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GCSE · Trigonometry and Transformations

G4 Trigonometric Graphs and Transformations

AQA 8365 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The graphs of y = sin(x), y = cos(x) and y = tan(x) are defined for 0 ≤ x ≤ 360.

(a)Which of these correctly gives the amplitude of y = 4cos(x)?(1)

A) 1
B) 4
C) 360
D) 90

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

(c)State the period, in degrees, of y = tan(x).(1)

(d)State the amplitude of y = sin(x).(1)

(Total for Question 1 is 4 marks)

Consider the graph of y = cos(x) for 0 ≤ x ≤ 360.

(a)State the coordinates of the y-intercept.(1)

(b)State the two values of x, with 0 ≤ x ≤ 360, at which the graph crosses the x-axis.(2)

(c)State the coordinates of the minimum point.(1)

(Total for Question 2 is 4 marks)

Each graph below is a transformation of y = sin(x). Describe fully the single transformation of y = sin(x) that produces each graph.

(a)y = sin(x) + 3(1)

(b)y = sin(x - 60)(1)

(c)y = 5sin(x)(1)

(d)y = sin(4x)(1)

(Total for Question 3 is 4 marks)

The graph of y = cos(x) is translated by vector (90, 0) and then stretched by scale factor 2 parallel to the y-axis, in that order.

(a)Write down the equation of the resulting graph.(2)

(Total for Question 4 is 2 marks)

Solve the following trigonometric equations for 0 ≤ x ≤ 360.

(a)sin(x) = 0.5(3)

(Total for Question 5 is 3 marks)

Solve cos(x) = -0.6 for 0 ≤ x ≤ 360, giving your answers correct to 1 decimal place.

(a)cos(x) = -0.6(3)

(Total for Question 6 is 3 marks)

Solve tan(x) = 2.5 for 0 ≤ x ≤ 360, giving your answers correct to 1 decimal place.

(a)tan(x) = 2.5(3)

(Total for Question 7 is 3 marks)

The graph of y = 2sin(x) - 1 is drawn for 0 ≤ x ≤ 360.

(a)Find the coordinates of the maximum point.(2)

(b)Find the coordinates of the minimum point.(2)

(c)State the coordinates of the y-intercept.(1)

(Total for Question 8 is 5 marks)

State the period, in degrees, of each of the following graphs.

(a)y = sin(3x)(1)

(b)y = cos(x/2)(1)

(Total for Question 9 is 2 marks)

The graph of y = -cos(x) is drawn for 0 ≤ x ≤ 360.

(a)Describe fully the single transformation that maps y = cos(x) onto y = -cos(x).(1)

(b)State the coordinates of the maximum point of y = -cos(x).(2)

(c)State the coordinates of one minimum point of y = -cos(x) in the given range.(1)

(Total for Question 10 is 4 marks)

The graph of y = sin(x) has a maximum point at (90, 1). Use this fact to find the coordinates of the stated point on each transformed graph.

(a)Find the coordinates of the maximum point of y = 3sin(x) + 2.(2)

(b)Find the coordinates of the maximum point of y = sin(x - 40) + 2.(2)

(c)Find the coordinates of the minimum point of y = -2sin(x).(2)

(Total for Question 11 is 6 marks)

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

(a)2sin(x) + 1 = 0(4)

(Total for Question 12 is 4 marks)

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

(a)cos(2x) = 0.5(5)

(Total for Question 13 is 5 marks)

Solve sin(x) = -0.3 for -180 ≤ x ≤ 180, giving your answers correct to 1 decimal place.

(a)sin(x) = -0.3(4)

(Total for Question 14 is 4 marks)

Using the symmetry properties of the sine, cosine and tangent graphs, find the exact value of each of the following without using a calculator.

(a)sin(150)(2)

(b)cos(210)(2)

(c)tan(315)(2)

(Total for Question 15 is 6 marks)

The graph of y = tan(x) is translated by vector (45, 0) and is then stretched by scale factor 1/2 parallel to the y-axis, in that order.

(a)Write down the equation of the resulting graph.(3)

(Total for Question 16 is 3 marks)

Given that sin(θ) = 0.28, where θ is obtuse (90 < θ < 180), use the identity sin²(θ) + cos²(θ) = 1 to find the value of cos(θ).

(a)Find the value of cos(θ).(4)

(Total for Question 17 is 4 marks)

Solve 3sin(x) = 2cos(x) for 0 ≤ x ≤ 360, giving your answers correct to 1 decimal place.

(a)3sin(x) = 2cos(x)(4)

(Total for Question 18 is 4 marks)

Solve sin(2x - 30) = 0.5 for 0 ≤ x ≤ 360, giving all solutions.

(a)sin(2x - 30) = 0.5(5)

(Total for Question 19 is 5 marks)

The graph of y = a*cos(x) + b, where a > 0, has a maximum point at (0, 7) and a minimum point at (180, -1).

