Ratios and Graphs of Trigonometric Functions: Mathematics Tingkatan 5 questions

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

What is the corresponding reference angle of 238°?

42°
58°
122°
32°

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238° is in quadrant III, so α = 238° − 180° = 58°. 122° comes from 360° − 238° (a quadrant IV formula), and 32° is not related to 238° by any formula.

For the teacher: Remembering, low difficulty, standard 6.1.1

Textbook p. 176

In quadrant III, which trigonometric ratio is positive?

Sine only
Cosine only
Tangent only
All three ratios

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A-S-T-C: All in quadrant I, Sin in II, Tan in III, Cos in IV. In quadrant III, sine and cosine are negative and tangent is positive.

For the teacher: Remembering, low difficulty, standard 6.1.1

Textbook p. 176

Which expression is equal to sin 160°?

sin 20°
−sin 20°
cos 20°
−cos 20°

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160° is in quadrant II with α = 180° − 160° = 20°. Sine is positive in quadrant II, so sin 160° = + sin 20°.

For the teacher: Understanding, low difficulty, standard 6.1.1

Textbook p. 177

Without using a calculator, sin 240° = ______

√32
−12
−√32
12

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240° is in quadrant III, α = 60°. Sine is negative there: sin 240° = −sin 60° = −√32. −12 would be −sin 30° or −cos 60°.

For the teacher: Applying, low difficulty, standard 6.1.2

Textbook p. 180

Given sin θ = 0.5 and 0° < θ < 360°, the values of θ are ______

30° and 210°
60° and 120°
30° and 330°
30° and 150°

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Sine is positive in quadrants I and II. sin⁻¹ 0.5 = 30°, so θ = 30° or 180° − 30° = 150°. 210° and 330° are the angles where sine is −0.5.

For the teacher: Applying, low difficulty, standard 6.1.3

Textbook p. 181

Given cos θ = −0.7071 and 0° < θ < 360°, the values of θ are ______

135° and 225°
45° and 135°
225° and 315°
45° and 315°

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Cosine is negative in quadrants II and III. cos⁻¹ 0.7071 = 45°, so θ = 180° − 45° = 135° or 180° + 45° = 225°.

For the teacher: Applying, low difficulty, standard 6.1.3

Textbook p. 181

Diagram 1 shows a unit circle. P is the point (−0.6, −0.8). What is the value of tan θ?

A unit circle with radius OP drawn into quadrant III; P is labelled (−0.6, −0.8) and θ is the anticlockwise angle from the positive x-axis — Diagram 1

−43
34
43
−34

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tan θ = y ÷ x = (−0.8) ÷ (−0.6) = 43. Both coordinates are negative, so the quotient is positive, as expected in quadrant III. 34 is x ÷ y.

For the teacher: Understanding, low difficulty, standard 6.1.1

Textbook p. 178

What are the amplitude and the period of y = 3 sin 2x + 1?

3 and 360°
3 and 720°
6 and 180°
3 and 180°

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Amplitude = a = 3. Period = 360° ÷ b = 360° ÷ 2 = 180°. The +1 only moves the graph up.

For the teacher: Remembering, low difficulty, standard 6.2.2

Textbook p. 188

What is the period of y = tan 3x?

30°
60°
120°
540°

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The period of a tangent function is 180° ÷ b = 180° ÷ 3 = 60°. 120° comes from using 360° ÷ 3.

For the teacher: Understanding, low difficulty, standard 6.2.2

Textbook p. 188

Diagram 2 shows the graph of a trigonometric function for 0° ≤ x ≤ 360°. Which function does it show?

A wave starting at y = 1 at x = 0, rising to a maximum of 3 at 90 degrees, crossing y = 1 at 180 degrees, reaching a minimum of −1 at 270 degrees and returning to 1 at 360 degrees — Diagram 2

y = 2 sin x + 1
y = 2 cos x + 1
y = 3 sin x − 1
y = sin 2x + 1

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Maximum 3, minimum −1: amplitude = (3 − (−1)) ÷ 2 = 2 and c = (3 + (−1)) ÷ 2 = 1. The graph starts on the centre line and rises, so it is a sine graph: y = 2 sin x + 1.

For the teacher: Applying, medium difficulty, standard 6.2.3

Textbook p. 188

How many times does the graph of y = cos 2x cut the x-axis for 0° ≤ x ≤ 360°?

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The period is 180°, so there are two cycles, each cutting the axis twice (cos 2x = 0 at x = 45°, 135°, 225°, 315°). y = cos x itself cuts the axis only 2 times.

For the teacher: Analysing, high difficulty, standard 6.2.1

Textbook p. 185

Farah writes sin 200° = sin 20° because 20° is the corresponding reference angle. Which statement is correct?

She is correct, because sin θ always equals sin α.
She is correct, because sine is positive in quadrant III.
She is wrong, because 200° is in quadrant II.
She is wrong, because sine is negative in quadrant III, so sin 200° = −sin 20°.

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200° is in quadrant III where only tangent is positive. The reference angle is right (20°) but the sign is missing: sin 200° = −sin 20° = −0.3420.

