4522 譚穎恩 Tam Wing Yan - Class Practice Plus 4A09 - Enhanced

Book 4B Ch9 More about Trigonometry
Name: ____________________
Class Practice Plus 9.1 Class: __________
Trigonometric Ratios of Any Angle
Key Points
◼ Let P(x, y) be a point on the terminal side of .
y
sin  =
r
x
cos  =
r
y
tan  =
x
where r = x 2 + y 2 þ 0
◼ The signs of sin, cos and tan can be summarized by the 8CAST9 diagram.
Example 1 Level 1
In the figure, P(–5, 12) is a point on the terminal side of an angle . Find the
values of sin  , cos and tan  .
Exercise Q3 – 6
Solution
OP = (−5) 2 + 122  OP = r = x 2 + y 2
= 169
= 13
12 y
6 sin  =  sin  =
r
13
5 x
cos = −  cos  =
13 r
12
tan  = −  tan  =
y
5 x
HKDSE Mastering Mathematics 113 © Pearson Education Asia Limited 2019

Quick Practice 1.1 Quick Practice 1.2
In the figure, P(3, 4) is a In the figure, P(24, –7) is
point on the terminal side a point on the terminal
of an angle . Find the side of an angle . Find
values of sin  , cos the values of sin  ,
and tan  . cos and tan  .
Solution Solution
Example 2 Level 1
21
Given that sin  = − , where 270 ü  ü 360 , find the values of cos and tan  .
29
Exercise Q7 – 9
Solution
7 270 ü  ü 360
6  lies in quadrant IV.
Let P(x, y) be a point on the terminal side of  and OP = r . Tips for Students
You may draw the following diagram
y − 21
7 sin  = = first.
r 29
6 Let y = −21 and r = 29 .
r = x2 + y2
29 = x 2 + (−21) 2
292 = x 2 + 212
x 2 = 400
x = 20 or x = −20 (rejected)  As  lies in quadrant IV, the x-coordinate must be positive.
20 21
6 cos = , tan  = −  cos  =
x
, tan  =
y
29 20 r x

3 8
Given that sin  = , where 0 ü  ü 90 , find Given that cos = − , where 180 ü  ü 270 ,
5 17
the values of cos and tan  . find the values of sin  and tan  .
Solution Solution
Example 3 Level 2
12
Given that cos = , find the values of sin  and tan  .
13
Exercise Q15, 18
Solution
12
7 cos = þ0
13
6  lies in quadrant I or quadrant IV.
Let P(x, y) be a point on the terminal side of  and OP = r .
Case 1:  lies in quadrant I.
6 x þ 0 and y þ 0
x
6 Let x = 12 and r = 13 .  cos  =
r
r = x2 + y2
13 = 122 + y 2
132 = 122 + y 2
y 2 = 25
y = 5 or y = −5 (rejected)
5 5
6 sin  = , tan  =
13 12

Case 2:  lies in quadrant IV.
6 x þ 0 and y ü 0
6 Let x = 12 and r = 13 .
r = x2 + y2
13 = 122 + y 2
132 = 122 + y 2
y 2 = 25
y = 5 (rejected) or y = −5
5 5
6 sin  = − , tan  = −
13 12
5 5 5 5
Combining cases 1 and 2, sin  = , tan  = or sin  = − , tan  = −
13 12 13 12

8 7
Given that tan  = , find the values of sin  Given that sin  = − , find the values of
15 25
and cos . cos and tan  .
Solution Solution

Exercise
(In this exercise, leave your answers in surd form if necessary.)
Level 1
1. Find  in each of the following figures.
(a) (b)
2. Sketch each of the following angles on a rectangular coordinate plane and determine in which
quadrant the angle lies.
(a) 153 (b) 216
(c) –29 (d) –165

In each of the following figures, P lies on the terminal side of . Find sin , cos and tan  .
(3 – 6)
3. 4.
5. 6.

9
7. Given that sin  = , where  lies in quadrant II, find cos and tan  .
41
35
8. Given that cos = , where  lies in quadrant I, find sin and tan  .
37
15
9. Given that tan  = − , where 270 ü  ü 360 , find sin and cos .
8

Level 2
In each of the following figures, P lies on the terminal side of . Find sin , cos and tan  .
(10 – 11)
10. 11.
2
12. Given that cos = and sin  ü 0 , find sin  and tan  .
7
13. Given that tan = −5 and cos þ 0 , find sin and cos .

9
14. Given that tan = and cos ü 0 , find sin  and cos .
8
11
15. Given that sin  = − , find cos and tan  .
61
Additional Questions
5
16. Given that tan = , where  lies in quadrant III, find sin and cos .
2
1
17. Given that cos = − and tan ü 0 , find sin and tan  .
6
11
18. Given that sin  = , find cos and tan  .
6

Name: ____________________
Trigonometric Identities
Key Points
◼ Trigonometric Identities
(a)
(b)
Example 4 Level 1
Express each of the following trigonometric ratios in terms of the same trigonometric ratio of an
acute angle.
(a) sin124 (b) cos 255 (c) tan 319
Exercise Q1 – 3
Solution
(a) sin 124 = sin(180 − 56) Tips for Students
Since 124 lies in quadrant II, express
= sin 56  sin(180 −  ) = sin
it in the form of (180 – ) first.
(b) cos 255 = cos(180 + 75)

= − cos75  cos(180 +  ) = − cos 
(c) tan 319 = tan(360 − 41)

= − tan 41  tan(360 −  ) = − tan 

Express each of the following trigonometric Express each of the following trigonometric
ratios in terms of the same trigonometric ratio of ratios in terms of the same trigonometric ratio
an acute angle. of an acute angle.
(a) sin150 (a) sin 325
(b) cos320 (b) cos 208
(c) tan 210 (c) tan167
Solution Solution
Example 5 Level 1 Reference: HKDSE 2014 Paper 2 Q19
Simplify the following expressions.

tan(− )
(a)
sin(360  −  )
(b) sin(180 +  ) cos(360 −  ) tan(180 −  )
Exercise Q7 – 9, 11, 13, 15 – 16
Solution
tan(− )
(a)
sin(360 −  )
− tan 
=
sin 
sin  sin 
−  tan  =
= cos
cos 
sin 
1
=−
cos
(b) sin(180 +  ) cos(360 −  ) tan(180 −  )

= (− sin  )(cos )(− tan  )
sin 
= sin  cos •
cos
= sin 
2

Simplify the following expressions. Simplify the following expressions.
(a) sin(180 −  ) sin(− ) tan(180  +  )
(a)
(b) cos(360 −  ) tan(180 +  ) sin(− )
(b) cos(180 +  ) tan(180 −  ) sin(360 +  )
Solution Solution
Example 6 Level 2 Reference: HKDSE 2015 Paper 2 Q19
Simplify the following expressions.

