Sec 3 A-Math Trigonometric functions and graphs worksheet with answers

A free Sec 3 A-Math worksheet on trigonometric functions and graphs: 10 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 3 A-Math worksheet: Trigonometric functions and graphs

Total: 40 marksSuggested time: 1 hour

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

(a)State the amplitude of

y = -2\sin 2x + 4

Answer (a)[1]

(b)State the period, in degrees, of

y = 3\cos 2x + 1

Answer (b)[1]

Without using a calculator, find the exact value of

\tan \dfrac{\pi}{3} - \tan \dfrac{\pi}{6}

Answer[2]

Without using a calculator, find the exact value of

\sin \dfrac{\pi}{3} \cos \dfrac{\pi}{6}

Answer[2]

(a)State the amplitude of

y = 2\cos x

Answer (a)[1]

(b)State the maximum value of

y = 5\sin x + 3

Answer (b)[2]

(a)Find the basic angle of

351^\circ

Answer (a)[2]

(b)Find the basic angle of

172^\circ

Answer (b)[2]

(a)Find the basic angle of

250^\circ

Answer (a)[2]

(b)Without using a calculator, find the exact value of

\tan 135^\circ

Answer (b)[2]

(a)Find the basic angle of

215^\circ

Answer (a)[2]

(b)Express

\tan 290^\circ

in terms of a trigonometric ratio of its basic angle.

Answer (b)[2]

Section B

(a)Without using a calculator, find the exact value of

\sin 45^\circ

Answer (a)[2]

(b)Find the basic angle of

160^\circ

Answer (b)[3]

The diagram shows the graph of

y = a\cos bx + c

for

0 \leq x \leq 2\pi

, where

a

b

and

c

are positive integers. The curve has a maximum point at

\left(0, 4\right)

and a minimum point at

\left(\dfrac{\pi}{3}, -2\right)

(a)State the value of

a

Answer (a)[1]

(b)State the value of

b

Answer (b)[1]

(c)State the value of

c

Answer (c)[1]

(d)Solve the equation

3\cos 3x + 1 = -\dfrac{1}{2}

for

0 \leq x \leq \pi

, giving your answers in terms of

\pi

Answer (d)[3]

(e)State the number of solutions of the equation

3\cos 3x + 1 = 4

for

0 \leq x \leq 2\pi

Answer (e)[1]

The diagram shows the graph of

y = a\cos bx + c

for

0^\circ \leq x \leq 360^\circ

, where

a

b

and

c

are positive integers. The curve has a maximum point at

(0^\circ, 4)

and a minimum point at

(60^\circ, 0)

(a)State the value of

a

Answer (a)[1]

(b)State the value of

b

Answer (b)[1]

(c)State the value of

c

Answer (c)[1]

(d)Find the values of

x

, for

{0^\circ \leq x \leq 360^\circ}

, for which

y = 1

Answer (d)°[3]

(e)State the number of solutions of the equation

a\cos bx + c = 0

for

0^\circ \leq x \leq 360^\circ

Answer (e)[1]

END OF PAPER

Answers and worked solutions

Question 1

(a)Answer:

\text{Amplitude} = 2

[1]

\text{Amplitude} = 2

(b)Answer:

\text{Period} = 180^\circ

[1]

\text{Period} = \dfrac{360^\circ}{2}

\text{Period} = 180^\circ

Question 2

Answer:

= \dfrac{2\sqrt{3}}{3}

[2]

\tan \dfrac{\pi}{3} = \sqrt{3}

\tan \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{3}

\tan \dfrac{\pi}{3} - \tan \dfrac{\pi}{6} = \left(\sqrt{3}\right) - \left(\dfrac{\sqrt{3}}{3}\right)

= \dfrac{2\sqrt{3}}{3}

Question 3

Answer:

= \dfrac{3}{4}

[2]

\sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}

\cos \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}

\sin \dfrac{\pi}{3} \times \cos \dfrac{\pi}{6} = \left(\dfrac{\sqrt{3}}{2}\right) \times \left(\dfrac{\sqrt{3}}{2}\right)

= \dfrac{3}{4}

Question 4

(a)Answer:

\text{Amplitude} = 2

[1]

\text{Amplitude} = 2

(b)Answer:

= 8

[2]

\text{Amplitude} = 5

\text{Maximum} = 3 + 5

= 8

Question 5

(a)Answer:

\alpha = 9^\circ

[2]

