What is 5.2: Small angle approximations?

For |θ| close to zero and measured in radians, sinθ≈θ, tanθ≈θ and cosθ1-θ^2/2. Replace each trigonometric function by its stated approximation, simplify algebraically, and retain only a solution whose magnitude is small enough for the approximation to be credible.

How many marks are 5.2 questions worth?

5.2 practice drills on this page are worth 1–6 marks each; marks in a real exam vary by question.

Which past exam papers has 5.2 appeared in?

We have not yet indexed a verified past-paper appearance for 5.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

5.2

Understand and use the standard small angle approximations of sine, cosine and tangent: sin θ ≈ θ, cos θ ≈ 1 − θ²/2, tan θ ≈ θ.

Draft — not yet indexed

Small angle approximations

Worked answers and methods for 5.2 on Edexcel A-level Maths 9MA0.

Explanation

For $|\theta|$ close to zero and measured in radians, $\sin\theta\approx\theta$ , $\tan\theta\approx\theta$ and $\cos\theta\approx1-\theta^2/2$ . Replace each trigonometric function by its stated approximation, simplify algebraically, and retain only a solution whose magnitude is small enough for the approximation to be credible.
For example, $1-\cos\theta\approx\theta^2/2$ , so $(1-\cos\theta)/\theta^2\approx1/2$ for a small non-zero $\theta$ .
The approximations are radian results, not degree results; another common error is to accept a large root created by the approximate polynomial.
The approximation sign matters: these are local approximations near zero, not identities.
Quote the approximation used, substitute the small radian angle, and reject any root whose magnitude is inconsistent with the small-angle assumption.

Worked example

Without using a calculator's trigonometric keys, estimate $\dfrac{1-\cos(0.08)}{(0.08)^2}$ by a small-angle approximation.

1.Use $\cos\theta\approx1-\theta^2/2$ .
2.Then $1-\cos(0.08)\approx(0.08)^2/2$ , so division by $(0.08)^2$ gives $1/2=0.5$ .

Answer: $0.5$

Common mistakes

Don't apply the small-angle approximations to an angle measured in degrees.
Don't accept a large algebraic root even though the approximation is valid only near zero.

Exam tip

State that the angle is in radians and reject any solution that is not small.

Worked practice

Tier 1 · Easy

Use a standard small-angle approximation to estimate

\sin(0.064)

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$0.064$	1
Notes
Since $0.064$ is a small angle in radians, use $\sin\theta\approx\theta$ to obtain $\sin(0.064)\approx0.064$ .

(1 mark)

Tier 2 · Standard

Use small-angle approximations to estimate the non-zero small positive solution of

\sin x=5(1-\cos x)

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$x\approx0.4$ radians	3
Notes
For small $x$ , use $\sin x\approx x$ and $1-\cos x\approx x^2/2$ . The equation becomes $x\approx5x^2/2$ . Since the required solution is non-zero, divide by $x$ to obtain $1\approx5x/2$ , hence $x\approx2/5=0.4$ radians.

(3 marks)

Tier 3 · Hard

A small positive angle

x

satisfies

\sin x+\cos x=1.08

. Use the standard small-angle approximations to estimate

x

, giving

4

decimal places, and explain which algebraic root is admissible.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$x\approx0.0835$ radians The other root is not a small angle.	4
Notes
Substitution gives $x+1-x^2/2\approx1.08$ , hence $x^2-2x+0.16\approx0$ . Therefore $x\approx1\pm\sqrt{0.84}$ , giving $0.08348\ldots$ or $1.9165\ldots$ . Only $0.08348\ldots$ is small, so $x\approx0.0835$ radians; the larger root lies outside the approximation's intended range.

(4 marks)

Tier 1 · Easy

For

x=0.06

radians, estimate

\cos x

using the appropriate small-angle approximation.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$0.9982$	2
Notes
For a small angle in radians, $\cos x\approx1-x^2/2$ . Hence $\cos(0.06)\approx1-(0.06)^2/2=0.9982$ .

(2 marks)

Tier 2 · Standard

A pendulum of length

2.5\,\text{m}

is displaced through

0.12

radians from the downward vertical. By considering the vertical component of the length, derive an expression for its vertical rise

h

. Then use a standard small-angle approximation to estimate

h

in centimetres, giving your answer to

3

significant figures.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$h=2.5(1-\cos(0.12))\,\text{m}$ $h\approx1.80\,\text{cm}$	3
Notes
Initially the bob is $2.5\,\text{m}$ vertically below the pivot. After the displacement its vertical distance below the pivot is $2.5\cos(0.12)\,\text{m}$ , so $h=2.5-2.5\cos(0.12)=2.5(1-\cos(0.12))\,\text{m}$ . For a small angle in radians, $1-\cos x\approx x^2/2$ . Hence $h\approx2.5(0.12)^2/2=0.018\,\text{m}=1.80\,\text{cm}$ to $3$ significant figures.