(a) cos(270 −  ) tan(90 +  ) cos(− )
(b) cos(360  −  ) sin(90  +  ) + sin 2 (− )
Exercise Q10, 12, 14, 17 – 18, 22 – 23
Solution
(a) cos(270 −  ) tan(90 +  ) cos(− )
ö 1 ö
= (− sin  )÷ − ÷(cos )
ø tan  ø
cos
sin 
= cos 
2
(b) cos(360 −  ) sin(90 +  ) + sin 2 (− )

= (cos )(cos ) + (− sin  ) 2
= cos2  + sin 2 
=1

Simplify the following expressions. Simplify the following expressions.
(a) cos(270 −  ) − sin(− ) tan(270  +  ) sin(− )
(a)
(b) sin(90 +  ) + cos(180 +  ) sin(90  +  )
(b) cos 2 (90  +  ) − cos(270  −  ) sin(180  +  )
Solution Solution
Example 7 Level 2
2 sin(90  +  )
Prove that  sin(180  −  ) − cos(90  +  ) .
tan(270  −  )
Exercise Q19 – 20, 24
Solution
2 sin(90 +  )
L.H.S. =
tan(270 −  )
2 cos
=
1
tan 
= 2 cos tan 
sin 
= 2 cos •
cos
= 2 sin 
R.H.S. = sin(180 −  ) − cos(90 +  )
= sin  − (− sin  )
= 2 sin 
6 L.H.S. = R.H.S.
2 sin(90  +  )
6  sin(180  −  ) − cos(90  +  )
tan(270  −  )

cos(90  +  ) Prove that
Prove that  tan  .
sin(270  +  ) cos(90  +  )
cos(270  −  ) tan(270  +  )  .
tan(180  −  )
Solution Solution
Exercise
Level 1
acute angle. (1 – 3)
1. (a) cos134 (b) sin167
2. (a) tan 199 (b) cos256
3. (a) sin 295 (b) tan 339

acute angle. (4 – 6)
4. (a) tan 362 (b) sin 444
5. (a) cos(−48) (b) tan(−60)
6. (a) sin(−112) (b) cos(−154)
Simplify the following expressions. (7 – 14)

7. sin(360 −  ) • sin(180 +  ) 8. cos(180 +  ) • sin(90 −  )
9. sin(180 −  ) • tan(90 −  ) 10. tan(270 −  ) • cos(270 −  )

tan(180  +  ) cos(270  −  )
11. 12.
sin(360  −  ) tan(90  +  )
13. 1 − sin(360 −  ) sin(180 +  ) 14. cos(180 +  ) sin(270 −  ) − 1
Level 2
Simplify the following expressions. (15 – 18)
tan(360  −  ) cos(360  +  )
15. tan(360 −  ) cos(180 +  ) + sin(− ) 16.
sin(180  −  )

sin(270  −  )
17. cos(90 +  ) − cos(270 −  ) tan(90 −  ) 18. + cos(180  −  )
tan(90  +  )
Prove the following identities. (19 – 20)

sin(90  −  ) sin(− )
19.  tan(270  −  ) 20. − cos 2 (180  +  )  sin 2 
cos(270  +  ) cos(270  −  )
21. Express each of the following trigonometric ratios in terms of the same trigonometric ratio of
an acute angle.
(a) sin 214 (b) cos 276
sin(270  −  )
22. Simplify .
cos(90  +  ) tan(270  +  )
23. Simplify sin 2 (− ) + cos(180  +  ) sin(270  +  ) .
sin 2 (180 −  ) 1
24. Prove that  .
cos(90 −  ) cos(180 +  ) tan(270 +  )

Name: ____________________
Solving Trigonometric Equations

by Algebraic Methods
Example 8 Level 1
Solve the following equations for 0  x  360 .

(Give your answers correct to 1 decimal place.)
(a) sin x = 0.25
(b) tan x = −1.27
Keying sequence: Exercise Q1 – 6, 21 – 22
Solution SHIFT sin 0.25 ) EXE
(a) sin x = 0.25 which gives 14.477... Tips for Students
6 x = 14.5 or 180 − 14.5 7 sin x > 0
6 Possible solutions of x lie in
x = 14.5 (cor. to 1 d.p.) or 165.5 (cor. to 1 d.p.) quadrant I or quadrant II.
Keying sequence:
SHIFT tan 1.27 ) EXE Tips for Students

(b) tan x = −1.27 7 tan x < 0
which gives 51.783...
6 x = 180 − 51.8 or 360 − 51.8 6 Possible solutions of x lie in
quadrant II or quadrant IV.
x = 128.2 (cor. to 1 d.p.) or 308.2 (cor. to 1 d.p.)

Solve the following equations for 0  x  360 . Solve the following equations for 0  x  360 .
(a) sin x = 0.5 (Give your answers correct to 1 decimal place.)
(b) tan x = 1 (a) cos x = 0.69
(b) sin x = −0.726
Solution Solution

NF Example 9 Level 1
Solve 9 cos x + 5 = 0 for 0  x  360 .

(Give your answers correct to the nearest degree.)
Exercise Q7 – 10
Solution
9 cos x + 5 = 0
9 cos x = −5 Keying sequence:
5 SHIFT cos 5 ab/c 9 ) EXE Tips for Students

cos x = − 7 cos x < 0
9 which gives 56.251... 6 Possible solutions of x lie in
6 x = 180 − 56 or 180 + 56 quadrant II or quadrant III.
x = 124 (cor. to the nearest degree) or 236 (cor. to the nearest degree)

Solve 2 cos x = 1 for 0  x  360 . Solve 7 sin x + 6 = 0 for 0  x  360 .
Solution Solution
Solve cosx = sin 40 for 0  x  360 .

Solution
cosx = sin 40
= cos(90 − 40)
= cos50
6 x = 50 or 360 − 50
x = 50 or 310

Solve sin x = sin 60 for 0  x  360 . Solve sin x = cos72 for 0  x  360 .
Solution Solution
Solve 4 sin x − 5 cos x = 0 for 0  x  360 .

Solution
4sin x − 5cos x = 0 Tips for Students
4sin x = 5cos x Using a suitable trigonometric identity,
express the given equation in terms of
sin x 5 only one kind of trigonometric ratio.
=
cos x 4
5 sin x
tan x =   tan x
4 cos x
6 x = 51 or 180 + 51
x = 51 (cor. to the nearest degree) or 231 (cor. to the nearest degree)

Solve sin x = cos x for 0  x  360 . Solve 2 sin x − 3cos x = 0 for 0  x  360 .
Solution Solution

NF Example 12 Level 2 Reference: HKDSE 2018 Paper 2 Q38
Solve the equation 9 cos2 x − 20 cos x + 11 = 0 for 0  x  360 .

Solution
9 cos 2 x − 20 cos x + 11 = 0  It is a quadratic equation in cos x.
(cos x − 1)(9 cos x − 11) = 0
cos x − 1 = 0 or 9 cos x − 11 = 0
11
6 cos x = 1 or cos x = (rejected)  −1  cosx  1
9
x = 0 or 360

Solve the equation sin 2 x + 7 sin x = 0 for Solve the equation 2 cos2 x − 13 cos x + 6 = 0
0  x  360 . for 0  x  360 .
Solution Solution
Exercise
(In this exercise, give your answers correct to 1 decimal place if necessary.)
Level 1
Solve the following equations for 0  x  360 . (1 – 6)
1. sin x = 0.65 2. cos x = 0.75
7 40.50 or 180040.50 X 41.40or 110041.40

119.50 118.60
3. tan x = −1.56 4. sin x = −0.29
x 180057.30 or 360057.30
122.70 or302.70
11 180416.90 or 360016.9
196.90 or143.10

2
5. tan x = −1 6. cos x = −
2
4 1800450or 1600
450
150or315 X 1800450 or 180945
1350 or 2250

NF 7. 6 sin x − 5 = 0 NF 8. 7 cos x − 2 = 0
sinx I
4 56.40 or 180056.4 7 71.40 or360071.40
121.60 286.60
NF 9. 3 tan x + 14 = 0 NF 10. 2 + 11sin x = 0
tanx I Sinx Y
X 18077.90 or 360077.90
x 180 10.5 or 160010.50
102.10or282.1
190.50or 349.50
NF 11. 1 − 5 cos x = cos x − 2 NF 12. 5 tan x − 9 = 11tan x + 3
box I stanklitany 3 9
lox t btanx 12
x 600 or3600boo tanx 2
sooo 11 180063.4 or310061.4
116.6 or296.60

NF 13. cosx = sin 29 NF 14. sin x = cos87
x 610013600 610 X 3 or 180030

1770
2990

Level 2
NF 15. 3sin x − 13cos x = 0 NF 16. 7 sin x + 5 cos x = 0
3sinx 1coax Minx 5 lox
t tanks
tanx I
4 77.00 or 1800170 21 180035.50 or 360015.50
257 144.50 or 324.50

