Pure: Trigonometric Graphs - 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

P18 Pure: Trigonometric Graphs

EDEXCEL 9MA0 · Calculator allowed · about 130 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question concerns the graph of y = sin(x) for 0 degrees ≤ x ≤ 360 degrees.

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

(b)Write down the coordinates of the maximum point and the minimum point of y = sin(x) for 0 degrees ≤ x ≤ 360 degrees.(2)

(c)State the x-intercepts of y = sin(x) for 0 degrees ≤ x ≤ 360 degrees.(2)

(d)Sketch y = sin(x) for 0 degrees ≤ x ≤ 360 degrees, labelling the coordinates of the maximum point, the minimum point and every x-intercept found in parts (b) and (c).(3)

(Total for Question 1 is 8 marks)

This question concerns the graph of y = cos(x) for 0 degrees ≤ x ≤ 360 degrees.

(a)State the period and the range of y = cos(x).(2)

(b)Write down the coordinates of the maximum point(s) and the minimum point of y = cos(x) for 0 degrees ≤ x ≤ 360 degrees.(2)

(c)State the x-intercepts of y = cos(x) for 0 degrees ≤ x ≤ 360 degrees.(2)

(d)Given that cos(x) = sin(x + 90), describe fully the single transformation that maps the graph of y = sin(x) onto the graph of y = cos(x).(2)

(Total for Question 2 is 8 marks)

This question concerns the graph of y = tan(x) for -180 degrees ≤ x ≤ 180 degrees.

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

(b)Sketch y = tan(x) for -180 degrees ≤ x ≤ 180 degrees, showing the vertical asymptotes and the points where the curve crosses the x-axis.(4)

(c)State the equations of the vertical asymptotes of y = tan(x) for -180 degrees ≤ x ≤ 180 degrees.(2)

(d)State the domain restriction that applies to y = tan(x) for all real x, using n to represent an integer.(2)

(Total for Question 3 is 9 marks)

This question concerns the graph of y = sin(x) + 1 for 0 degrees ≤ x ≤ 360 degrees.

(a)Sketch y = sin(x) + 1 for 0 degrees ≤ x ≤ 360 degrees.(3)

(b)State the period of y = sin(x) + 1, and the coordinates of its maximum and minimum points for 0 degrees ≤ x ≤ 360 degrees.(3)

(c)Find the coordinates of the point where the curve y = sin(x) + 1 touches the x-axis, and explain why the curve touches the x-axis at this point rather than crossing it.(2)

(Total for Question 4 is 8 marks)

This question concerns the graph of y = cos(2x) for 0 degrees ≤ x ≤ 360 degrees.

(a)State the period, in degrees, of y = cos(2x).(1)

(b)Sketch y = cos(2x) for 0 degrees ≤ x ≤ 360 degrees.(3)

(c)Find the exact coordinates of every point where y = cos(2x) crosses the x-axis for 0 degrees ≤ x ≤ 360 degrees.(3)

(Total for Question 5 is 7 marks)

The diagram shows part of the curve y = a + b cos(cx) for 0 degrees ≤ x ≤ 180 degrees, where a, b and c are positive constants.

Figure (to be drawn): A smooth cosine-shaped curve is drawn for 0 ≤ x ≤ 180. The curve starts at a labelled maximum point (0, 7). It decreases smoothly to a labelled minimum point at (90, 1). It then increases smoothly back up to a second labelled maximum point at (180, 7). The horizontal axis is labelled x (degrees), marked from 0 to 180. The vertical axis is labelled y, marked from 0 to 8. A dashed horizontal guide line is shown from each of the three labelled points to the y-axis, and a dashed vertical guide line is shown from each labelled point down to the x-axis. No other points on the curve are labelled.

(a)Using the information shown in the diagram, find the values of the constants a, b and c.(6)

(b)State the range of y = a + b cos(cx), using your values from part (a).(1)

(Total for Question 6 is 7 marks)

This question concerns the graph of y = cosec(x) for 0 degrees ≤ x ≤ 360 degrees.

(a)Write cosec(x) in terms of sin(x).(1)

(b)State the values of x in the range 0 degrees ≤ x ≤ 360 degrees for which cosec(x) is undefined, explaining why cosec(x) is undefined at these values.(3)

(c)Sketch y = cosec(x) for 0 degrees < x < 360 degrees, x not equal to 180 degrees, showing the vertical asymptotes and the coordinates of the two turning points.(4)

(d)State the range of y = cosec(x).(1)

(Total for Question 7 is 9 marks)

This question concerns the graph of y = sec(x) for 0 degrees ≤ x ≤ 360 degrees.

