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

20 original exam-style questions - 3 pages of questions with a full mark scheme - free printable PDF.

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A-Level · Pure Mathematics

P13D Pure: Trigonometry Depth: Fluency and Exam Drill

EDEXCEL 9MA0 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

State the exact value of sin(30 degrees).

(Total for Question 1 is 1 mark)

State the exact value of cos(60 degrees).

(Total for Question 2 is 1 mark)

State the exact value of tan(45 degrees).

(Total for Question 3 is 1 mark)

Simplify sin²(θ) + cos²(θ).

(Total for Question 4 is 1 mark)

Write sec(θ) in terms of cos(θ).

(Total for Question 5 is 1 mark)

Write cosec(θ) in terms of sin(θ).

(Total for Question 6 is 1 mark)

State the identity for cos(2theta) in terms of cos(θ) only.

(Total for Question 7 is 1 mark)

State the identity for sin(2theta) in terms of sin(θ) and cos(θ).

(Total for Question 8 is 1 mark)

Simplify sin(θ) / cos(θ).

(Total for Question 9 is 1 mark)

Solve, for 0 degrees ≤ x < 90 degrees, the equation cos x = 0.5, giving your answer in degrees.

(Total for Question 10 is 1 mark)

Show that, for cos θ not equal to 0, sec² θ - tan² θ = 1.

(Total for Question 11 is 2 marks)

Solve, for 0 degrees ≤ θ < 360 degrees, the equation sin θ = -0.5, giving all solutions in degrees.

(Total for Question 12 is 2 marks)

Solve, for 0 degrees ≤ x < 360 degrees, the equation cos(2x) = 0.5, giving all solutions.

(Total for Question 13 is 2 marks)

Given that sin A = 3/5, where A is acute, find the exact value of cos A.

(Total for Question 14 is 2 marks)

Show that, for cos θ not equal to 0 and sin θ not equal to 1, (1 - sin θ)/cos θ + cos θ/(1 - sin θ) = 2 sec θ.

(Total for Question 15 is 4 marks)

Solve, for 0 degrees ≤ θ < 360 degrees, the equation 2cos² θ + sin θ = 2, giving your answers in degrees.

(Total for Question 16 is 4 marks)

Prove that cos(3theta) = 4cos³(θ) - 3cos(θ), using the addition formula for cos(A+B) together with the double angle identities for cos(2theta) and sin(2theta).

(Total for Question 17 is 4 marks)

Solve, for 0 degrees ≤ x < 360 degrees, the equation 3 sin x = 2 cos x, giving your answers to 1 decimal place.

(Total for Question 18 is 3 marks)

f(θ) = 5 sin θ + 12 cos θ, for 0 degrees ≤ θ < 360 degrees.

(a)Express f(θ) in the form R sin(θ + α), where R > 0 and 0 < α < 90, giving the value of α to 1 decimal place.(2)

(b)Hence solve, for 0 degrees ≤ θ < 360 degrees, the equation 5 sin θ + 12 cos θ = 6.5, giving your answers to 1 decimal place.(2)

(Total for Question 19 is 4 marks)

The height, H metres, of a point on a large observation wheel above the ground, t minutes after the ride begins, is modelled by H(t) = 65 + 60 cos(12t), where the angle 12t is measured in degrees.

(a)State the maximum possible height given by the model.(1)

(b)Find the smallest positive value of t, in minutes, for which the height is at its minimum possible value.(2)

(Total for Question 20 is 3 marks)

Mark scheme · P13D Pure: Trigonometry Depth: Fluency and Exam Drill

Question 1

B1 1/2 cao
Answer: 1/2

Question 2

B1 1/2 cao
Answer: 1/2

Question 3

B1 1 cao
Answer: 1

Question 4

B1 1 cao
Answer: 1

Question 5

B1 1/cos(θ) cao
Answer: 1/cos(θ)

Question 6

B1 1/sin(θ) cao
Answer: 1/sin(θ)

Question 7

B1 2cos²(θ) - 1, oe
Answer: 2cos²(θ) - 1

Question 8

B1 2sin(θ)cos(θ) cao
Answer: 2sin(θ)cos(θ)

Question 9

B1 tan(θ) cao
Answer: tan(θ)

Question 10

B1 x = 60 cao
Answer: x = 60

Question 11

M1 writes sec² θ - tan² θ = (1 - sin² θ)/cos² θ = cos² θ / cos² θ
A1 cso: = 1
Answer: Proof: sec² θ - tan² θ = 1

Question 12

M1 finds the reference angle arcsin(0.5) = 30 and identifies the 3rd and 4th quadrants
A1 θ = 210 or θ = 330, both values
Answer: θ = 210 or θ = 330

Question 13

M1 solves for 2x over the extended range 0 ≤ 2x < 720, giving 2x = 60, 300, 420, 660
A1 x = 30, 150, 210, 330, all four values
Answer: x = 30, 150, 210, 330

Question 14

M1 uses sin² A + cos² A = 1 to find cos² A = 1 - 9/25 = 16/25
A1 cos A = 4/5 (positive, since A is acute)
Answer: cos A = 4/5

Question 15

M1 combines the two fractions over the common denominator cos θ (1 - sin θ)
A1 forms the numerator (1-sin θ)² + cos² θ = 2 - 2sin θ, using sin²+cos²=1
M1 factorises the numerator as 2(1 - sin θ)
A1 cso: cancels (1 - sin θ) to leave 2/cos θ = 2 sec θ
Answer: Proof: (1-sin θ)/cos θ + cos θ/(1-sin θ) = 2 sec θ

Question 16

M1 uses cos² θ = 1 - sin² θ to form the equation -2sin² θ + sin θ = 0
A1 factorises to sin θ(1 - 2sin θ) = 0, giving sin θ = 0 or sin θ = 0.5
A1 θ = 0 or θ = 180 (from sin θ = 0)
A1 θ = 30 or θ = 150 (from sin θ = 0.5), all four values found
Answer: θ = 0, 30, 150 or 180

Question 17

M1 writes cos(3theta) = cos(2theta + θ) and expands using the addition formula to cos2theta costheta - sin2theta sintheta
M1 substitutes cos2theta = 2cos²θ - 1 and sin2theta = 2sinthetacostheta
A1 simplifies to 2cos³θ - costheta - 2sin²θ costheta
A1 cso: uses sin²θ = 1 - cos²θ to obtain 4cos³θ - 3costheta
Answer: Proof: cos(3theta) = 4cos³(θ) - 3cos(θ)

Question 18

M1 rearranges to tan x = 2/3
A1 reference angle awrt 33.7
A1 x = 33.7 or x = 213.7 (both values, since tan is positive in the 1st and 3rd quadrants)
Answer: x = 33.7 or x = 213.7

Question 19

(a) M1 R = √5² + 12² = 13
(a) A1 α = awrt 67.4, from tan α = 12/5
(a) Answer: f(θ) = 13 sin(θ + 67.4)
(b) M1 uses the form from (a): 13 sin(θ + α) = 6.5, so sin(θ + α) = 0.5, giving θ + α = 150 or 390
(b) A1 θ = awrt 82.6 or θ = awrt 322.6 (ft from (a))
(b) Answer: θ = 82.6 or θ = 322.6

Question 20

(a) B1 125 m
(a) Answer: 125 m
(b) M1 recognises the minimum occurs when cos(12t) = -1, so 12t = 180
(b) A1 t = 15 minutes cao
(b) Answer: t = 15 minutes

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

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P13D Pure: Trigonometry Depth: Fluency and Exam Drill

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