NF 17. tan 2 x − tan x = 0 NF 18. 6 sin 2 x − 17 sin x + 7 = 0
tank I or o
sink0.5 or rejected
450
100011800100
00,450,180 2250,1600 1500
NF 19. 7 cos2 x + 6 cos x − 1 = 0 NF 20. 3 cos2 x = cos x
case 1 or's
800,818,360081.80 278.2 I
70.50 9092700,160070.50289.5
21. sin x = −0.21 1
22. tan x = −
11 180012.10or 360012.10 3
192.100134790 x 180030 orsoooo
sooo.no
NF 23. sin x = − cos25 NF 24. sin x − 1 = 1 + 3 sin x
1806500,16065
24500.295 24,4 19100
NF 25. 9 sin x = 4 cos x NF 26. 6 sin x − 5 sin x + 1 = 0
2
int on
tanx I 11 19.5930 1500,180 19.50 160.50
x 24.0or1800240
say
Name: ____________________
Graphs of Trigonometric Functions
Key Points
◼ Graphs of Trigonometric Functions
Trigonometric Function Properties
(i) −1  sin x  1 for all values of x.
(i.e. max. value of sin x = 1, min. value of
sin x = −1 )
(ii) y = sin x is a periodic function with period
360.
(i) −1  cos x  1 for all values of x.

(i.e. max. value of cos x = 1 , min. value of
cos x = −1)
(ii) y = cos x is a periodic function with period
360.
(i) tan x has neither a maximum nor a minimum.

(ii) y = tan x is a periodic function with period
180.
Example 13 Level 1
The figure shows the graph of a periodic function

y = sin 4 x for 0  x  360 .
(a) Find the maximum and the minimum values
of y.
(b) Find the period of the function y = sin 4 x
from its graph.
Solution
(a) From the graph, the maximum and the
minimum values of y are 1 and –1
respectively.

(b) 7 The graph repeats itself every 90.
Rough Work
6 The period of the function y = sin 4 x
is 90.

The figure shows the graph of a periodic The figure shows the graph of a periodic
function y = cos2 x for 0  x  360 . x
function y = sin for − 720  x  720 .
2
Find the maximum and the minimum values of

the function y and its period.
Find the maximum and the minimum values of
the function y and its period.
Solution Solution

Example 14 Level 1
Find the maximum and the minimum values of the following functions.
(a) y = 2 + 4 cos x
(b) y = 3 sin 2 x
Exercise Q5 – 8, 15 – 16
Solution
(a) 7 − 1  cos x  1
6 The maximum value of y = 2 + 4(1)
=6 7 −1  cosx  1
The minimum value of y = 2 + 4(−1) 6 −4  4 cosx  4
= −2  6 −2  2 + 4 cos x  6
(b) 7 − 1  sin x  1
6 The maximum value of y = 3(1)
=3
The minimum value of y = 3(0)  The minimum value of sin 2 x is 0.
=0

Find the maximum and the minimum values of Find the maximum and the minimum values of
the following functions. the following functions.
(a) y = 8 sin x 1
(a) y = 1+ sin x
(b) y = cos x
2
3
(b) y = 7 cos 2 x
Solution Solution

Example 15 Level 2
Find the maximum and the minimum values of the following functions.
(a) y = 6 − 4 sin x
1
(b) y=
6 − 4 sin x
Solution
(a) 7 − 1  sin x  1
6 The maximum value of y = 6 − 4(−1)  Maximum value of y = 6 – 4 (minimum value of sin x)
= 10
6 The minimum value of y = 6 − 4(1)  Minimum value of y = 6 – 4 (maximum value of sin x)
=2
1 1
(b) The maximum value of y =  Maximum value of y =
minimum value of (6 − 4 sin x)
2
1 1
The minimum value of y =  Minimum value of y =
10 maximum value of (6 − 4 sin x)

Find the maximum and the minimum values of Find the maximum and the minimum values of
the following functions. the following functions.
(a) y = cos x + 2 (a) y = 9 − 3 sin x
1 1
(b) y= (b) y=
cos x + 2 9 − 3 sin x
Solution Solution

Exercise
Level 1
Each of the following figures shows the graph of a periodic function y = f (x) . Find the maximum
and the minimum values of the function and its period. (1 – 4)
1. 2.
Maxvalue 2 Maxvalue I
Mini value 2
Mini value 1
Period 1800
period 1200
3. 4.
Maxvalue 1 Maxvalue 6
Mini value 3 Mini value 2
Period 900 Period 1200

Find the maximum and the minimum values of the following functions. (5 – 8)
5. y = 7 sin x 6. y = −5 cos x
Max value Mini value Max value Mini value
YI ll
YI y Stl
I
7. y = 9 cos x + 7 8. y = 5 − 6 sin x
Max value Mini value

Ma value Mini value
4 941 7 4 9611 7
2 4 56611 E
Level 2
9. The figure shows the graph of y = 3 sin 2 x − 2 for
0  x  360 .
of y.
(b) Find the period of y = 3 sin 2 x − 2 .
a Max value Mini value

L 5
Cbl Period 1800

10. y = 7 + 3 cos 2 x 11. y = 6 − 2 sin 2 x

4 7 341 4 7 110
o 4 62101 4 6 21
L 4
2 8
12. y = 13. y =
7 + 4 sin x 5 − 2 cos x
Max value Mini value Max value Mini value
Y E Y É 4 27
5 y sty
14. The figure shows the graph of y = f (x) for
0  x  1440 . y = f(x)

of y.
(b) Find the period of y = f (x) .
191 Max value Mini value

4 o
b Period 7200
15. y = 1− 3 sin x Maxvalue Minivalue
1114 I
16. y = 2 cos x
2
Maxvalue Minivalue
Y I
E
2
17. y= Maxvalue Minivalue
3 cos x + 8
I
Name: ____________________
Graphical Solutions of Trigonometric Equations
Example 16 Level 1
The figure shows the graph of y = cos x .

Solve the following equations for 0  x  360
graphically.
(a) cos x = −0.5
(b) cos x = 1
(c) cos x = 1.5
Solution
Tips for Students

The precision of the solutions depends on
the scales of the axes. The scales in this
figure allow us to read the values of x
correct to the nearest 6.
(a) Draw the straight line y = −0.5 on the graph of y = cos x .

7 The two graphs intersect at x =120 and x = 240 for 0  x  360 .
6 The solutions of cos x = −0.5 are x =120 or 240 for 0  x  360 .
(b) Draw the straight line y = 1 on the graph of y = cos x .

7 The two graphs intersect at x = 0 and x = 360 for 0  x  360 .
6 The solutions of cos x = 1 are x = 0 or 360 for 0  x  360 .
(c) Draw the straight line y = 1.5 on the graph of y = cos x .

7 The two graphs do not intersect for 0  x  360 .
6 There are no real solutions for cos x = 1.5 .

The figure shows the graph of y = sin x . The figure shows the graph of y = sin 2 x .
Solve the following equations for 0  x  360 Solve the following equations for
graphically. 0  x  360 graphically.
(a) sin x = 0
(b) sin x = −1
(c) sin x = −2
(a) sin 2 x = 0
(b) sin 2x = 1
(c) sin 2x = −1
Solution Solution

The figure shows the graph of y = 4 sin x .

(a) Rewrite the equation 4 sin x + 1 = 0 into the form
4 sin x = k , where k is a constant.
(b) Hence, solve 4 sin x + 1 = 0 for 0  x  360
graphically.
Exercise Q4
Solution
(a) 4 sin x + 1 = 0
4 sin x = −1
(b) Draw the straight line y = −1 on the graph of y = 4 sin x .
7 The two graphs intersect at x = 193.5 and x = 346.5 for 0  x  360 .