(a)Find the values of a and b.(4)

(Total for Question 20 is 4 marks)

The graph of y = 2sin(x) + k, for 0 ≤ x ≤ 360, has a minimum value of -5.

(a)Find the value of k.(2)

(b)Using your value of k, explain why the equation 2sin(x) + k = 0 has no solutions for 0 ≤ x ≤ 360.(1)

(Total for Question 21 is 3 marks)

The height of the tide, H metres, in a small harbour on the Cornish coast is modelled by H(t) = 6 + 2sin(30t), where t is the number of hours after midnight and 0 ≤ t ≤ 24.

(a)State the maximum and minimum heights of the tide predicted by the model.(2)

(b)Find the two smallest positive values of t at which the tide height is exactly 6 m.(3)

(c)Calculate the height of the tide at t = 4, giving your answer correct to 2 decimal places.(2)

(Total for Question 22 is 7 marks)

The graphs of y = sin(x) and y = cos(x) are both defined for 0 ≤ x ≤ 360.

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

(b)Hence write sin(x) in terms of cos, in the form sin(x) = cos(x - a) for some constant a.(1)

(Total for Question 23 is 3 marks)

Mark scheme · G4 Trigonometric Graphs and Transformations

Question 1

(a) B1 B) 4
(a) Answer: B) 4
(b) B1 360 cao
(b) Answer: 360
(c) B1 180 cao
(c) Answer: 180
(d) B1 1 cao
(d) Answer: 1

Question 2

(a) B1 (0, 1) oe
(a) Answer: (0, 1)
(b) B1 x = 90
(b) B1 x = 270
(b) Answer: x = 90, 270
(c) B1 (180, -1) oe
(c) Answer: (180, -1)

Question 3

(a) B1 translation by vector (0, 3) oe
(a) Answer: Translation by vector (0, 3)
(b) B1 translation by vector (60, 0) oe
(b) Answer: Translation by vector (60, 0)
(c) B1 stretch, scale factor 5, parallel to the y-axis oe
(c) Answer: Stretch, scale factor 5, parallel to the y-axis
(d) B1 stretch, scale factor 1/4, parallel to the x-axis oe
(d) Answer: Stretch, scale factor 1/4, parallel to the x-axis

Question 4

(a) M1 applies the translation correctly to give cos(x - 90)
(a) A1 y = 2cos(x - 90) oe cao
(a) Answer: y = 2cos(x - 90)

Question 5

(a) M1 sin^-1(0.5) = 30 (principal value)
(a) A1 x = 30
(a) A1 x = 150 (using 180 - 30) cao
(a) Answer: x = 30, 150

Question 6

(a) M1 cos^-1(-0.6) = 126.9 (awrt, principal value)
(a) A1 x = 126.9 (awrt 1 d.p.)
(a) A1 x = 233.1 (awrt 1 d.p., using 360 - 126.9)
(a) Answer: x = 126.9, 233.1 (to 1 d.p.)

Question 7

(a) M1 tan^-1(2.5) = 68.2 (awrt, principal value)
(a) A1 x = 68.2 (awrt 1 d.p.)
(a) A1 x = 248.2 (awrt 1 d.p., using 68.2 + 180)
(a) Answer: x = 68.2, 248.2 (to 1 d.p.)

Question 8

(a) M1 maximum of sin(x) is 1, giving y = 2(1) - 1 = 1
(a) A1 (90, 1) cao
(a) Answer: (90, 1)
(b) M1 minimum of sin(x) is -1, giving y = 2(-1) - 1 = -3
(b) A1 (270, -3) cao
(b) Answer: (270, -3)
(c) B1 (0, -1) oe
(c) Answer: (0, -1)

Question 9

(a) B1 120 cao
(a) Answer: 120
(b) B1 720 cao
(b) Answer: 720

Question 10

(a) B1 reflection in the x-axis oe
(a) Answer: Reflection in the x-axis
(b) M1 cos(x) is a minimum (-1) at x = 180, so -cos(x) = -(-1) = 1
(b) A1 (180, 1) cao
(b) Answer: (180, 1)
(c) B1 (0, -1) or (360, -1) oe, one required
(c) Answer: (0, -1)

Question 11

(a) M1 3(1) + 2 = 5, with x-coordinate unchanged at 90
(a) A1 (90, 5) cao
(a) Answer: (90, 5)
(b) M1 x-coordinate shifts to 90 + 40 = 130
(b) A1 (130, 3) cao
(b) Answer: (130, 3)
(c) M1 the original maximum of sin(x) becomes a minimum after reflection: -2(1) = -2
(c) A1 (90, -2) cao
(c) Answer: (90, -2)

Question 12

(a) M1 rearrange to sin(x) = -0.5
(a) M1 reference angle sin^-1(0.5) = 30
(a) A1 x = 210 (using 180 + 30)
(a) A1 x = 330 (using 360 - 30) cao
(a) Answer: x = 210, 330

