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GCSE · Trigonometry

G3 Further Trigonometric Identities and Equations

AQA 8365 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question tests recall of exact trigonometric values. Do not use a calculator for these, even though a calculator is allowed for the rest of this paper.

(a)Write down the exact value of sin 30 degrees.(1)

(b)Write down the exact value of cos 60 degrees.(1)

(c)Write down the exact value of tan 45 degrees.(1)

(Total for Question 1 is 3 marks)

Trigonometric identities.

(a)Which of the following is a correct trigonometric identity, true for all values of x?(1)

A) sin²(x) - cos²(x) = 1
B) sin²(x) + cos²(x) = 1
C) sin(x) + cos(x) = 1
D) tan(x) = cos(x)/sin(x)

(Total for Question 2 is 1 mark)

Angle x is acute, and sin(x) = 3/5.

(a)Show that cos(x) = 4/5.(2)

(b)Hence find the exact value of tan(x).(1)

(Total for Question 3 is 3 marks)

Solving trigonometric equations.

(a)Solve tan(x) = 1.732 for 0 ≤ x ≤ 360, giving your answers to 1 decimal place.(3)

(Total for Question 4 is 3 marks)

Solving trigonometric equations.

(a)Solve 5 sin(x) = 3 for 0 ≤ x ≤ 360, giving your answers to 1 decimal place.(3)

(Total for Question 5 is 3 marks)

In triangle ABC, angle A = 52 degrees, angle B = 71 degrees, and side a (opposite angle A) = 8.4 cm.

Diagram NOT accurately drawn

(a)Calculate the length of side b (opposite angle B), giving your answer correct to 3 significant figures.(3)

(Total for Question 6 is 3 marks)

In triangle ABC, a = 7 cm, b = 9 cm and c = 12 cm, where side c is opposite angle C.

Diagram NOT accurately drawn

(a)Calculate the size of angle C, giving your answer correct to 1 decimal place.(4)

(Total for Question 7 is 4 marks)

Triangle PQR has PQ = 10 cm and PR = 14 cm. The area of triangle PQR is 55 cm², and angle QPR is acute.

Diagram NOT accurately drawn

(a)Calculate the size of angle QPR, giving your answer correct to 1 decimal place.(3)

(Total for Question 8 is 3 marks)

In triangle ABC, angle A = 35 degrees, a = 6 cm (opposite angle A) and b = 9 cm (opposite angle B).

Diagram NOT accurately drawn

(a)Find the two possible values of angle B, giving your answers correct to 1 decimal place.(4)

(Total for Question 9 is 4 marks)

Solving trigonometric equations.

(a)Solve 2 sin²(x) - 1 = 0 for 0 ≤ x ≤ 360, giving all solutions.(4)

(Total for Question 10 is 4 marks)

Solving trigonometric equations.

(a)Solve 2 sin²(x) + sin(x) - 1 = 0 for 0 ≤ x ≤ 360, giving all solutions.(5)

(Total for Question 11 is 5 marks)

Trigonometric identities.

(a)Show that (1 - cos²(x)) / sin(x) = sin(x), for sin(x) not equal to 0.(2)

(Total for Question 12 is 2 marks)

The graph of y = sin(x) is transformed to give the graph of y = 3 sin(x) - 2.

(a)State the amplitude of y = 3 sin(x) - 2.(1)

(b)State the maximum and minimum values of y = 3 sin(x) - 2.(2)

(c)State the period of y = 3 sin(x) - 2, in degrees.(1)

(Total for Question 13 is 4 marks)

The graph of y = cos(x) is transformed to give the graph of y = cos(2x).

(a)Describe fully the single transformation that maps the graph of y = cos(x) onto the graph of y = cos(2x).(2)

(b)State the period of y = cos(2x), in degrees.(1)

(c)Find all solutions of cos(2x) = 0.5 for 0 ≤ x ≤ 360.(4)

(Total for Question 14 is 7 marks)

ABCDEFGH is a cuboid, with base ABCD and top EFGH. AB = 6 cm, BC = 8 cm and the vertical edge CG = 5 cm. AG is the diagonal from vertex A to vertex G.