6 The solutions of 4 sin x + 1 = 0 are x = 193.5 or 346.5 for 0  x  360 .

The figure shows the graph of y = tan x. The figure shows the graph of y = 4 cos x.
(a) Rewrite the equation tan x − 1 = 0 into the

form tan x = k , where k is a constant.
(b) Hence, solve tan x −1 = 0 for (a) Rewrite the equation 4 cos x − 1 = 0 into
0  x  360 graphically. the form 4 cos x = k , where k is a constant.
(b) Hence, solve 4 cos x − 1 = 0 for
0  x  360 graphically.
Solution Solution

Exercise
Level 1
1. The figure shows the graph of y = sin x .
Solve the following equations for
(a) sin x = 1
4 900
(b) sin x = 0.8
4 540or 1260
(c) sin x = −0.4
4 2040 or 3360
2. The figure shows the graph of y = cos x .

(a) cos x = 0
900or 2700
(b) cos x = 0.6
X 540013060
(c) cos x = −0.9
1560 or 2040

3. The figure shows the graph of y = tan x . Solve the
following equations for 0  x  360 graphically.
(a) tan x = 1
4 450 or 2250
(b) tan x = 2.8
4 720 or 2520
(c) tan x = −0.5
11 1530 or 3310
NF 4. The figure shows the graph of y = 1+ cos x .
(a) Rewrite the equation cos x = 0.6 into the form

1 + cos x = k , where k is a constant.
(b) Hence, solve cos x = 0.6 for 0  x  360
graphically.
a 111.6 b X 540 or 1060

ItCox 1.6
cost 0.6

Level 2
NF 5. The figure shows the graph of y = 3 sin x . Solve
the following equations for 0  x  360
graphically.
(a) 3 sin x = 1.2
X 22.5001 151.50
(b) sin x = 0.3
X 180 or 1620
x
NF 6. The figure shows the graph of y = tan . Solve the
2
following equations for 0  x  360 graphically.
x
(a) tan = 0.8
2
4 76.50
x
(b) 2 tan +6=0
2
tan I
7 2160

NF 7. The figure shows the graph of y = sin 2 x .
(a) 1 − sin 2x = 1.5
Sina0.5
since0.5
103.50 166.5 283.50 346.50
or
(b) 2 + sin 2x = 2.3
sih2X 0
X 90,810,1890 or 2610
8. The figure shows the graph of y = cos x .
(a) cos x = −0.4
(b) cos x = 1.2
al Cox 0.4
1140or2460
b cost 1.2
i Minivalue 1 thereis norealsolutions
NF 9. The figure shows the graph of y = − sin 4 x .

Solve the following equations for 0  x  180
graphically.
(a) 1 − sin 4x = 0.4
(b) 2 + sin 4x = 2.4
al 1sin4 0.4 bl 2tsin4x 2.4

sin4x o.o sin4x o
sinax o.ci
90,360,990or1260
x 60,190,960or1290

4B 冊第9章續O角
姓]：____________________
課堂練習 9.1 班w：__________
任意角的O角比
要點提示
◼ 設 P(x, y) 為  的終邊P的一點2
y
sin  =
r
x
cos  =
r
y
tan  =
x
其中 r = x 2 + y 2 þ 0
◼ sin1cos 和 tan 的k負值可綜合r<CAST=圖2
例Ü 1 程度 1
在圖中，P(–5, 12) 是  的終邊P的一點2求 sin  1cos 和 tan  的

值2
練習第3 – 6題
解
OP = (−5) 2 + 122  OP = r = x 2 + y 2
= 169
= 13
12 y
6 sin  =  sin  =
r
13
5 x
cos = −  cos  =
13 r
12
tan  = −  tan  =
y
5 x
香港中學文憑活用數學 113 © ÿ生教育出版亞洲有限公司 2019

即時練習 1.1 即時練習 1.2
在圖中，P(3, 4) 是  的在圖中，P(24, –7) 是 

終邊P的一點2求的終邊P的一點2求
sin  1 cos 和 tan  sin  1 cos 和 tan 
的值2 的值2
解解
例Ü 2 程度 1
21
已知 sin  = − ，其中 270 ü  ü 360 ，求 cos 和 tan  的值2
29
練習第7 – 9題
解
7 270 ü  ü 360
6  屬於象限 IV2
設 P(x, y) 是  的終邊P的一點，而 OP = r 2 小錦囊
y − 21 s們可先繪畫Q圖2
7 sin  = =
r 29
6 設 y = −21 及 r = 29 2
r = x2 + y2
29 = x 2 + (−21) 2
292 = x 2 + 212
x 2 = 400
x = 20 或 x = −20 (捨去)  由於  屬於象限 IV，因l P 的 x 坐標必定是一個k值2
20 21
6 cos = , tan  = −  cos  =
x
, tan  =
y
29 20 r x

3 8
已知 sin  = ，其中 0 ü  ü 90 ，求 cos 已知 cos = − ，其中 180 ü  ü 270 ，求
5 17
和 tan  的值2 sin  和 tan  的值2
解解
例Ü 3 程度 2
12
已知 cos = ，求 sin  和 tan  的值2
13
練習第 15, 18 題
解
12
7 cos = þ0
13
6  屬於象限 I 或象限 IV2
設 P(x, y) 是  的終邊P的一點，而 OP = r 2
情況 1：  屬於象限 I2
6 xþ0 及 y þ0
x
6 設 x = 12 及 r = 13 2  cos  =
r
r = x2 + y2
13 = 122 + y 2
132 = 122 + y 2
y 2 = 25
y = 5 或 y = −5 (捨去)
5 5
6 sin  = , tan  =
13 12

情況 2：  屬於象限 IV2
6 xþ0 及 yü0
6 設 x = 12 及 r = 13 2
r = x2 + y2
13 = 122 + y 2
132 = 122 + y 2
y 2 = 25
y = 5 (捨去) 或 y = −5
5 5
6 sin  = − , tan  = −
13 12
5 5 5 5
綜合情況 1 和情況 2， sin  = , tan  = 或 sin  = − , tan  = −
13 12 13 12

8 7
已知 tan  = ，求 sin  和 cos 的值2 已知 sin  = − ，求 cos 和 tan  的值2
15 25
解解

練習
(在本練習中，如有需要，答案以根式表示2)
程度 1
1. 求Q列各圖中的 2
(a) (b)
2. 在直角坐標平面P繪畫Q列各角，然後v斷它所屬的象限2
(a) 153 (b) 216
(c) –29 (d) –165

在Q列各圖中，P 位於  的終邊P2求 sin 1 cos 和 tan  2(3 – 6)
3. 4.
5. 6.

9
7. 已知 sin  = ，其中  屬於象限 II，求 cos 和 tan  2
41
35
8. 已知 cos = ，其中  屬於象限 I，求 sin 和 tan  2
37
15
9. 已知 tan  = − ，其中 270 ü  ü 360 ，求 sin 和 cos 2
8

程度 2
在Q列各圖中，P 位於  的終邊P2求 sin 1 cos 和 tan  2(10 – 11)
10. 11.
2
12. 已知 cos = 及 sin  ü 0 ，求 sin  和 tan  2
7
13. 已知 tan = −5 及 cos þ 0 ，求 sin 和 cos 2

9
14. 已知 tan = 及 cos ü 0 ，求 sin  和 cos 2
8
11
15. 已知 sin  = − ，求 cos 和 tan  2
61
Ý 外 Ü 目
5
16. 已知 tan = ，其中  屬於象限 III，求 sin 和 cos 2
2
1
17. 已知 cos = − 及 tan ü 0 ，求 sin 和 tan  2
6
11
18. 已知 sin  = ，求 cos 和 tan  2
6

姓]：____________________
課堂練習 9.2 班w：__________
O角恆等式
要點提示
◼ O角恆等式
(a)
(b)
例Ü 4 程度 1
試以\類的O角比來表示Q列各Ü，其中各角均須為銳角2
(a) sin124 (b) cos 255 (c) tan 319
練習第1 – 3題
解
(a) sin 124 = sin(180 − 56) 小錦囊
= sin 56  sin(180 −  ) = sin 由於 124屬於象限 II，故先將它表
示r (180 – ) 的形式2
(b) cos 255 = cos(180 + 75)

