What is 5.6: Compound and double angle formulae?

The compound-angle formulae are sin(A± B)=sin Acos B±cos Asin B, cos(A± B)=cos Acos B Asin B and tan(A± B)=tan A±tan B/1 Atan B; note the sign reversal in cosine.

How many marks are 5.6 questions worth?

5.6 practice drills on this page are worth 3–5 marks each; marks in a real exam vary by question.

Which past exam papers has 5.6 appeared in?

We have not yet indexed a verified past-paper appearance for 5.6 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

5.6

Understand and use double angle formulae; formulae for sin(A ± B), cos(A ± B), tan(A ± B) with geometrical proofs; express a cos θ + b sin θ in the form r cos(θ ± α) or r sin(θ ± α).

Draft — not yet indexed

Compound and double angle formulae

Worked answers and methods for 5.6 on Edexcel A-level Maths 9MA0.

Definition

Specification 5.6: understand and use double angle formulae; use of formulae for $\sin(A\pm B)$ , $\cos(A\pm B)$ and $\tan(A\pm B)$ ; understand geometrical proofs of these formulae; understand and use expressions for $a\cos\theta+b\sin\theta$ in the equivalent forms of $r\cos(\theta\pm\alpha)$ or $r\sin(\theta\pm\alpha)$ .

Explanation

The compound-angle formulae are $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ , $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ and $\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}$ ; note the sign reversal in cosine.
Putting $A=B$ gives $\sin2\theta=2\sin\theta\cos\theta$ , $\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta$ and $\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$ , and rearranging the cosine forms gives $\cos^2\theta=\frac12(1+\cos2\theta)$ for integration.
To write $a\cos\theta+b\sin\theta$ as $R\cos(\theta\pm\alpha)$ , expand the target, match coefficients to get $R\cos\alpha=a$ and $R\sin\alpha=b$ , so $R=\sqrt{a^2+b^2}$ and $\tan\alpha=\dfrac ba$ with $\alpha$ in the stated range and in the stated units.
The $R$ form then gives the maximum $R$ , the minimum $-R$ , the values of $\theta$ where they occur, and solves $a\cos\theta+b\sin\theta=c$ in one step.

Worked example

Express

5\cos\theta-12\sin\theta

in the form

R\cos(\theta+\alpha)

, where

R>0

and

0<\alpha<\dfrac{\pi}{2}

. Give the value of

\alpha

to 3 decimal places.

(3)

(Total for Question 1 is 3 marks)

Working

1.Expand: $R\cos(\theta+\alpha)=R\cos\theta\cos\alpha-R\sin\theta\sin\alpha$ .
2.Match coefficients: $R\cos\alpha=5$ and $R\sin\alpha=12$ , so $R=\sqrt{5^2+12^2}=13$ .
3.Divide: $\tan\alpha=\dfrac{12}{5}$ , so $\alpha=\arctan\dfrac{12}{5}=1.176$ (radians, in the required range).

Answer: $13\cos(\theta+1.176)$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	$R=13$	B1
	$\tan\alpha=\pm\dfrac{12}{5}$ or $\tan\alpha=\pm\dfrac{5}{12}$	M1
	$\alpha=$ awrt $1.176$	A1
Notes
M1: Condone $\cos\alpha=\frac{5}{13}$ or $\sin\alpha=\frac{12}{13}$
A1: Degrees ( $67.4^\circ$ ) do not earn this mark

(3 marks)

Common mistakes

Don't use the same sign in $\cos(A\pm B)$ instead of reversing it in the expansion.
Don't choose the sign of $\alpha$ in an $R$ -form without expanding and comparing both coefficients.

Mark-scheme wording to use

“K = 500 B1 tan α = 480/140 ⇒ α = ... M1 α = awrt 73.74° or 500cos(θ + 73.74)° A1 … B1: Correct value for K. Condone R = 500 M1: Award for tan α = ±480/140 ⇒ α = ..., tan α = ±140/480 ⇒ α = ..., sin α = ±480/"500" ⇒ α = ... or cos α = ±140/"500" ⇒ α = ... … A1: α = awrt 73.74{°} or correct expression 500cos(θ + 73.74){°}” — Pearson 9MA0, June 2024 Paper 1, Q12(a)
“R = 1000 + 500cos(30t + 73.74)° or R = 1000 + 140cos(30t)° − 480sin(30t)° B1ft (b)(ii) {R_min =}500 B1ft … B1ft: Correct equation of the model in either form including the R = following through on their numerical K (0 < K ≤ 750) and their numerical α. … B1ft: 500 or follow through on (their A – their K) or (1500 – 2 × their K) provided it is non-negative and less than 1500. It must be clear this is their answer to (b)(ii) so expect to see e.g. (b) or R_min = or an indication it is the minimum.” — Pearson 9MA0, June 2024 Paper 1, Q12(b)
“B1 for the exact value R = sqrt(109) … M1 for tan(alpha) = 3/10 or an equivalent comparison. A1 for alpha = 16.70 degrees and the correct completed form” — Pearson 9MA0, Sample Paper 2, Q13(a)