(3 marks)

Tier 3 · Hard

A sensor hangs from a straight cable of length

24\,\text{m}

. The cable makes a small positive angle

x

radians with the downward vertical, and the sensor is

23.88\,\text{m}

vertically below the support. Use standard small-angle approximations to estimate

x

and the sensor's horizontal displacement from the support.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$x\approx0.1$ radians Horizontal displacement $\approx2.4\,\text{m}$	4
Notes
The vertical component gives $24\cos x=23.88$ , so $\cos x=0.995$ . Using $\cos x\approx1-x^2/2$ gives $1-x^2/2\approx0.995$ , hence $x^2\approx0.01$ and the positive angle is $x\approx0.1$ radians. The horizontal displacement is $24\sin x\approx24x=2.4\,\text{m}$ .

(4 marks)

Tier 2 · Standard

For

x=0.05

radians, use standard small-angle approximations to estimate

\dfrac{3\sin(2x)-2\tan x}{1-\cos x}

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$160$	3
Notes
For small $x$ in radians, $\sin(2x)\approx2x$ , $\tan x\approx x$ and $1-\cos x\approx x^2/2$ . The expression is therefore approximately $\dfrac{6x-2x}{x^2/2}=\dfrac{8}{x}$ . At $x=0.05$ , this is $8/0.05=160$ .

(3 marks)

Tier 3 · Hard

For small

x

F(x)=a\sin x+b\cos x+\tan(2x)

, where

a

and

b

are positive constants. It is known that

F(x)\approx5+7x-\dfrac52x^2

. Use standard small-angle approximations to find

a

and

b

. Hence estimate the smaller positive solution of

F(x)=5.24

, giving your answer to

3

significant figures.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$a=5$ and $b=5$ $x\approx0.0347$ radians	5
Notes
Using $\sin x\approx x$ , $\cos x\approx1-x^2/2$ and $\tan(2x)\approx2x$ gives $F(x)\approx b+(a+2)x-bx^2/2$ . Comparing coefficients with $5+7x-5x^2/2$ gives $b=5$ and $a=5$ . The equation is then $5+7x-2.5x^2\approx5.24$ , or $2.5x^2-7x+0.24\approx0$ . Its roots are $0.034716\ldots$ and $2.76528\ldots$ ; the smaller positive solution is $x\approx0.0347$ radians.

(5 marks)

Tier 3 · Hard

A rigid boom of length

26\,\text{m}

is raised at an angle of

0.18

radians above the horizontal. Use standard small-angle approximations to estimate the horizontal and vertical coordinates of its end relative to the pivot, giving each coordinate to

4

significant figures. Calculate the corresponding coordinates using trigonometric functions, and show that the distance between the approximate and calculated positions is less than

2.6\,\text{cm}

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Approximate coordinates $(25.58,\ 4.680)\,\text{m}$ Calculated coordinates $(25.58,\ 4.655)\,\text{m}$ Position error $<2.6\,\text{cm}$	5
Notes
Using $\cos x\approx1-x^2/2$ and $\sin x\approx x$ , the approximate coordinates are $\bigl(26(1-0.18^2/2),26(0.18)\bigr)=(25.5788,4.68)$ , giving $(25.58,4.680)\,\text{m}$ to $4$ significant figures. Direct calculation gives $(26\cos0.18,26\sin0.18)=(25.579936\ldots,4.654768\ldots)$ , giving $(25.58,4.655)\,\text{m}$ . Using the unrounded coordinates, the distance between the positions is $\sqrt{(25.5788-25.579936\ldots)^2+(4.68-4.654768\ldots)^2}=0.0252566\ldots\,\text{m}=2.52566\ldots\,\text{cm}<2.6\,\text{cm}$ .

(5 marks)

Q10

Tier 3 · Hard

10.

A shallow circular arc has chord length

30\,\text{m}

and maximum height

0.60\,\text{m}

above the chord. The chord subtends a small angle

2u

radians at the centre. By resolving the radius along and perpendicular to the chord, show that

15=r\sin u

and

0.60=r(1-\cos u)

. Use standard small-angle approximations to obtain the values of

u

r

and

2u

predicted by the approximation model; in particular, report the approximate value of

r

, not the value from exact circle geometry.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$15=r\sin u$ and $0.60=r(1-\cos u)$ $u\approx0.080$ radians $r\approx187.5\,\text{m}$ Angle subtended by the chord $\approx0.160$ radians	6
Notes
The perpendicular from the centre bisects the chord, so the half-chord gives $15=r\sin u$ . The distance from the centre to the chord is $r\cos u$ , hence the height is $r-r\cos u=0.60$ . For small $u$ , use $\sin u\approx u$ and $1-\cos u\approx u^2/2$ . Thus $15\approx ru$ and $0.60\approx ru^2/2$ . Dividing the second relation by the first gives $0.60/15\approx u/2$ , so the approximation model gives $u\approx0.080$ . It then gives $r\approx15/0.080=187.5\,\text{m}$ and $2u\approx0.160$ radians. (Exact circle geometry would give $r=187.8\,\text{m}$ , but that is not the value requested.)

(6 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 5.2. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in 5 Trigonometry

Want help turning this into marks?

Bring 5.2 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call