Semester 1 Mathematics STPM (Teacher's Edition)

96

EXERCISE 1.10

1. For each of the following, express in terms of and ,

a) = 2(3 + )

b) ln( − 2) − ln( + 2) =

2. Solve the equation (3 − log3 ) log3 3 = 1

3. Given that log = 3 log 2 − log ( − 2 ) , express in terms of .
2

4. Find the value of such that (9 − log2 ) log2 8 = 2

5. Solve the equation ln + ln( + 2) = 1

6. Solve the simultaneous equations,

log3 = 5 and 2 = 2
log9

7. Solve the simultaneous equation

log4 = 1 and (log2 )(log2 ) = −2
2

8. Given that log (3 − 4 ) + log 3 = 2 + log (1 − 2 ) , where 0 < < 1,
log2
2

Find the value of .

ANSWER 1(a)
Section = 2(3 + )
1.3 2 (3 + ) =

Exercise 2 − = −6
1.10
(2 − ) = −6

= −6 = 6
2 − − 2
6
= ln [ − 2 ]

1(b) ln( − 2) − ln( + 2) =

− 2
ln [ + 2] =

− 2
+ 2 =

− 2 = ( + 2)

− = 2 + 2
2 + 2
= 1 −

97

2 (3 − log3 ) log3 3 = 1
1

(3 − log3 ) log3 3 = 1
3 − log3 = log3 3 + log3
2 log3 = 2
= 3

3
log 2 = 3 log 2 − log ( − 2 )
23
log 2 = log ( − 2 )

8
2 = ( − 2 )

( − 2 ) = 8 2

2 − 2 − 8 2 = 0

2 ± √(2 )2 − 4(1)(−8 2)
= 2(1)
= 0 or 4
Since > 0, ∴ = 4

OR (other method)
2 − 2 − 8 2 = 0

( + 2 )( − 4 ) = 0
= −2 or 4
Since > 0 and > 0, ∴ = 4

4 (9 − log2 ) log2 8 = 2
log2 8
(9 − log2 ) log2 2 = 2

3
(9 − log2 ) log2 2 + log2 = 2
3(9 − log2 ) = 2(1 + log2 )

5 log2 = 25

log2 = 5
= 25

= 32

5 ln ( + 2) = 1
( + 2) =
2 + 2 − = 0

−2 ± √22 − 4(1)( )
= 22
= −1 ± √1 −
since > 0, ∴ = −1 + √1 −

98

6 log3 = 5

log3 + log3 = 5 − − − − − −( )
2

log9 = 2

log3 2

= 2
log3 9

2
log3 = 4

2 log3 − log3 = 4 − − − − − −( )
( ) + ( ):

3 log3 = 9

log3 = 3
= 33 = 27

log3 = 2
= 32 = 9

7 1
log4 = 2

log2 = 1
log2 4 2

log2 = 1

log2 + log2 = 1

log2 = 1 − log2 − − − − − −( )

(log2 )(log2 ) = −2 − − − − − −( )

Substitute ( ) into ( )

(1 − log2 )(log2 ) = −2
log2 − (log2 )2 = −2
(log2 )2 − log2 − 2 = 0

(log2 − 2)(log2 + 1) = 0
log2 = 2 or log2 = −1

For log2 = 2
= 22 = 4

log2 = 1 − 2 = −1

= 2−1 = 1
2

For log2 = −1

= 2−1 1
=2

log2 = 1 − (−1) = 2

= 22 = 4

99

8 2
log (3 − 4 ) + log 3 = 1 + log (1 − 2 )

log 2

log (3 − 4 ) + log 3 = 2 log 2 + log (1 − 2 )
log 3 (3 − 4 ) = log 22(1 − 2 )
3 (3 − 4 ) = 22(1 − 2 )

9 2 − 12 − 4 + 8 = 0

12 ± √(−12 )2 − 4(9)(−4 + 8 )
= 2(9)

12 ± √144 2 + 144 − 288
= 2(9)

12 ± 12( − 1)
= 2(9)

= 2 or 4 − 2
3 3

since > 0 and 0 < < 1 , ∴ = 2
2 3

OR (other method)

9 2 − 12 − 4 + 8 = 0

9 2 − 4 − 12 + 8 = 0

(3 + 2)(3 − 2) − 4 (3 − 2) = 0

(3 − 2)[3 + 2 − 4 ] = 0

= 2 or 4 − 2
3 3

since > 0 and 0 < < 1 , ∴ = 2
2 3

100 Learning Outcome:
a) Relate the periodicity and symmetries of sine, cosine and tangent
1.4(a) functions to their graphs and identify the inverse sine, inverse cosine
TRIGONOMETRIC and inverse tangent function and their graphs
FUNCTIONS b) Use basics trigonometric identities and the formulae for sin (A ± B),
cos (A ± B) and tan (A ± B) including sin 2A, cos 2A and tan 2A

1.4 TRIGONOMETRIC FUNCTIONS

1. Graphs of basic trigonometric functions

Functions Graph Examples
Period : 2
= sin Amplitude : 1
Domain : (−∞, ∞)
Range : [-1, 1]

= cos Period : 2
Amplitude : 1
Domain : (−∞, ∞)
Range : [-1, 1]

= tan Period :

Domain : All real numbers except
+ ,
an integer
2

Range : [−∞, ∞]

= cot Period :

Domain : All real numbers except
, an integer
Range : [−∞, ∞]

Amplitude : defined

101 Period : 2
= csc
Domain : All real numbers except
= sec , an integer
Range : ≤ −1 ≥ 1

Amplitude : defined

Period :

