Pure Mathematics 1
Coursebook, Chapter 5
Sketch graphs of the sine, cosine and
tangent functions.
1 Sketch the graphs of these functions
for 0° ø x ø 360°.
2021 A LEVELS MATHEMATICS
52
a
b
c
Pure Mathematics 1
Coursebook, Chapter 5
Find exact values of the sine, cosine and
tangent of 30° , 45° and 60° and related
angles.
Pure Mathematics 1
Solve trigonometric equations using
y = 2 sin x + 1
y = tan( x − 90°)
y = cos(2 x + 90°)
2 Find the exact values of:
a sin120°
b cos 225°
c tan150°
M.Austin: No. 25-01, 25-02, Jalan Austin Heights 8/3, Taman Mount Austin (Tel: 019-7732389)
Name :
Lecturer :
sin θ
Coursebook, Chapter 5
and :
Subject : A levels Mathematics the identities tan θ ≡
Class
cos θ
2
2
Chapter : Pure Maths 3
Lesson
No :
sin θ + cos θ ≡ 1.
Topic :
Date :
Day/Time:
Chapter 3
Trigonometry
3 Solve for 0° ø x ø 360° .
Cheng Wui Leap
a 5 sin x = 3 cos x
A2 Maths
b 2 cos2 x − sin x = 1
1
Why do we study trigonometry?
REWIND
addition
toyou
the three
mainhow
trigonometrical
functions (sine, cosine and tangent) there
In In
this
chapter
will learn
to:
are three more commonly used trigonometrical functions (cosecant, secant and cotangent)
understand
the relationship
of the
cosecant
and cotangent
functions to
cosine, sine
that
we will learn
about. These
aresecant,
sometimes
referred
to as the reciprocal
trigonometrical
and tangent, and use properties and graphs of all six trigonometric functions for angles of any
functions.
magnitude
We
also learn about
the compound
angle
formulae,
which
form a basis
for numerous
usewill
trigonometrical
identities
for the simplifi
cation
and exact
evaluation
of expressions
and in
important
mathematical
techniques.
These
techniques
include
solving
equations,
the course of solving equations, and select an identity or identities appropriate to thederiving
context,
further
the addition
of different sine and cosine functions and
showingtrigonometric
familiarity in identities,
particular with
the use of:
deriving
rules for differentiating and integrating trigonometric functions, which we will
sec 2 θ = 1 + tan2 θ and cosec 2 θ = 1 + cot 2 θ
cover later in this book.
the expansion of sin( A ± B ), cos( A ± B ) and tan( A ± B )
As you
know, scientists
engineers
the formulae
for sin 2and
A, cos
2 A and represent
tan 2 A oscillations and waves using trigonometric
functions.
These functions
have
uses in
navigation,
engineering
the expression
of a sin θ also
+ b cos
θ further
in the forms
R sin(
θ ± α ) and
R cos(θ ±and
α ). physics.
■
■
•
•
•
•
3.1 The cosecant, secant and cotangent ratios
There are a total of six trigonometric ratios. You have already learnt about the ratios sine,
cosine and tangent. In this section you will learn about the other three ratios, which are
cosecant (cosec), secant (sec) and cotangent (cot).
KEY POINT 3.1
The three reciprocal trigonometric ratios are defined as:
cosec θ =
1
sin θ


tan θ 
sin θ 
cos θ
1
1
Copyright
- Review
cot Only
θ = - Not
sec θ Material
=
 = for Redistribution

cos θ
Consider the right-angled triangle:
y
x
y
sin θ =
cos θ =
tan θ =
r
r
x
r
r
x
cosec θ =
sec θ =
cot θ =
y
x
y
r
y
θ
x
Copyright Material - Review Only - Not for Redistribution
By: WL Cheng (2021)
Page 1 of 13
This chapter builds on
the work we covered in
the Pure Mathematics 1
Coursebook, Chapter 5.
WEB LINK
Explore the
Trigonometry:
Compound angles
station on the
Underground
Mathematics website.
Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3
Cambridge
International AS & A Level Mathematics: Pure Mathematics 2 & 3
2021
A LEVELS MATHEMATICS
Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3
The following identities can be found from this triangle:
The following identities can be found from this triangle:
x 2 2+ y2 2 = r 2 2
Divide both sides by x 2 .2
x
+
y
=
r
Divide both sides by x .
The following
identities
can be found from this triangle:
2
2
y 2  r  2
r
y

1+  y  =  r 
Use y= tan θ and r = sec θ .
2+ x 2
2 x 
2
= sec θ .
1
=
tan
and
Use
=
θ
x
x
x + y = r  
Divide both sides by x .
x2
x2
x
x
2
2
1 + tan
y  2θ ≡
rsec
r
y
 2θθ
= sec θ .
1 +1 + tan
=θ ≡ sec
Use
= tan θ and
 x
 x
x
x
2
2
2
Divide both sides by y2 2.
1 + xtan2+2 θy ≡2 =
secr2 θ
Divide both sides by y .
x + y = r2
2
2
 x2  22


r
x
r
2
2
2
cotby
θ yand
Use
Divide
bothx=
sides
.
r = cosec θ .
xyx+y++1=1=r=yr
Use y = cot θ and y = cosec θ .
 y2 
 2y 
y
y
 x 2
 r
x
r
2
= cot θ and
= cosec θ .
Use
θ+ +1 1= ≡ cosec
θ
cot