= − cos75  cos(180 +  ) = − cos 
(c) tan 319 = tan(360 − 41)

= − tan 41  tan(360 −  ) = − tan 

試以\類的O角比來表示Q列各Ü，其中各角試以\類的O角比來表示Q列各Ü，其中各
均須為銳角2 角均須為銳角2
(a) sin150 (a) sin 325
(b) cos320 (b) cos 208
(c) tan 210 (c) tan167
解解
例Ü 5 程度 1 相關試題：香港中學文憑 2014 卷二第 19 題
化簡Q列各式2
tan(− )
(a)
sin(360  −  )
(b) sin(180 +  ) cos(360 −  ) tan(180 −  )
練習第 7 – 9, 11, 13, 15 – 16 題
解
tan(− )
(a)
sin(360 −  )
− tan 
=
sin 
sin  sin 
−  tan  =
= cos
cos 
sin 
1
=−
cos
(b) sin(180 +  ) cos(360 −  ) tan(180 −  )

= (− sin  )(cos )(− tan  )
sin 
cos
= sin 
2

化簡Q列各式2 化簡Q列各式2
(a) sin(180 −  ) sin(− ) tan(180  +  )
(a)
(b) cos(360 −  ) tan(180 +  ) sin(− )
(b) cos(180 +  ) tan(180 −  ) sin(360 +  )
解解
例Ü 6 程度 2 相關試題：香港中學文憑 2015 卷二第 19 題
化簡Q列各式2
(a) cos(270 −  ) tan(90 +  ) cos(− )
(b) cos(360  −  ) sin(90  +  ) + sin 2 (− )
練習第 10, 12, 14, 17 – 18, 22 – 23 題
解
(a) cos(270 −  ) tan(90 +  ) cos(− )
ö 1 ö
= (− sin  )÷ − ÷(cos )
ø tan  ø
cos
sin 
= cos 
2
(b) cos(360 −  ) sin(90 +  ) + sin 2 (− )

= (cos )(cos ) + (− sin  ) 2
= cos2  + sin 2 
=1

化簡Q列各式2 化簡Q列各式2
(a) cos(270 −  ) − sin(− ) tan(270  +  ) sin(− )
(a)
(b) sin(90 +  ) + cos(180 +  ) sin(90  +  )
(b) cos 2 (90  +  ) − cos(270  −  ) sin(180  +  )
解解
例Ü 7 程度 2
2 sin(90  +  )
證明  sin(180  −  ) − cos(90  +  ) 2
tan(270  −  )
練習第 19 – 20, 24 題
2 sin(90 +  )
左方 =
tan(270 −  )
2 cos
=
1
tan 
= 2 cos tan 
sin 
= 2 cos •
cos
= 2 sin 
右方 = sin(180 −  ) − cos(90 +  )
= sin  − (− sin  )
= 2 sin 
6 左方 = 右方
2 sin(90  +  )
6  sin(180  −  ) − cos(90  +  )
tan(270  −  )

cos(90  +  ) 證明
證明  tan  2
sin(270  +  ) cos(90  +  )
cos(270  −  ) tan(270  +  )  2
tan(180  −  )
解解
練習
程度 1
試以\類的O角比來表示Q列各Ü，其中各角均須為銳角2(1 – 3)
1. (a) cos134 (b) sin167
2. (a) tan 199 (b) cos256
3. (a) sin 295 (b) tan 339

試以\類的O角比來表示Q列各Ü，其中各角均須為銳角2(4 – 6)
4. (a) tan 362 (b) sin 444
5. (a) cos(−48) (b) tan(−60)
6. (a) sin(−112) (b) cos(−154)
化簡Q列各式2(7 – 14)
7. sin(360 −  ) • sin(180 +  ) 8. cos(180 +  ) • sin(90 −  )
9. sin(180 −  ) • tan(90 −  ) 10. tan(270 −  ) • cos(270 −  )

tan(180  +  ) cos(270  −  )
11. 12.
sin(360  −  ) tan(90  +  )
13. 1 − sin(360 −  ) sin(180 +  ) 14. cos(180 +  ) sin(270 −  ) − 1
程度 2
化簡Q列各式2(15 – 18)
tan(360  −  ) cos(360  +  )
15. tan(360 −  ) cos(180 +  ) + sin(− ) 16.
sin(180  −  )

sin(270  −  )
17. cos(90 +  ) − cos(270 −  ) tan(90 −  ) 18. + cos(180  −  )
tan(90  +  )
證明Q列各恆等式2(19 – 20)
sin(90  −  ) sin(− )
19.  tan(270  −  ) 20. − cos 2 (180  +  )  sin 2 
cos(270  +  ) cos(270  −  )
Ý 外 Ü 目
21. 試以\類的O角比來表示Q列各Ü，其中各角均須為銳角2
(a) sin 214 (b) cos 276
sin(270  −  )
22. 化簡 2
cos(90  +  ) tan(270  +  )
23. 化簡 sin 2 (− ) + cos(180  +  ) sin(270  +  ) 2
sin 2 (180 −  ) 1
24. 證明  2
cos(90 −  ) cos(180 +  ) tan(270 +  )

姓]：____________________
課堂練習 9.3 班w：__________
利用代數方法解O角方程
例Ü 8 程度 1
解Q列各方程，其中 0  x  360 2
(答案須準確至一位小數2)
(a) sin x = 0.25
(b) tan x = −1.27
練習第 1 – 6, 21 – 22 題
解按鍵次序：
SHIFT sin 0.25 ) EXE

(a) sin x = 0.25 小錦囊
可得 14.477...
6 x = 14.5 或 180 − 14.5 7 sin x > 0
6 x 的解可能屬於象限 I 或
x = 14.5 (準確至一位小數) 或 165.5 (準確至一位小數) 象限 II2
按鍵次序：
(b) tan x = −1.27 SHIFT tan 1.27 ) EXE

小錦囊
6 x = 180 − 51.8 或 360 − 51.8 可得 51.783... 7 tan x < 0
6 x 的解可能屬於象限 II 或
x = 128.2 (準確至一位小數) 或 308.2 (準確至一位小數) 象限 IV2

解Q列各方程，其中 0  x  360 2 解Q列各方程，其中 0  x  360 2
(a) sin x = 0.5 (答案須準確至一位小數2)
(b) tan x = 1 (a) cos x = 0.69
(b) sin x = −0.726
解解

非Ā礎例Ü 9 程度 1
解 9 cos x + 5 = 0 ，其中 0  x  360 2

(答案須準確至最接近的度2)
練習第 7 – 10 題
解
9 cos x + 5 = 0
9 cos x = −5 按鍵次序：
5 SHIFT cos 5 ab/c 9 ) EXE 小錦囊

cos x = − 7 cos x < 0
9 可得 56.251... 6 x 的解可能屬於象限 II 或
6 x = 180 − 56 或 180 + 56 象限 III2
x = 124 (準確至最接近的度) 或 236 (準確至最接近的度)

解 2 cos x = 1，其中 0  x  360 2 解 7 sin x + 6 = 0 ，其中 0  x  360 2
解解
解 cosx = sin 40 ，其中 0  x  360 2

練習第 13 – 14, 23 題
解
cosx = sin 40
= cos(90 − 40)
= cos50
6 x = 50 或 360 − 50
x = 50 或 310

解 sin x = sin 60 ，其中 0  x  360 2 解 sin x = cos72 ，其中 0  x  360 2
解解
解 4 sin x − 5 cos x = 0 ，其中 0  x  360 2