Domain : All real numbers except

2 + , an integer

Range : ≤ −1 ≥ 1

Amplitude : defined

2. Graphing y = k + A sin (Bx + C) and y = k + A cos (Bx + C)

| | = Amplitude

Solve Bx + C = 0 and Bx + C = 2π for period and these shift

Bx + C = 0 Bx + C = 2π

= − = − + 2



Phase shift period

∴ period = 2



Phase shift (horizontal translation)

If − is positive : shift to the right



If − is negative : shift to the left



| | is vertical translation

If k positive : translate vertically up k units

If k negative : translate vertically down k units

102

3. Graphing = tan( + ) or = cot( + )

Amplitude is undefined

Solve + = 0 and + = 2 for period and these shift

+ = 0 + = 2

= − = − +



Phase shift period

4. Graphing = sec ( + ) or = csc ( + )

Amplitude is undefined

Solve + = 0 and + = 2

= − = − + 2



5. Inverse trigonometric functions

(a) The inverse sine function sin-1 or arc sin is defined as the inverse of the restricted sine function
− ≤
y = sin x for ≤ , −1 ≤ ≤ 1 .
2 2

Thus y = sin-1 x = arc sin x is equivalent to x = sin y for −1 ≤ ≤ 1, − ≤ ≤

• sin ( −1 ) = , −1 ≤ ≤ 1 22

sin−1(sin ) = , − ≤ ≤

22

(b) The inverse cos function cos-1 or arc cos is defined as the inverse of the restricted cosine
function = cos for 0 ≤ ≤ , −1 ≤ ≤ 1 .

Thus = cos−1 =
cos is equivalent to = cos for −1 ≤ ≤ 1, − ≤ ≤

22

• cos ( −1 ) = , −1 ≤ ≤ 1

−1(cos ) = , 0 ≤ ≤

6. Basic Trigonometric Identities

Reciprocal Identities sec = 1 cot = 1
csc = 1
cos tan
sin
tan(− ) = − tan
Quotient Identities cot = cos
tan = sin
sin
cos

Identities for Negatives
sin(− ) = − sin cos(− ) = cos

Pythagorean Identities
sin2 + cos2 = 1 tan2 + 1 = sec2 1 + cot2 = csc2

103

Sum Identities and difference Identities
sin (x ± y) = sin x cos y ± cos x sin y
cos (x ± y) = cos x cos y ∓ sin x sin y
tan ± tan
tan( ± ) = 1 ∓ tan tan

7. Cofunction Identities sin ( − ) = cos tan ( − ) = cot
cos ( − ) = sin
2 2
2
Shrink the graph of
8. Double-angle Identities y = sin x horizontally by a
sin 2 = 2 sin cos factor of 2, shrink it
vertically also by a factor
cos 2x = cos2 x – sin2 x of ½ and shift it unit to
= 1 – 2 sin2 x
= 2 cos2 x – 1 8

2 tan the right
tan 2 = 1 − tan2

9. Half-angle Identities
sin = ±√1−cos

2

cos = ±√1+s2in

tan = ±√11 − cos
+ cos

= sin = 1−cos
1+cos sin

EXAMPLE 1
Graph the function = 1 sin(2 − )

24

Solution:
= 1 sin(2 − )

24

1
= 2 sin [2 ( − 8)]

11
Amplitude = |2| = 2

104

2
Period = | 2 | =


Phase shift = 4 = 8

2

EXAMPLE 2

Graph the function = −1+ 3 sin(2 + ).

2 4

Solution:

Amplitude = 3 Period = |2 | =
4
2

2x + π = 0 2x + π = 2π The graph completes one full cycle as x

x = − x = − + = varies over the interval [− , ]
2 2 2 2 2

phase shift period

∴ phase shift = −
2

We graph = 3 sin(2 + ) , then vertically translate the graph down ½ unit.

4

Shrink the graph of y = sin x

horizontally by a factor of 2,

shrink it vertically also by a
factor of ½ and shift it unit to

8

the right

EXAMPLE 3 (b) 2 − 1
Simplify the following 1 − 2

(a) sin

cos tan(− )

Solution: (b) = 2 − 1 2 −1
2
sin sin 1 − 2 2
(a) = cos (− tan ) 1− 2
cos tan(− )

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

= 1 ⋅ 2
2
= 1

= − sin = −1 = 2
2
sin

= 2

105 1. = sin

WORKSHEET 1.9

Sketch the graph of = sin for
0 ≤ ≤ 3600. Hence sketch the graph of

(a) = 1 + sin
(b) = sin ( + 30°)
(c) = sin 2

(a) = 1 + sin (b) = sin ( + 30°)

(c) = sin 2

EXERCISE 1.11

1. State the amplitude A and period P of each function and graph the function over the indicated
interval.

(a) = 3 sin , − 2 ≤ ≤ 2

(b) = − 1 cos , − 2 ≤ ≤ 2

2

106

2. State the amplitude and period of the function and graph the function over the indicated
interval.

(a) = 2 + 2 sin , −4 ≤ ≤4

2

(b) = 4 − 2 cos , −4 ≤ ≤ 4

2

3. Find the period and phase shift than graph each function.

(a) = tan(2 + ), −3 < < 3

44

(b) = −2 tan ( − ) , −1 < <7

44

4. Evaluate cos (sin−1 √2 + cos−1 35).

2

5. Solve = sin−1 √2

2

6. Simplify each of the following trigonometric expressions.

(a) 2 sin2 + sin − 3
1 − cos2 − sin

(b) sin + cot

1 + cos

7. Simplify the following

(a) sin2 − 9 ∙ 10 cos + 5
2 cos + 1
3 sin + 9

(b) 1 − 2
sin2 − cos2 cos sin

(c) csc4 − 1
csc2

ANSWER 1(a) = 3 sin = |3| = 3 = 2
Section
1.4

Exercise
1.11

107

1(b) The graph of − 1 cos 2 is the graph of − 1 cos 2 reflected across
2 2

the x-axis.