2
2
ycot
y
y
 θ + 1≡y cosec

θ
2
2
We θwill
these
cot
+ 1need
≡ cosec
θ important identities to solve trigonometric equations later in this section.
We will need these important identities to solve trigonometric equations later in this section.
We will need these important identities to solve trigonometric equations later in this section.
KEY POINT 3.2
KEY POINT 3.2
KEY POINT
3.2
1 + tan2 2θ ≡ sec 2 2θ
1 + tan θ ≡ sec θ
2 2 θ ≡ sec 2 θ
1+
tan
cot
θ + 1 ≡ cosec 2 2θ
2
cot
θ + 1 ≡ cosec
θ
2
2
cot θ + 1 ≡ cosec θ
The graphs of y = sin θ and y = cosec θ
y = cosec
θ
The
graphs
= θsinand
θ and
y = cosec
θ
The
graphs
of of
y =ysin
y
y
y
y = cosec θ
y = cosec θ y = cosec θ
54
5454
y
1 y y
y
1 1
OO
O
sin θ
1
y = siny=θ sin θ
=
90 90
90
180 180
180
270
270
270
θ
360
θ
360 θ
360
–1
–1
O
–1
–1
1
1
O
O 90
90
180
90
–1
–1
The function y = cosec θ is the reciprocal of the function y = sin θ .
The function y = cosec θ is the reciprocal of the function y = sin θ .
The function y = cosec θ is the reciprocal of the function y = sin θ .
The function y = sin θ is zero when θ = 0°, 180°, 360°, … , this means that y = cosec θ
The function y = sin θ is zero when θ = 10°, 180°, 360°, … , this means that y = cosec θ
y = of
sinthese
θ ispoints
zero when
θ = 0is°,not
180defined.
°, 360°,Hence
… , this
means
that y = cosec θ
function
vertical
asymptotes
is The
not defined
at each
because
1
0
is not defined at each of these points because
1is not defined. Hence vertical asymptotes
is not
defined
thesepoints.
points because0 is not defined. Hence vertical asymptotes
need
to be
drawnat
ateach
each of these
need to be drawn at each of these points.
need to be drawn at each of these points.
0
Copyright Material - Review Only - Not for Redistribution
By: WL Cheng (2021)
Page 2 of 13
Copyright Material - Review Only - Not for Redistribution
Copyright Material - Review Only - Not for Redistribution
180
270
180
270θ
360
270
360 θ
360 θ
Chapter 3: Trigonometry
2021 A LEVELS MATHEMATICS
The graphs of y = cos θ and y = sec θ
We can find the graph of y = sec θ in a similar way from the graph of y = cos θ .
y
y = sec θ
Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3
y
1
O
1
y = cos θ
WORKED EXAMPLE 3.2
90
–1
180
270
O
360 θ
180
90
360 θ
270
–1
Solve sec 2 x − tan x − 3 = 0 for 0° ø x ø 360°.
Answer
sec 2 x − tan x − 3 = 0
Use 1 + tan2 x ≡ sec 2 x.
tan2 x − tan x − 2 = 0
Factorise.
The graphs of y = tan θ and y = cot θ
(tan x − 2)(tan x + 1) = 0
We can find the graph of y = cot θ from the graph of y = tan θ .
tan x = 2 or tan x = −1
y
The values of x are 63.4°, 135°, 243.4°, 315°.
y
y=2
1
y = tan θ
O
y = tan x
2
tan x = 2 ⇒ x = 63.4° (calculator)
or x = 63.4° + 180° = 243.4°
tan x = −1 ⇒ x = − 45° (not in required range)
or x = − 45° + 180° = 135°
or180
+ 180° = 315
x = 135° 270
90
360 ° θ
y
3
y = cot θ
–45
–90
63.4 90
O
180
270
180
90
x
360
O
–1
270
y = –1
360 θ
–2
–3
WORKED EXAMPLE 3.1
EXERCISE 3A
56
Find the exact value of cosec 240° .
1 Find the exact values of:
Answer a sec 60°
b cosec 45°
e cosec135
f cot 330°
1 °
cosec 240° =
sin 240°
2 Find the exact values of:
1
=
π
π
b cot
a cosec
–sin 60°
6
3
51π
7π
=
f cosec
e sec 3
4
− 6
2
3 Solve each
2 equation for 0° ø x ø 360°.
=−
a sec x 3= 3
b cot x = 0.8
c
cot120°
d sec 300°
g sec 150
y °
h cot( −30°).
A
S
240°
c
g
T
π
4
60° O
4π
cot
3
sec
2π
3
 π
h sec  −  .
 6
d cot
x
C
c
cosec x = −3
d
3sec x − 4 = 0
c
cot x = 2
d
2 cot x + 5 = 0
c
cot 2 x = 1
d
2 sec 2 x = 3
4 Solve each equation for 0 ø x ø 2π .
a cosec x = 2
b
sec x = −1
5 Solve each equation for 0° ø x ø 180°.
a cosec 2 x = 1.2
b
sec 2 x = 4
6 Solve each equation for the given domains.
a
c
Material
Redistribution
cosec( x − 30°) Copyright
= 2 for 0° ø
x ø 360°- Review Only - Not bfor sec(2
x + 60°) = −1.5 for 0° ø x ø 180°
π

