Pure: Trigonometry Depth - Worksheets, Questions and Revision

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

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A-Level · Pure Mathematics

P13 Pure: Trigonometry Depth

EDEXCEL 9MA0 · Calculator allowed · about 135 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Show that, for sin x not equal to 0,

sin x / (1 + cos x) + (1 + cos x) / sin x = 2 / sin x

(Total for Question 1 is 4 marks)

Show that, for sin θ cos θ not equal to 0,

sec² θ + cosec² θ = sec² θ cosec² θ

(Total for Question 2 is 4 marks)

Solve, for 0 degrees ≤ θ < 360 degrees, the equation

sin 2theta = cos θ

giving all four solutions as exact values. Take care not to lose any solutions.

(Total for Question 3 is 5 marks)

Solve, for 0 degrees ≤ x < 360 degrees, the equation

2cos² x - 3 sin x = 3

giving all solutions as exact values.

(Total for Question 4 is 5 marks)

Solve, for 0 degrees ≤ x < 360 degrees, the equation

sin(x + 45 degrees) = 2 sin(x - 45 degrees)

giving your answers to 1 decimal place.

(Total for Question 5 is 5 marks)

Starting from the compound angle formula cos(θ + θ) = cos θ cos θ - sin θ sin θ:

(a)Prove that cos 2theta = 1 - 2sin² θ.(3)

(b)Hence show that sin² θ = (1 - cos2theta)/2, and use this result to find the exact value of sin²(15 degrees), giving your answer in the form (a - √b)/c where a, b, c are integers.(3)

(Total for Question 6 is 6 marks)

Solve, for 0 degrees ≤ θ < 360 degrees, the equation

4cos 2theta + 8cos θ + 1 = 0

giving your answers to 1 decimal place.

(Total for Question 7 is 6 marks)

Given that cos A = 3/5, where A is acute, and sin B = 5/13, where B is obtuse.

(a)Find the exact value of sin(A + B).(3)

(b)Find the exact value of cos(A + B), and hence show that 180 degrees < A + B < 270 degrees.(4)

(Total for Question 8 is 7 marks)

For cos θ not equal to 0 and sin θ not equal to -1:

(a)Show that (1 + sin θ)/cos θ + cos θ/(1 + sin θ) = 2 sec θ.(4)

(b)Hence, or otherwise, solve for 0 degrees ≤ θ < 360 degrees, the equation

(1 + sin θ)/cos θ + cos θ/(1 + sin θ) = 4(4)

(Total for Question 9 is 8 marks)

For tan θ not equal to plus or minus 1:

(a)Prove that tan 2theta = 2 tan θ / (1 - tan² θ).(3)

(b)Hence solve, for 0 degrees ≤ θ ≤ 180 degrees, the equation tan2theta = 3tan θ, giving non-exact answers to 1 decimal place. You should find four solutions, including θ = 0 and θ = 180.(5)

(Total for Question 10 is 8 marks)

Starting from tan(A - B) = sin(A - B)/cos(A - B):

(a)Prove that tan(A - B) = (tanA - tanB)/(1 + tanA tanB), for cosA cosB not equal to 0 and 1 + tanA tanB not equal to 0.(4)

(b)Hence, using exact values for tan60 degrees and tan45 degrees, find the exact value of tan15 degrees, giving your answer in the form a + b*√c, where a, b and c are integers.(4)

(Total for Question 11 is 8 marks)

f(x) = 7 sin x - 24 cos x, for 0 degrees ≤ x < 360 degrees.

(a)Express f(x) in the form R sin(x - α), where R > 0 and 0 < α < 90 degrees, giving the value of α to 1 decimal place.(4)

(b)Hence solve, for 0 degrees ≤ x < 360 degrees, the equation 7 sin x - 24 cos x = 10, giving your answers to 1 decimal place.(5)

(Total for Question 12 is 9 marks)

g(θ) = 4cos θ + 3sin θ, for 0 degrees ≤ θ < 360 degrees.

(a)Express g(θ) in the form R cos(θ - α), where R > 0 and 0 < α < 90 degrees, giving α to 2 decimal places.(4)

(b)Using the form found in part (a), find the maximum value of [g(θ)]² - 2g(θ), and state a value of θ, 0 degrees ≤ θ < 360 degrees, at which this maximum occurs.(5)

(Total for Question 13 is 9 marks)

The number of hours of daylight, H, in a UK town on a day t months after 1 January is modelled by

H(t) = 12 + 2.5 sin(30t) - 1.5 cos(30t), for 0 ≤ t < 12

where the angle is measured in degrees.

