Why are there three versions of 2A?

Because ^2 A - ^2 A can be rewritten with the Pythagorean identity into 2 ^2 A - 1 or 1 - 2 ^2 A. You pick the form that leaves only the ratio you want — very useful when integrating.

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Addition and Double-Angle Formulae

Q: Why does the cosine addition formula have a sign flip?

It falls out of the geometry/derivation: (A+B) = A B - A B. The sign inside the bracket becomes the opposite sign in the formula, unlike sine, which keeps the same sign.

Q: Where do the double-angle formulae come from?

From the addition formulae with B = A. For example (A+A) = A A + A A = 2 A A, giving 2A.

Q: What is the R-form used for?

It rewrites a + b as one wave R ( - ). This makes the maximum (R) and minimum (-R) immediate and turns two-term trig equations into single-ratio ones.

The addition formulae tell you $sin$ and $cos$ of a sum of angles. Setting the two angles equal gives the double-angle formulae, and combining a sine and cosine wave gives the powerful $R$ -form used for maxima and equations.

32 min Video by TLMaths Trigonometry

The big picture

So far you could only handle trig of a single angle. The addition formulae unlock $combinations$ of angles, which is where trigonometry becomes genuinely powerful: they let you find exact values of unusual angles, they generate the double-angle formulae that are essential for integration, and they let you rewrite $a cos θ + b sin θ$ as a single wave $R cos (θ - α)$ . That $R$ -form instantly reveals the maximum and minimum of a combined oscillation — the mathematics behind resonance, tides and signal processing. This lesson is dense, but almost everything in it flows from the three addition formulae at the top.

A-Level Maths E6-04: Introducing the Double Angle FormulaeWatch the walkthrough, then work through the notes below.

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What you'll be able to do

State and use the addition formulae for $sin$ , $cos$ and $tan$
Derive and apply the double-angle formulae
Use these to find exact values and solve equations
Write $a cos θ + b sin θ$ in the form $R cos (θ \pm α)$

The addition formulae

These give the trig ratios of $A \pm B$ . Watch the signs carefully — cosine’s formula swaps the sign, which is the most-tested detail in the whole lesson.

sin (A \pm B) = sin A cos B \pm cos A sin B

Sine keeps the sign.

cos (A \pm B) = cos A cos B \mp sin A sin B

Cosine flips the sign (note the

\mp

tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B}

Tangent, from the other two.

cos (4 5^{\circ} + 3 0^{\circ}) = cos 4 5^{\circ} cos 3 0^{\circ} - sin 4 5^{\circ} sin 3 0^{\circ}

= \frac{1}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{1}{2}

= \frac{3 - 1}{2 2}

Answer

cos 7 5^{\circ} = \frac{3 - 1}{2 2}

Tip — For cosine, the sign inside flips outside: $cos (A + B)$ uses a $minus$ , $cos (A - B)$ uses a $plus$ . Sine keeps the same sign.

Double-angle formulae

Set $B = A$ in the addition formulae and they collapse into the $double-angle$ formulae. Cosine gives three equivalent forms — pick whichever suits the problem (the $2 cos^{2} A - 1$ and $1 - 2 sin^{2} A$ versions are invaluable for integration).

sin 2 A = 2 sin A cos A

From

sin (A + A)

cos 2 A = cos^{2} A - sin^{2} A = 2 cos^{2} A - 1 = 1 - 2 sin^{2} A

Three equivalent forms.

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

From

tan (A + A)

1Since

A

is acute and

sin A = \frac{3}{5}

cos A = \frac{4}{5}

(3–4–5 triangle).

sin 2 A = 2 sin A cos A = 2 \cdot \frac{3}{5} \cdot \frac{4}{5}

= \frac{24}{25}

Answer

sin 2 A = \frac{24}{25}

Why this works

The three forms of $cos 2 A$ are not clutter — each is chosen to leave only $cos^{2}$ or only $sin^{2}$ , which is exactly what you need to integrate $cos^{2} x$ or $sin^{2} x$ later.