cot x +
= 2 for 0 ø x ø 2 π
d 2 cosec(2 x − 1) = 3 for − π ø x ø π

4
By: WL Cheng (2021)
Page 3 of 13
Copyright Material - Review Only - Not for Redistribution
55
6
56
56
y=2
2
tan xEXAMPLE
= 2 ⇒ 3.2
x = 63.4° (calculator)
WORKED
1 Find the
exact
values
of:
or x = 63.4° + 180° = 243.4°
1
2
tan
=
−
1
⇒
=
−
45
°
(not
in
required
range)
x
x
a
sec
60
°
b
c cot120°
cosec
45
°
–45
2021
Solve
secA xLEVELS
− tan xMATHEMATICS
− 3 = 0 for 0° ø x ø 360°.
63.4 90
O
–90
180
or
=
−
45
°
+
180
°
=
135
°
x
e cosec135°
f cot 330°
g sec 150°
–1
Answer
or x = 135° + 180° = 315°
–2
22 Find the exact values of:
Use 1 + tan2 x ≡ sec 2 x.
secvalues
3=
0
x − tan
The
of x −are
π 63.4°, 135°, 243.4°, 315°.
π
π
–3
b cot
c sec
a cosec
2
Factorise.
6
3
4
tan x − tan x − 2 = 0
7π
4π
y
(tan x − 2)(tan
1) = 0
x +5 π
f cosec
g coty = tan x
e sec
6
4
3
3
tan x = 2 or tan x = −1
EXERCISE
3A
for 0° ø x ø 360°.
2
tan x3 = Solve
2 ⇒each
63.4° (calculator)
x =equation
a sec
= 3= 63.4
b ° cot x = 0.8
c cosec x = −3
orxexact
° +of:
180° = 243.4
x
1 Find
the
values
1
tan x = −1 ⇒ x = − 45° (not in required range)
sec 60
° equation for 0 ø bx øcosec
4 a Solve
each
2π . 45°
–90
or x = − 45° + 180° = 135°
e a cosec135
°= 2
f bcotsec
330x° = −1
cosec
x
or x = 135° + 180° = 315°
2 5 Find
exact
values
Solve
each
equation
0°°ø
x ø°.180°.
The values
of the
x are
63.4
°, 135°of:
,for
243.4
, 315
π
π
b bcotsec 2 x = 4
a a cosec
cosec6 2 x = 1.2
3
5π
7π
sec each equation for the given
f cosec
6 e Solve
domains.
6
4
EXERCISE 3A
x − 30°) =for
2 0for
ø°360
a cosec(
° ø0x° ø
360
. °
øx
3 Solve
each equation
π
 x +  = 2 for 0 ø x ø 2 π
cot
a c sec
xexact
b cot x = 0.8
1 Find
the
= 3 values
4  of:
sec 60
° equation for 0 ø xbø 2π
cosec
4 a Solve
each
. 45°
e a cosec135
° 2
cosec x =
f b cot
sec330
x =° −1
–45
O
–1
c
cot120
°
63.4
90
g csec 150
cot°x = 2
d sec 300°
270
x
360
h cot( −y30
°).
= –1
2π
3
 π
h sec  −  .
 6
d cot
y=2
d
3sec x − 4 = 0
d sec 300°
180
270
360
x
–1
=5
h cot(
°). xy +
d −230cot
=0
–2
–3
2π
π
c csec cot 2 x = 1
d cot
d 32 sec 2 x = 3
4
 π
4π
g cot
h sec  −  .
 6
3
b sec(2 x + 60°) = −1.5 for 0° ø x ø 180°
2 cosec(2
c dcosec
x = −3 x − 1) = 3
c
cot120°
for
π øxx−ø4 π= 0
d −3sec
d sec 300°
gc sec
cot150
x =° 2
dh 2cot(
cot−x30
+°5).= 0
2 5 Find
theeach
exact
values for
of: 0° ø x ø 180°.
Solve
equation
2π
π
π
π
sec
bb cot
a a cosec
Copyright
Material
cosec 2 x = 1.2
sec
2 x = 4 - Review Only cc- Not
cotfor
2 x Redistribution
=1
dd 2cot
sec 2 x = 3
6
3
4
3
 π
5
7
4
π
π
π
6 e Solve
domains.
sec each equation for the given
f cosec
g cot
h sec  −  .
 6
6
4
3
b sec(2 x + 60°) = −1.5 for 0° ø x ø 180°
a cosec( x − 30°) = 2 for 0° ø x ø 360°
3 Solve each equation for 0° ø x ø 360°.
π
d 2 cosec(2 x − 1) = 3 for − π ø x ø π
c cot  x +  = 2 for 0 ø x ø 2 π
a sec x = 3 4 
b cot x = 0.8
c cosec x = −3
d 3sec x − 4 = 0
4 Solve each equation for 0 ø x ø 2π .
a cosec x = 2
b
sec x = −1
c
cot x = 2
5 Solve each equation for 0° ø x ø 180°.
Copyright Material - Review Only - Not for Redistribution
a cosec 2 x = 1.2
b sec 2 x = 4
c cot 2 x = 1
d
2 cot x + 5 = 0
d
2 sec 2 x = 3
6 Solve each equation for the given domains.
a cosec( x − 30°) = 2 for 0° ø x ø 360°
π
c cot  x +  = 2 for 0 ø x ø 2 π

