Trigonometric Functions: Additional Mathematics questions

Section A: objective questions

13 marks

Choose the correct answer. Each question carries 1 mark.

In trigonometry, a negative angle is an angle measured

anticlockwise from the positive x-axis
clockwise from the positive x-axis
anticlockwise from the negative x-axis
clockwise from the positive y-axis

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Positive angles go anticlockwise and negative angles go clockwise, both from the positive x-axis.

For the teacher: Remembering, low difficulty, standard 6.1.1

Textbook p. 190

In which quadrant does −735° lie?

Quadrant I
Quadrant II
Quadrant III
Quadrant IV

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−735° + 3(360°) = 345°, which is between 270° and 360°, so Quadrant IV. Equivalently −735° = −2(360°) − 15°, which is 15° clockwise from the x-axis.

For the teacher: Applying, low difficulty, standard 6.1.1

Textbook p. 192

In which quadrant does −11π/6 rad lie?

Quadrant I
Quadrant II
Quadrant III
Quadrant IV

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−11π/6 + 2π = π/6, which is between 0 and π/2, so Quadrant I.

For the teacher: Applying, low difficulty, standard 6.1.1

Textbook p. 192

Which of the following is equal to sec θ?

1/sin θ
1/tan θ
1/cos θ
cos θ/sin θ

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sec θ = 1/cos θ. 1/sin θ is cosec θ and 1/tan θ is cot θ.

For the teacher: Remembering, low difficulty, standard 6.2.1

Textbook p. 193

Using a calculator, cot 40° = ______ (correct to 3 decimal places).

0.839
1.192
1.305
0.766

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cot 40° = 1/tan 40° = 1/0.8391 = 1.192. 0.839 is tan 40° itself.

For the teacher: Applying, low difficulty, standard 6.2.1

Textbook p. 194

sin 25° is equal to

cos 25°
sin 65°
cos 65°
tan 65°

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sin θ = cos (90° − θ), so sin 25° = cos 65°.

For the teacher: Understanding, low difficulty, standard 6.2.1

Textbook p. 194

For an angle θ, sin θ is positive and cos θ is negative. In which quadrant does θ lie?

Quadrant I
Quadrant II
Quadrant III
Quadrant IV

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Sine is positive in Quadrants I and II. Cosine is negative in Quadrants II and III. Only Quadrant II satisfies both.

For the teacher: Understanding, low difficulty, standard 6.2.2

Textbook p. 197

The exact value of cos 240° is

−1/2
1/2
−√3/2
√3/2

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240° is in Quadrant III with reference angle 60°. cos 60° = 1/2 and cosine is negative there, so −1/2.

For the teacher: Applying, low difficulty, standard 6.2.2

Textbook p. 198

What is the period of the graph y = 2 sin 3x?

120°
360°
60°
1 080°

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Period = 360°/3 = 120°. 1 080° comes from multiplying 360° by 3.

For the teacher: Understanding, medium difficulty, standard 6.3.1

Textbook p. 203

What is the maximum value of y = 3 − 2 cos 3x?

−1
1
3
5

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cos 3x lies between −1 and 1, so −2 cos 3x lies between −2 and 2. The maximum is 3 + 2 = 5.

For the teacher: Applying, medium difficulty, standard 6.3.1

Textbook p. 205

(sec²x − 1)/tan x can be simplified to

sin x
sec x
tan x
cot x

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1 + tan²x = sec²x, so sec²x − 1 = tan²x. Then tan²x/tan x = tan x.

For the teacher: Applying, medium difficulty, standard 6.4.2

Textbook p. 212

The exact value of sin 75° is

(√6 − √2)/4
(√3 + 1)/2
√6/4
(√6 + √2)/4

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sin 75° = sin 45° cos 30° + cos 45° sin 30° = (√6 + √2)/4 = 0.966. The first option is cos 75°.

For the teacher: Applying, high difficulty, standard 6.5.1

Textbook p. 216

Given sin A = 3/5 and A is acute, the value of sin 2A is

6/5
9/25
24/25
7/25

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cos A = 4/5. sin 2A = 2 sin A cos A = 2(3/5)(4/5) = 24/25. 7/25 is cos 2A.

For the teacher: Applying, medium difficulty, standard 6.5.2

Textbook p. 218

Section B: structured questions

30 marks

Answer all questions. Show your working.

In a right-angled triangle ABC, the right angle is at B, AB = 12 cm, BC = 5 cm and angle BAC = θ.

[1 mark] (a)Find the length of AC.

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AC = √(12² + 5²) = 13 cm [1]1

Tick each point you got.

[2 marks] (b)Find the value of cosec θ, sec θ and cot θ as fractions.

