Math Core

If you know $\sin\theta$ , can you find $\sin 2\theta$ ? Doubling the answer won't work: $\sin 30^\circ = \frac{1}{2}$ , but $\sin 60^\circ$ is $\frac{\sqrt{3}}{2}$ , not $1$ . The double-angle formulas tell you exactly how the trig values of $2\theta$ depend on those of $\theta$ . They show up constantly in calculus, physics and in solving trig equations.

Deriving the formulas

There is nothing new to memorize at first. Set $A = B = \theta$ in the sum formulas from the last lesson.

\begin{aligned} \sin 2\theta &= \sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta \\ \cos 2\theta &= \cos(\theta + \theta) = \cos\theta\cos\theta - \sin\theta\sin\theta = \cos^2\theta - \sin^2\theta \\ \tan 2\theta &= \tan(\theta + \theta) = \frac{\tan\theta + \tan\theta}{1 - \tan\theta\tan\theta} = \frac{2\tan\theta}{1 - \tan^2\theta} \end{aligned}

The cosine formula has two more useful forms. Replace $\cos^2\theta$ with $1 - \sin^2\theta$ , or replace $\sin^2\theta$ with $1 - \cos^2\theta$ :

\cos 2\theta = (1 - \sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta, \qquad \cos 2\theta = \cos^2\theta - (1 - \cos^2\theta) = 2\cos^2\theta - 1.

Double-angle formulas

\sin 2\theta = 2\sin\theta\cos\theta \qquad\qquad \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}

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

Pick the form of $\cos 2\theta$ that matches the information you have: only sine, only cosine, or both.

The graphs make the difference between $\sin 2x$ and $2\sin x$ clear. Doubling the angle squeezes the wave so it repeats twice as fast. Doubling the output stretches it taller.

y = sin 2x (period π, height 1) and y = 2 sin x (period 2π, height 2) are very different functions.Open in grapher →

Common mistake

$\sin 2\theta \ne 2\sin\theta$ . The $2$ is inside the function, so it changes the angle, not the output. Always use $2\sin\theta\cos\theta$ .

Values from one known value

Worked example: All three double-angle values

Suppose $\sin\theta = \dfrac{5}{13}$ and $\theta$ is in Quadrant II. Find $\sin 2\theta$ , $\cos 2\theta$ and $\tan 2\theta$ .

First, $\cos\theta = -\dfrac{12}{13}$ (negative in QII). Then

\begin{aligned} \sin 2\theta &= 2\sin\theta\cos\theta = 2\cdot\frac{5}{13}\cdot\left(-\frac{12}{13}\right) = -\frac{120}{169} \\ \cos 2\theta &= \cos^2\theta - \sin^2\theta = \frac{144}{169} - \frac{25}{169} = \frac{119}{169} \\ \tan 2\theta &= \frac{\sin 2\theta}{\cos 2\theta} = -\frac{120}{119} \end{aligned}

Check the signs: $\theta$ is between $90^\circ$ and $180^\circ$ , so $2\theta$ is between $180^\circ$ and $360^\circ$ . A negative sine and a positive cosine put $2\theta$ in Quadrant IV, which is consistent.

Recognizing the pattern

Expressions like $2\sin x\cos x$ or $\cos^2 x - \sin^2 x$ can be collapsed into a single term.

Worked example: Collapse and evaluate

Find the exact values of (a) $2\sin 15^\circ \cos 15^\circ$ and (b) $1 - 2\sin^2\dfrac{\pi}{8}$ .

(a) This is $\sin(2\cdot 15^\circ) = \sin 30^\circ = \dfrac{1}{2}$ .

(b) This is $\cos\left(2\cdot\dfrac{\pi}{8}\right) = \cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$ .

Building bigger formulas

Combine the double-angle and sum formulas to reach triple angles and beyond.

Worked example: A triple-angle formula

Write $\cos 3\theta$ in terms of $\cos\theta$ only.

\begin{aligned} \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\theta\cos\theta)\sin\theta \\ &= 2\cos^3\theta - \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 \end{aligned}

Check with $\theta = 60^\circ$ : $\cos 180^\circ = -1$ , and $4\left(\tfrac{1}{2}\right)^3 - 3\left(\tfrac{1}{2}\right) = \tfrac{1}{2} - \tfrac{3}{2} = -1$ . It matches.

Verifying identities with double angles

When an identity mixes $2x$ and $x$ , rewrite everything in terms of $x$ first.

Worked example: Verify an identity

Verify: $\dfrac{\sin 2x}{1 + \cos 2x} = \tan x$ .

Use $\cos 2x = 2\cos^2 x - 1$ so the $1$ in the denominator cancels.

\begin{aligned} \frac{\sin 2x}{1 + \cos 2x} &= \frac{2\sin x\cos x}{1 + (2\cos^2 x - 1)} && \text{double-angle formulas} \\ &= \frac{2\sin x \cos x}{2\cos^2 x} && \text{simplify} \\ &= \frac{\sin x}{\cos x} = \tan x && \text{cancel } 2\cos x \end{aligned}

Tip

To choose a form of $\cos 2\theta$ , look at what should cancel. Next to $1 + \cos 2\theta$ , use $2\cos^2\theta - 1$ . Next to $1 - \cos 2\theta$ , use $1 - 2\sin^2\theta$ .

Where double angles show up

Double angles appear naturally in science. If you launch a ball at speed $v$ and angle $\theta$ above level ground (ignoring air resistance), it lands a horizontal distance

R = \frac{v^2 \sin 2\theta}{g}

away, where $g$ is the acceleration due to gravity. The formula is usually derived with the product $2\sin\theta\cos\theta$ and then collapsed using the double-angle formula. Now you can read off facts that are hard to see otherwise. The range is largest when $\sin 2\theta = 1$ , that is, when $2\theta = 90^\circ$ , so $\theta = 45^\circ$ . And since $\sin 2\theta = \sin(180^\circ - 2\theta)$ , the angles $30^\circ$ and $60^\circ$ give exactly the same range, as do any two launch angles that add to $90^\circ$ .

When you work a double-angle problem, a short routine keeps you out of trouble. First find both $\sin\theta$ and $\cos\theta$ , with correct signs from the quadrant of $\theta$ . Then substitute into the formula you need. Finally, decide which quadrant $2\theta$ lands in and make sure the signs of your answers agree with it.

Practice

Practice 1

Which expression is equal to $\sin 2x$ for all $x$ ?

Practice 2

If $\cos\theta = \dfrac{3}{5}$ and $\theta$ is in Quadrant IV, find $\sin 2\theta$ .