4
b
sec(2 x + 60°) = −1.5 for 0° ø x ø 180°
d
2 cosec(2 x − 1) = 3
for − π ø x ø π
Copyright Material - Review Only - Not for Redistribution
By: WL Cheng (2021)
Page 4 of 13
6
tan x = −1 ⇒ x = − 45° (not in required range)
a sec or
x =x3 = − 45° + 180° = 135
b ° cot x = 0.8 –90
or x = 135° + 180° = 315°
4 Solve each equation for 0 ø x ø 2π .
2021 A LEVELS MATHEMATICS
The values of x are 63.4°, 135°, 243.4°, 315°.
b sec x = −1
a cosec x = 2
–45
O
c
63.4 90
cosec x
= −3
180
270d
3sec360
x − 4 =x 0
y = –1
–1
–2
–3
c
cot x = 2
d
2 cot x + 5 = 0
c
cot 2 x = 1
d
2 sec 2 x = 3
5 Solve each equation for 0° ø x ø 180°.
a cosec 2 x = 1.2
EXERCISE 3A
sec 2 x = 4
b
6 Solve each equation for the given domains.
1 Find the exact values of:
a cosec( x − 30°) = 2 for 0° ø x ø 360°
a sec 60°
b cosec 45°
π
c cot x +  = 2 for 0 ø x ø 2 π
 ° 4
e cosec135
f cot 330°
sec(2 x + 60°) = −1.5 for 0° ø x ø 180°
cot120°
d sec 300°
d 2 cosec(2 x − 1) = 3 for − π ø x ø π
g sec 150°
h cot( −30°).
b
c
2 Find the exact values of:
π
π
π
b cot
c sec
a cosec
6
3
4
5π
7π
4π
f cosec
g cot
e sec
Copyright Material
6
4 - Review Only - Not for3Redistribution
2π
3
 π
h sec  −  .
 6
d cot
3 Solve each equation for 0° ø x ø 360°.
a sec x = 3
b
cot x = 0.8
c
cosec x = −3
d
3sec x − 4 = 0
c
cot x = 2
d
2 cot x + 5 = 0
c
cot 2 x = 1
d
2 sec 2 x = 3
b
sec(2 x + 60°) = −1.5 for 0° ø x ø 180°
d
2 cosec(2 x − 1) = 3
4 Solve each equation for 0 ø x ø 2π .
a cosec x = 2
b
sec x = −1
5 Solve each equation for 0° ø x ø 180°.
a cosec 2 x = 1.2
b
sec 2 x = 4
6 Solve each equation for the given domains.
a
c
cosec( x − 30°) = 2 for 0° ø x ø 360°
π
cot  x +  = 2 for 0 ø x ø 2 π

4
for − π ø x ø π
Copyright Material - Review Only - Not for Redistribution
By: WL Cheng (2021)
Page 5 of 13
2021 A LEVELS MATHEMATICS
Chapter 3: Trigonometry
7 Solve each equation for −180° ø x ø 180°.
cosec 2 x = 4
b
sec 2 x = 9
c
d sec x = cos x
e
cosec x = sec x
f
a
1
x=4
2
2 tan x = 3 cosec x
9 cot 2
8 Solve each equation for 0° ø θ ø 360° .
a 2 tan2 θ − 1 = sec θ
2 cot 2 θ − cosec θ = 13
c
e tan2 θ + 3sec θ = 0
b
3 cosec 2 θ = 13 − cot θ
d
cosec θ + cot θ = 2 sin θ
3 sec 2 θ = cosec θ
f
9 Solve each equation for 0° ø θ ø 180°.
a sec θ = 3 cos θ − tan θ
sec 4 θ + 2 = 6 tan2 θ
c
b
2 sec 2 2θ = 3tan 2θ + 1
d
2 cot 2 2θ + 7 cosec 2θ = 2
Chapter 3: Trigonometry
10 Solve each equation for 0 ø θ ø 2 π .
tan2 θ + 3sec θ + 3 = 0
a
3 cot 2 θ + 5 cosec θ + 1 = 0
b
eacheach
equation
−180° ø
x ø 180°for
. the interval 0 ø x ø 2π .
117 a Solve
Sketch
of the for
following
functions
2
4 x
a i cosec
y = 1x+=sec
= 92x
b ii secy2 =x cot
2 1
c iii9 cot
x = 4 x − π 
y = 2 cosec
2

