What is G21: Exact trig values?

Know exact sine and cosine values for 0°, 30°, 45°, 60° and 90°, and exact tangent values through 60°. The sequences are sinθ:0,1/2,√(2)/2,√(3)/2,1 and cosθ:1,√(3)/2,√(2)/2,1/2,0.

How many marks are G21 questions worth?

In verified past-paper appearances, G21 questions on Edexcel GCSE Maths 1MA1 have been worth 1–5 marks.

Which past exam papers has G21 appeared in?

G21 has appeared in 11 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

G21

Know the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60° and 90°; know the exact value of tan θ for θ = 0°, 30°, 45° and 60°

Exact trig values

Worked answers, methods and verified real exam appearances for G21 on Edexcel GCSE Maths 1MA1.

Explanation

Know exact sine and cosine values for $0^\circ$ , $30^\circ$ , $45^\circ$ , $60^\circ$ and $90^\circ$ , and exact tangent values through $60^\circ$ . The sequences are $\sin\theta:0,\frac12,\frac{\sqrt2}{2},\frac{\sqrt3}{2},1$ and $\cos\theta:1,\frac{\sqrt3}{2},\frac{\sqrt2}{2},\frac12,0$ .
Also $\tan0^\circ=0$ , $\tan30^\circ=\frac1{\sqrt3}$ , $\tan45^\circ=1$ and $\tan60^\circ=\sqrt3$ . The $1:\sqrt3:2$ and $1:1:\sqrt2$ triangles explain these values by pairing opposite, adjacent and hypotenuse sides with the marked angle.
If a value is forgotten, reconstruct it from the appropriate special triangle rather than using a decimal.
Use exact values throughout multi-step area or algebra calculations.
Examiners require fractions and surds.

The

1:\sqrt3:2

and

1:1:\sqrt2

triangles generate the exact trigonometric values.

Worked example

Work out the exact value of $3\sin60^\circ-2\cos45^\circ$ .

1.Use $\sin60^\circ=\frac{\sqrt3}{2}$ and $\cos45^\circ=\frac{\sqrt2}{2}$ .
2.Substitute: $3\left(\frac{\sqrt3}{2}\right)-2\left(\frac{\sqrt2}{2}\right)$ .
3.Simplify without decimals.

Answer: $\frac{3\sqrt3}{2}-\sqrt2$ .

Common mistakes

Don't enter the values into a calculator and give decimals when the question asks for exact form.
Don't swap $\sin30^\circ=\frac12$ with $\sin60^\circ=\frac{\sqrt3}{2}$ .

Exam tip

Write each exact trig value before substitution; the unsimplified exact form usually secures the method.

Worked practice

Tier 1 · Easy

Work out the exact value of

\sin30^\circ+\cos60^\circ

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$1$	1	$\sin30^\circ=\frac12$ and $\cos60^\circ=\frac12$ . Therefore the sum is $\frac12+\frac12=1$ .

Tier 2 · Standard

Work out the exact value of

2\sin45^\circ\cos45^\circ+\cos60^\circ

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$\frac32$	3	$\sin45^\circ=\cos45^\circ=\frac{\sqrt2}{2}$ and $\cos60^\circ=\frac12$ . Therefore $2\sin45^\circ\cos45^\circ+\cos60^\circ=2(\frac{\sqrt2}{2})(\frac{\sqrt2}{2})+\frac12=1+\frac12=\frac32$ .

Tier 3 · Hard

The hypotenuse of a right-angled triangle is

12\text{ cm}

and one acute angle is

30^\circ

. Work out its exact area.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$18\sqrt3\text{ cm}^2$	4	The side opposite $30^\circ$ is $12\sin30^\circ=12\times\frac12=6\text{ cm}$ . The adjacent side is $12\cos30^\circ=12\times\frac{\sqrt3}{2}=6\sqrt3\text{ cm}$ . Hence the area is $\frac12\times6\times6\sqrt3=18\sqrt3\text{ cm}^2$ .