Amplitude = |1| = 1

22

2 (a) = 2 + 2 sin

2

A = 2, period = 2 = 4



4

The graph is the same as

the graph of = 2 2

shifted 2 unit up

2(b)
= 4 − 2 cos 2
2
= 4, period = = 4

2
The graph is the same as the graph of = 2 cos reflected about x-

2

axis and shifted 4 units up.


= 4 − 2 cos 2

108

3(a) = tan(2 + )

2 + = 0 2 + = Note
One cycle of y =
2 = − 2 = − + tan(2x + ) is
= − = − + completed as 2 +
varies from 0 to
2 22

phase shift period


∴ ℎ ℎ = − 2 , = 2

Sketch the graph for one

period [− , 0] then extend
2
3 3 ]
over the interval [− 4 ,
4

3(b)

109

4 −√2
10

5
= 2

6 (a) 2+ 3

sin

(b) cosec x

7 (a) 5 (sin − 3)
3
1+2 sin +2 cos
(b) 2 − 2

(c) 2 + 2

CLONE STPM
1. Sketch the graph of = 2 cos ( − ) for 0 ≤ ≤ 2 .

3

2. Sketch the graph of = √13 cos( − 56.3°) for 0 ≤ ≤ 2 .

STPM PAST YEAR

STPM MT 2015 (Section B Q7(c) )
1. Sketch = 13 cos( + 0.4) for 0 ≤ ≤ 2 .

STPM MT 2018 (Section B Q7(b) )
= 2 sin( − )
2. Sketch the graph of for 0 ≤ ≤ 2 .
6

ANSWER 1
Section
1.4

Clone
STPM


= 2 cos ( − 3)

110
2

ANSWER 1 = √13 cos( − 56.3°)
Section = 13 cos( + 0.4)
1.4

STPM = 2 sin( − 6)
Past
Year

2

111

1.4 (b) Learning Outcome:
TRIGONOMETRIC Find the solution, within specified intervals, of trigonometric equations
EQUATIONS

Trigonometric Equations

Simple trigonometric equations

The simple trigonometric equations like sin = 1 or cos = − √3 or tan(3 + ) = 1 can
2 2

be solved in few steps:

1. Determine the quadrants, the angle should be in, based on the given trigonometric equation.

2. Find the basic angle by using a scientific calculator

3. Determine the range of values of the required angles, for examples the range of values for

or 3 .

4. Determine the values of angles in those quadrants.

EXAMPLE 1

Solve the following trigonometric equations, for 0° ≤ ≤ 360°

a) tan = 1.0 b) cos = − 1

2

Solution: Solution:
tan = 1.0 1

Basic angle: cos = − 2
Basic angle:
.
= cos−1 1 = 60°
= tan−11 = 45o 2

tan x has positive sign in the first But cos x has negative sign and thus the angles
and third quadrants . are in the second and third quadrants

∴ = 45°, -135o or 225o ∴ = 120°, −120° 240

EXAMPLE 2
Solve sin − √3 cos = 0 for 0° ≤ ≤ 360°

Solution:
sin − √3 cos = 0
sin = √3 cos
tan = √3
= tan−1√3 = 60°

Since tan > 0, thus x falls in the I and III quadrants. Hence, = 60° , 240°

112

EXAMPLE 3
Solve the equation 3 2 − 17 cos + 10 = 0 for 0° ≤ ≤ 360°.

Solution:

3cos2 − 17 cos + 10 = 0

Let = cos , 3 2 − 17 + 10 = 0

(3 − 2)( − 5) = 10

= 2 = 5

3

Substitute back:

cos = 2 cos = 5 (undefined)
3

2
cos = 3 ∶ = 48.19° , 311.81

EXERCISE 1.12

1. Find the solution for the following trigonometric equations in the given interval
a) tan = 0.2 , 0 ≤ ≤ 360°
b) sin 2 = 0.5 , 0° ≤ ≤ 360°
c) cos(2 + 30°) = 0.5 , − 180° ≤ ≤ 180°

2. Solve the equation exactly for 0≤ ≤ 2
a) 2 cos − 3 = −5
b) cosec = −2
c) 2sin2 − 1 = 0

3. Solve 2(cos + 1) = 1 for 0≤ ≤ 2

4. Solve 4 tan + 2 = 2 tan

5. Solve the equation 2sin2 − 5 sin + 3 = 0 for 0 ≤ ≤ 2

6. Solve 5cos2 − 2 cos 2 = 3 for 0≤ ≤ 2

7. 5 tan − cot 2 + 5 = 0 for 0° ≤ ≤ 360°

8. Solve 5sin2 − 4 sin − 1 = 0 for 0° ≤ ≤ 360°

113

ANSWER 1(a) tan = 0.2
Section
1.4 Quadrant I and III

Exercise Basic angle
1.12 = −10.2 = 11.31

= 11.31° 191.31°

1(b) sin 2 = 0.5
Quadrant I and IV
Basic angle
= −10.5 = 30°

2 = 30°, 150°, 390°, 510°
= 15°, 75°, 195°, 255°
1(c) cos(2 + 30°) = 0.5
Quadrant I and IV
Basic angle
= −10.5 = 60°

2 + 30° = −60° , 60°
2 = −90° , 30°
= −45° , 15°
2(a) 2 − 3 = −5
2 cos = −2
cos = −1
=
2(b) = −2