2
f 2 tan x = 3 cosec x

π
vi y = sec  2 x + 

4
P
d sec x = cos x
e cosec x = sec x
1
v y = 1 + cosec x
iv y = 1 − sec x
2
8 Solve each equation for 0° ø θ ø 360° .
b Write down the equation of each of the asymptotes for your graph for part a vi.
b 3 cosec 2 θ = 13 − cot θ
a 2 tan2 θ − 1 = sec θ
12 Prove each
of these identities.
d cosec θ + cot θ = 2 sin θ
c 2 cot 2 θ − cosec θ = 13
b cosec x −2 sin x ≡ cos x cot x
a sin x2+ cos x cot x ≡ cosec x
f
e tan θ + 3sec θ = 0
3 sec θ = cosec θ
c sec x cosec x − cot x ≡ tan x
d (1 + sec x )(cosec x − cot x ) ≡ tan x
9 Solve each equation for 0° ø θ ø 180°.
13 Prove each of these identities.
b 2 sec 2 2θ = 3tan 2θ + 1
a sec θ = 3 cos θ − tan θ
1
1
2
sin2 x cos x
b d sec
a c sec 4 θ + 2 = 6≡tan
x 2+2sec
x ≡2θ = 2
θ
2 cot
θ +x7tan
cosec
tan x + cot x
1 − sin x
2
1
+
tan
x
1 − cos2 x
2
≡ sec x cosec x
≡ 1 − sinfor
x 0 ø θ ø 2π .
d
10c Solve each
equation
sec 2 x − 1
tan x
2
2
a tan
b 1 3+cot
sin xθ + 5 cosec θ + 1 = 20
sinθx + 3sec θ + 3 = 0
≡
cosec
x
f
≡ ( tan x + sec x )
e
1 − cos2 x
1 − sin x
11 a Sketch
2πx.
cos x
1 each of1the following 2functions for the interval 0 ø x øcos
+
≡ 2 cosec x
g
h
+
≡ 2 sec x

π
1 i+ cos
− cos
1 + sin x 1 − sin x iii y = 2 cosec  x − 
y =x 1 +1sec
x x
ii y = cot 2 x

2
PS
14 Solveiveach
y =equation
1 − sec x for 0° ø θ ø 180°.
P
P
P
1

π
x
vi y = sec  2 x + 
2

4
2
2
sec 3 θ down
− 5 secthe
θ−
8 sec θ +of
3 =each
0 of the asymptotes for your
b 2graph
cot 3 θfor
+ 3part
cosec
θ − 8 cot θ = 0
a b 6Write
equation
a vi.
v
y = 1 + cosec
12 Prove each of these identities.
a
sin x + cos x cot x ≡ cosec x
b
cosec x − sin x ≡ cos x cot x
c
sec x cosec x − cot x ≡ tan x
d
(1 + sec x )(cosec x − cot x ) ≡ tan x
b
sec 2 x + sec x tan x ≡
13 Prove each of these identities.
1
≡ sin x cos x
a
tan x + cot x
1
1 − sin x
1 + tan2 x
1 − cos2 x
2
≡ sec x cosec x
≡
1
−
sin
x
d
sec 2(2021)
x −1
tan x
By: WL Cheng
Page 6 of 13
Copyright Material - Review Only - Not for Redistribution
1 + sin x
sin x
2
≡ cosec x
f
≡ ( tan x + sec x )
e
2
1 − cos x
1 − sin x
cos x
cos x
1
1
+
≡ 2 cosec 2 x
g
h
+
≡ 2 sec x
1 + cos x 1 − cos x
1 + sin x 1 − sin x
c
57
57
8 Solve each equation for 0° ø θ ø 360° .
a 2 tan2 θ − 1 = sec θ
2
2 cot MATHEMATICS
θ − cosec θ = 13
2021 A cLEVELS
e tan2 θ + 3sec θ = 0
b
3 cosec 2 θ = 13 − cot θ
d
cosec θ + cot θ = 2 sin θ
f
3 sec 2 θ = cosec θ
9 Solve each equation for 0° ø θ ø 180°.
a sec θ = 3 cos θ − tan θ
c
sec 4 θ + 2 = 6 tan2 θ
b
2 sec 2 2θ = 3tan 2θ + 1
d
2 cot 2 2θ + 7 cosec 2θ = 2
b
3 cot 2 θ + 5 cosec θ + 1 = 0
10 Solve each equation for 0 ø θ ø 2 π .
a
tan2 θ + 3sec θ + 3 = 0
Chapter 3: Trigonometry
11 a Sketch each of the following functions for the interval 0 ø x ø 2π .
y =equation
1 + sec xfor −180° ø x ø 180°.
7 Solve ieach
ii
y = cot 2 x
b sec 2 x = 9
c
a cosec 2 x = 4
1
v y = 1 + cosec x
iv y = 1 − sec x
2
d sec x = cos x
e cosec x = sec x
f
b Write down the equation of each of the asymptotes for your graph for part a vi.
8 Solve each equation for 0° ø θ ø 360° .
12 Prove each of these identities.
a 2 tan2 θ − 1 = sec θ
a sin x + cos x cot x ≡ cosec x
c 2 cot 2 θ − cosec θ = 13
c sec x cosec x − cot x ≡ tan x
e tan2 θ + 3sec θ = 0
13 Prove each of these identities.
9 Solve each equation
for 0° ø θ ø 180°.
1
≡ sin x cos x
a
cotθx− tan θ
θ =x 3+cos
a sectan
P
P