練習第 15 – 16, 25 題
解
4sin x − 5cos x = 0 小錦囊
4sin x = 5cos x 利用適當的O角恆等式，把給定方程表
示r只含有一類O角比的方程2
sin x 5
=
cos x 4
5 sin x
tan x =   tan x
4 cos x
6 x = 51 或 180 + 51
x = 51 (準確至最接近的度) 或 231 (準確至最接近的度)

解 sin x = cos x ，其中 0  x  360 2 解 2 sin x − 3cos x = 0 ，其中 0  x  360 2
解解

非Ā礎例Ü 12 程度 2 相關試題：香港中學文憑 2018 卷二第 38 題
解 9 cos2 x − 20 cos x + 11 = 0 ，其中 0  x  360 2

練習第 17 – 20, 26 題
解
9 cos 2 x − 20 cos x + 11 = 0  這是一個以 cos x 為變數的二次方程2
(cos x − 1)(9 cos x − 11) = 0
cos x − 1 = 0 或 9 cos x − 11 = 0
11
6 cos x = 1 或 cos x = (捨去)  −1  cosx  1
9
x = 0 或 360

解 sin 2 x + 7 sin x = 0 ，其中 0  x  360 2 解 2 cos2 x − 13 cos x + 6 = 0 ，其中
0  x  360 2
解解
練習
(在本練習中，如有需要，取答案準確至一位小數2)
程度 1
解Q列各方程，其中 0  x  360 2(1 – 6)
1. sin x = 0.65 2. cos x = 0.75
3. tan x = −1.56 4. sin x = −0.29

2
5. tan x = −1 6. cos x = −
2
解Q列各方程，其中 0  x  360 2(7 – 12)

非Ā礎 7. 6 sin x − 5 = 0 非Ā礎 8. 7 cos x − 2 = 0
非Ā礎 9. 3 tan x + 14 = 0 非Ā礎 10. 2 + 11sin x = 0
非Ā礎 11. 1 − 5 cos x = cos x − 2 非Ā礎 12. 5 tan x − 9 = 11tan x + 3
解Q列各方程，其中 0  x  360 2(13 – 14)

非Ā礎 13. cosx = sin 29 非Ā礎 14. sin x = cos87

程度 2
解Q列各方程，其中 0  x  360 2(15 – 16)
非Ā礎 15. 3sin x − 13cos x = 0 非Ā礎 16. 7 sin x + 5 cos x = 0
解Q列各方程，其中 0  x  360 2(17 – 20)

非Ā礎 17. tan 2 x − tan x = 0 非Ā礎 18. 6 sin 2 x − 17 sin x + 7 = 0
非Ā礎 19. 7 cos2 x + 6 cos x − 1 = 0 非Ā礎 20. 3 cos2 x = cos x
Ý 外 Ü 目
解Q列各方程，其中 0  x  360 2(21 – 26)
1
21. sin x = −0.21 22. tan x = −
3
非Ā礎 23. sin x = − cos25 非Ā礎 24. sin x − 1 = 1 + 3 sin x
非Ā礎 25. 9 sin x = 4 cos x 非Ā礎 26. 6 sin 2 x − 5 sin x + 1 = 0

姓]：____________________
課堂練習 9.4 班w：__________
O角函數的圖像
要點提示
◼ O角函數的圖像
O角函數性質
(i) 對於所有 x 值， −1  sin x  1 2
(即 sin x 的極大值 = 1，sin x 的極小值 = –1)
(ii) y = sin x 是一個週期函數，它的週期為 3602
(i) 對於所有 x 值， −1  cos x  1 2

(即 cos x 的極大值 = 1，cos x 的極小值 = –1)
(ii) y = cos x 是一個週期函數，它的週期為 3602
(i) tan x 沒有極大值，也沒有極小值2

(ii) y = tan x 是一個週期函數，它的週期為 1802
例Ü 13 程度 1
圖中所示為週期函數 y = sin 4 x 的圖像，其中

0  x  360 2
(a) 求 y 的極大值和極小值2
(b) 從函數 y = sin 4 x 的圖像，求它的週期2
練習第 1 – 4, 14 題
解
(a) 從圖像可得，y 的極大值和極小值分w為
1 和 –12

(b) 7 圖像每隔 90 便重複一次2
草稿
6 函數 y = sin 4 x 的週期為 902

圖中所示為週期函數 y = cos2 x 的圖像， x
圖中所示為週期函數 y = sin 的圖像，
其中 0  x  360 2 2
其中 − 720  x  720 2
求函數 y 的極大值和極小值，及它的週期2
求函數 y 的極大值和極小值，及它的週期2
解解

例Ü 14 程度 1
求Q列各函數的極大值和極小值2
(a) y = 2 + 4 cos x
(b) y = 3 sin 2 x
練習第 5 – 8, 15 – 16 題
解
(a) 7 − 1  cos x  1
6 y 的極大值 = 2 + 4(1)
=6 7 −1  cosx  1
y 的極小值 = 2 + 4(−1) 6 −4  4 cosx  4
= −2  6 −2  2 + 4 cos x  6
(b) 7 − 1  sin x  1
6 y 的極大值 = 3(1)
=3
y 的極小值 = 3(0)  sin 2 x 的極小值是 02
=0

求Q列各函數的極大值和極小值2 求Q列各函數的極大值和極小值2
(a) y = 8 sin x 1
(a) y = 1+ sin x
(b) y = cos x
2
3
(b) y = 7 cos 2 x
解解

例Ü 15 程度 2
求Q列各函數的極大值和極小值2
(a) y = 6 − 4 sin x
1
(b) y=
6 − 4 sin x
練習第 10 – 13, 17 題
解
(a) 7 − 1  sin x  1
6 y 的極大值 = 6 − 4(−1)  y 的極大值 = 6 – 4 (sin x 的極小值)
= 10
6 y 的極小值 = 6 − 4(1)  y 的極小值 = 6 – 4 (sin x 的極大值)
=2
1 1
(b) y 的極大值 =  y 的極大值 =
(6 − 4 sin x) 的極小值
2
1 1
y 的極小值 =  y 的極小值 =
10 (6 − 4 sin x) 的極大值

求Q列各函數的極大值和極小值2 求Q列各函數的極大值和極小值2
(a) y = cos x + 2 (a) y = 9 − 3 sin x
1 1
(b) y= (b) y=
cos x + 2 9 − 3 sin x
解解

練習
程度 1
Q列各圖所示為週期函數 y = f (x) 的圖像2求各函數的極大值1極小值及週期2(1 – 4)
1. 2.
3. 4.

求Q列各函數的極大值和極小值2(5 – 8)
5. y = 7 sin x 6. y = −5 cos x
7. y = 9 cos x + 7 8. y = 5 − 6 sin x
程度 2
9. 圖中所示為 y = 3 sin 2 x − 2 的圖像，其中
0  x  360 2
(b) 求 y = 3 sin 2 x − 2 的週期2

10. y = 7 + 3 cos 2 x 11. y = 6 − 2 sin 2 x
2 8
12. y = 13. y =
7 + 4 sin x 5 − 2 cos x
Ý 外 Ü 目
14. 圖中所示為 y = f (x) 的圖像，其中
0  x  1440 2 y = f(x)
(b) 求 y = f (x) 的週期2
15. y = 1− 3 sin x
16. y = 2 cos 2 x
2
17. y=
3 cos x + 8

姓]：____________________
課堂練習 9.5 班w：__________
O角方程的圖解法
例Ü 16 程度 1
圖中所示為 y = cos x 的圖像2

利用圖解法解Q列各方程，其中
0  x  360 2
(a) cos x = −0.5
(b) cos x = 1
(b) cos x = 1.5
練習第 1 – 3, 8 題
小錦囊
方格紙的格線決定了所得的解的準確
度2在這圖像中，x 的值可準確至最接
近的 62
(a) 在 y = cos x 的圖像P加P直線 y = −0.5 2