1
sin = −2

1
sin = − 2

7 11
= 6 , 6

114

2(c) 2 2 − 1 = 0
2 2 = 1

√ 2 = ±√12

sin = ± 1 = ± √2
√2 2

3 5 7
= 4 , 4 , 4 , 4

3 2(cos + 1) = 1
2 cos + 2 = 1
2 cos = −1
1
cos = − 2

2 4
= 3 , 3

4 4 tan + 2 = 2 tan
2 tan = −2
tan = −1
< 0 ,
tan = 1
= 135° , 315

5 2 2 − 5 sin + 3 = 0

(2 sin − 3)(sin − 1) = 0

2 sin − 3 = 0

2sin = 3 sin = 1

sin = 3 ∴ = 1
∴ =
2
2
Ignore

6 5 2 − 2(2 2 − 1) = 3

5 2 − 4 2 + 2 = 3

2 − 1 = 0

(cos − 1)(cos + 1) = 0

cos = 1 cos = −1

= 0 , 2 = ,

∴ = , 0 , 2

7 5 tan − cot 2 + 5 = 0
1

5 tan − tan 2 + 5 = 0
(1 − 2 )

5 tan − 2 tan + 5 = 0
10 2 − (1 − 2 ) + 10 tan = 0

11 2 + 10 tan − 1 = 0

(11 tan − 1)(tan + 1) = 0

115

1
tan = 11 tan = −1


Basic angle:

= −1 1 = −1(1)
= 45o
11

= 5.19°

∴ = 5.19°, 185.19° ∴ = 135°, 315°

8 5 2 − 4 sin − 1 = 0

(5 sin + 1)(sin − 1) = 0

5 sin + 1 = 0 sin − 1 = 0

5 sin = −1 sin = 1

1 = 90°
sin = − 5
sin < 0 in Quadrant III,VI

= − 12°

∴ = 90°, 192°, 348°

CLONE STPM

1. Solve the following trigonometric equations for −180° ≤ ≤ 180°

a) sin = 1

5

b) cos = − √3

5

2. Find the solutions for cos ( + ) = sin ( + ) in [− , ].

6 3

3. By using t = tan , solve the following equation in the given interval

2

4 sin − 3 cos = 3 ; [− , )

4. Given that 4sin + 5cos = 0, find the value of tan .
Hence, solve the equation (1 − tan )(4sin + 5cos ) = 0 in the interval
0o ≤ ≤ 360o, giving your values of to the nearest 0.1o.

5. Find the solutions of 9sin2 + 10 sin cos − 2cos2 = 1 in the interval [0° , 360°).

116

ANSWER 1(a) 1
Section Quadrant I , II sin = 5
1.4 1
= −1 (5) = 11.54°
Clone
STPM

= 11.54°
= 180° − 11.54° = 168.46°

1(b) √3
cos = − 5

= −1 √3 = 69.73°
(5)

Quadrant II, III
= 180° − 69.73° = 110.27°

= −(180 − 69.73°)
= -110.27°

2
( + 6) = ( + 3)

cos 6 − sin 6 = sin cos 3 + cos sin 3

√3 1 1 √3
2 cos − 2 sin = 2 sin + 2 cos

sin = 0

= − , 0,

3 4 sin − 3 cos = 3

(2 ) (1 − 2)
4 1 + 2 − 3 1 + 2 = 3
8 − 3 + 3 2 = 3 + 3 2

8 = 6

3
= 4

3
∴ tan 2 = 4
3
2 = −1 (4)


2 = 36.87°

= 73.74°

117 4sin + 5cos = 0
4
4tan + 5 = 0
5 5

= − 4
(1 − tan )(4sin + 5cos ) = 0

1 − tan = 0 4sin + 5cos = 0
5

tan = 1 tan = − 4
= 450, 2250, 128. 70, 308. 70

9 2 + 10 cos − 2 2 = 1

9 2 + 10 cos − 2 2 = 2 + 2

8 2 + 10 cos − 3 2 = 0

(4 sin − cos )(2 sin + 3 cos ) = 0

4 sin − cos = 0 2 sin + 3 cos = 0

4 sin = cos 2 sin = −3 cos

1 3
tan = 4 tan = − 2
1 3
= −1 (4) = −1 (2)

= 14.04° = 56.31°

∴ = 14.04°, 123.69°, 194.04°, 303.69°

1.4(c) EQUATION Learning outcome:
Express a sin Ѳ ± b cos Ѳ in the form r sin (Ѳ ± α) and
INVOLVING THE r cos (Ѳ ± α)

FORMS
a sin Ѳ ± b cos Ѳ

Expressing sin + cos in the form sin( ± ) or cos ( ± )
c)

For r > 0 and 0 < < 2

sin + cos = sin( + )

sin − cos = sin( − )

cos + sin = cos( − )

− sin = cos ( + )

Where = √ 2 + 2 = tan−1 ( )



118

EXAMPLE 1
Solve 4 cos − 3 sin = 1 for 0 ≤ ≤ 360°

Solution:
Let 4 cos − 3 sin = cos ( + )

4 cos − 3 sin = cos cos − sin sin

Compare both sides
4 = cos …….. (1)
3 = sin ……….(2)

(1)2 + (2)2 42 + 32 = 2 Remember to take
(2)(2) ÷ (1) 25 = R2 only the numerical
values. Ignore the
R=5 sign
3 = tan

4

= 36.87°

Then 5cos ( + 36.87°)=1
cos( + 36.87) = 1

5

Basic angle : 78.46o
+ 36.87° = 78.46° 281.54°

Thus, = 41.59° 244.67°.