Diagram NOT accurately drawn

(a)Calculate the size of the angle between AG and the plane ABCD, giving your answer correct to 1 decimal place.(3)

(b)Calculate the length of AG, giving your answer correct to 3 significant figures.(2)

(Total for Question 15 is 5 marks)

A ship sails from port A on a bearing of 072 degrees for 15 km to reach point B. It then changes course and sails on a bearing of 155 degrees for 21 km to reach point C.

Diagram NOT accurately drawn

(a)Show that angle ABC = 97 degrees.(2)

(b)Calculate the distance AC, giving your answer correct to 3 significant figures.(3)

(c)Find the bearing of C from A, giving your answer to the nearest degree.(4)

(Total for Question 16 is 9 marks)

x is an angle measured in degrees.

(a)Prove that (sin(x) + cos(x))² = 1 + 2 sin(x) cos(x) for all values of x.(3)

(b)Given that sin(x) + cos(x) = 1.2, use your result from part (a) to find the value of sin(x) cos(x).(2)

(c)Given also that (sin(x) - cos(x))² = 1 - 2 sin(x) cos(x), find the value of (sin(x) - cos(x))^2.(2)

(Total for Question 17 is 7 marks)

Solving trigonometric equations.

(a)Solve 3 cos²(x) + cos(x) - 2 = 0 for 0 ≤ x ≤ 360, giving your answers to 1 decimal place where necessary.(5)

(Total for Question 18 is 5 marks)

Mark scheme · G3 Further Trigonometric Identities and Equations

Question 1

(a) B1 1/2 oe
(a) Answer: 1/2
(b) B1 1/2 oe
(b) Answer: 1/2
(c) B1 1
(c) Answer: 1

Question 2

(a) B1 B) sin²(x) + cos²(x) = 1
(a) Answer: B) sin²(x) + cos²(x) = 1

Question 3

(a) M1 use sin²(x) + cos²(x) = 1 to get cos²(x) = 1 - (3/5)² = 16/25
(a) A1 cos(x) = 4/5 (positive root, since x is acute), cso
(a) Answer: 4/5
(b) B1 tan(x) = 3/4 oe ft from part (a)
(b) Answer: 3/4

Question 4

(a) M1 arctan(1.732) = 60.0 (principal value, awrt)
(a) A1 x = 60.0
(a) A1 x = 240.0 (ft: principal value + 180, since tan has period 180)
(a) Answer: x = 60.0 or x = 240.0

Question 5

(a) M1 sin(x) = 0.6 (rearrange)
(a) A1 x = 36.9 (awrt)
(a) A1 x = 143.1 (ft: 180 - first value, awrt)
(a) Answer: x = 36.9 or x = 143.1

Question 6

(a) M1 correct sine rule stated: b/sin(B) = a/sin(A)
(a) M1 correct substitution: b = 8.4 x sin(71) / sin(52)
(a) A1 b = 10.1 cm awrt (3sf)
(a) Answer: 10.1 cm

Question 7

(a) M1 cosine rule rearranged: cos(C) = (a² + b² - c²) / (2ab)
(a) M1 correct substitution: cos(C) = (49 + 81 - 144) / (2 x 7 x 9)
(a) A1 cos(C) = -1/9 oe (-0.111...)
(a) A1 C = 96.4 degrees awrt
(a) Answer: 96.4 degrees

Question 8

(a) M1 rearrange Area = 1/2 x PQ x PR x sin(QPR) to make sin(QPR) the subject
(a) A1 sin(QPR) = 55/70 = 11/14 oe (0.786)
(a) A1 QPR = 51.8 degrees awrt
(a) Answer: 51.8 degrees

Question 9

(a) M1 correct sine rule stated: sin(B)/b = sin(A)/a
(a) A1 sin(B) = 9 x sin(35) / 6 = 0.860(4) oe
(a) A1 B = 59.4 degrees awrt
(a) A1 B = 120.6 degrees (180 - first value, awrt)
(a) Answer: B = 59.4 degrees or B = 120.6 degrees

Question 10

(a) M1 rearrange to sin²(x) = 0.5
(a) M1 sin(x) = +1/√2 or sin(x) = -1/√2 (awrt ±0.707)
(a) A1 x = 45 and x = 135 (from positive root)
(a) A1 x = 225 and x = 315 (from negative root) cao
(a) Answer: x = 45, 135, 225, 315