7 該兩個圖像相交於 x =120 及 x = 240 ，其中 0  x  360 2
6 對於 0  x  360 ， cos x = −0.5 的解是 x =120 或 2402
(b) 在 y = cos x 的圖像P加P直線 y = 1 2

7 該兩個圖像相交於 x = 0 及 x = 360 ，其中 0  x  360 2
6 對於 0  x  360 ， cos x = 1 的解是 x = 0 或 3602
(c) 在 y = cos x 的圖像P加P直線 y = 1.5 2

7 該兩個圖像在 0  x  360 中並沒有交點2
6 cos x = 1.5 沒有實數解2

圖中所示為 y = sin x 的圖像2利用圖解法解圖中所示為 y = sin 2 x 的圖像2利用圖解法
Q列各方程，其中 0  x  360 2 解Q列各方程，其中 0  x  360 2
(a) sin x = 0
(b) sin x = −1
(c) sin x = −2
(a) sin 2 x = 0
(b) sin 2x = 1
(c) sin 2x = −1
解解

圖中所示為 y = 4 sin x 的圖像2

(a) 把方程 4 sin x + 1 = 0 重寫r 4 sin x = k 的形
式，其中 k 是常數2
(b) 由l，利用圖解法解 4 sin x + 1 = 0 ，其中
0  x  360 2
練習第4題
解
(a) 4 sin x + 1 = 0
4 sin x = −1
(b) 在 y = 4 sin x 的圖像P加P直線 y = −1 2
7 該兩個圖像相交於 x = 193.5 及 x = 346.5 ，其中 0  x  360 2

6 對於 0  x  360 ， 4 sin x + 1 = 0 的解是 x = 193.5 或 346.52

圖中所示為 y = tan x 的圖像2 圖中所示為 y = 4 cos x 的圖像2
(a) 把方程 tan x − 1 = 0 重寫r tan x = k

的形式，其中 k 是常數2
(b) 由l，利用圖解法解 tan x − 1 = 0 ，其中 (a) 把方程 4 cos x − 1 = 0 重寫r 4 cos x = k
0  x  360 2 的形式，其中 k 是常數2
(b) 由l，利用圖解法解 4 cos x − 1 = 0 ，其中
0  x  360 2
解解

練習
程度 1
1. 圖中所示為 y = sin x 的圖像2利用圖解
法解Q列各方程，其中 0  x  360 2
(a) sin x = 1
(b) sin x = 0.8
(c) sin x = −0.4
2. 圖中所示為 y = cos x 的圖像2利用圖解

法解Q列各方程，其中 0  x  360 2
(a) cos x = 0
(b) cos x = 0.6
(c) cos x = −0.9

3. 圖中所示為 y = tan x 的圖像2利用圖解法
解Q列各方程，其中 0  x  3602
(a) tan x = 1
(b) tan x = 2.8
(c) tan x = −0.5
非Ā礎 4. 圖中所示為 y = 1+ cos x 的圖像2

(a) 把方程 cos x = 0.6 重寫r 1 + cos x = k 的
形式，其中 k 是常數2
(b) 由l，利用圖解法解 cos x = 0.6 ，其中
0  x  360 2

程度 2
非Ā礎 5. 圖中所示為 y = 3 sin x 的圖像2利用圖解法
解Q列各方程，其中 0  x  360 2
(a) 3 sin x = 1.2
(b) sin x = 0.3
x
非Ā礎 6. 圖中所示為 y = tan
的圖像2利用圖解法
2
解Q列各方程，其中 0  x  360 2
x
(a) tan = 0.8
2
x
(b) 2 tan +6=0
2

非Ā礎 7. 圖中所示為 y = sin 2 x 的圖像2利用圖解
法解Q列各方程，其中 0  x  360 2
(a) 1 − sin 2x = 1.5
(b) 2 + sin 2x = 2.3
Ý 外 Ü 目
8. 圖中所示為 y = cos x 的圖像2利用圖解法
解Q列各方程，其中 0  x  360 2
(a) cos x = −0.4

(b) cos x = 1.2
非Ā礎 9. 圖中所示為 y = − sin 4 x 的圖像2利用圖解法

解Q列各方程，其中 0  x  180 2
(a) 1 − sin 4x = 0.4

(b) 2 + sin 4x = 2.4

Answers
Answers
9 More about Trigonometry 12. sin = −

45
, tan  = −
45
7 2
Class Practice Plus 9.1
5 ö ö 1 ö 26 ö
Quick Practice 13. sin  = − ÷ or − 5 26 ÷ , cos  = ÷ or ÷
÷
26 ø 26 ÷ø 26 ÷ø 26 ÷
4 3 4 ø
1.1 sin  = , cos  = , tan  =
5 5 3 ö ö
9 ÷ or − 9 145 ÷ ,
14. sin  = −
7 24 7 ÷
145 ø 145 ÷ø
1.2 sin  = − , cos  = , tan  = −
25 25 24
ö
8 ö
4 3 cos  = − ÷ or − 8 145 ÷
2.1 cos  = , tan  = ÷
145 ø 145 ÷
5 4 ø
15 15 60 11 60 11
2.2. sin  = − , tan  = 15. cos  = − , tan  = or cos  = , tan  = −
17 8 61 60 61 60
8 15 8 15 Additional Questions
3.1 sin  = , cos  = or sin  = − , cos  = −
17 17 17 17
5 2
16. sin = − , cos  = −
24 7 24 7 3 3
3.2 cos  = − , tan  = or cos  = , tan  = −
25 24 25 24
35
17. sin = , tan  = − 35
6
Exercise
5 11 5 11
Level 1 18. cos  = , tan  = or cos  = − , tan  = −
6 5 6 5
1. (a) –240 (b) 315
2. (a) II (b) III
(c) IV (d) III
Quick Practice
15 8 15
3. sin  = , cos  = , tan  =
17 17 8 4.1 (a) sin 30 (b) cos40
12 5 12 (c) tan30
4. sin  = , cos  = − , tan  = −
13 13 5 4.2 (a) −sin35 (b) − cos28
5. sin  = −
20
, cos  = −
21
, tan  =
20 (c) − tan13
29 29 21
5.1 (a) − sin 2  (b) sin
9 40 9
6. sin  = − , cos  = , tan  = − 1
41 41 40 5.2 (a) − (b) sin 2 
cos 
40 9
7. cos  = − , tan  = − 6.1 (a) 0 (b) 0
41 40
6.2 (a) 1 (b) 0
12 12
8. sin  = , tan  =
37 35
Exercise
15 8 Level 1
9. sin  = − , cos  =
17 17
1. (a) − cos46 (b) sin13
Level 2
2. (a) tan19 (b) − cos76
5 ö 5 29 ö ö ö
10. sin  = ÷ or ÷ , cos  = − 2 ÷ or − 2 29 ÷ , 3. (a) −sin 65 (b) − tan21
÷
29 ø 29 ÷ 29 ÷ 29 ÷
ø ø ø
4. (a) tan2 (b) sin 84
5
tan  = − 5. (a) cos48 (b) − tan60
2
6. (a) −sin 68 (b) − cos26
ö 7 ö ö ö
11. sin  = − ÷ or − 7 50 ÷ , cos  = 1 ÷ or 50 ÷, 7. sin  2
8. − cos 2 
÷
50 ø 50 ÷ø 50 ÷ 50 ÷
ø ø
9. cos 10. − cos
tan = −7

