Learn9MA0Paper 1 · Pure Mathematics 1TrigonometryDouble-angle formulae

Trigonometry · Subtopic 12 of 13

Double-Angle Formulae

Setting $B=A$ in the compound-angle formulae gives formulae for $\sin2A$ , $\cos2A$ and $\tan2A$ . Three versions of $\cos2A$ let you match the function in an equation, and they are essential for integration later.

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Section 1 of 14

Learning objectives

Derive and use $\sin2A$ , $\cos2A$ (three forms) and $\tan2A$ .
Find exact values of $\sin2\theta$ and $\cos2\theta$ from one ratio of $\theta$ .
Choose the right form of $\cos2A$ to turn an equation into a quadratic in one function.
Rewrite $\sin^2x$ and $\cos^2x$ in terms of $\cos2x$ .

Pause and think

Put $B=A$ in $\sin(A+B)=\sin A\cos B+\cos A\sin B$ . What do you get?

Section 2 of 14

Core idea

Start hereDouble the angle, not double the value

Double an angle and you don't double its sine: $\sin60^\circ\neq2\sin30^\circ$ . But put $B=A$ into the compound formulas and exact rules for $\sin2A$ , $\cos2A$ and $\tan2A$ appear for free. They're essential for solving equations and for integrating $\sin^2x$ .

The big idea

$\sin2A=2\sin A\cos A$ ; $\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A$ ; $\tan2A=\dfrac{2\tan A}{1-\tan^2A}$ .

1. Derivations

From the compound formulas with B = A

$\sin2A=\sin(A+A)=\sin A\cos A+\cos A\sin A=2\sin A\cos A$
$\sin(A+B)$ .
$\cos2A=\cos A\cos A-\sin A\sin A=\cos^2A-\sin^2A$
$\cos(A+B)$ .
$=\cos^2A-(1-\cos^2A)=2\cos^2A-1$
Replace $\sin^2A$ .
$=(1-\sin^2A)-\sin^2A=1-2\sin^2A$
Or replace $\cos^2A$ .
$\tan2A=\frac{\tan A+\tan A}{1-\tan A\tan A}=\frac{2\tan A}{1-\tan^2A}$
$\tan(A+B)$ .

sin 2x is not 2 sin x

\sin2x

squashes the wave horizontally;

2\sin x

stretches it vertically. Different functions.

TrickPick the right cos 2A Equation has

\sin x

? Use

1-2\sin^2x

. Has

\cos x

? Use

2\cos^2x-1

. Rearranged:

\sin^2A=\frac12(1-\cos2A)

\cos^2A=\frac12(1+\cos2A)

(vital for integration).

2. Solving equations

Worked example $\sin2x=\sin x$ , $0\le x\le360^\circ$

Solve.

Goal: All $x$ .
Strategy: Double angle; factorise (never cancel $\sin x$ ).

$2\sin x\cos x-\sin x=0$
Expand; rearrange.
$\sin x(2\cos x-1)=0$
Factorise.
$\sin x=0:\ 0^\circ,180^\circ,360^\circ;\quad\cos x=\tfrac12:\ 60^\circ,300^\circ$
Solve each.

Answer

0^\circ,60^\circ,180^\circ,300^\circ,360^\circ

✓ Check

x=300^\circ

\sin600^\circ=\sin240^\circ=-\frac{\sqrt3}2=\sin300^\circ

✓.

Worked example $\cos2x+3\sin x=2$

Solve for $0\le x\le360^\circ$ .

Goal: All $x$ .
Strategy: Sin version of $\cos2x$ .

$1-2\sin^2x+3\sin x=2$
Substitute.
$2\sin^2x-3\sin x+1=0\Rightarrow(2\sin x-1)(\sin x-1)=0$
Rearrange; factorise.
$\sin x=\tfrac12:\ 30^\circ,150^\circ;\quad\sin x=1:\ 90^\circ$
Solve.

Answer

30^\circ,90^\circ,150^\circ

✓ Check

x=90

\cos180+3=-1+3=2

✓.

3. Identities

Worked example $\cos3x$ in terms of $\cos x$

Show $\cos3x\equiv4\cos^3x-3\cos x$ .

Goal: Proof.
Strategy: $3x=2x+x$ .

$\cos(2x+x)=\cos2x\cos x-\sin2x\sin x$
Compound.
$=(2\cos^2x-1)\cos x-2\sin^2x\cos x$
Double angles.
$=2\cos^3x-\cos x-2(1-\cos^2x)\cos x=4\cos^3x-3\cos x$
Replace $\sin^2x$ ; collect.

AnswerProved ∎.

✓ Check

x=0

1=4-3

✓;

x=60^\circ

\cos180=-1

4\cdot\frac18-\frac32=-1

✓.

Pause and think

If $\cos x=\frac13$ , find $\cos2x$ .

Summary

$B=A$ in compound formulas.
Three forms of $\cos2A$ ; choose to match.
Factorise, don't cancel.

Double-angle formula: Expresses a function of $2A$ in terms of $A$ .

HistoryA 45th-degree challenge. In 1593 the Dutch mathematician Adriaan van Roomen challenged the world to solve an equation of degree

45

. François Viète solved it within days by recognising it as the formula for

\sin45\theta

written in terms of

\sin\theta

, built by repeatedly applying multiple-angle formulae.

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