EXAMPLE 2
Find the minimum and maximum values of

a) 8 cos
b) 5 cos 2

Solution:
(a) −8 ≤ 8 cos ≤ 8
Minimum value = - 8
Maximum value = 8

(b) −5 ≤ 5 cos 2 ≤ 5
Minimum value = - 5
Maximum value = 5

119

EXAMPLE 3
Express sin 3 − cos 3 in the form sin (3 − ) with R > 0 and 0 ≤ ≤ 90°.
Solve the equation sin 3 − cos 3 = 1 for 0° ≤ ≤ 180°.

Solution:

sin 3 − cos 3 = R sin (3 − )

sin 3 − cos 3 = R sin 3 cos − sin cos 3

Compare both sides, 1 = cos ______(1)
1 = sin ______(2)

(1)2 + (2)2, 2 = 2( 2 + 2 )
2 = 2

= √2

(2) , tan = 1
(1)

= 45°

∴ sin 3 − cos 3 = √2 sin(3 − 45°)

Solve √2 sin (3 − 45°) = 1

sin(3 − 45°) = 1 Basic angle :45o

√2

So, 3 − 45° = 45°, 135°, 405°, 495°

3 = 90°, 180°, 450°, 540°

= 30°, 60°, 150°, 180

EXERCISE 1.13

1. By reducing the following trigonometric equations into cos( ∓ ) = or

sin( ± ) = , find the solutions in the given intervals.

(a) 4 sin − 3 cos = 2 [−360°, 360°]

(b) 24 sin + 7 cos = 5 [0°, 360°]

2. Each of the following expressions can be written in the form R cos (x − α) with −π < α < π.
In each case determine the values of R and α (in radians) correct to 3 decimal places

(a) 5 cos x + 12 sin x

(b) 6 cos x + 5 sin x

3. Solve the following equations for 0 < x < 2π
(a) cos x − 2 sin x = 1

(b) cos x + 2 sin x = 1

4. Express 7 sin 2 − 24 cos 2 = 15 in the form sin( − ) = , find the solution for the
interval 0° ≤ ≤ 360°.

120

5. Express the equation 4 sin 2 + 3 cos 2 in the form cos (2 − ).
(a) Hence solve the equation 4 sin 2 + 3 cos 2 = 1 for 0 ≤ ≤ 2 .
(b) Find the maximum and minimum values of 4 sin 2 + 3 cos 2 + 3.

ANSWER 1(a) 4 sin − 3 cos = 2
Section
1.4 Let sin( − ) = 4 sin − 3 cos

Exercise cos − cos sin = 4 sin − 3 cos
1.4(c)
cos = 4 − − − −(1)

sin = 3 − − − −(2)

(2) 3
(1) , tan = 4
3
= −1 (4) = 36.87°

2 2 + 2 2 = 32+42

2 = 25

= 5

5 sin( − 36.87°) = 2

2
sin( − 36.87°) = 5
2
= −1 (5) = 23.58°

∴ − 36.87° = −336.42°, −203.58°, 23.58°, 156.42°

= −166.71°, −299.55°, 60.45°, 193.29°

1(b) 24 sin + 7 cos = 5

Let sin( + ) = 24 sin + 7 cos

cos + cos sin = 24 sin + 7 cos

cos = 24 − − − (1)

sin = 7 − − − −(2)

(2) 7
(1) , tan = 24
7
= −1 (24) = 16.26°

2 2 + 2 2 = 242+72

2 = 625

= 25

25 sin( + 16.26°) = 5

sin( + 16.26°) = 1
5
1
= −1 (5) = 11.54°

∴ + 16.26° = 168.46°, 371.54°

= 152.2°, 355.28°

121

2(a) 5 cos + 12 sin

5 cos + 12 sin = sin ( + )

cos = 5 − − − − − (1)

sin = 12 − − − − − (2)

Squaring and adding (1) and (2)
2 = 52 + 122

= √169 = 13

(2)/(1):

sin 12
cos = 5

12
tan = 5
= 1.176

∴ = 13 = 1.176

2(b) 6 cos x + 5 sin x

6 cos x + 5 sin x = r sin ( + )

cos = 6 − − − − − (1)

sin = 5 − − − − − (2)

Squaring and adding (1) and (2)
2 = 62 + 52

= √61

(2)/(1)

sin 5
cos = 6

5
tan = 6
= 0.6947

∴ = √61 = 0.6947

3(a) cos − 2 sin = 1

cos − 2 sin = cos ( + )

= √12 + 22 = √5

= −1 2 = 1.1071
(1)

√5 cos ( + 1.1071) = 1
1

cos ( + 1.1071) =
√5

+ 1.1071 = 1.1071 , 5.176

= 4.069

122

3(b) cos + 2 sin = 1
4 cos + 2 sin = cos ( − )

5 = √12 + 22 = √5
Then,
= −1 2 = 1.1071
(1)

√5 cos ( − 1.1071) = 1

cos ( − 1.1071) = 1
√5

− 1.1071 = 1.1071 , 5.176

= 2.2142

7 sin 2 − 24 cos 2 = 15
sin (2 − ) = 7 sin 2 − 24 cos 2

cos = 7 − − − − − (1)

sin = 24 − − − − − (2)

2 = 72 + 242

= √625 = 25

24
tan = 7
24
= −1 (7) = 73.74°

25 sin (2 − 73.74°) = 15

3
sin (2 − 73.74°) = 5
3
= −1 (5) = 36.87°

2 − 73.74° = 36.87° , 143.13°, 396.81°, 503.13°

2 = 110.61°, 216.87°, 470.55°, 576.87°

= 55.31°, 108.44°, 235.28°, 288.41°

4 sin 2 + 3 cos 2 = cos (2 − )