π
iii y = 2 cosec  x − 

2
1
9 cot 2 x = 4
π
vi y2= sec  2 x + 

2 tan x = 3 cosec x 4 
57
3 cosec 2 θ = 13 − cot θ
cosec x − sin x ≡ cos x cot x
d cosec θ + cot θ = 2 sin θ
d (1 + sec x )(cosec x − cot x ) ≡ tan x
f
3 sec 2 θ = cosec θ
b
b
1
sec 2 x + sec x tan x ≡
2
1
−
sin
x
b 2 sec 2θ = 3tan 2θ + 1
2
2
1 + tan x
1 − cos x
2 sin2 x
≡ sec 2xθcosec
1−
θ
d d 2 cot 2 2θ + 7 cosec
=2 x
c c sec 4 θ 2+ 2 = 6 ≡tan
sec x − 1
tan x
1 + sin x
sin x
2
10 Solve
forx0 ø θ ø 2 π .
≡ cosec
f
≡ ( tan x + sec x )
e each equation
1 − cos2 x
1 − sin x
2
a tan2 θ1+ 3sec θ + 31 = 0
b 3 cotcos
θx+ 5 cosec
cosθx+ 1 = 0
+
≡ 2 cosec 2 x
g
h
+
≡ 2 sec x
1 + cos x 1 − cos x
1 + sin x 1 − sin x
11 a Sketch each of the following functions for the interval 0 ø x ø 2π .
PS
P
b

π
iii y = 2 cosec  x − 

2
3
2
3
2
a 6 sec θ − 5 sec θ − 8 sec θ + 3 = 0
1 b 2 cot θ + 3 cosec θ − 8 cot θ =0
π
vi y = sec  2 x + 
v y = 1 + cosec x
iv y = 1 − sec x
2

4
b Write down the equation of each of the asymptotes for your graph for part a vi.
y = 1 equation
+ sec x for 0° ø θ ø 180°.
i each
14 Solve
ii
y = cot 2 x
12 Prove each of these identities.
a
sin x + cos x cot x ≡ cosec x
b
cosec x − sin x ≡ cos x cot x
c
sec x cosec x − cot x ≡ tan x
d
(1 + sec x )(cosec x − cot x ) ≡ tan x
P
13 Prove each of these identities.
1
1
≡ sin x cos x
b sec 2 x + sec x tan x ≡
a
tan x + cot x
1 − sin x
2
Copyright Material - Review Only - Not for1 +
Redistribution
tan
x
1 − cos2 x
≡ sec x cosec x
≡ 1 − sin2 x
d
c
sec 2 x − 1
tan x
1 + sin x
sin x
2
≡ cosec x
f
≡ ( tan x + sec x )
e
2
1 − cos x
1 − sin x
cos x
cos x
1
1
+
≡ 2 cosec 2 x
g
h
+
≡ 2 sec x
1 + cos x 1 − cos x
1 + sin x 1 − sin x
PS
14 Solve each equation for 0° ø θ ø 180°.
a
6 sec 3 θ − 5 sec 2 θ − 8 sec θ + 3 = 0
By: WL Cheng (2021)
b
2 cot 3 θ + 3 cosec 2 θ − 8 cot θ = 0
Page 7 of 13
57
a sec θ = 3 cos θ − tan θ
c
sec θ + 2 = 6 tan θ
4
2
b
2 sec 2θ = 3tan 2θ + 1
d
2 cot 2 2θ + 7 cosec 2θ = 2
Chapter 3: Trigonometry
10 Solve each equation for 0 ø θ ø 2 π .
2021 A LEVELS MATHEMATICS
a
tan2 θ + 3sec θ + 3 = 0
b
3 cot 2 θ + 5 cosec θ + 1 = 0
7 11Solve
each equation
forfollowing
−180° øfunctions
x ø 180°.for the interval 0 ø x ø 2π .
a Sketch
each of the
a
2
cosec
i y x= 1=+4sec x
2 1
c iii9 cot
x = 4 x − π 
y = 2 cosec
2