What the examiner wants

The right ratio in the right units. The June 2024 Paper 1 report says the $R$ -form part was mostly done correctly, and that the common mistakes among those who dropped marks were using radians instead of degrees and calculating $\arctan\dfrac{140}{480}$ instead of $\arctan\dfrac{480}{140}$ .

Exam tip

Expand the proposed $R$ -form, compare coefficients, then state $R>0$ and the required range for $\alpha$ .

Worked practice

Tier 1 · Easy

Use a compound-angle formula to find the exact value of

\sin(75^\circ)

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\dfrac{\sqrt6+\sqrt2}{4}$	3
Notes
Write $75^\circ=45^\circ+30^\circ$ . Then $\sin75^\circ=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ=(1/\sqrt2)(\sqrt3/2)+(1/\sqrt2)(1/2)=(\sqrt6+\sqrt2)/4$ .

(3 marks)

Tier 2 · Standard

Given that

\sin\theta=\dfrac35

and

0<\theta<\dfrac{\pi}{2}

, find the exact values of

\sin2\theta

\cos2\theta

and

\tan2\theta

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\sin2\theta=\dfrac{24}{25}$ $\cos2\theta=\dfrac{7}{25}$ $\tan2\theta=\dfrac{24}{7}$	4
Notes
Because $\theta$ is acute, $\cos\theta=4/5$ . Hence $\sin2\theta=2\sin\theta\cos\theta=2(3/5)(4/5)=24/25$ . Also $\cos2\theta=\cos^2\theta-\sin^2\theta=16/25-9/25=7/25$ . Therefore $\tan2\theta=(\sin2\theta)/(\cos2\theta)=24/7$ .

(4 marks)

Tier 3 · Hard

Use two points on the unit circle and the cosine rule to prove geometrically that

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

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Equate the coordinate and cosine-rule expressions for the squared chord joining $(\cos A,\sin A)$ and $(\cos B,\sin B)$ .	5
Notes
Let $P=(\cos A,\sin A)$ and $Q=(\cos B,\sin B)$ on the unit circle. By coordinates, $PQ^2=(\cos A-\cos B)^2+(\sin A-\sin B)^2=2-2(\cos A\cos B+\sin A\sin B)$ . In triangle $OPQ$ , $OP=OQ=1$ . If $\phi\in[0,\pi]$ is the smaller central angle, then $\cos\phi=\cos(A-B)$ , so the cosine rule gives $PQ^2=2-2\cos(A-B)$ . Equating the two expressions for $PQ^2$ and dividing by $-2$ proves $\cos(A-B)=\cos A\cos B+\sin A\sin B$ .

(5 marks)

Tier 1 · Easy

By writing

15^\circ

as the difference of two standard angles, find

\tan15^\circ

exactly.

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$2-\sqrt3$	3
Notes
Write $15^\circ=45^\circ-30^\circ$ . Then $\tan15^\circ=(1-1/\sqrt3)/(1+1/\sqrt3)=(\sqrt3-1)/(\sqrt3+1)=2-\sqrt3$ .

(3 marks)

Tier 2 · Standard

Write

7\cos\theta+24\sin\theta

as the single sine expression

R\sin(\theta+\alpha)

, choosing

R

positive and

\alpha

acute. Hence find the exact range of

f(\theta)=\dfrac{1}{26+7\cos\theta+24\sin\theta}

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$25\sin(\theta+\alpha)$ , where $\alpha=\arctan(7/24)$ $\dfrac{1}{51}\leq f(\theta)\leq1$	4
Notes
Expanding gives $R\sin(\theta+\alpha)=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha$ . Thus $R\cos\alpha=24$ and $R\sin\alpha=7$ , so $R=\sqrt{24^2+7^2}=25$ and $\tan\alpha=7/24$ . Therefore $1\leq26+25\sin(\theta+\alpha)\leq51$ . All values are positive, so taking reciprocals reverses the bounds and gives $1/51\leq f(\theta)\leq1$ .

