Pure: Trigonometry - Worksheets, Questions and Revision

12 original exam-style questions - 4 pages of questions with a full mark scheme - free printable PDF.

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A-Level · Pure Mathematics

P5 Pure: Trigonometry

EDEXCEL 9MA0 · Calculator allowed · about 140 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A sector OAB of a circle, centre O and radius 8 cm, has angle AOB = θ radians, where θ = 5pi/12.

Diagram NOT accurately drawn

(a)Convert θ = 5pi/12 radians to degrees.(1)

(b)Find the arc length AB, giving your answer as an exact multiple of π.(2)

(c)Find the area of the sector OAB, giving your answer to 3 significant figures.(2)

(d)Find the perimeter of the segment cut off by the chord AB, giving your answer to 3 significant figures.(4)

(Total for Question 1 is 9 marks)

Solve the following trigonometric equations, giving all solutions in the given ranges.

(a)Solve 2 sin(3 θ - 20 degrees) = 1.2 for 0 degrees ≤ θ ≤ 360 degrees, giving your answers to 1 decimal place.(5)

(b)Solve cos(x/2) = -0.85 for 0 degrees ≤ x ≤ 720 degrees, giving your answers to 1 decimal place.(4)

(Total for Question 2 is 9 marks)

In a yacht race, three buoys A, B and C are positioned such that AB = 3.6 km, BC = 5.1 km and angle ABC = 112 degrees. The bearing of B from A is 072 degrees, and C is on a bearing greater than that of B from A.

Diagram NOT accurately drawn

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

(b)Calculate angle BCA, giving your answer to 1 decimal place.(3)

(c)Calculate the bearing of C from A, giving your answer to the nearest degree.(3)

(Total for Question 3 is 9 marks)

(a) Show that cosec(θ) - sin(θ) = cos(θ) cot(θ), for 0 degrees < θ < 180 degrees, θ not equal to 90 degrees.

(a)Show that cosec(θ) - sin(θ) = cos(θ) cot(θ), for 0 degrees < θ < 180 degrees, θ not equal to 90 degrees.(4)

(b)Hence solve cosec(θ) - sin(θ) = cos(θ) for 0 degrees ≤ θ ≤ 360 degrees, θ not equal to 0, 180, 360 degrees.(4)

(Total for Question 4 is 8 marks)

(a) Show that the equation 2cot²(θ) + cosec(θ) = 1 can be written in the form 2cosec²(θ) + cosec(θ) - 3 = 0.

(a)Show that the equation 2cot²(θ) + cosec(θ) = 1 can be written in the form 2cosec²(θ) + cosec(θ) - 3 = 0.(3)

(b)Hence solve 2cot²(θ) + cosec(θ) = 1 for 0 degrees ≤ θ ≤ 360 degrees, giving non-exact answers to 1 decimal place.(6)

(Total for Question 5 is 9 marks)

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

(a)Express f(θ) in the form R cos(θ + α), where R > 0 and 0 < α < 90 degrees, giving α to 2 decimal places.(3)

(b)Hence write down the maximum value of f(θ), and find the smallest positive value of θ at which this maximum occurs, to 1 decimal place.(2)

(c)Solve f(θ) = 6 for 0 degrees ≤ θ ≤ 360 degrees, giving your answers to 1 decimal place.(5)

(Total for Question 6 is 10 marks)

(a) Prove that sin(3theta) = 3sin(θ) - 4sin³(θ).

(a)Prove that sin(3theta) = 3sin(θ) - 4sin³(θ).(4)

(b)Hence, or otherwise, solve 4sin³(θ) - 3sin(θ) + 0.5 = 0 for 0 degrees ≤ θ ≤ 360 degrees.(5)

(Total for Question 7 is 9 marks)

Given that θ is small and measured in radians.

(a)Show that, for small θ, (2theta - sin(2theta)) / θ³ is approximately equal to 4/3.(4)

(b)Using your calculator, find the exact-value estimate from part (a) and the true value of (2theta - sin(2theta))/θ³ when θ = 0.1 radians, and hence find the percentage error in the estimate, giving your answer to 2 significant figures.(4)

(Total for Question 8 is 8 marks)

This question concerns the inverse trigonometric functions arcsin and arccos.