(a)Write sec(x) in terms of cos(x).(1)

(b)State the values of x in the range 0 degrees ≤ x ≤ 360 degrees for which sec(x) is undefined, giving a reason.(3)

(c)Sketch y = sec(x) for 0 degrees ≤ x ≤ 360 degrees, x not equal to 90 degrees or 270 degrees, showing the vertical asymptotes and the coordinates of the turning points.(4)

(d)State the range of y = sec(x).(1)

(Total for Question 8 is 9 marks)

This question concerns the graph of y = cot(x) for 0 degrees ≤ x ≤ 360 degrees.

(a)Write cot(x) in terms of tan(x), and also in terms of sin(x) and cos(x).(2)

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

(c)State the values of x in the range 0 degrees ≤ x ≤ 360 degrees for which cot(x) is undefined.(2)

(d)State the x-intercepts of y = cot(x) for 0 degrees < x < 360 degrees.(2)

(e)Determine, with justification, whether y = cot(x) is an odd function or an even function.(3)

(Total for Question 9 is 10 marks)

This question concerns the graph of y = cosec(2x).

(a)State the period, in degrees, of y = cosec(2x).(1)

(b)Find the values of x in the range 0 degrees ≤ x ≤ 360 degrees for which y = cosec(2x) is undefined.(4)

(c)Sketch y = cosec(2x) for 0 degrees < x < 90 degrees, showing the equations of the vertical asymptotes and the coordinates of the turning point.(3)

(Total for Question 10 is 8 marks)

This question uses the graphs of y = sec(x) and y = cot(x).

(a)Explain, using the definition sec(x) = 1 / cos(x), why the range of y = sec(x) cannot include any value of y satisfying -1 < y < 1.(3)

(b)State, with a reason, the number of values of x in the interval 0 degrees ≤ x ≤ 360 degrees for which sec(x) = 0.5.(2)

(c)By considering the graph of y = cot(x) for 0 degrees < x < 360 degrees, determine the number of solutions of the equation cot(x) = k in this range, where k is any real number.(4)

(Total for Question 11 is 9 marks)

The function g is defined by g(x) = 3 cosec(x) + 2, for 0 degrees < x < 360 degrees, x not equal to 180 degrees.

(a)State the range of g(x).(2)

(b)Sketch y = g(x) for 0 degrees < x < 360 degrees, x not equal to 180 degrees, showing the equations of the vertical asymptotes and the coordinates of the turning points.(4)

(c)Show that the equation g(x) = 8 can be written in the form sin(x) = 1/2.(2)

(d)Hence solve g(x) = 8 for 0 degrees < x < 360 degrees, giving your answers in degrees.(2)

(Total for Question 12 is 10 marks)

Mark scheme · P18 Pure: Trigonometric Graphs

Question 1

(a) B1 360 degrees cao
(a) Answer: 360 degrees
(b) B1 maximum point (90, 1)
(b) B1 minimum point (270, -1)
(b) Answer: Maximum (90, 1); minimum (270, -1)
(c) B1 at least two of x = 0, 180, 360 correctly identified
(c) B1 all three values x = 0, 180, 360 cao, no extra values
(c) Answer: x = 0, 180, 360
(d) B1 single smooth wave, correct overall shape, starting at the origin
(d) B1 maximum (90, 1) and minimum (270, -1) both correctly labelled on the curve
(d) B1 x-intercepts at (0, 0), (180, 0) and (360, 0) all correctly labelled
(d) Answer: Sine wave from (0,0) up to (90,1), down through (180,0) to (270,-1), back up to (360,0)

Question 2

(a) B1 period = 360 degrees
(a) B1 range -1 ≤ y ≤ 1 oe
(a) Answer: Period 360 degrees; range -1 ≤ y ≤ 1
(b) B1 maximum points (0, 1) and (360, 1), both stated
(b) B1 minimum point (180, -1)
(b) Answer: Maxima (0, 1) and (360, 1); minimum (180, -1)
(c) B1 one of x = 90, 270 correctly identified
(c) B1 both values x = 90, 270 cao, no extra values
(c) Answer: x = 90, 270
(d) M1 recognises that y = sin(x + 90) is a horizontal translation of y = sin(x) by (-90, 0)
(d) A1 correctly states: translation by the vector (-90, 0), i.e. 90 degrees in the negative x-direction, cao
(d) Answer: Translation by the vector (-90, 0) (a shift of 90 degrees to the left)

