\).

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

Proof

[3 marks]

Prove that \(\cos(x + 60^\circ) + \cos(x - 60^\circ) \equiv \cos x\).

LHS \(= (\cos x\cos 60^\circ - \sin x\sin 60^\circ) + (\cos x\cos 60^\circ + \sin x\sin 60^\circ)\). ✓
The \(\sin x\) terms cancel: \(= 2\cos x\cos 60^\circ = 2\cos x \times \dfrac{1}{2} = \cos x\) = RHS. ✓

Calculation

[4 marks]

Solve the equation \(\cos(x - 60^\circ) = 3\sin x\) for \(0^\circ \le x \le 360^\circ\).

\(\dfrac{1}{2}\cos x + \dfrac{\sqrt{3}}{2}\sin x = 3\sin x\). ✓
\(\dfrac{1}{2}\cos x = \left(3 - \dfrac{\sqrt{3}}{2}\right)\sin x\), so \(\tan x = \dfrac{1/2}{3 - \sqrt{3}/2} = \dfrac{1}{6 - \sqrt{3}} = 0.2343\). ✓
\(x = 13.2^\circ\) or \(13.2^\circ + 180^\circ = 193.2^\circ\). ✓

Analysis

[3 marks]

Given that \(\tan(x + y) = 3\) and \(\tan x = 2\), find the exact value of \(\tan y\).

Let \(t = \tan y\): \(\dfrac{2 + t}{1 - 2t} = 3\). ✓
\(2 + t = 3 - 6t\), so \(7t = 1\). ✓
\(\tan y = \dfrac{1}{7}\). ✓

Analysis

[6 marks]

(a) Show that \(\sin(\theta + 60^\circ) + \cos(\theta + 30^\circ) \equiv \sqrt{3}\cos\theta\).

(b) Hence solve the equation \(\sin(\theta + 60^\circ) + \cos(\theta + 30^\circ) = 1\) for \(0^\circ < \theta < 360^\circ\).

(a) \(\sin(\theta + 60^\circ) = \dfrac{1}{2}\sin\theta + \dfrac{\sqrt{3}}{2}\cos\theta\). ✓
\(\cos(\theta + 30^\circ) = \dfrac{\sqrt{3}}{2}\cos\theta - \dfrac{1}{2}\sin\theta\). ✓
Add: the \(\sin\theta\) terms cancel, leaving \(\sqrt{3}\cos\theta\), as required. ✓
(b) \(\sqrt{3}\cos\theta = 1\), so \(\cos\theta = \dfrac{1}{\sqrt{3}} = 0.5774\). ✓
\(\theta = 54.7^\circ\) or \(360^\circ - 54.7^\circ = 305.3^\circ\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.