= cos 2 cos + sin 2 sin

cos = 3 − − − − − (1)

sin = 4 − − − − − (2)

2 = 32 + 42

= √25 = 5
4

tan = 3
∴ 4 sin 2 + 3 cos 2 = 5 cos (2 − 0.9273)

5 cos (2 − 0.9273) = 1

123

Basic angle : 1.369
Since 0 ≤ ≤ 2

2 − 0.9273 = 1.369, 4.914, 7.652, 11.2
= 1.15, 2.92, 4.29, 6.06

b) 4 sin 2 + 3 cos 2 + 3 = 5 cos (2 − 0.9273) + 3
Since −5 5 cos (2 − 0.9273) ≤ 5
Therefore, The maximum value is 5 + 3 = 8
and the minimum value is – 5 + 3 = - 2

CLONE STPM

1. By first expanding cos ( + 45°), then express cos ( + 45°) − √2 sin in the
form cos( +∝), where R  0 and 0° < ∝< 90°. Give the value of R correct to 4 significant
figures and the value of ∝ correct to 2 decimal places.
Hence, solve the equation cos ( + 45°) − √2 sin = 2, for 0° < < 360°.

2. Express 24cos − 7sin in the form r cos( +  ) where r  0 and 0     .

2

Hence,

(a) State the minimum and maximum values of
1

30 + 24 − 7
(b) Solve the equation 24 cos − 7 sin = 0 , 0 ≤ ≤ 2 .

3. Given that 5 cos 2 − 8 sin 2 ≡ cos (2 + ), where > 0 and 0 < < , state the

2

value of R and find the value of α in radians to three decimal places.

4. Express 4 sin − 3 cos in the form sin ( − ) where r > 0 and 0 < < 90°. Hence,
find the maximum and the minimum value of 4 sin − 3 cos and the corresponding values
of in the interval 0° ≤ ≤ 360°.

5. If ( ) = 5 − 4 cos , express ( ) in the form sin ( − ), where R > 0 and 0≤
≤ 4 . Hence
(a) find the minimum and maximum values of ( ) ,

(b) solve ( ) = √41 for 0° ≤ ≤ 360°

2

124

STPM PAST YEAR
1. Express 12 cos − 5 sin in the form cos ( + ) where r > 0 and 0 < < .

2

Hence,
(a) state the minimum and maximum values of 12 cos − 5 sin for real values of ,
(b) solve the equation 12 cos − 5 sin = 0, 0 ≤ ≤ 2

[STPM 2015]

ANSWER 1 cos( + 45°) = cos cos 45° − sin sin 45°
Section
1.4 = √2 cos − √2 sin
2 2
Clone
STPM cos( + 45°) − √2 sin = √2 cos − √2 sin − √2 sin
2 2

= √2 cos − 3√2 sin
2 2

Let √2 cos − 3√2 sin  cos( + )

2 2

 cos cos − sin sin

cos = √2 --------------①

2

sin = 3√2 ---------------②
2

①2 + ②2 : 2 = ( √2 ) 2 + ( 3√2 ) 2

22

R = √5 = 2.236

② 3√2
①
: tan = 2 =3
√2

2

= 71.57°

ii)cos( + 45°) − √2 sin = 2.236 cos( + 71.57° )

cos( + 45°) − √2 sin = 2

2.236 cos( + 71.57° ) = 2

cos ( + 71.57° ) = 500
559

Basic angle = 26.56°

( + 71.57° ) = 333.44° , 386.56°

= 261.87° , 314.99°

125

2 24cos − 7sin = r cos cos − r sin sin

r cos = 24

r sin = 7 B1 together
Let r=25 M1

 =0.284 M1

24cos − 7sin = 25cos( + 0.284) A1

a) M1

−25  25cos( + 0.284)  25

5  30 + 24cos − 7sin  55
1 1  1
5 30 + 24cos − 7sin 55

11

Maximum value = , Minimum value =

5 55

b) B1

24 cos − 7 sin = 0 M1
A1
25cos ( + 0.284) = 0
cos ( + 0.284) = 0

0    2
0.284    2 + 0.284

 + 0.284 =  or 3

22

 = 1.287or 4.428

3 5 cos 2 − 8 sin 2 ≡ cos (2 + )

cos = 5 − − − (1)

sin = 8 − − − (2)
(1)2 + (2)2 : 2 = 52 + 82

= √89

(2) / (1) :

sin 8
cos = 5

8
tan = 5
= 1.012

126 Let ( ) = 4 sin − 3 cos
4
4 sin − 3 cos = sin ( − )
5
cos = 4 − − − (1)
ANSWER
sin = 3 − − − (2)
2 = 42 + 32
(1)2 + (2)2 :

= 5

(2) / (1):

sin 3
cos = 4

3
tan = 4
= 36.9°

∴ 4 sin − 3 cos = 5 sin ( − 36.9°)

Maximum value = 5 when − 36.9° = 90°

≫ = 126.9°

Minimum value = - 5 when − 36.9° = 270°

≫ = 306.9°

5 sin − 4 cos = sin ( − )

cos = 5 − − − (1)

sin = 4 − − − (2)

(1)2 + (2)2: : 2 = 52 + 42

(2) / (1) : = √41
tan = 4

5

= 38.66°

5 sin − 4 cos = √41 sin ( − 38.66°)

a) Minimum value of ( ) is −√41.
Maximum value of ( ) √41.