Question 3

(a) B1 180 degrees cao
(a) Answer: 180 degrees
(b) B1 three separate increasing branches drawn, each going from -infinity to +infinity
(b) B1 two vertical asymptotes drawn as dashed lines at x = -90 and x = 90
(b) B1 x-intercepts correctly plotted at x = -180, 0, 180
(b) B1 curve shows correct symmetry about the origin (odd function), passing through (0, 0)
(b) Answer: Three increasing branches, asymptotes at x = -90 and x = 90, crossing the x-axis at x = -180, 0, 180
(c) B1 x = -90
(c) B1 x = 90
(c) Answer: x = -90, x = 90
(d) M1 recognises the asymptote pattern repeats with period 180 degrees from a base value of 90 degrees
(d) A1 x is not equal to 90 + 180n, n an integer, oe cao
(d) Answer: x is not equal to 90 + 180n, for n an integer

Question 4

(a) B1 single smooth wave with the same period as y = sin(x)
(a) B1 curve lies entirely on or above the x-axis, touching it at exactly one point
(a) B1 maximum value of 2 correctly shown on the curve
(a) Answer: Sine wave translated up by 1, from (0,1) up to (90,2), down to touch (270,0), back up to (360,1)
(b) B1 period = 360 degrees (unchanged by a vertical translation)
(b) B1 maximum point (90, 2)
(b) B1 minimum point (270, 0)
(b) Answer: Period 360 degrees; maximum (90, 2); minimum (270, 0)
(c) B1 coordinates (270, 0) stated
(c) B1 correct explanation: sin(x) ≥ -1 for all x, with equality only at x = 270 in this range, so sin(x) + 1 ≥ 0 with equality only there, meaning the curve cannot go below the x-axis and only just reaches it
(c) Answer: (270, 0); the curve touches rather than crosses because sin(x) + 1 ≥ 0 always, with equality only at x = 270

Question 5

(a) B1 180 degrees cao
(a) Answer: 180 degrees
(b) B1 correct shape showing two complete cosine cycles across the given range
(b) B1 maximum points at (0, 1), (180, 1) and (360, 1) all correctly shown
(b) B1 minimum points at (90, -1) and (270, -1) correctly shown
(b) Answer: Two full cosine cycles: max at x = 0, 180, 360; min at x = 90, 270
(c) M1 sets 2x = 90 + 180k and finds at least one correct value of 2x or x
(c) A1 at least three of the four values x = 45, 135, 225, 315 correct
(c) A1 all four x-intercepts correct cao, no extra values: (45, 0), (135, 0), (225, 0), (315, 0)
(c) Answer: x = 45, 135, 225, 315

Question 6

(a) M1 uses a = (maximum + minimum) / 2, e.g. a = (7 + 1) / 2
(a) A1 a = 4
(a) M1 uses b = (maximum - minimum) / 2, e.g. b = (7 - 1) / 2
(a) A1 b = 3
(a) M1 uses the period shown (180 degrees, from maximum to maximum) with period = 360 / c
(a) A1 c = 2 cao
(a) Answer: a = 4, b = 3, c = 2
(b) B1 1 ≤ y ≤ 7 oe, ft from values of a and b in (a)
(b) Answer: 1 ≤ y ≤ 7

Question 7

(a) B1 cosec(x) = 1 / sin(x)
(a) Answer: cosec(x) = 1 / sin(x)
(b) M1 recognises cosec(x) = 1/sin(x) is undefined where sin(x) = 0
(b) A1 x = 0, 180, 360 all correctly identified, no extra values
(b) B1 correct explanation: division by zero is undefined, so cosec(x) has no value wherever sin(x) = 0
(b) Answer: x = 0, 180, 360; undefined because sin(x) = 0 there, and division by zero is not defined
(c) B1 correct branch shapes: a U-shaped branch opening upwards on (0, 180) and an inverted branch opening downwards on (180, 360)
(c) B1 vertical asymptotes drawn as dashed lines at x = 0, 180, 360
(c) B1 turning point (90, 1) correctly labelled
(c) B1 turning point (270, -1) correctly labelled
(c) Answer: Two branches with asymptotes at x = 0, 180, 360; turning points (90, 1) and (270, -1)
(d) B1 y ≤ -1 or y ≥ 1 oe
(d) Answer: y ≤ -1 or y ≥ 1

