LearnYMA01WMA13 · Pure Mathematics 3TrigonometryDouble-angle formulae

Trigonometry · Subtopic 4 of 5

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.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 45 minRevision 5 · updated 2026-10-27

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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