2
f 2 tan x = 3 cosec x

π
vi y = sec  2 x + 

4
9 2x
b iisec 2yx==cot
d sec x = cos x
e cosec x = sec x
1
v y = 1 + cosec x
iv y = 1 − sec x
2
8 Solve each equation for 0° ø θ ø 360° .
b Write down the equation of each of the asymptotes for your graph for part a vi.
b 3 cosec 2 θ = 13 − cot θ
a 2 tan2 θ − 1 = sec θ
P
12 Prove each
of
these
identities.
d cosec θ + cot θ = 2 sin θ
c 2 cot 2 θ − cosec θ = 13
b cosec x 2− sin x ≡ cos x cot x
a sin2 x + cos x cot x ≡ cosec x
f
e tan θ + 3sec θ = 0
3 sec θ = cosec θ
c sec x cosec x − cot x ≡ tan x
d (1 + sec x )(cosec x − cot x ) ≡ tan x
9 Solve each equation for 0° ø θ ø 180°.
P
13 Prove each of these identities.
b 2 sec 2 2θ = 3tan 2θ + 1 1
a sec θ = 31 cos θ − tan θ
≡
sin
cos
x
x
b sec 2 x + sec x tan x ≡
a
4 x + cot x
θ + 2 = 6 tan2 θ
d 2 cot 2 2θ + 7 cosec 2θ1 −
= sin
2 x
c sectan
2
2
1 + tan x
1 − cos x
≡ sec x cosec x
≡ 1 − sin2 x
d
c
2 equation for 0 ø θ ø 2 π .
10 Solve sec
each
x −1
tan x
1 + sin2 x
ae tan2sin
θ +x3sec≡θcosec
+ 3 =x 0
+ 1 x=)20
fb 3 cot θ +≡ 5( cosec
tan x +θ sec
1 − cos2 x
1 − sin x
cos x
1 each of the
1 following functions
11 a Sketch
for the interval 0 ø x øcos
2π .x
+
≡ 2 cosec 2 x
g
h
+
≡ 2 sec x

π
1 + cos x 1 − cos x
1 + sin x 1 − sin x
ii y = cot 2 x
iii y = 2 cosec  x − 
i y = 1 + sec x