(4 marks)

Tier 3 · Hard

Starting from the addition formulae for sine and cosine, derive

\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}

for values of

A

and

B

for which the expressions are defined.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Divide the expansions of $\sin(A+B)$ and $\cos(A+B)$ by $\cos A\cos B$ .	4
Notes
Using the addition formulae, $\tan(A+B)=\dfrac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}$ . Divide numerator and denominator by $\cos A\cos B$ . The numerator becomes $\tan A+\tan B$ and the denominator becomes $1-\tan A\tan B$ , giving the required formula wherever all the displayed expressions are defined.

(4 marks)

Tier 2 · Standard

Express

5\cos^2x-2\sin^2x

in the form

a+b\cos(2x)

, where

a

and

b

are constants. Hence state its maximum and minimum values.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$a=3/2$ and $b=7/2$ Maximum $=5$ and minimum $=-2$	4
Notes
Use $\cos^2x=(1+\cos2x)/2$ and $\sin^2x=(1-\cos2x)/2$ . Then $5\cos^2x-2\sin^2x=\tfrac52(1+\cos2x)-(1-\cos2x)=\tfrac32+\tfrac72\cos2x$ . Since $-1\leq\cos2x\leq1$ , the maximum is $3/2+7/2=5$ and the minimum is $3/2-7/2=-2$ .

(4 marks)

Tier 3 · Hard

Using double-angle formulae, prove that

\tan(\theta/2)=\dfrac{\sin\theta}{1+\cos\theta}

for values of

\theta

for which both sides are defined. Hence find

\tan(\pi/8)

exactly.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$\tan(\theta/2)=\dfrac{\sin\theta}{1+\cos\theta}$ $\tan(\pi/8)=\sqrt2-1$	5
Notes
Write $u=\theta/2$ . Then $\sin\theta=2\sin u\cos u$ and $1+\cos\theta=1+2\cos^2u-1=2\cos^2u$ . Where both sides are defined, $\cos u\neq0$ , so division gives $\sin\theta/(1+\cos\theta)=\tan u=\tan(\theta/2)$ . Taking $\theta=\pi/4$ gives $\tan(\pi/8)=\dfrac{\sqrt2/2}{1+\sqrt2/2}=\dfrac{\sqrt2}{2+\sqrt2}=\sqrt2-1$ .

(5 marks)

Tier 3 · Hard

Given that

\cos(2\theta)=-7/25

and

\pi/2<\theta<3\pi/4

, find

\sin\theta

and

\cos\theta

exactly. Hence find the exact value of

\sin(\theta+\pi/4)

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\sin\theta=4/5$ and $\cos\theta=-3/5$ $\sin(\theta+\pi/4)=\sqrt2/10$	5
Notes
From $\cos(2\theta)=2\cos^2\theta-1=-7/25$ , we obtain $\cos^2\theta=9/25$ . The stated interval is in quadrant II, so $\cos\theta=-3/5$ and $\sin\theta=4/5$ . Using the sine addition formula, $\sin(\theta+\pi/4)=\sin\theta\cos(\pi/4)+\cos\theta\sin(\pi/4)=(4/5-3/5)/\sqrt2=\sqrt2/10$ .

(5 marks)

Q10

Tier 3 · Hard

10.

Show, by repeated use of the double-angle formula for sine, that

\sin(8x)=8\sin x\cos x\cos(2x)\cos(4x)

. Hence find the exact value of

\cos20^\circ\cos40^\circ\cos80^\circ

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$\sin(8x)=8\sin x\cos x\cos(2x)\cos(4x)$ $\cos20^\circ\cos40^\circ\cos80^\circ=1/8$	5
Notes
Apply $\sin(2u)=2\sin u\cos u$ three times: $\sin(8x)=2\sin(4x)\cos(4x)=4\sin(2x)\cos(2x)\cos(4x)=8\sin x\cos x\cos(2x)\cos(4x)$ . Set $x=20^\circ$ . Since $\sin160^\circ=\sin20^\circ$ and $\sin20^\circ\neq0$ , the identity gives $\sin20^\circ=8\sin20^\circ\cos20^\circ\cos40^\circ\cos80^\circ$ . Cancelling $\sin20^\circ$ gives the exact product $1/8$ .

(5 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 5.6. Browse the Edexcel A-level Maths 9MA0 past papers directly.

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