(a)Express 2.5 sin(30t) - 1.5 cos(30t) in the form R sin(30t - α), where R > 0 and 0 < α < 90 degrees, giving R to 3 significant figures and α to 1 decimal place.(4)

(b)Hence write down the maximum number of daylight hours given by the model, to 3 significant figures, and find the smallest positive value of t (to 2 decimal places) at which this maximum occurs.(3)

(c)Find the total length of time during the year (0 ≤ t < 12) for which there are at least 13.5 hours of daylight, giving your answer in months to 2 decimal places.(6)

(Total for Question 14 is 13 marks)

Mark scheme · P13 Pure: Trigonometry Depth

Question 1

M1 combines the two fractions over a common denominator sin x (1 + cos x)
A1 correct expanded numerator sin² x + (1 + cos x)², oe
M1 uses sin² x + cos² x = 1 to simplify the numerator to 2 + 2cos x, or 2(1 + cos x)
A1 cso: cancels (1 + cos x) to obtain 2 / sin x
Answer: Identity shown (proof)

Question 2

M1 writes sec² θ and cosec² θ in terms of sin and cos, and combines over the common denominator sin² θ cos² θ
A1 correct combined numerator sin² θ + cos² θ
M1 uses sin² θ + cos² θ = 1 to simplify to 1 / (sin² θ cos² θ)
A1 cso: rewrites 1/(sin² θ cos² θ) as sec² θ cosec² θ
Answer: Identity shown (proof)

Question 3

M1 uses sin2theta = 2 sin θ cos θ to rewrite the equation
M1 factorises correctly as cos θ (2 sin θ - 1) = 0 (does not divide through by cos θ)
A1 cos θ = 0 giving θ = 90 and θ = 270 (both)
A1 sin θ = 1/2 giving θ = 30 and θ = 150 (both)
A1 all four solutions given, with none lost or repeated, cao
Answer: θ = 30, 90, 150, 270 degrees

Question 4

M1 uses cos² x = 1 - sin² x to form an equation in sin x only
M1 rearranges to the 3-term quadratic 2sin² x + 3sin x + 1 = 0, oe
M1 correct factorisation (2sin x + 1)(sin x + 1) = 0, or correct use of the quadratic formula
A1 sin x = -1/2 leading to x = 210 and x = 330 (both), oe
A1 sin x = -1 leading to x = 270, with no extra solutions given
Answer: x = 210, 270, 330 degrees

Question 5

M1 expands both sides using the compound angle formulae, sin(x+45)=sinx cos45+cosx sin45 and sin(x-45)=sinx cos45-cosx sin45
M1 uses cos45 = sin45 to cancel the common factor from every term
M1 rearranges the resulting equation sinx+cosx = 2sinx-2cosx to 3cosx = sinx, i.e. tanx = 3
A1 x = awrt 71.6
A1 x = awrt 251.6
Answer: x = 71.6, 251.6 degrees (awrt)

Question 6

(a) M1 uses the compound angle formula with A = B = θ to write cos2theta = cos² θ - sin² θ
(a) M1 uses cos² θ = 1 - sin² θ
(a) A1 cso: cos2theta = 1 - 2sin² θ
(a) Answer: Identity proved: cos2theta = 1 - 2sin² θ
(b) M1 rearranges the result from (a) to give sin² θ = (1-cos2theta)/2
(b) M1 substitutes θ = 15 and uses cos30 = √3/2
(b) A1 cso: sin²(15) = (2 - √3)/4
(b) Answer: sin²(15 degrees) = (2 - √3)/4

Question 7

M1 uses cos2theta = 2cos² θ - 1 to form an equation in cos θ only
M1 rearranges to the 3-term quadratic 8cos² θ + 8cos θ - 3 = 0, oe
dM1 solves the quadratic (formula or completing the square), dependent on the previous method mark
A1 rejects cos θ = awrt -1.29 as impossible (with a reason given), retaining cos θ = awrt 0.291
A1 θ = awrt 73.1
A1 θ = awrt 286.9
Answer: θ = 73.1, 286.9 degrees (awrt)

Question 8

(a) M1 finds cos B = -12/13, using the 5-12-13 triangle with the correct negative sign for the obtuse angle B
(a) M1 uses sin(A+B) = sinA cosB + cosA sinB with correct substitution
(a) A1 sin(A+B) = -33/65, oe
(a) Answer: sin(A+B) = -33/65
(b) M1 uses cos(A+B) = cosA cosB - sinA sinB
(b) M1 correct substitution of values
(b) A1 cos(A+B) = -56/65, oe
(b) A1 cso: since sin(A+B) and cos(A+B) are both negative (from part (a) and above), A+B lies in the third quadrant, so 180 degrees < A+B < 270 degrees
(b) Answer: cos(A+B) = -56/65; A+B lies between 180 and 270 degrees

Question 9

(a) M1 combines the two fractions over the common denominator cos θ (1 + sin θ)
(a) A1 correct expanded numerator (1+sin θ)² + cos² θ
(a) M1 uses sin² θ + cos² θ = 1 to simplify the numerator to 2 + 2sin θ, or 2(1+sin θ)
(a) A1 cso: cancels (1+sin θ) to obtain 2/cos θ = 2 sec θ
(a) Answer: Identity proved: (1+sin θ)/cos θ + cos θ/(1+sin θ) = 2 sec θ
(b) M1 uses part (a) to write 2 sec θ = 4 (ft)
(b) M1 sec θ = 2, so cos θ = 1/2
(b) A1 θ = 60
(b) A1 θ = 300
(b) Answer: θ = 60, 300 degrees