Answers
1 sin 2  14.1 (a) max. value = 8, min. value = –8

11. − 12.
cos  cos  (b) max. value = 1, min. value = 0
13. cos  2
14. − sin 2 4 2
14.2 (a) max. value = , min. value =
3 3
Level 2
(b) max. value = 7, min. value = 0
15. 0 16. − 1
15.1 (a) max. value = 3, min. value = 1
17. cos − sin 18. sin − cos
1
Additional Questions (b) max. value = 1, min. value =
3
21. (a) −sin34 (b) cos84
15.2 (a) max. value = 12, min. value = 6
22. − 1 23. 1
1 1
(b) max. value = , min. value =
6 12
Exercise
Quick Practice
Level 1
8.1 (a) 30 or 150 (b) 45 or 225
1. max. value = 2 , min. value = –2, period = 180
8.2 (a) 46.4 or 313.6 (b) 226.6 or 313.4
2. max. value = 3 , min. value = 1, period = 120
9.1 60 or 300 9.2 239 or 301
3. max. value = –1 , min. value = –3, period = 90
10.1 60 or 120 10.2 18 or 162
4. max. value = 6 , min. value = –2, period = 120
11.1 45 or 225 11.2 56 or 236
5. max. value = 7 , min. value = –7
12.1 0, 180 or 360 12.2 60 or 300
Exercise
Level 1
Level 2
1. 40.5 or 139.5 2. 41.4 or 318.6
9. (a) max. value = 1 , min. value = –5
3. 122.7 or 302.7 4. 196.9 or 343.1
(b) 180
5. 135 or 315 6. 135 or 225
10. max. value = 10 , min. value = 7
7. 56.4 or 123.6 8. 73.4 or 286.6
9. 102.1 or 282.1 10. 190.5 or 349.5
2 2
11. 60 or 300 12. 116.6 or 296.6 12. max. value = , min. value =
3 11
13. 61 or 299 14. 3 or 177
8 8
Level 2 13. max. value = , min. value =
3 7
15. 77.0 or 257.0 16. 144.5 or 324.5
17. 0, 45, 180, 225 or 360
14. (a) max. value = 4 , min. value = 0
18. 30 or 150
(b) 720
19. 81.8, 180 or 278.2
20. 70.5, 90, 270 or 289.5
2 2
21. 192.1 or 347.9 22. 150 or 330 17. max. value = , min. value =
5 11
23. 245 or 295 24. 270
25. 24.0 or 204.0 26. 19.5, 30, 150 or 160.5
Quick Practice
16.1 (a) 0, 180 or 360
Quick Practice
(b) 270
13.1 (a) max. value = 1, min. value = –1
(c) no real solutions
(b) period = 180
16.2 (a) 0, 90, 180, 270 or 360
13.2 (a) max. value = 1, min. value = –1
(b) 45 or 225
(b) period = 720
(c) 135 or 315

Answers
17.1 (a) tanx = 1 (b) 45 or 225
17.2 (a) 4cosx = 1 (b) 76.5 or 283.5
Exercise
Level 1
1. (a) 90 (b) 54 or 126
(c) 204 or 336
2. (a) 90 or 270 (b) 54 or 306
(c) 156 or 204
3. (a) 45 or 225 (b) 72 or 252
(c) 153 or 333
4. (a) 1+ cosx = 1.6 (b) 54 or 306
Level 2
5. (a) 22.5 or 157.5 (b) 18 or 162
6. (a) 76.5 (b) 216
7. (a) 103.5, 166.5, 283.5 or 346.5
(b) 9, 81, 189 or 261
8. (a) 114 or 246 (b) no real solutions
9. (a) 9, 36, 99 or 126
(b) 6, 39, 96 or 129

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Book 4B Ch9 More about Trigonometry

Trigonometric Ratios of Any Angle

HKDSE Mastering Mathematics 113 © Pearson Education Asia Limited 2019

HKDSE Mastering Mathematics 114 © Pearson Education Asia Limited 2019

HKDSE Mastering Mathematics 115 © Pearson Education Asia Limited 2019

Quick Practice 3.1 Quick Practice 3.2

HKDSE Mastering Mathematics 116 © Pearson Education Asia Limited 2019

(c) –29 (d) –165

HKDSE Mastering Mathematics 117 © Pearson Education Asia Limited 2019

HKDSE Mastering Mathematics 118 © Pearson Education Asia Limited 2019

HKDSE Mastering Mathematics 119 © Pearson Education Asia Limited 2019

HKDSE Mastering Mathematics 120 © Pearson Education Asia Limited 2019

HKDSE Mastering Mathematics 121 © Pearson Education Asia Limited 2019

(b) cos 255 = cos(180 + 75)

(c) tan 319 = tan(360 − 41)

HKDSE Mastering Mathematics 122 © Pearson Education Asia Limited 2019

Example 5 Level 1 Reference: HKDSE 2014 Paper 2 Q19

Simplify the following expressions.

(b) sin(180 +  ) cos(360 −  ) tan(180 −  )

HKDSE Mastering Mathematics 123 © Pearson Education Asia Limited 2019

Example 6 Level 2 Reference: HKDSE 2015 Paper 2 Q19

Simplify the following expressions.

(b) cos(360 −  ) sin(90 +  ) + sin 2 (− )

HKDSE Mastering Mathematics 124 © Pearson Education Asia Limited 2019

HKDSE Mastering Mathematics 125 © Pearson Education Asia Limited 2019

2. (a) tan 199 (b) cos256

3. (a) sin 295 (b) tan 339

HKDSE Mastering Mathematics 126 © Pearson Education Asia Limited 2019

5. (a) cos(−48) (b) tan(−60)

6. (a) sin(−112) (b) cos(−154)

Simplify the following expressions. (7 – 14)

9. sin(180 −  ) • tan(90 −  ) 10. tan(270 −  ) • cos(270 −  )

HKDSE Mastering Mathematics 127 © Pearson Education Asia Limited 2019

13. 1 − sin(360 −  ) sin(180 +  ) 14. cos(180 +  ) sin(270 −  ) − 1

HKDSE Mastering Mathematics 128 © Pearson Education Asia Limited 2019

Prove the following identities. (19 – 20)

23. Simplify sin 2 (− ) + cos(180  +  ) sin(270  +  ) .

HKDSE Mastering Mathematics 129 © Pearson Education Asia Limited 2019

Solving Trigonometric Equations

Solve the following equations for 0  x  360 .

SHIFT tan 1.27 ) EXE Tips for Students

Quick Practice 8.1 Quick Practice 8.2

HKDSE Mastering Mathematics 130 © Pearson Education Asia Limited 2019

Solve 9 cos x + 5 = 0 for 0  x  360 .

5 SHIFT cos 5 ab/c 9 ) EXE Tips for Students

Quick Practice 9.1 Quick Practice 9.2

Solve cosx = sin 40 for 0  x  360 .

HKDSE Mastering Mathematics 131 © Pearson Education Asia Limited 2019

Solve 4 sin x − 5 cos x = 0 for 0  x  360 .

Quick Practice 11.1 Quick Practice 11.2

HKDSE Mastering Mathematics 132 © Pearson Education Asia Limited 2019

Solve the equation 9 cos2 x − 20 cos x + 11 = 0 for 0  x  360 .

Quick Practice 12.1 Quick Practice 12.2

7 40.50 or 180040.50 X 41.40or 110041.40

3. tan x = −1.56 4. sin x = −0.29

HKDSE Mastering Mathematics 133 © Pearson Education Asia Limited 2019

Solve the following equations for 0  x  360 . (7 – 12)

NF 9. 3 tan x + 14 = 0 NF 10. 2 + 11sin x = 0

NF 11. 1 − 5 cos x = cos x − 2 NF 12. 5 tan x − 9 = 11tan x + 3

香港中學文憑活用數學 113 © ÿ生教育出版亞洲有限公司 2019

在圖中，P(3, 4) 是  的在圖中，P(24, –7) 是 

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香港中學文憑活用數學 115 © ÿ生教育出版亞洲有限公司 2019

香港中學文憑活用數學 116 © ÿ生教育出版亞洲有限公司 2019

香港中學文憑活用數學 117 © ÿ生教育出版亞洲有限公司 2019

香港中學文憑活用數學 118 © ÿ生教育出版亞洲有限公司 2019

香港中學文憑活用數學 119 © ÿ生教育出版亞洲有限公司 2019