(a)State the domain and range of the function f(x) = arcsin(x).(2)

(b)Sketch the graph of y = arccos(x) for -1 ≤ x ≤ 1, indicating the coordinates of the endpoints of the curve and the point where the curve crosses the y-axis.(3)

(c)Solve arccos(2x - 1) = π/3, giving x as an exact fraction.(3)

(d)Given that arcsin(x) = arccos(x) + π/6, and using the identity arcsin(x) + arccos(x) = π/2, find the exact value of x.(4)

(Total for Question 9 is 12 marks)

The depth of water, D metres, in a harbour, t hours after midnight on a particular day, is modelled by D(t) = 6 + 2.5 sin(30t) degrees, for 0 ≤ t ≤ 24, where the angle is measured in degrees.

(a)State the maximum depth of water given by the model, and find the times during the 24 hour period at which it occurs.(3)

(b)Find the depth of water at 08:30, giving your answer to 3 significant figures.(2)

(c)Find the times during the 24 hour period at which the depth of water is exactly 7 m, giving your answers to the nearest minute.(6)

(Total for Question 10 is 11 marks)

Consider the equation 2cos²(x) + 7sin(x) - 5 = 0, for -180 degrees ≤ x ≤ 180 degrees.

(a)Show that the equation 2cos²(x) + 7sin(x) - 5 = 0 can be written in the form 2sin²(x) - 7sin(x) + 3 = 0.(2)

(b)Hence solve the equation 2cos²(x) + 7sin(x) - 5 = 0 for -180 degrees ≤ x ≤ 180 degrees, giving a reason why any additional roots of the quadratic in sin(x) must be rejected.(5)

(Total for Question 11 is 7 marks)

(a) Prove that cos(A + B) cos(A - B) = cos²(A) - sin²(B).

(a)Prove that cos(A + B) cos(A - B) = cos²(A) - sin²(B).(4)

(b)Using the result in part (a), or otherwise, show that cos(75 degrees) cos(15 degrees) = 1/4.(3)

(c)Hence find the exact value of cos²(105 degrees) - sin²(15 degrees).(3)

(Total for Question 12 is 10 marks)

Mark scheme · P5 Pure: Trigonometry

Question 1

(a) B1 75 degrees cao
(a) Answer: 75 degrees
(b) M1 uses s = r*θ with θ in radians, e.g. 8 * 5pi/12
(b) A1 10pi/3 cm oe cao (exact form)
(b) Answer: 10pi/3 cm (approx 10.5 cm)
(c) M1 uses A = 0.5 * r² * θ, e.g. 0.5 * 8² * 5pi/12
(c) A1 awrt 41.9 cm² cao
(c) Answer: 41.9 cm² (3 s.f.)
(d) M1 correct method for chord AB, e.g. AB = 2*8*sin(37.5 deg) or cosine rule AB² = 8²+8²-2*8*8*cos75
(d) A1 AB = awrt 9.74 cm
(d) dM1 (dep on both M marks) perimeter = arc length + chord AB
(d) A1 awrt 20.2 cm cao
(d) Answer: 20.2 cm (3 s.f.)

Question 2

(a) M1 rearranges to sin(3theta - 20) = 0.6
(a) M1 finds a reference/principal angle, awrt 36.9 degrees
(a) M1 correct method to generate further values of (3theta - 20) using 180 - x and/or +360k, and converts back by adding 20 and dividing by 3
(a) A1 at least 3 of the 6 correct values, awrt 1 d.p.
(a) A1 all six correct: θ = 19.0, 54.4, 139.0, 174.4, 259.0, 294.4 (awrt), cao, no extra values in range
(a) Answer: θ = 19.0, 54.4, 139.0, 174.4, 259.0, 294.4 degrees
(b) M1 sets u = x/2 with correct range 0 ≤ u ≤ 360
(b) M1 finds principal value arccos(-0.85), awrt 148.2 degrees
(b) M1 correct second value 360 - 148.2 = 211.8 degrees, and doubles both to find x
(b) A1 x = 296.4 and x = 423.6 (awrt 1 d.p.), cao, no extra values
(b) Answer: x = 296.4 degrees, 423.6 degrees