Tier 1 · Easy

The angle

x

is one of

0^\circ

30^\circ

45^\circ

60^\circ

90^\circ

. Given that

\sin x=\cos x

, write down

x

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$45^\circ$	1	$\sin45^\circ=\cos45^\circ=\frac{\sqrt2}{2}$ , so $x=45^\circ$ .

Tier 2 · Standard

The acute angle

x

is one of

30^\circ

45^\circ

60^\circ

. Given that

\sin x=\frac12

, write down

x

and the exact value of

\cos x

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$x=30^\circ$ $\cos x=\frac{\sqrt3}{2}$	2	$\sin30^\circ=\frac12$ , so $x=30^\circ$ . The corresponding exact cosine value is $\cos30^\circ=\frac{\sqrt3}{2}$ .

Tier 3 · Hard

A right-angled triangle has an acute angle of

60^\circ

. The side between the

60^\circ

angle and the right angle is

4\text{ cm}

. Work out the exact area of the triangle.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$8\sqrt3\text{ cm}^2$	3	Let the opposite side be $h$ . Then $\tan60^\circ=\frac{h}{4}$ , so $h=4\sqrt3\text{ cm}$ . The area is $\frac12\times4\times4\sqrt3=8\sqrt3\text{ cm}^2$ .

Tier 2 · Standard

Solve

2x\sin30^\circ+3\cos60^\circ=\frac{7}{2}

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$x=2$	3	Use $\sin30^\circ=\frac12$ and $\cos60^\circ=\frac12$ . The equation becomes $x+\frac32=\frac72$ , so $x=2$ .

Tier 3 · Hard

Hassan says

2\sin30^\circ+\cos60^\circ=\tan45^\circ

. Is Hassan correct? You must show all your working.

(3)

(Total for Question 8 is 3 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	No, the left side is $\frac32$ but the right side is $1$ .	3	$2\sin30^\circ+\cos60^\circ=2(\frac12)+\frac12=\frac32$ . However, $\tan45^\circ=1$ . Since $\frac32\ne1$ , Hassan is not correct.

Tier 3 · Hard

Solve

x^2\tan45^\circ-8x\sin30^\circ+3\cos0^\circ=0

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$x=1$ or $x=3$	4	Use $\tan45^\circ=1$ , $\sin30^\circ=\frac12$ and $\cos0^\circ=1$ . The equation becomes $x^2-4x+3=0$ . Factorising gives $(x-1)(x-3)=0$ , so $x=1$ or $x=3$ .

Q10

Tier 3 · Hard

Higher only: A right-angled triangle has an acute angle of

60^\circ

and area

48\sqrt3\text{ cm}^2

. Work out the exact perimeter of the triangle.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$12\sqrt6+12\sqrt2\text{ cm}$	4	Let the side adjacent to $60^\circ$ be $x\text{ cm}$ . The opposite side is $x\tan60^\circ=x\sqrt3$ . Hence $\frac12x(x\sqrt3)=48\sqrt3$ , so $x^2=96$ and $x=4\sqrt6$ . The opposite side is $4\sqrt6\times\sqrt3=12\sqrt2$ , and the hypotenuse is $x\div\cos60^\circ=8\sqrt6$ . The perimeter is $4\sqrt6+12\sqrt2+8\sqrt6=12\sqrt6+12\sqrt2\text{ cm}$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-06	1H	Q22	4	Non-calculator	Higher	QP MS
2019-06	1H	Q14	2	Non-calculator	Higher	QP MS
2023-11	1H	Q20	4	Non-calculator	Higher	QP MS
2022-06	1F	Q30	1	Non-calculator	Foundation	QP MS
2023-06	1H	Q22	2	Non-calculator	Higher	QP MS
2021-11	1H	Q18	5	Non-calculator	Higher	QP MS
2024-06	3H	Q21	4	Allowed	Higher	QP MS
2024-11	1H	Q14	3	Non-calculator	Higher	QP MS
2023-06	1F	Q30	1	Non-calculator	Foundation	QP MS
2022-11	1H	Q22	4	Non-calculator	Higher	QP MS
2019-06	3H	Q22	4	Allowed	Higher	QP MS

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