\text{Quadrant} = 4

\mathstrut \alpha = 360^\circ - 351^\circ

\alpha = 9^\circ

(b)Answer:

\alpha = 8^\circ

[2]

\text{Quadrant} = 2

\mathstrut \alpha = 180^\circ - 172^\circ

\alpha = 8^\circ

Question 6

(a)Answer:

\alpha = 70^\circ

[2]

\text{Quadrant} = 3

\mathstrut \alpha = 250^\circ - 180^\circ

\alpha = 70^\circ

(b)Answer:

= -1

[2]

\text{Quadrant} = 2

\mathstrut \alpha = 180^\circ - 135^\circ

\alpha = 45^\circ

\tan 135^\circ = -\tan 45^\circ

= -1

Question 7

(a)Answer:

\alpha = 35^\circ

[2]

\text{Quadrant} = 3

\mathstrut \alpha = 215^\circ - 180^\circ

\alpha = 35^\circ

(b)Answer: N[2]

\text{Quadrant} = 4

\mathstrut \alpha = 360^\circ - 290^\circ

\alpha = 70^\circ

\tan 290^\circ = -\tan 70^\circ

Question 8

(a)Answer:

\sin 45^\circ = \dfrac{\sqrt{2}}{2}

[2]

\sin 45^\circ = \dfrac{\sqrt{2}}{2}

(b)Answer:

\alpha = 20^\circ

[3]

\text{Quadrant} = 2

\mathstrut \alpha = 180^\circ - 160^\circ

\alpha = 20^\circ

Question 9

(a)Answer:

a = 3

[1]

a = \dfrac{4 - (-2)}{2}

\phantom{a} = 3

(b)Answer:

b = 3

[1]

One period is

\dfrac{2\pi}{3}

(from a maximum to the next).

\dfrac{2\pi}{b} = \dfrac{2\pi}{3}

, so

b = 3

(c)Answer:

c = 1

[1]

c = \dfrac{4 + (-2)}{2}

\phantom{c} = 1

(d)Answer:

x = \dfrac{2\pi}{9}, \dfrac{4\pi}{9}, \dfrac{8\pi}{9}

[3]

\cos 3x = -\dfrac{1}{2}

Basic angle

\dfrac{\pi}{3}

; for

0 \leq 3x \leq 3\pi

3x = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{8\pi}{3}

x = \dfrac{2\pi}{9}, \dfrac{4\pi}{9}, \dfrac{8\pi}{9}

(e)Answer:

4

[1]

y = 4

is the greatest value: the line touches the curve at each maximum point.There are

4

maximum points for

0 \leq x \leq 2\pi

, including both ends.

Question 10

(a)Answer:

a = 2

[1]

a = \dfrac{4 - (0)}{2}

= 2

(b)Answer:

b = 3

[1]

Period

= 2 \times (60^\circ - 0^\circ)

= 120^\circ

b = \dfrac{360^\circ}{120^\circ}

= 3

(c)Answer:

c = 2

[1]

c = \dfrac{4 + (0)}{2}

= 2

(d)Answer:

40^\circ

80^\circ

160^\circ

200^\circ

280^\circ

320^\circ

°[3]

2\cos 3x + 2 = 1

\cos 3x = -\dfrac{1}{2}

Basic angle

= 60^\circ

3x = 120^\circ, 240^\circ, 480^\circ, 600^\circ, 840^\circ, 960^\circ

x = 40^\circ, 80^\circ, 160^\circ, 200^\circ, 280^\circ, 320^\circ

(e)Answer:

3

[1]

y = 0

is the least value of

y

: the line

y = 0

touches the curve at each minimum point.There are

3

such points for

0^\circ \leq x \leq 360^\circ

Practice questions by skill

Sec 3 A-Math Trigonometric functions and graphs worksheet with answers

In the print window, choose Save as PDF to download it.