Question 3

(a) M1 correct statement of the cosine rule, AC² = AB² + BC² - 2*AB*BC*cos(ABC)
(a) M1 correct substitution: AC² = 3.6² + 5.1² - 2*3.6*5.1*cos(112)
(a) A1 AC = awrt 7.26 km cao
(a) Answer: AC = 7.26 km (3 s.f.)
(b) M1 correct statement of the sine rule, sin(BCA)/AB = sin(ABC)/AC
(b) M1 correct rearrangement: sin(BCA) = 3.6*sin(112)/7.2612 (ft AC from (a))
(b) A1 angle BCA = awrt 27.4 degrees cao (acute value taken since angle ABC is obtuse)
(b) Answer: angle BCA = 27.4 degrees (1 d.p.)
(c) M1 angle BAC = 180 - 112 - 27.4 (ft) = 40.6 degrees
(c) M1 bearing of C from A = bearing of B from A + angle BAC = 072 + 40.6 (since C is on a greater bearing)
(c) A1 113 degrees (awrt, cao), i.e. bearing 113 degrees
(c) Answer: Bearing of C from A = 113 degrees (to nearest degree)

Question 4

(a) M1 writes cosec(θ) as 1/sin(θ)
(a) M1 combines into a single fraction: (1 - sin²(θ))/sin(θ)
(a) M1 uses 1 - sin²(θ) = cos²(θ)
(a) A1 cso completes to cos(θ)*(cos(θ)/sin(θ)) = cos(θ) cot(θ)
(a) Answer: Identity shown (proof)
(b) M1 uses part (a) to form cos(θ)*cot(θ) = cos(θ), i.e. cos(θ)*(cot(θ) - 1) = 0
(b) M1 cos(θ) = 0 leading to θ = 90, 270
(b) M1 cot(θ) = 1, i.e. tan(θ) = 1, leading to θ = 45, 225
(b) A1 all four values correct: θ = 45, 90, 225, 270 degrees cao, no extras
(b) Answer: θ = 45, 90, 225, 270 degrees

Question 5

(a) M1 uses the identity cot²(θ) = cosec²(θ) - 1
(a) M1 substitutes to give 2(cosec²(θ) - 1) + cosec(θ) = 1
(a) A1 cso correct rearrangement to 2cosec²(θ) + cosec(θ) - 3 = 0
(a) Answer: Shown (identity rearrangement)
(b) M1 lets c = cosec(θ) and factorises 2c² + c - 3 = 0 as (2c+3)(c-1) = 0
(b) A1 cosec(θ) = 1 or cosec(θ) = -3/2 (i.e. sin(θ) = 1 or sin(θ) = -2/3), both stated
(b) M1 sin(θ) = 1 leads to θ = 90 degrees
(b) M1 finds reference angle for sin(θ) = -2/3, awrt 41.8 degrees, and correct method for third/fourth quadrant solutions
(b) A1 θ = 221.8 and 318.2 (awrt), both correct
(b) A1 all three final solutions correct: θ = 90, 221.8, 318.2 degrees, cao, no extras
(b) Answer: θ = 90, 221.8, 318.2 degrees

Question 6

(a) B1 R = 13, from √5² + 12²
(a) M1 correct trig ratio for α, e.g. tan(α) = 12/5
(a) A1 α = 67.38 degrees (awrt 2 d.p.), cao
(a) Answer: f(θ) = 13 cos(θ + 67.38 degrees)
(b) B1 maximum value = 13 (ft R from (a))
(b) B1 θ = 292.6 degrees (awrt, ft from α), from θ + α = 360
(b) Answer: Maximum value = 13, occurring at θ = 292.6 degrees
(c) M1 forms 13cos(θ + 67.38) = 6, i.e. cos(θ + 67.38) = 6/13 (ft from (a))
(c) M1 finds principal value arccos(6/13), awrt 62.5 degrees
(c) M1 correct method for second value within the shifted range (using 360 - 62.5 and/or 360 + 62.5)
(c) A1 θ = 230.1 degrees (awrt), ft
(c) A1 θ = 355.1 degrees (awrt), cao
(c) Answer: θ = 230.1 degrees, 355.1 degrees

Question 7

(a) M1 writes sin(3theta) = sin(2theta + θ) = sin(2theta)cos(θ) + cos(2theta)sin(θ)
(a) M1 substitutes sin(2theta) = 2sin(θ)cos(θ) and cos(2theta) = 1 - 2sin²(θ)
(a) M1 uses cos²(θ) = 1 - sin²(θ) to write the expression entirely in sin(θ)
(a) A1 cso completes correctly to 3sin(θ) - 4sin³(θ)
(a) Answer: Identity proved
(b) M1 recognises from (a) that 4sin³(θ) - 3sin(θ) = -sin(3theta)
(b) M1 forms sin(3theta) = 0.5
(b) M1 correct method for all solutions of 3theta in [0, 1080] using 30+360k and 150+360k
(b) A1 at least 3 of the 6 correct values of θ
(b) A1 all six correct: θ = 10, 50, 130, 170, 250, 290 degrees, cao
(b) Answer: θ = 10, 50, 130, 170, 250, 290 degrees