Question 11

(a) M1 let s = sin(x) and form 2s² + s - 1 = 0, then factorise (2s - 1)(s + 1) = 0 (or use the quadratic formula)
(a) A1 s = 0.5
(a) A1 s = -1
(a) A1 x = 30 and x = 150 (from s = 0.5)
(a) A1 x = 270 (from s = -1) cao
(a) Answer: x = 30, 150, 270

Question 12

(a) M1 use sin²(x) + cos²(x) = 1 to write 1 - cos²(x) = sin²(x)
(a) A1 sin²(x) / sin(x) = sin(x), fully shown, cso
(a) Answer: sin(x) (shown)

Question 13

(a) B1 3
(a) Answer: 3
(b) B1 maximum = 1
(b) B1 minimum = -5
(b) Answer: maximum = 1, minimum = -5
(c) B1 360 degrees
(c) Answer: 360 degrees

Question 14

(a) B1 (horizontal) stretch
(a) B1 scale factor 1/2, parallel to the x-axis oe
(a) Answer: Horizontal stretch, scale factor 1/2, parallel to the x-axis
(b) B1 180 degrees, ft from part (a)
(b) Answer: 180 degrees
(c) M1 let u = 2x and find the principal value: arccos(0.5) = 60
(c) M1 use periodicity/symmetry of cos to find all values of u in 0 ≤ u ≤ 720: u = 60, 300, 420, 660
(c) A1 two of x = 30, 150, 210, 330 correct (x = u/2)
(c) A1 all four of x = 30, 150, 210, 330 correct cao
(c) Answer: x = 30, 150, 210, 330

Question 15

(a) B1 AC = 10 cm (Pythagoras in right-angled triangle ABC: √6² + 8²)
(a) M1 tan(angle) = CG/AC = 5/10, using right-angled triangle ACG (right angle at C)
(a) A1 angle = 26.6 degrees awrt
(a) Answer: 26.6 degrees
(b) M1 AG = √AC² + CG² = √10² + 5², or full 3D Pythagoras √6² + 8² + 5²
(b) A1 AG = 11.2 cm awrt
(b) Answer: 11.2 cm

Question 16

(a) M1 back bearing from B to A = 072 + 180 = 252 degrees
(a) A1 angle ABC = 252 - 155 = 97 degrees, cso
(a) Answer: 97 degrees (shown)
(b) M1 correct cosine rule stated: AC² = AB² + BC² - 2 x AB x BC x cos(ABC)
(b) M1 correct substitution: AC² = 15² + 21² - 2 x 15 x 21 x cos(97)
(b) A1 AC = 27.3 km awrt (3sf)
(b) Answer: 27.3 km
(c) M1 correct sine rule stated: sin(BAC)/BC = sin(ABC)/AC
(c) A1 angle BAC = 49.9 degrees awrt
(c) M1 correct strategy: bearing of C from A = bearing of B from A + angle BAC = 072 + BAC
(c) A1 bearing = 122 degrees awrt (accept 121 to 122)
(c) Answer: 122 degrees

Question 17

(a) M1 expand the bracket: sin²(x) + 2 sin(x) cos(x) + cos²(x)
(a) M1 use sin²(x) + cos²(x) = 1 to simplify
(a) A1 fully shown: = 1 + 2 sin(x) cos(x), cso
(a) Answer: 1 + 2 sin(x) cos(x) (shown)
(b) M1 substitute 1.2 into the identity: 1.2² = 1 + 2 sin(x) cos(x)
(b) A1 sin(x) cos(x) = 0.22 cao
(b) Answer: 0.22
(c) M1 substitute sin(x) cos(x) = 0.22 into 1 - 2 sin(x) cos(x)
(c) A1 = 0.56 cao
(c) Answer: 0.56

Question 18

(a) M1 let c = cos(x) and form 3c² + c - 2 = 0
(a) M1 solve the quadratic (formula or factorisation) to get c = 2/3 or c = -1
(a) A1 x = 180 (from c = -1)
(a) A1 x = 48.2 (from c = 2/3, awrt)
(a) A1 x = 311.8 (ft: 360 - previous value, awrt)
(a) Answer: x = 48.2, 180, 311.8

Further Trigonometric Identities and Equations - Worksheets, Questions and Revision

G3 Further Trigonometric Identities and Equations

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18