Question 8

(a) B1 sec(x) = 1 / cos(x)
(a) Answer: sec(x) = 1 / cos(x)
(b) M1 recognises sec(x) = 1/cos(x) is undefined where cos(x) = 0
(b) A1 x = 90, 270 both correctly identified, no extra values
(b) B1 correct reason: division by zero is undefined, so sec(x) has no value wherever cos(x) = 0
(b) Answer: x = 90, 270; undefined because cos(x) = 0 there, and division by zero is not defined
(c) B1 correct branch shapes: an upward U-shaped branch on each side of x = 180, and a downward branch between x = 90 and x = 270
(c) B1 vertical asymptotes drawn as dashed lines at x = 90 and x = 270
(c) B1 turning points (0, 1) and (360, 1) both correctly labelled
(c) B1 turning point (180, -1) correctly labelled
(c) Answer: Branches with asymptotes at x = 90, 270; turning points (0, 1), (180, -1), (360, 1)
(d) B1 y ≤ -1 or y ≥ 1 oe
(d) Answer: y ≤ -1 or y ≥ 1

Question 9

(a) B1 cot(x) = 1 / tan(x)
(a) B1 cot(x) = cos(x) / sin(x)
(a) Answer: cot(x) = 1/tan(x) = cos(x)/sin(x)
(b) B1 180 degrees cao
(b) Answer: 180 degrees
(c) B1 recognises cot(x) is undefined where sin(x) = 0
(c) B1 x = 0, 180, 360 all correctly identified, no extra values
(c) Answer: x = 0, 180, 360
(d) B1 recognises cot(x) = 0 where cos(x) = 0
(d) B1 x = 90, 270 both correctly identified, no extra values
(d) Answer: x = 90, 270
(e) M1 considers cot(-x) = cos(-x) / sin(-x)
(e) A1 uses cos(-x) = cos(x) and sin(-x) = -sin(x) to simplify to cot(-x) = -cot(x)
(e) A1 correct conclusion: cot(x) is an odd function, since cot(-x) = -cot(x) for all valid x
(e) Answer: y = cot(x) is an odd function, since cot(-x) = -cot(x)

Question 10

(a) B1 180 degrees cao
(a) Answer: 180 degrees
(b) M1 recognises cosec(2x) undefined where sin(2x) = 0
(b) M1 sets 2x = 0, 180, 360, 540, 720 (using the correct range 0 ≤ 2x ≤ 720)
(b) A1 at least three of the five values x = 0, 90, 180, 270, 360 correct
(b) A1 all five values correct cao, no extra values: x = 0, 90, 180, 270, 360
(b) Answer: x = 0, 90, 180, 270, 360
(c) B1 single U-shaped branch opening upwards, unbounded near both ends of the interval
(c) B1 vertical asymptotes drawn as dashed lines at x = 0 and x = 90
(c) B1 turning point (45, 1) correctly labelled
(c) Answer: One U-shaped branch, asymptotes at x = 0 and x = 90, turning point (45, 1)

Question 11

(a) M1 states -1 ≤ cos(x) ≤ 1 for all x, with cos(x) not equal to 0 where sec(x) is defined
(a) A1 deduces |sec(x)| = 1/|cos(x)| ≥ 1 for all x where sec(x) is defined
(a) A1 correctly concludes sec(x) ≤ -1 or sec(x) ≥ 1, so no value of sec(x) can lie strictly between -1 and 1
(a) Answer: Since |cos(x)| ≤ 1 (and nonzero where sec is defined), |sec(x)| = 1/|cos(x)| ≥ 1, so sec(x) is never strictly between -1 and 1
(b) B1 0 solutions
(b) B1 correct reason: 0.5 lies strictly between -1 and 1, and the range of sec(x) is y ≤ -1 or y ≥ 1, so sec(x) = 0.5 has no solutions
(b) Answer: 0 solutions, since 0.5 is not in the range of sec(x)
(c) M1 identifies that y = cot(x) consists of two separate branches on (0, 180) and (180, 360)
(c) A1 recognises each branch is continuous and strictly decreasing, taking every real value exactly once (unbounded above near the lower end, unbounded below near the upper end)
(c) A1 deduces the horizontal line y = k meets each branch exactly once, for any real value of k
(c) B1 correct conclusion: cot(x) = k has exactly 2 solutions in 0 < x < 360, for every real value of k (unlike sec(x) = k or cosec(x) = k, whose number of solutions depends on k)
(c) Answer: Exactly 2 solutions, for every real value of k