Answers and worked solutions

Question 1

(a)Answer:

\text{Amplitude} = 2

[1]

\text{Amplitude} = 2

(b)Answer:

\text{Period} = 180^\circ

[1]

\text{Period} = \dfrac{360^\circ}{2}

\text{Period} = 180^\circ

Question 2

Answer:

= \dfrac{2\sqrt{3}}{3}

[2]

\tan \dfrac{\pi}{3} = \sqrt{3}

\tan \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{3}

\tan \dfrac{\pi}{3} - \tan \dfrac{\pi}{6} = \left(\sqrt{3}\right) - \left(\dfrac{\sqrt{3}}{3}\right)

= \dfrac{2\sqrt{3}}{3}

Question 3

Answer:

= \dfrac{3}{4}

[2]

\sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}

\cos \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}

\sin \dfrac{\pi}{3} \times \cos \dfrac{\pi}{6} = \left(\dfrac{\sqrt{3}}{2}\right) \times \left(\dfrac{\sqrt{3}}{2}\right)

= \dfrac{3}{4}

Question 4

(a)Answer:

\text{Amplitude} = 2

[1]

\text{Amplitude} = 2

(b)Answer:

= 8

[2]

\text{Amplitude} = 5

\text{Maximum} = 3 + 5

= 8

Question 5

(a)Answer:

\alpha = 9^\circ

[2]

\text{Quadrant} = 4

\mathstrut \alpha = 360^\circ - 351^\circ

\alpha = 9^\circ

(b)Answer:

\alpha = 8^\circ

[2]

\text{Quadrant} = 2

\mathstrut \alpha = 180^\circ - 172^\circ

\alpha = 8^\circ

Question 6

(a)Answer:

\alpha = 70^\circ

[2]

\text{Quadrant} = 3

\mathstrut \alpha = 250^\circ - 180^\circ

\alpha = 70^\circ

(b)Answer:

= -1

[2]

\text{Quadrant} = 2

\mathstrut \alpha = 180^\circ - 135^\circ

\alpha = 45^\circ

\tan 135^\circ = -\tan 45^\circ

= -1

Question 7

(a)Answer:

\alpha = 35^\circ

[2]

\text{Quadrant} = 3

\mathstrut \alpha = 215^\circ - 180^\circ

\alpha = 35^\circ

(b)Answer: N[2]

\text{Quadrant} = 4

\mathstrut \alpha = 360^\circ - 290^\circ

\alpha = 70^\circ

\tan 290^\circ = -\tan 70^\circ

Question 8

(a)Answer:

\sin 45^\circ = \dfrac{\sqrt{2}}{2}

[2]

\sin 45^\circ = \dfrac{\sqrt{2}}{2}

(b)Answer:

\alpha = 20^\circ

[3]

\text{Quadrant} = 2

\mathstrut \alpha = 180^\circ - 160^\circ

\alpha = 20^\circ

Question 9

(a)Answer:

a = 3

[1]

a = \dfrac{4 - (-2)}{2}

\phantom{a} = 3

(b)Answer:

b = 3

[1]

One period is

\dfrac{2\pi}{3}

(from a maximum to the next).

\dfrac{2\pi}{b} = \dfrac{2\pi}{3}

, so

b = 3

(c)Answer:

c = 1

[1]

c = \dfrac{4 + (-2)}{2}

\phantom{c} = 1

(d)Answer:

x = \dfrac{2\pi}{9}, \dfrac{4\pi}{9}, \dfrac{8\pi}{9}

[3]

\cos 3x = -\dfrac{1}{2}

Basic angle

\dfrac{\pi}{3}

; for

0 \leq 3x \leq 3\pi

3x = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{8\pi}{3}

x = \dfrac{2\pi}{9}, \dfrac{4\pi}{9}, \dfrac{8\pi}{9}

(e)Answer:

4

[1]

y = 4

is the greatest value: the line touches the curve at each maximum point.There are

4

maximum points for

0 \leq x \leq 2\pi

, including both ends.

Question 10

(a)Answer:

a = 2

[1]

a = \dfrac{4 - (0)}{2}

= 2

(b)Answer:

b = 3

[1]

Period

= 2 \times (60^\circ - 0^\circ)

= 120^\circ

b = \dfrac{360^\circ}{120^\circ}

= 3

(c)Answer:

c = 2

[1]

c = \dfrac{4 + (0)}{2}

= 2

(d)Answer:

40^\circ

80^\circ

160^\circ

200^\circ

280^\circ

320^\circ

°[3]

2\cos 3x + 2 = 1

\cos 3x = -\dfrac{1}{2}

Basic angle

= 60^\circ

3x = 120^\circ, 240^\circ, 480^\circ, 600^\circ, 840^\circ, 960^\circ

x = 40^\circ, 80^\circ, 160^\circ, 200^\circ, 280^\circ, 320^\circ

(e)Answer:

3

[1]

y = 0

is the least value of

y

: the line

y = 0

touches the curve at each minimum point.There are

3

such points for

0^\circ \leq x \leq 360^\circ