AS & A Level · Practice questions · Double angle formulae

Double the angle, halve the work.

Six original Cambridge-style questions on the double angle formulae: exact values, equations, identities and triple angles, 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

Pick the form of cos 2x that fits.

sin: \(\sin 2A = 2\sin A\cos A\).
cos: \(\cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A\).
tan: \(\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}\).
Equations: one angle, one function; factorise rather than divide.
Proofs: \(1 - \cos 2x = 2\sin^2 x\) and \(1 + \cos 2x = 2\cos^2 x\) remove the 1.

Recall

[2 marks]

Given that \(\sin A = \dfrac{1}{4}\), find the exact value of \(\cos 2A\).

\(\cos 2A = 1 - 2\sin^2 A = 1 - 2 \times \dfrac{1}{16}\). ✓
\(= \dfrac{7}{8}\). ✓

Calculation

[3 marks]

Without a calculator, find the exact values of (a) \(\cos^2 75^\circ - \sin^2 75^\circ\) (b) \(\sin 75^\circ\cos 75^\circ\).

(a) \(\cos^2 75^\circ - \sin^2 75^\circ = \cos 150^\circ = -\dfrac{\sqrt{3}}{2}\). ✓
(b) \(\sin 75^\circ\cos 75^\circ = \dfrac{1}{2}\sin 150^\circ = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\). ✓

Calculation

[4 marks]

Solve the equation \(3\cos 2x + 7\cos x = 0\) for \(0^\circ \le x \le 360^\circ\).

\(3(2\cos^2 x - 1) + 7\cos x = 0\), so \(6\cos^2 x + 7\cos x - 3 = 0\). ✓
\((3\cos x - 1)(2\cos x + 3) = 0\). ✓
\(\cos x = -\dfrac{3}{2}\) is impossible. ✓
\(\cos x = \dfrac{1}{3}\): \(x = 70.5^\circ\) or \(360^\circ - 70.5^\circ = 289.5^\circ\). ✓

Proof

[3 marks]

Prove that \(\dfrac{\sin 2x}{1 + \cos 2x} \equiv \tan x\).

\(1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x\). ✓
LHS \(= \dfrac{2\sin x\cos x}{2\cos^2 x} = \dfrac{\sin x}{\cos x} = \tan x\) = RHS. ✓

Proof

[4 marks]

Show that \(\cos 3x \equiv 4\cos^3 x - 3\cos x\).

\(\cos 3x = \cos(2x + x) = \cos 2x\cos x - \sin 2x\sin x\). ✓
\(= (2\cos^2 x - 1)\cos x - 2\sin^2 x\cos x\). ✓
\(= 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x\). ✓
\(= 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x = 4\cos^3 x - 3\cos x\). ✓

Analysis

[6 marks]

(a) Prove that \(\tan x + \cot x \equiv 2\operatorname{cosec} 2x\).

(b) Hence solve the equation \(\tan x + \cot x = 8\) for \(0^\circ < x < 180^\circ\).

(a) LHS \(= \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \dfrac{1}{\sin x\cos x}\). ✓
\(\sin x\cos x = \dfrac{1}{2}\sin 2x\), so LHS \(= \dfrac{2}{\sin 2x} = 2\operatorname{cosec} 2x\) = RHS. ✓
(b) \(2\operatorname{cosec} 2x = 8\), so \(\sin 2x = \dfrac{1}{4}\). ✓
For \(0^\circ < 2x < 360^\circ\): \(2x = 14.48^\circ\) or \(180^\circ - 14.48^\circ = 165.52^\circ\). ✓
\(x = 7.2^\circ,\ 82.8^\circ\). ✓

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.