Question 13

(a) M1 let u = 2x, so 0 ≤ u ≤ 720
(a) M1 first two solutions for u: 60 and 300 (using cos^-1(0.5) = 60 and 360 - 60)
(a) A1 further two solutions for u using the 360-degree period: 420 and 660
(a) M1 divide all four values of u by 2 to return to x
(a) A1 x = 30, 150, 210, 330 (all four required) cao
(a) Answer: x = 30, 150, 210, 330

Question 14

(a) M1 sin^-1(-0.3) = -17.5 (awrt, principal value, which lies in the given range)
(a) A1 x = -17.5 (awrt 1 d.p.)
(a) M1 use symmetry for the range -180 ≤ x ≤ 180 to find the second solution: -180 - (-17.5)
(a) A1 x = -162.5 (awrt 1 d.p.)
(a) Answer: x = -17.5, -162.5 (to 1 d.p.)

Question 15

(a) M1 sin(150) = sin(180 - 150) = sin(30)
(a) A1 1/2 oe cao
(a) Answer: 1/2
(b) M1 210 is in the third quadrant, where cosine is negative; reference angle = 210 - 180 = 30
(b) A1 -√3/2 oe cao
(b) Answer: -√3/2
(c) M1 315 is in the fourth quadrant, where tangent is negative; reference angle = 360 - 315 = 45
(c) A1 -1 cao
(c) Answer: -1

Question 16

(a) M1 applies the translation correctly to give tan(x - 45)
(a) M1 applies the vertical stretch of scale factor 1/2 to give (1/2)tan(x - 45)
(a) A1 y = 0.5tan(x - 45) oe cao
(a) Answer: y = 0.5tan(x - 45)

Question 17

(a) M1 correctly quotes and rearranges sin²(θ) + cos²(θ) = 1 to cos²(θ) = 1 - sin²(θ)
(a) M1 cos²(θ) = 1 - 0.28² = 1 - 0.0784 = 0.9216
(a) A1 √0.9216 = 0.96 (awrt)
(a) A1 cos(θ) = -0.96 (negative root selected since θ is obtuse) cao
(a) Answer: cos(θ) = -0.96

Question 18

(a) M1 divide both sides by cos(x) to form 3tan(x) = 2, hence tan(x) = 2/3
(a) M1 tan^-1(2/3) = 33.7 (awrt, principal value)
(a) A1 x = 33.7 (awrt 1 d.p.)
(a) A1 x = 213.7 (awrt 1 d.p., using 33.7 + 180)
(a) Answer: x = 33.7, 213.7 (to 1 d.p.)

Question 19

(a) M1 let u = 2x - 30, so the range for u is -30 ≤ u ≤ 690
(a) M1 first two solutions for u: 30 and 150 (using sin^-1(0.5) = 30 and 180 - 30)
(a) M1 further two solutions for u using the 360-degree period: 390 and 510
(a) A1 x = 30 and x = 90 (from u = 30 and u = 150, using x = (u + 30)/2)
(a) A1 x = 210 and x = 270 (from u = 390 and u = 510) cao, all four values required
(a) Answer: x = 30, 90, 210, 270

Question 20

(a) M1 forms the equation a + b = 7 (from cos(0) = 1)
(a) M1 forms the equation -a + b = -1 (from cos(180) = -1)
(a) A1 b = 3 (by adding the two equations: 2b = 6)
(a) A1 a = 4 cao
(a) Answer: a = 4, b = 3

Question 21

(a) M1 the minimum of 2sin(x) is -2, so -2 + k = -5
(a) A1 k = -3 cao
(a) Answer: k = -3
(b) B1 rearranges to sin(x) = 1.5 and states that this is impossible because sin(x) can only take values between -1 and 1 (1.5 is outside this range) oe ft from part (a)
(b) Answer: 2sin(x) - 3 = 0 gives sin(x) = 1.5, which is impossible since -1 ≤ sin(x) ≤ 1 for all x.

Question 22

(a) B1 maximum height = 8 m
(a) B1 minimum height = 4 m
(a) Answer: Maximum = 8 m, minimum = 4 m
(b) M1 sets up 6 + 2sin(30t) = 6, i.e. sin(30t) = 0
(b) A1 t = 6 (from 30t = 180)
(b) A1 t = 12 (from 30t = 360) cao
(b) Answer: t = 6, t = 12
(c) M1 substitutes t = 4 to give H = 6 + 2sin(120)
(c) A1 H = 7.73 (awrt 2 d.p.)
(c) Answer: H = 7.73 m

Question 23

(a) M1 translation
(a) A1 by vector (-90, 0), i.e. 90 units in the negative x-direction (left) oe cao
(a) Answer: Translation by vector (-90, 0), i.e. 90 to the left
(b) B1 sin(x) = cos(x - 90) oe cao
(b) Answer: sin(x) = cos(x - 90)

Trigonometric Graphs and Transformations - Worksheets, Questions and Revision