For the teacher: Evaluating, high difficulty, standard 6.1.1

Textbook p. 176

Section B: structured questions

33 marks

Answer all questions. Show your working.

Use the corresponding reference angle to answer this question.

[2 marks] (a)State the quadrant and the corresponding reference angle of 147°.

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Quadrant II [1]1
α = 180° − 147° = 33° [1]1

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[2 marks] (b)Find the value of cos 251°.

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Quadrant III, α = 251° − 180° = 71°; cos is negative: −cos 71° [1]1
= −0.3256 [1]1

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[2 marks] (c)Find the value of tan 305°.

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Quadrant IV, α = 360° − 305° = 55°; tan is negative: −tan 55° [1]1
= −1.4281 [1]1

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For the teacher: Applying, low difficulty, standard 6.1.2

Textbook p. 179

Calculate the values of θ for 0° < θ < 360°. Give answers to 1 decimal place where needed.

[2 marks] (a)sin θ = −0.5

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Negative: quadrants III and IV, α = sin⁻¹ 0.5 = 30° [1]1
θ = 180° + 30° = 210° or 360° − 30° = 330° [1]1

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[2 marks] (b)tan θ = −1.5

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Negative: quadrants II and IV, α = tan⁻¹ 1.5 = 56.3° [1]1
θ = 180° − 56.3° = 123.7° or 360° − 56.3° = 303.7° [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 6.1.3

Textbook p. 181

In Diagram 3, EABC is a straight line, AB = BC = 5 cm and BD = 12 cm. Angle y is the obtuse angle EAD.

Straight line EABC with D directly above B. AB and BC are each 5 cm and BD is 12 cm. The obtuse angle y is at A between AE and AD — Diagram 3

[1 mark] (a)Calculate the length of AD.

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AD = √(5² + 12²) = 13 cm [1]1

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[2 marks] (b)Find cos y.

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Reference angle is ∠DAB, cos ∠DAB = 513 [1]1
y is in quadrant II, cos is negative: cos y = −513 [1]1

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[2 marks] (c)Find tan y.

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tan ∠DAB = 125 [1]1
tan y = −125 [1]1

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For the teacher: Applying, medium difficulty, standard 6.1.4

Textbook p. 182

A function is y = 2 sin 3x + 1 for 0° ≤ x ≤ 360°.

[1 mark] (a)State the amplitude.

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2 [1]1

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[1 mark] (b)Find the period of the function.

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360° ÷ 3 = 120° [1]1

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[2 marks] (c)State the maximum value and the minimum value.

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Maximum = 1 + 2 = 3 [1]1
Minimum = 1 − 2 = −1 [1]1

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For the teacher: Understanding, low difficulty, standard 6.2.2

Textbook p. 188

Diagram 4 shows the depth of water, y metres, in a harbour x hours after midnight. The depth is modelled by y = a sin bx + c.

A wave of water depth against time for 24 hours. It starts at 4 m, reaches a maximum of 6 m at 3 hours, a minimum of 2 m at 9 hours and repeats every 12 hours — Diagram 4

[2 marks] (a)State the values of a and c.

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a = (6 − 2) ÷ 2 = 2 [1]1
c = (6 + 2) ÷ 2 = 4 [1]1

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[2 marks] (b)Find the value of b.

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Period = 12 hours, so 360° ÷ b = 12 [1]1
b = 30 [1]1

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[1 mark] (c)Calculate the depth of water at 1 a.m.

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y = 2 sin 30° + 4 = 5 m [1]1

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For the teacher: Applying, medium difficulty, standard 6.2.3

Textbook p. 191

The height above the ground of a rider on a Ferris wheel is recorded as in part (a). The wheel turns steadily.

[1 mark] (a)The heights are 20, 31, 20, 9, 20 m at times 0, 2, 4, 6, 8 minutes. The height starts at the centre line and then goes up. The graph is that of a ______ function.

sinecosinetangent

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sine

A sine graph starts on its centre line and rises. A cosine graph would start at its maximum.

[2 marks] (b)State the amplitude and the value of c.

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Amplitude = (31 − 9) ÷ 2 = 11 [1]1
c = (31 + 9) ÷ 2 = 20 [1]1

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[2 marks] (c)Find the period and the value of b.

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Period = 8 minutes [1]1
b = 360° ÷ 8 = 45 [1]1

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[1 mark] (d)Write the function for the height, y m, after x minutes.

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y = 11 sin 45x + 20 [1]1

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For the teacher: Applying, medium difficulty, standard 6.2.3

Textbook p. 190

Sketch the graph of y = 3 cos 2x for 0° ≤ x ≤ 360°.

[3 marks] (a)Sketch the graph. Mark the maximum and minimum values and the x-intercepts.

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Amplitude 3: maximum 3 and minimum −3, no vertical shift [1]1
Period = 360° ÷ 2 = 180°, so two complete cycles, starting at (0°, 3) [1]1
Cuts x-axis at 45°, 135°, 225°, 315°; minimum at 90° and 270°; maximum at 0°, 180°, 360° [1]1

Tick each point you got.

For the teacher: Creating, high difficulty, standard 6.2.2

Textbook p. 189

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Section B: structured questions

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19