The R-formula: combining two waves

An expression like $a cos θ + b sin θ$ is actually a single sinusoid in disguise. You can write it as $R cos (θ - α)$ (or a sine version), where $R = a^{2} + b^{2}$ and $α$ is found from $tan α = \frac{b}{a}$ . This $\textbf{$ R $-form}$ makes the maximum ( $R$ ) and minimum ( $- R$ ) obvious, and turns a two-term equation into a single-ratio one you can solve.

a cos θ + b sin θ = R cos (θ - α), R = a^{2} + b^{2}

Combine two waves into one; maximum value is

R

R = 3^{2} + 4^{2} = 25 = 5

tan α = \frac{b}{a} = \frac{4}{3}

, so

α = 53. 1^{\circ}

3So

3 cos θ + 4 sin θ = 5 cos (θ - 53. 1^{\circ})

4The maximum of

cos

is 1, so the maximum value is

R = 5

Answer

5 cos (θ - 53. 1^{\circ})

; maximum value

5

Think like an examiner

The sign flip in $cos (A \pm B)$ is the most common error — cosine’s formula uses the opposite sign.
Choose the $cos 2 A$ form ( $2 cos^{2} A - 1$ or $1 - 2 sin^{2} A$ ) that matches what the question needs, especially for integration set-ups.
For the $R$ -form, compute $R = a^{2} + b^{2}$ exactly and find $α$ in the correct quadrant; the maximum is $R$ .
When finding exact values, split the awkward angle into a sum/difference of standard angles ( $7 5^{\circ} = 4 5^{\circ} + 3 0^{\circ}$ ).

Common misconceptions

Writing

cos (A + B) = cos A cos B + sin A sin B

It is a minus:

cos (A + B) = cos A cos B - sin A sin B

Thinking

sin 2 A = 2 sin A

sin 2 A = 2 sin A cos A

— the cosine factor is essential.

Taking

R = a + b

in the

R

-form.

R = a^{2} + b^{2}

(Pythagoras of the coefficients).

Using

cos 2 A = 2 cos A - 1

It is

2 cos^{2} A - 1

— the cosine is squared.

Compound-angle formulae

sin (A \pm B) = sin A cos B \pm cos A sin B

Sine addition.

cos (A \pm B) = cos A cos B \mp sin A sin B

Cosine addition (sign flips).

sin 2 A = 2 sin A cos A

Double angle (sine).

cos 2 A = 2 cos^{2} A - 1

Double angle (cosine).

R = a^{2} + b^{2}

R-form amplitude.

Stretch yourself

Use $cos 2 A = 1 - 2 sin^{2} A$ to solve $cos 2 x + 3 sin x = 2$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .

Hint — Substitute the double-angle form to get a quadratic in $sin x$ , then factorise and solve each simple equation.

Questions students ask

Why does the cosine addition formula have a sign flip?

It falls out of the geometry/derivation:

cos (A + B) = cos A cos B - sin A sin B

. The sign inside the bracket becomes the opposite sign in the formula, unlike sine, which keeps the same sign.

Where do the double-angle formulae come from?

From the addition formulae with

B = A

. For example

sin (A + A) = sin A cos A + cos A sin A = 2 sin A cos A

, giving

sin 2 A

Why are there three versions of

cos 2 A

Because

cos^{2} A - sin^{2} A

can be rewritten with the Pythagorean identity into

2 cos^{2} A - 1

1 - 2 sin^{2} A

. You pick the form that leaves only the ratio you want — very useful when integrating.

What is the R-form used for?

It rewrites

a cos θ + b sin θ

as one wave

R cos (θ - α)

. This makes the maximum (

R

) and minimum (

- R

) immediate and turns two-term trig equations into single-ratio ones.

Key takeaways

Addition formulae give trig of $A \pm B$ ; cosine’s version flips the sign.
Double-angle formulae come from setting $B = A$ ; $cos 2 A$ has three forms.
Use them for exact values and to turn equations into quadratics.
The $R$ -form writes $a cos θ + b sin θ = R cos (θ - α)$ , $R = a^{2} + b^{2}$ .
The $R$ -form reveals the maximum ( $R$ ) and minimum ( $- R$ ) at a glance.

How this fits the course

Build on

Integration (later)soon

Leads to

Exponentials and Logarithms

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