b) ( ) = √41

2

√41 sin ( − 38.66°) = √41
√2

1
sin( − 38.66°) =

√2
− 38.66° = 45°, 135°

= 78.66°, 173.66°

127 1a 12 cos − 5 sin = cos ( + )

Section 12 cos − 5 sin = cos cos − sin sin
1.4
By equating the corresponding elements
STPM
Past Year cos = 12 sin = 5

→ 2 2 = 122 2 2 = 52

sin 52
cos = 122

5
tan = 12
= 0.3948

2 = 52 + 122

2 = 132

∴ = 13

∴ 12 cos − 5 sin = 13 cos ( + 0.3948)

Minimum value occurs when cos ( + 0.3948) = −1

∴ − 13

Maximum value occurs when cos ( + 0.3948) = 1

∴ 13

1b 12 cos − 5 sin = 0
13 cos ( + 0.3948) = 0

cos ( + 0.3948) = 0
3

θ + 0.3948 = 2 , 2
= 1.1760 , 4.3176

= 1.18 , 4.32 (3 )

1.4(d) Learning outcome:
TRIGONOMETRY Sketch the graph of the trigonometric functions and solve the inequality
INEQUALITIES

A trigonometry inequality is an inequality in standard form: R(x) > 0 (or < 0) that contains one or a
few trigonometry functions of the variable arc x. Solving the inequality R(x) means finding all the
values of the variable arc x whose trigonometry functions make the inequality R(x) true. All these
values of x constitute the solution set of the trigonometry inequality R(x). Solution sets of trigonometry
inequalities are expressed in intervals.

There are 4 main common types of basic trigonometry inequalities: sin x < a (or > a) cos x < a (or > a)
a is a given number tan x < a (or > a) cot x < a (or > a) Solving basic trigonometry inequalities
proceeds by using trigonometry conversion tables (or calculators), then by considering the various
positions of the variable arc x that rotates on the trigonometry circle.

EXAMPLE 1

128

1. Find the set of values of x which satisfy the inequality sin < √3 cos , for 0 ≤ ≤ 2 .

Solution:

sin < √3 cos , 0 ≤ ≤ 2
0 ≤ ≤
(i) when , cos > 0 , tan ≥0
2
sin
cos < √3

tan < √3


0 ≤ < 3

(ii) when < < , cos < 0 , tan < 0

sin 2

cos > √3

tan > √3
in 2
no values of ≤ ≤

(iii) when < < 3 , cos < 0 , tan > 0

2

sin
cos > √3

tan > √3
4 3
3 < < 2

129

(iv) when 3 < ≤ 2 , cos > 0 , tan ≤ 0

sin 2

cos < √3

tan < √3

3
2 < ≤ 2

Combining the above results, the set of values of x satisfying sin < √3 cos , is 4 < ≤ 2

3

EXAMPLE 2

Solve sin < 1 .
2

Solution:

1
sin = 2

5
= 6 , 6
5
∴ (0 , 6) ∪ ( 6 , 2 )

EXERCISE 1.14

1. Solve tan ≥ √3
2. Solve 2 cos ≤ 1 in the interval 0 ≤ ≤ 2 .
3. Solve the trigonometric inequality cos 2 > 3 sin + 2 for − < ≤ .
4. Sketch the graph of = sin 2 in the range 0 ≤ ≤ . Hence, solve the inequality

|sin 2 | < 21, where 0 ≤ ≤ 2 .

130

5. Sketch the graph of = 12 cos − 5 sin for 0 ≤ ≤ 2 , and determine −5 ≤
12 cos − 5 sin ≤ 0.

ANSWER 1
Section
1.4

Exercise
1.14

tan = √3
4

= 3 , 3
For tan ≥ √3

4
∴ [3 , 2) ∪ [ 3 , 2 )

2

2 cos ≤ 1

consider 2 cos = 1
1

cos = 2
5

= 3 , 3
For 2 cos ≤ 1 ,

5
[3 , 3 ]

131 cos 2 > 3 sin + 2
3
Using cos 2 ≡ 1 − 2 2
4
We have 1 − 2 2 > 3 sin + 2

2 2 + 3 sin + 1 < 0

(2 sin + 1)(sin + 1) < 0

∴ 2 sin + 1 < 0 sin + 1 > 0

2 sin + 1 > 0 sin + 1 < 0

1 sin > −1
sin < − 2 sin < −1

1
sin > − 2

Not possible

1
∴ −1 < sin < − 2
= − 1 = − − 5
When ,
2 6 6
= −
When = −1 ,
2
5
Hence, − 6 < < − 6

The solution to the inequality cos 2 > 3 sin + 2 is

5
{ : − 6 < < − 6 }

0 ≤ ≤ 2 → 0 ≤ 2 ≤ 4

1
sin 2 = 2 2 ≥ 0
|sin 2 | = 1 { 1 2 < 0
2
− sin 2 = 2
1 5 13 17
sin 2 = 2 , 2 = 6 , 6 , 6 , 6
11 19 21
= 12 , 12 , 12 , 12
=1 = −1
For − sin 2 → sin 2
2 2
7 11 19 21
2 = 6 , 6 , 6 , 6

7 11 19 21
= 12 , 12 , 12 , 12

132
5

−5 = 12 cos − 5 sin

−5 = 13 cos ( + 0.395)

cos ( + 0.395) = 5
− 13
0 ≤ ≤ 2

55
− 13 ≤ ≤ 2 − 13
+ 0.395 = 1.966 4.318

= 1.571 3.923

From the graph, the range satisfying
−5 ≤ 12 cos − 5 sin ≤ 0

are 1.176 ≤ ≤ 1.571 3.923 ≤ ≤ 4.317

CLONE STPM

1. Sketch the graphs of y = cos, y = sin 2 for  − ,  on the same coordinate axes. Hence
solve the inequality cos  sin 2 .

