Pure: Trigonometry: Fluency and Exam Drill - Worksheets, Questions and Revision

20 original exam-style questions - 2 pages of questions with a full mark scheme - free printable PDF.

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A-Level · Pure Mathematics

P5D Pure: Trigonometry: Fluency and Exam Drill

EDEXCEL 9MA0 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Convert π/6 radians to degrees.

(Total for Question 1 is 1 mark)

Convert 60 degrees to radians, giving your answer in terms of π.

(Total for Question 2 is 1 mark)

A sector has radius 5 cm and angle 2 radians at the centre. Find the arc length.

(Total for Question 3 is 1 mark)

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

(Total for Question 4 is 1 mark)

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

(Total for Question 5 is 1 mark)

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

(Total for Question 6 is 1 mark)

A right-angled triangle has hypotenuse 10 cm and one angle of 30 degrees. Find the length of the side opposite this angle.

(Total for Question 7 is 1 mark)

Solve sin(x) = 0.5 for 0 degrees ≤ x ≤ 90 degrees.

(Total for Question 8 is 1 mark)

State the value of cos²(θ) + sin²(θ) for any angle θ.

(Total for Question 9 is 1 mark)

A sector has radius 6 cm and angle 1.5 radians at the centre. Find the area of the sector.

(Total for Question 10 is 1 mark)

A sector has radius 8 cm and arc length 20 cm. Find the angle at the centre, in radians.

(Total for Question 11 is 2 marks)

Solve cos(x) = 0.5 for 0 degrees ≤ x ≤ 360 degrees, giving all solutions.

(Total for Question 12 is 2 marks)

A triangle has sides a = 7 cm and b = 9 cm with included angle C = 50 degrees. Find the area of the triangle, giving your answer to 3 significant figures.

(Total for Question 13 is 2 marks)

Solve 2sin(x) - 1 = 0 for 0 degrees ≤ x ≤ 360 degrees, giving all solutions.

(Total for Question 14 is 3 marks)

In triangle ABC, AB = 8 cm, BC = 11 cm and angle ABC = 95 degrees. Calculate the length of AC, giving your answer to 3 significant figures.

(Total for Question 15 is 3 marks)

Solve tan(x) = -1 for 0 degrees ≤ x ≤ 360 degrees, giving all solutions.

(Total for Question 16 is 3 marks)

Solve 3cos(x) + 1 = 0 for -180 degrees ≤ x ≤ 180 degrees, giving your answers to 1 decimal place.

(Total for Question 17 is 3 marks)

Prove the identity sin(θ)/(1+cos(θ)) + (1+cos(θ))/sin(θ) = 2/sin(θ), for sin(θ) not equal to 0.

(Total for Question 18 is 4 marks)

The angle θ is such that sin(θ) = 3/5 and θ is obtuse. Find the exact value of cos(θ) and the exact value of tan(θ).

(Total for Question 19 is 4 marks)

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

(Total for Question 20 is 4 marks)

Mark scheme · P5D Pure: Trigonometry: Fluency and Exam Drill

Question 1

B1 30 degrees cao
Answer: 30 degrees

Question 2

B1 π/3 cao
Answer: π/3

Question 3

B1 10 cm cao
Answer: 10 cm

Question 4

B1 1/2 cao
Answer: 1/2

Question 5

B1 1/2 cao
Answer: 1/2

Question 6

B1 1 cao
Answer: 1

Question 7

B1 5 cm cao
Answer: 5 cm

Question 8

B1 x = 30 degrees cao
Answer: x = 30 degrees

Question 9

B1 1 cao
Answer: 1

Question 10

B1 27 cm² cao
Answer: 27 cm²

Question 11

M1 uses θ = s/r
A1 2.5 radians cao
Answer: 2.5 radians

Question 12

M1 finds the principal value, 60 degrees
A1 x = 60 and x = 300 degrees, both correct, no extras
Answer: x = 60 degrees, 300 degrees

Question 13

M1 uses Area = 0.5 * a * b * sin(C)
A1 awrt 24.1 cm² cao
Answer: 24.1 cm² (3 s.f.)

Question 14

M1 rearranges to sin(x) = 0.5
A1 x = 30 degrees
A1 x = 150 degrees, no extra values in range
Answer: x = 30 degrees, 150 degrees

Question 15

M1 correct statement of the cosine rule, AC² = AB² + BC² - 2(AB)(BC)cos(ABC)
M1 correct substitution: AC² = 8² + 11² - 2(8)(11)cos(95)
A1 AC = awrt 14.2 cm cao
Answer: AC = 14.2 cm (3 s.f.)

Question 16

M1 finds the reference angle 45 degrees
A1 x = 135 degrees
A1 x = 315 degrees, no extra values in range
Answer: x = 135 degrees, 315 degrees

Question 17

M1 rearranges to cos(x) = -1/3
M1 finds the principal value, awrt 109.5 degrees
A1 x = 109.5 and x = -109.5 degrees (awrt 1 d.p.), both values, no extras in range
Answer: x = 109.5 degrees, -109.5 degrees

Question 18

M1 combines the two fractions over a common denominator sin(θ)(1+cos(θ))
M1 expands the numerator sin²(θ) + (1+cos(θ))²
A1 uses sin²(θ)+cos²(θ)=1 to simplify the numerator to 2+2cos(θ)
A1 cancels the common factor (1+cos(θ)) to complete the proof, cso
Answer: Identity proved

Question 19

M1 uses sin²(θ) + cos²(θ) = 1 to find cos²(θ) = 16/25
A1 cos(θ) = -4/5 (negative, since θ is obtuse)
M1 uses tan(θ) = sin(θ)/cos(θ)
A1 tan(θ) = -3/4
Answer: cos(θ) = -4/5; tan(θ) = -3/4

Question 20

M1 lets s = sin(x) and factorises 2s²+s-1 as (2s-1)(s+1)
A1 sin(x) = 1/2 or sin(x) = -1, both stated
M1 correct method for all solutions from sin(x)=1/2 (30, 150) and sin(x)=-1 (270)
A1 x = 30, 150, 270 degrees, cao, no extras
Answer: x = 30, 150, 270 degrees

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Pure: Trigonometry: Fluency and Exam Drill - Worksheets, Questions and Revision

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P5D Pure: Trigonometry: Fluency and Exam Drill

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Question 20

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