For the graph of y = sin(x), 0 ≤ x ≤ 360, state (i) the coordinates of the y-intercept, and (ii) the coordinates of the maximum point.
(Total for Question 3 is 2 marks)
4
Describe fully the single transformation that maps y = sin(x) onto y = sin(x) - 4.
(Total for Question 4 is 1 mark)
5
Describe fully the single transformation that maps y = cos(x) onto y = cos(x + 30).
(Total for Question 5 is 1 mark)
6
Describe fully the single transformation that maps y = sin(x) onto y = 6sin(x).
(Total for Question 6 is 1 mark)
7
Describe fully the single transformation that maps y = cos(x) onto y = cos(5x).
(Total for Question 7 is 1 mark)
8
State the period, in degrees, of y = sin(3x), and the amplitude of y = 2sin(3x).
(Total for Question 8 is 2 marks)
9
The graph of y = tan(x) is translated by vector (0, 5). Write down the equation of the resulting graph, and state its period in degrees.
(Total for Question 9 is 2 marks)
10
Find the coordinates of the maximum point of y = 4sin(x) + 1, for 0 ≤ x ≤ 360.
(Total for Question 10 is 2 marks)
11
Find the coordinates of the minimum point of y = 3cos(x) - 2, for 0 ≤ x ≤ 360.
(Total for Question 11 is 2 marks)
12
State the equation of the graph obtained when y = sin(x) is reflected in the x-axis.
(Total for Question 12 is 1 mark)
13
The graph of y = cos(x) is stretched by scale factor 3 parallel to the y-axis, then translated by vector (0, 2). Write down the equation of the resulting graph.
(Total for Question 13 is 2 marks)
14
Solve sin(x) = -0.7 for 0 ≤ x ≤ 360, giving your answers correct to 1 decimal place.
(Total for Question 14 is 3 marks)
15
The graph of y = a*sin(x) + b, where a > 0, has a maximum value of 11 and a minimum value of -3. Find the values of a and b.
(Total for Question 15 is 3 marks)
16
The depth of water, D metres, at the entrance to a marina is modelled by D(t) = 5 + 3sin(30t), where t is the number of hours after midnight and 0 ≤ t ≤ 24. Find the maximum depth predicted by the model, and the smallest positive value of t at which this maximum occurs.
(Total for Question 16 is 3 marks)
17
Solve 2cos(x) + 1 = 0 for 0 ≤ x ≤ 360, giving all solutions.
(Total for Question 17 is 3 marks)
18
The graph of y = sin(x) undergoes a stretch of scale factor 1/2 parallel to the x-axis, followed by a translation by vector (0, -3).
(a)Write down the equation of the resulting graph.(1)
(b)State the period, in degrees, of the resulting graph.(1)
(c)Find the coordinates of one minimum point of the resulting graph, for 0 ≤ x ≤ 360.(2)
(Total for Question 18 is 4 marks)
19
The graph of y = p*cos(qx), for 0 ≤ x ≤ 360, where p > 0 and q is a positive integer, has amplitude 6 and completes exactly 4 full cycles in the range 0 ≤ x ≤ 360.
(a)Find the value of p.(1)
(b)Find the value of q.(2)
(c)Hence find the coordinates of the first maximum point of the graph for x > 0.(2)
(Total for Question 19 is 5 marks)
Mark scheme · G4D Trigonometric Graphs and Transformations: Fluency and Exam Drill
Question 1
B1 5 cao
Answer: 5
Question 2
B1 360 cao
Answer: 360
Question 3
B1 (0, 0) oe
B1 (90, 1) oe
Answer: (0, 0) and (90, 1)
Question 4
B1 translation by vector (0, -4) oe
Answer: Translation by vector (0, -4)
Question 5
B1 translation by vector (-30, 0) oe
Answer: Translation by vector (-30, 0)
Question 6
B1 stretch, scale factor 6, parallel to the y-axis oe
Answer: Stretch, scale factor 6, parallel to the y-axis
Question 7
B1 stretch, scale factor 1/5, parallel to the x-axis oe
Answer: Stretch, scale factor 1/5, parallel to the x-axis
Question 8
B1 period = 120 cao
B1 amplitude = 2 cao
Answer: period = 120, amplitude = 2
Question 9
B1 y = tan(x) + 5 oe cao
B1 180, ft from the equation
Answer: y = tan(x) + 5, period = 180
Question 10
M1 maximum of 4sin(x) is 4, giving y = 4(1) + 1 = 5
A1 (90, 5) cao
Answer: (90, 5)
Question 11
M1 minimum of 3cos(x) is -3, giving y = 3(-1) - 2 = -5
A1 (180, -5) cao
Answer: (180, -5)
Question 12
B1 y = -sin(x) cao
Answer: y = -sin(x)
Question 13
M1 applies the stretch to give 3cos(x)
A1 y = 3cos(x) + 2 oe cao
Answer: y = 3cos(x) + 2
Question 14
M1 reference angle arcsin(0.7) = 44.4 (awrt)
A1 x = 224.4 (awrt, using 180 + 44.4)
A1 x = 315.6 (awrt, using 360 - 44.4)
Answer: x = 224.4 or x = 315.6
Question 15
M1 forms the equation a + b = 11 (from the maximum, sin(x) = 1)
M1 forms the equation -a + b = -3 (from the minimum, sin(x) = -1)
A1 a = 7 and b = 4 (both required) cao
Answer: a = 7, b = 4
Question 16
M1 maximum of 3sin(30t) is 3, giving D = 5 + 3 = 8
A1 maximum depth = 8 m
A1 t = 3 (from 30t = 90) cao
Answer: Maximum depth = 8 m, at t = 3
Question 17
M1 rearrange to cos(x) = -0.5
A1 x = 120 cao
A1 x = 240 cao
Answer: x = 120, 240
Question 18
(a) B1 y = sin(2x) - 3 oe cao
(a) Answer: y = sin(2x) - 3
(b) B1 180, ft from part (a)
(b) Answer: 180
(c) M1 identifies a minimum of sin(2x) where 2x = 270, i.e. x = 135 (or x = 315), ft from part (a)
(c) A1 (135, -4) oe, e.g. (315, -4), cao
(c) Answer: (135, -4)
Question 19
(a) B1 p = 6 cao
(a) Answer: p = 6
(b) M1 4 cycles over 360 degrees means the period is 360/4 = 90
(b) A1 q = 4, from period = 360/q = 90, cao
(b) Answer: q = 4
(c) M1 cos(4x) is a maximum when 4x = 360 (the smallest positive solution, since 4x = 0 gives x = 0), ft from parts (a) and (b)