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cosec θ = 13/5 [1]1
sec θ = 13/12 and cot θ = 12/5 [1]1

Opposite to θ is BC = 5, adjacent is AB = 12.

Tick each point you got.

For the teacher: Applying, low difficulty, standard 6.2.1

Textbook p. 193

Without using a calculator, find the exact value of

[1 mark] (a)sin 210°

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Quadrant III, reference angle 30°; sin 210° = −1/2 [1]1

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[1 mark] (b)cos 315°

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Quadrant IV, reference angle 45°; cos 315° = √2/2 [1]1

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[1 mark] (c)tan 120°

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Quadrant II, reference angle 60°; tan 120° = −√3 [1]1

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[1 mark] (d)cosec 150°

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sin 150° = sin 30° = 1/2, so cosec 150° = 2 [1]1

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For the teacher: Applying, low difficulty, standard 6.2.2

Textbook p. 197

Diagram 1 shows the graph of y = a sin bx for 0° ≤ x ≤ 360°, where a and b are constants and a, b > 0.

A sine curve with maximum value 3 and minimum value −3 that completes two full cycles between 0 and 360 degrees — Diagram 1

[1 mark] (a)State the amplitude of the graph.

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3 [1]1

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

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Two cycles in 360°, so period = 180° [1]1

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[1 mark] (c)Write the equation of the graph.

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b = 360°/180° = 2, so y = 3 sin 2x [1]1

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[1 mark] (d)Find the value of y when x = 45°.

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y = 3 sin 90° = 3 [1]1

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For the teacher: Applying, low difficulty, standard 6.3.1

Textbook p. 203

Diagram 2 shows the graph of y = 3 cos 2x + 2 for 0 ≤ x ≤ π.

A cosine curve starting at 5 when x = 0, falling to −1 at x = pi/2, then rising to 5 at x = pi — Diagram 2

[2 marks] (a)State the maximum value and the minimum value of y.

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Maximum value = 3 + 2 = 5 [1]1
Minimum value = −3 + 2 = −1 [1]1

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[2 marks] (b)The equation 3 cos 2x = 2x/π − 1 is to be solved using the graph. State the equation of the straight line that must be drawn on the same axes.

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Add 2 to both sides: 3 cos 2x + 2 = 2x/π + 1 [1]1
y = 2x/π + 1 [1]1

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[1 mark] (c)Hence state the number of solutions of the equation in (b) for 0 ≤ x ≤ π.

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2 [1]1

The line goes from (0, 1) to (π, 3). It is below the curve at x = 0, above it near x = π/2, and below it again at x = π, so it crosses twice.

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For the teacher: Analysing, medium difficulty, standard 6.3.2

Textbook p. 208

Prove that tan²x − sin²x = tan²x sin²x.

[3 marks] (a)Prove the identity, working on the left-hand side.

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tan²x − sin²x = sin²x/cos²x − sin²x [1]1
= sin²x (1 − cos²x)/cos²x [1]1
= sin²x · sin²x/cos²x = tan²x sin²x [1]1

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For the teacher: Analysing, medium difficulty, standard 6.4.2

Textbook p. 213

Solve the equation 2 sin²x + cos x − 1 = 0 for 0° ≤ x ≤ 360°.

[1 mark] (a)Rewrite the equation as a quadratic equation in cos x.

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2(1 − cos²x) + cos x − 1 = 0, so 2cos²x − cos x − 1 = 0 [1]1

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[3 marks] (b)Hence find all the values of x.

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(2 cos x + 1)(cos x − 1) = 0 [1]1
cos x = −1/2: x = 120°, 240° [1]1
cos x = 1: x = 0°, 360° [1]1

Tick each point you got.

For the teacher: Applying, high difficulty, standard 6.6.1

Textbook p. 223

Given that sin A = 4/5 and cos B = 5/13, where A and B are acute angles, find the exact value of

[2 marks] (a)sin (A + B)

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cos A = 3/5 and sin B = 12/13 [1]1
(4/5)(5/13) + (3/5)(12/13) = 56/65 [1]1

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[2 marks] (b)cos (A − B)

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cos A cos B + sin A sin B [1]1
(3/5)(5/13) + (4/5)(12/13) = 63/65 [1]1

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For the teacher: Applying, high difficulty, standard 6.5.1

Textbook p. 217

Given that tan A = 3/4 and A is an acute angle, find

[1 mark] (a)sin 2A

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Sides 3, 4, 5: sin A = 3/5, cos A = 4/5; sin 2A = 2(3/5)(4/5) = 24/25 [1]1

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[1 mark] (b)cos 2A

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(4/5)² − (3/5)² = 7/25 [1]1

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[1 mark] (c)tan 2A

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sin 2A/cos 2A = 24/7 [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 6.5.2

Textbook p. 218

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

Question 13

Section B: structured questions

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21