Question 8

(a) M1 uses the small angle approximation sin(2theta) = 2theta - (2theta)³/6 + ...
(a) A1 correctly simplifies (2theta)³/6 to 4theta³/3
(a) M1 substitutes to give 2theta - sin(2theta) = 4theta³/3
(a) A1 cso divides by θ³ to obtain 4/3
(a) Answer: Shown that (2theta - sin2theta)/θ³ = 4/3 (approx)
(b) M1 estimate = 4/3 = 1.333 (awrt)
(b) M1 true value = (0.2 - sin(0.2))/0.001, using calculator in radians
(b) M1 true value = awrt 1.33 (1.3307), and correct percentage error method: abs(estimate - true)/true * 100
(b) A1 percentage error = awrt 0.20% cao
(b) Answer: Estimate = 1.33 (4/3); true value = 1.3307; percentage error = 0.20%

Question 9

(a) B1 domain: -1 ≤ x ≤ 1
(a) B1 range: -90 degrees ≤ f(x) ≤ 90 degrees (or -π/2 ≤ f(x) ≤ π/2), oe
(a) Answer: Domain: -1 ≤ x ≤ 1; Range: -90 ≤ f(x) ≤ 90 degrees (-π/2 to π/2)
(b) B1 smooth decreasing curve shape, no turning points, correct general shape from top-left to bottom-right
(b) B1 endpoints correctly labelled: (-1, 180 degrees) [or (-1, π)] and (1, 0)
(b) B1 y-intercept correctly labelled: (0, 90 degrees) [or (0, π/2)]
(b) Answer: Decreasing curve from (-1, 180 deg) through (0, 90 deg) to (1, 0 deg)
(c) M1 takes cosine of both sides: 2x - 1 = cos(π/3)
(c) A1 cos(π/3) = 1/2 (oe) used
(c) A1 x = 3/4 cao (exact)
(c) Answer: x = 3/4
(d) M1 uses the identity to write arccos(x) = π/2 - arcsin(x)
(d) M1 substitutes into given equation and rearranges to 2 arcsin(x) = 2pi/3, i.e. arcsin(x) = π/3
(d) A1 x = sin(π/3)
(d) A1 x = √3/2 cao (exact form)
(d) Answer: x = √3/2

Question 10

(a) B1 maximum depth = 8.5 m
(a) M1 sets 30t = 90 (+360k) from sin(30t) = 1
(a) A1 t = 3 and t = 15, i.e. 03:00 and 15:00, both correct
(a) Answer: Maximum depth 8.5 m, at 03:00 and 15:00
(b) M1 substitutes t = 8.5 to give D = 6 + 2.5sin(255)
(b) A1 D = awrt 3.59 m cao
(b) Answer: D = 3.59 m (3 s.f.)
(c) M1 forms sin(30t) = 0.4
(c) M1 finds principal value arcsin(0.4), awrt 23.6 degrees
(c) M1 correct method to find all values of 30t within [0, 720] using 23.6+360k and (180-23.6)+360k
(c) A1 two correct times in the first 12 hours (00:47 and 05:13, ft)
(c) A1 two correct times in the second 12 hours (12:47 and 17:13, ft)
(c) A1 all four times correct to the nearest minute cao: 00:47, 05:13, 12:47, 17:13
(c) Answer: 00:47, 05:13, 12:47, 17:13

Question 11

(a) M1 substitutes cos²(x) = 1 - sin²(x)
(a) A1 cso correct rearrangement (with sign change) to 2sin²(x) - 7sin(x) + 3 = 0
(a) Answer: Shown (rearrangement)
(b) M1 factorises 2sin²(x) - 7sin(x) + 3 = 0 as (2sin(x) - 1)(sin(x) - 3) = 0
(b) A1 sin(x) = 1/2, with sin(x) = 3 rejected because sin(x) is never greater than 1
(b) M1 finds principal value x = 30 degrees
(b) A1 finds second value x = 150 degrees (using 180 - 30)
(b) B1 confirms these are the only two solutions in -180 ≤ x ≤ 180 (no solutions from the negative part of the range, since sin(x) = 1/2 > 0)
(b) Answer: x = 30 degrees, 150 degrees