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WMA11/01 Jun 2019, Question 9: Trig Graphs: Amplitude and Period

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 7 marks

The question, in short

Using the given sine curve plot, find a transformed minimum point, sketch two related curves on a copy, and count solutions of tan θ = 1 + sin θ over extended ranges.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 2 marks

Give the coordinates of the lowest point of y = 4 sin θ for 0 ≤ θ ≤ 360°.

  1. 1y = 4 sin θ is y = sin θ stretched by factor 4 parallel to the y-axis
  2. 2The minimum of sin θ is at (270°, −1)
  3. 3Minimum of 4 sin θ is at (270°, −4)

Answer(270°, −4)

Part (b) · 2 marks

On Diagram 1, draw and label y = 1 + sin θ and y = tan θ for 0 ≤ θ ≤ 360°.

  1. 1y = 1 + sin θ is y = sin θ translated up by 1: (0, 1), (90°, 2), (180°, 1), (270°, 0), (360°, 1)
  2. 2y = tan θ has vertical asymptotes at θ = 90° and θ = 270°
  3. 3tan θ passes through (0, 0), (180°, 0), (360°, 0) and goes beyond y = 1 and y = −1

Answery = 1 + sin θ: the sine curve shifted up 1, through (0, 1), (90°, 2), (180°, 1), (270°, 0), (360°, 1). y = tan θ: through (0, 0), rising to +∞ at the asymptote θ = 90°, a branch from −∞ to +∞ through (180°, 0) between asymptotes θ = 90° and θ = 270°, then from −∞ up to (360°, 0)

Part (ci) · 2 marks

Use your sketches to count how many solutions tan θ = 1 + sin θ has for 0 ≤ θ ≤ 2160°.

  1. 1From the sketch the curves meet twice in 0 ≤ θ ≤ 360°
  2. 22160° is 6 full periods of 360°
  3. 3Number of solutions = 6 × 2 = 12

Answer12

Part (cii) · 1 mark

Use your sketches to count how many solutions tan θ = 1 + sin θ has for 0 ≤ θ ≤ 1980°.

  1. 11980° = 2160° − 180°
  2. 2The last 180° (1980° to 2160°) contains one of the 12 solutions (the one between 180° and 270° of that cycle)
  3. 3Number of solutions = 12 − 1 = 11

Answer11

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Also searched as: WMA11 June 2019 · WMA11/01 · wma11 jun19 · IAS Maths Pure Mathematics 1 June 2019 · question 9.
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