Double Angles

A2 · P3 · 3 min

Put $B = A$ in the compound-angle formulae and you get the double-angle formulae. They connect an angle with twice that angle, which is exactly what is needed to solve equations mixing $\theta$ and $2\theta$ , to integrate $\sin^2 x$ and $\cos^2 x$ , and to simplify expressions with squares.

The formulae

Key result

\sin 2A \equiv 2\sin A\cos A

\cos 2A \equiv \cos^2 A - \sin^2 A \equiv 2\cos^2 A - 1 \equiv 1 - 2\sin^2 A

\tan 2A \equiv \frac{2\tan A}{1 - \tan^2 A}

Rearranged, the cosine versions give the power-reducing forms used in integration:

\cos^2 A \equiv \tfrac{1}{2}(1 + \cos 2A), \qquad \sin^2 A \equiv \tfrac{1}{2}(1 - \cos 2A).

Choose the version of $\cos 2A$ that leaves you with only the function already present in the rest of the equation.

Exact values from a given ratio

Given $\cos\theta = \tfrac{1}{4}$ and $\theta$ is acute, find the exact values of $\sin 2\theta$ , $\cos 2\theta$ and $\tan 2\theta$ .

Solution

$\sin\theta = \sqrt{1 - \tfrac{1}{16}} = \tfrac{\sqrt{15}}{4}$ .

$\sin 2\theta = 2 \cdot \tfrac{\sqrt{15}}{4} \cdot \tfrac{1}{4} = \tfrac{\sqrt{15}}{8}$ . $\cos 2\theta = 2\left(\tfrac{1}{16}\right) - 1 = -\tfrac{7}{8}$ . $\tan 2\theta = \dfrac{\sqrt{15}/8}{-7/8} = -\tfrac{\sqrt{15}}{7}$ .

Solving equations

Mixed sin 2x and sin x

Solve $\sin 2\theta = \sin\theta$ for $0^\circ \leq \theta \leq 360^\circ$ .

Solution

$2\sin\theta\cos\theta - \sin\theta = 0 \Rightarrow \sin\theta(2\cos\theta - 1) = 0$ .

$\sin\theta = 0$ : $\theta = 0^\circ, 180^\circ, 360^\circ$ . $\cos\theta = \tfrac{1}{2}$ : $\theta = 60^\circ, 300^\circ$ .

Choosing the right cos 2A

Solve $\cos 2\theta + 3\sin\theta = 2$ for $0 \leq \theta \leq 2\pi$ .

Solution

The other term is $\sin\theta$ , so use $\cos 2\theta = 1 - 2\sin^2\theta$ :

$1 - 2\sin^2\theta + 3\sin\theta - 2 = 0 \Rightarrow 2\sin^2\theta - 3\sin\theta + 1 = 0 \Rightarrow (2\sin\theta - 1)(\sin\theta - 1) = 0$ .

$\sin\theta = \tfrac{1}{2}$ : $\theta = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$ . $\sin\theta = 1$ : $\theta = \tfrac{\pi}{2}$ .

An equation in tan

Solve $\tan 2\theta = 3\tan\theta$ for $0^\circ < \theta < 180^\circ$ , $\theta \neq 90^\circ$ .

Solution

$\dfrac{2t}{1 - t^2} = 3t$ with $t = \tan\theta$ . Either $t = 0$ , which has no solution for $0^\circ < \theta < 180^\circ$ , or $2 = 3(1 - t^2) \Rightarrow t^2 = \tfrac{1}{3} \Rightarrow t = \pm\tfrac{1}{\sqrt{3}}$ .

$\theta = 30^\circ, 150^\circ$ .

Identities and simplification

Proving with double angles

Prove that $\dfrac{1 - \cos 2\theta}{\sin 2\theta} \equiv \tan\theta$ .

Solution

\frac{1 - (1 - 2\sin^2\theta)}{2\sin\theta\cos\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta.

Triple angle

Show that $\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta$ .

Solution

$\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta\cos\theta - \sin 2\theta\sin\theta = (2\cos^2\theta - 1)\cos\theta - 2\sin^2\theta\cos\theta$ .

$= 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta = 4\cos^3\theta - 3\cos\theta$ .

In integration

$\displaystyle\int \sin^2 x \, dx = \int \tfrac{1}{2}(1 - \cos 2x) \, dx = \tfrac{1}{2}x - \tfrac{1}{4}\sin 2x + c$ . See Integration Rules.

Watch out

$\cos 2\theta$ has three forms; using $\cos^2\theta - \sin^2\theta$ when the equation contains $\sin\theta$ leaves two functions and gets you nowhere. Match the form to the rest of the equation.

Exam tip

Half-angle versions are the same identities with $A = \tfrac{\theta}{2}$ : $\cos\theta = 1 - 2\sin^2\tfrac{\theta}{2}$ . Questions with $\tfrac{1}{2}x$ and $x$ together are double-angle questions in disguise.

Practice

Question

Given $\sin\theta = \tfrac{5}{13}$ with $\theta$ obtuse, find $\sin 2\theta$ and $\cos 2\theta$ .
Solve $\cos 2x = \cos x$ for $0^\circ \leq x \leq 360^\circ$ .
Solve $2\sin 2\theta = 3\cos\theta$ for $0 \leq \theta \leq 2\pi$ .
Prove $\dfrac{\sin 2\theta}{1 + \cos 2\theta} \equiv \tan\theta$ .
Express $\sin^2 x\cos^2 x$ in terms of $\cos 4x$ .

Answers

$\cos\theta = -\tfrac{12}{13}$ : $\sin 2\theta = -\tfrac{120}{169}$ , $\cos 2\theta = \tfrac{119}{169}$ .
$2\cos^2 x - \cos x - 1 = 0 \Rightarrow \cos x = 1, -\tfrac{1}{2}$ : $x = 0^\circ, 120^\circ, 240^\circ, 360^\circ$ .
$\cos\theta(4\sin\theta - 3) = 0$ : $\theta = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, 0.848, 2.29$ .
$\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \tan\theta$ .
$\tfrac{1}{4}\sin^2 2x = \tfrac{1}{8}(1 - \cos 4x)$ .

The formulae

Solving equations

Identities and simplification

In integration

Practice

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