Question 10

(a) M1 writes tan2theta = sin2theta/cos2theta and uses sin2theta=2sintheta costheta, cos2theta=cos²θ-sin²θ
(a) M1 divides the numerator and denominator by cos² θ
(a) A1 cso: tan2theta = 2tan θ/(1-tan² θ)
(a) Answer: Identity proved: tan2theta = 2tan θ/(1-tan² θ)
(b) M1 uses the result from (a) to form 2tan θ/(1-tan² θ) = 3tan θ
(b) M1 rearranges (letting t = tan θ) to t(3t² - 1) = 0, or equivalent, and factorises
(b) A1 tan θ = 0 giving θ = 0 and θ = 180
(b) A1 tan θ = plus or minus 1/√3 giving θ = 30 and θ = 150
(b) A1 all four solutions stated, with no extra or repeated values, cao
(b) Answer: θ = 0, 30, 150, 180 degrees

Question 11

(a) B1 states tan(A-B) = sin(A-B)/cos(A-B)
(a) M1 expands the numerator and denominator using the compound angle formulae, sinA cosB - cosA sinB and cosA cosB + sinA sinB
(a) M1 divides both the numerator and the denominator by cosA cosB
(a) A1 cso: tan(A-B) = (tanA - tanB)/(1 + tanA tanB)
(a) Answer: Identity proved: tan(A-B) = (tanA-tanB)/(1+tanA tanB)
(b) M1 uses tan(A-B) with A=60, B=45, substituting tan60=√3 and tan45=1
(b) M1 correct unsimplified expression (√3-1)/(1+√3)
(b) M1 rationalises the denominator by multiplying the numerator and denominator by (√3-1)
(b) A1 cao: tan15 degrees = 2 - √3
(b) Answer: tan15 degrees = 2 - √3

Question 12

(a) M1 R = √7² + 24²
(a) A1 R = 25
(a) M1 tan α = 24/7 (from R cos α = 7, R sin α = 24), or equivalent
(a) A1 α = 73.7 (awrt)
(a) Answer: f(x) = 25 sin(x - 73.7 degrees) (awrt)
(b) M1 uses the result from (a) to write 25 sin(x - 73.7...) = 10 (ft on their R, α)
(b) M1 sin(x - α) = 0.4, leading to a reference angle of awrt 23.6 degrees
(b) M1 correct method for the second solution: (180 - their reference angle) + their α
(b) A1 x = awrt 97.3
(b) A1 x = awrt 230.2
(b) Answer: x = 97.3, 230.2 degrees (awrt)

Question 13

(a) M1 R = √4² + 3²
(a) A1 R = 5
(a) M1 tan α = 3/4
(a) A1 α = 36.87 (awrt)
(a) Answer: g(θ) = 5cos(θ - 36.87 degrees) (awrt)
(b) M1 lets u = g(θ) = R cos(θ - α) (ft) and states the range -5 ≤ u ≤ 5
(b) M1 writes the expression as u² - 2u = (u-1)² - 1, or evaluates the expression at both endpoints u=5 and u=-5
(b) A1 correct values at both endpoints: value 15 (at u=5) and value 35 (at u=-5)
(b) A1 maximum value = 35, cao (identifying u=-5, the endpoint further from the vertex u=1, as the maximum)
(b) A1 θ = awrt 216.87 (ft), from cos(θ - α) = -1
(b) Answer: Maximum value = 35, occurring at θ = 216.87 degrees (awrt)

Question 14

(a) M1 R = √2.5² + 1.5²
(a) A1 R = awrt 2.92 (exact value √34/2)
(a) M1 tan α = 1.5/2.5 (= 0.6)
(a) A1 α = awrt 31.0
(a) Answer: H(t) = 12 + 2.92 sin(30t - 31.0) (awrt), R = √34/2, α = awrt 31.0 degrees
(b) B1 maximum = 12 + R = awrt 14.9 hours
(b) M1 sets 30t - α = 90 (using their α, ideally unrounded for accuracy)
(b) A1 t = awrt 4.03 (ft)
(b) Answer: Maximum = 14.9 hours (3sf), occurring at t = 4.03 months after 1 January (awrt)
(c) M1 forms the inequality 12 + R sin(30t - α) ≥ 13.5, i.e. sin(30t - α) ≥ 1.5/R (ft their R, α)
(c) M1 recognises that the boundary sin(30t - α) = 1.5/R = sin(α) is satisfied when 30t - α = α or 30t - α = 180 - α
(c) A1 t = 6 (exact), from the second boundary equation
(c) M1 t = 2 x α/30, using their unrounded α, dependent on the previous M1
(c) A1 t = awrt 2.06 (ft), giving the interval [2.06, 6] during which the daylight is at least 13.5 hours
(c) A1 duration = 6 - 2.06 = awrt 3.94 months, cao
(c) Answer: There are at least 13.5 hours of daylight for 3.94 months (between t = 2.06 and t = 6)

Question 1

4 marks

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Question 2

4 marks

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