2
PS
14 Solve each equation for 0° ø θ ø 180°.
1

π
vi y = sec  2 x + 
v y = 1 + cosec x
iv y = 1 − sec x
3
2
3
2
4
b2 2 cot θ + 3 cosec θ − 8 cot θ = 0 
a 6 sec θ − 5 sec θ − 8 sec θ + 3 = 0
b Write down the equation of each of the asymptotes for your graph for part a vi.
P
P
12 Prove each of these identities.
a
sin x + cos x cot x ≡ cosec x
b
cosec x − sin x ≡ cos x cot x
c
sec x cosec x − cot x ≡ tan x
d
(1 + sec x )(cosec x − cot x ) ≡ tan x
b
sec 2 x + sec x tan x ≡
13 Prove each of these identities.
1
≡ sin x cos x
a
tan x + cot x
c
e
g
PS
1
1 − sin x
1 + tan2 x
1 − cos2 x
≡ sec x cosec x
≡ 1 − sin2 x
d
2
Copyright Material - Review Only - Not for Redistribution
sec x − 1
tan x
1 + sin x
sin x
2
≡ cosec x
f
≡ ( tan x + sec x )
1 − cos2 x
1 − sin x
cos x
cos x
1
1
+
≡ 2 cosec 2 x
h
+
≡ 2 sec x
1 + cos x 1 − cos x
1 + sin x 1 − sin x
14 Solve each equation for 0° ø θ ø 180°.
a
6 sec 3 θ − 5 sec 2 θ − 8 sec θ + 3 = 0
By: WL Cheng (2021)
b
2 cot 3 θ + 3 cosec 2 θ − 8 cot θ = 0
Copyright Material - Review Only - Not for Redistribution
Page 8 of 13
57
57
b Write down the equation of each of the asymptotes for your graph for part a vi.
P
12 Prove each of these identities.
sin x +MATHEMATICS
cos x cot x ≡ cosec x
2021 aA LEVELS
b
cosec x − sin x ≡ cos x cot x
sec x cosec x − cot x ≡ tan x
d
(1 + sec x )(cosec x − cot x ) ≡ tan x
b
sec 2 x + sec x tan x ≡
c
P
13 Prove each of these identities.
1
≡ sin x cos x
a
tan x + cot x
c
e
g
PS
57
1 − cos2 x
≡ 1 − sin2 x
sec 2 x − 1
sin x
≡ cosec x
1 − cos2 x
1
1
+
≡ 2 cosec 2 x
1 + cos x 1 − cos x
d
f
h
1
1 − sin x
1 + tan2 x
≡ sec x cosec x
tan x
1 + sin x
2
≡ ( tan x + sec x )
1 − sin x
cos x
cos x
+
≡ 2 sec x
1 + sin x 1 − sin x
14 Solve each equation for 0° ø θ ø 180°.
a
6 sec 3 θ − 5 sec 2 θ − 8 sec θ + 3 = 0
b
2 cot 3 θ + 3 cosec 2 θ − 8 cot θ = 0
Copyright Material - Review Only - Not for Redistribution
By: WL Cheng (2021)
Page 9 of 13
sec 2 x − 1
sin x
≡ cosec x
e
1 − cos2 x
1 MATHEMATICS
1
2021gA LEVELS
+
≡ 2 cosec 2 x
1 + cos x 1 − cos x
PS
f
h
tan x
1 + sin x
2
≡ ( tan x + sec x )
1 − sin x
cos x
cos x
+
≡ 2 sec x
1 + sin x 1 − sin x
14 Solve each equation for 0° ø θ ø 180°.
a
6 sec 3 θ − 5 sec 2 θ − 8 sec θ + 3 = 0
b
2 cot 3 θ + 3 cosec 2 θ − 8 cot θ = 0
Copyright Material - Review Only - Not for Redistribution
By: WL Cheng (2021)
Page 10 of 13
Mathematics:
Pure
Mathematics
2&3
2021
A LEVELS
MATHEMATICS
c −
Exercise 3A
1
b
2
c −
d 2
e
2
f
− 3
c
2
f
− 2
g −
2
g
4
5
6
2
3
a 2
d −
3
1
3
a 2
1
3
1
3
h − 3
1
b
3
2
e −
3
2
h
3
a 70.5°, 289.5°
b 51.3°, 231.3°
c 199.5°, 340.5°
d 41.4°, 318.6°
π 5π
,
6
6
c 0.464, 3.61
a
b π
d 2.76, 5.90
a 28.2°, 61.8°
b 37.8°, 142.2°
c 22.5°, 112.5°
d 24.1°, 155.9°
a 60°, 180°
b 35.9°, 84.1°
c 2.82, 5.96
d −2.28, −1.44, 0.865, 1.71
7
a −150°, − 30°, 30°, 150°
b −109.5°, − 70.5°, 70.5°, 109.5°
c −112.6°, 112.6°
d −180°, 0°, 180°
e −135°, 45°
f
8
−60°, 60°
a 48.2°, 180°, 311.8°
b
31.0°, 153.4°, 211.0°, 333.4°
c 19.5°, 160.5°, 203.6°, 336.4°
d
60°, 180°, 300°
e 107.6°, 252.4°
3
3
f
9
27.2°, 152.8°
a 41.8°, 138.2°
b
13.3°, 22.5°, 103.3°, 112.5°
c 45°, 60°, 120°, 135°
d 97.2°, 172.8°
10
a
2π
4π
, π,
3
3
b
7 π 11π
,
6
6
By: WL Cheng (2021)
aterial - Review Only - Not for Redistribution
Page 11 of 13
Answers
2021 A LEVELS MATHEMATICS
11
vi
y
a i
y
6
3
4
2
2
1
O
–2
π
–
2
π
3π
–
2
2π x
O
–1
π
–
2
π
2π x
3π
–
2
–4
–2
–6
–3
ii
y
π
5π
9π
13π
, x=
, x=
, x=
8
8
8
8
a Proof
b Proof
b x=
12
c Proof
O
π
–
2
π
3π
–
2
2π x
13
14
y
iii
1
4
2
–2
a Proof
b Proof
c Proof
d Proof
e Proof
f
g Proof
h Proof
a 48.2°, 180°
2
π
–
2
π
3π
–
2
2π x
a 1
d
3
–6
y
6
4
2
O
–2
π
–
2
π
3π
–
2
2π x
3
2
6+ 2
4
7
b 2+ 3
c
g −2 + 3
h
a
33
65
77
a
85
6− 2
4
6+ 2
4
16
65
36
b
85
b
4
2
O
By: WL Cheng (2021)
π
–
2
π
3π
–
2
c
f
e
6
2
2
e 1
d −2 + 3
5
–6
y
6
a
b
4+3 3
10
Proof
4
–4
v
b 45°, 63.4°, 161.6°
3
1
cos x − sin x
2
2
–4
iv
Proof
Exercise 3B
6
O
d Proof
2π x
Copyright Material - Review Only - Not for Redistribution
Page 12 of 13
1
2
3
3
f
2− 6
4
6− 2
4
33
56
13
c
84
c −
Answers
2021 A LEVELS MATHEMATICS
vi
y
3
2
1
π x
O
–1
π
–
2
π
2π x
3π
–
2
–2
–3
π
5π
9π
13π
, x=
, x=
, x=
8
8
8
8
a Proof
b Proof
b x=
12
c Proof
x
13
14
d Proof
a Proof
b Proof
c Proof
d Proof
e Proof
f
g Proof
h Proof
a 48.2°, 180°
Proof
b 45°, 63.4°, 161.6°
Exercise 3B
1
2
2π x
a 1
d
3
a
3
2
6+ 2
4
b 2+ 3
c
g −2 + 3
h
a
33
65
77
a
85
c
f
e
6
2
2
e 1
d −2 + 3
5
7
317
b
4+3 3
10
Proof
4
2π x
3
1
cos x − sin x
2
2
6− 2
4
6+ 2
4
16
65
36
b
85
b
1
2
3
3
f
2− 6
4
6− 2
4
33
56
13
c
84
c −
2π x
By: WL Cheng (2021)
- Review Only - Not for Redistribution
Page 13 of 13