2. Sketch the graph of = |2 cos 2 | in the range of 0 ≤ ≤ . Hence, solve the inequality

|cos 2 |  1 , where 0 ≤ ≤ .
2

3. Sketch the graph of y = 24cos − 7sin for 0    2 and determine the range of  in

this interval satisfying the inequality −5  24cos − 7sin  0 .

133 1

ANSWER
Section
1.4

Clone
STPM

cos −2 sin cos = 0

cos (1 − 2 sin ) = 0

cos = 0 , = − 2 , 2

1 5
sin = 2 = 6 , 6

5
∴ ∈ { : − ≤ ≤ , 6 ≤ ≤ 2 , 6 ≤ ≤ }

2

1
|cos 2 | = 2
2 = , − , + , 2 −
333 3
2 5
= 6 , 3 , 3 , 6

Solution set of is

or < 2 or 5 < < }
{ : 0 < < 6 3 <3 6

134
3

24 cos − 7 sin = −5
25 cos( + 0.284) = −5
cos ( + 0.284) = −0.2

+ 0.284 = 1.7722 4.510

= 1.4882 4.226
∴ { ∶ 1.287 ≤ ≤ 1.488 or 4.226 ≤ ≤ 4.428}

STPM PAST YEAR

1. Sketch the graph of 12 cos − 5 sin for 0 ≤ ≤ 2 and determine the range of values
of in this interval satisfying the inequality −5 ≤ 12 cos − 5 sin ≤ 0. [STPM 2015]

2. Sketch the graph of = sin 2 in the range 0 ≤ ≤ . Hence, solve the inequality

|sin 2 | < 21, where 0 ≤ ≤ . [STPM2013]

ANSWER 1 12 cos − 5 sin = cos ( + )
12 cos − 5 sin = cos cos − sin sin
Section
1.4 By equating the corresponding elements

STPM cos = 12 sin = 5
Past
Year → 2 2 = 122 2 2 = 52

sin 52
cos = 122

5
tan = 12
= 0.3948

135

2 = 52 + 122
2 = 132
∴ = 13

∴ 12 cos − 5 sin = 13 cos ( + 0.3948)

Minimum value occurs when cos ( + 0.3948) = −1
∴ Minimum value is − 13

Maximum value occurs when cos ( + 0.3948) = 1
∴ Maximum value is 13

12 cos − 5 sin = 0
13 cos ( + 0.3948) = 0

cos ( + 0.3948) = 0
3

θ + 0.3948 = 2 , 2
= 1.1760 , 4.3176
= 1.18 , 4.32 (3 )

−5 ≤ 12 cos − 5 sin ≤ 0
−5 ≤ 13 cos ( + 0.3948) ≤ 0

When 13 cos ( + 0.3948) = −5

cos ( + 0.3948) = 5
− 13
+ 0.3948 = 1.9656 , 4.3176

= 1.5708 , 3.9228

= 1.57 , 3.92 (3 )

The range of values of
= {1.18 ≤ ≤ 1.57 , 3.92 ≤ ≤ 4.32}

136
2

|sin 2 | = 1
2
5 7 11
2 = 6 , 6 , 6 , 6
5 7 11
= 12 , 12 , 12 , 12

For |sin 2 | < 1 , 0 ≤ ≤ , the solution set is

2

5 7 11
[0 , 12] ∪ (12 , 12) ∪ [ 12 , ]

137

2.1 SEQUENCES Learning Outcome:
a) Use an explicit formula and a recursive formula for a sequence
Sequences b) Find the limit of a convergent sequence

Learning outcome: Use an explicit formula and a recursive formula for a sequence.

A sequence is a list of numbers, stated in a particular order, such that each number can be derived
from the previous number according to a certain rule.

Each number of a sequence is called a term.
1, 3, 5, 7, …

This is a sequence of odd numbers. If 1, 2, 3, … represent the terms, then

1 = 1,
2 = 3,
3 = 5,

⋮
= 2 − 1.

The sequence 1, 3, 5, 7, … can be generated using the formula = 2 − 1, where = 1, 2, 3, … .
Since can be expressed in terms of , we say that = 2 − 1, an explicit formula.

The above sequence of odd numbers can also be generated by a recursive formula = −1 + 2
with 1 = 1. To see this, observe the description below:

1 = 1,
2 = 3 = 1 + 2 = 1 + 2,
3 = 5 = 1 + 2 + 2 = 2 + 2,

⋮
= −1 + 2.

As we can see that the next term, can be generated by using the previous term, −1 plus 2.
That is why this type of formula is called recursive formula.

138

EXAMPLE 1
Write down the first six terms of each of these sequences.
+1 = 2 + 1, 1 = 4

Solution:
Given, 1 = 4,
then 2 = 2 1 + 1 = 2(4) + 1 = 9,

3 = 2 2 + 1 = 2(9) + 1 = 19,
4 = 2 3 + 1 = 2(19) + 1 = 39,
5 = 2 4 + 1 = 2(39) + 1 = 79,
6 = 2 5 + 1 = 2(79) + 1 = 159.
The first six terms of the sequence are 4, 9, 19, 39, 79, 159.

EXAMPLE 2
Write down the first six terms of each of these sequences.
+1 = 2 − 5, 1 = 2

Solution:
Given, 1 = 2,
then 2 = ( 1)2 − 5 = (2)2 − 5 = −1

3 = ( 2)2 − 5 = (−1)2 − 5 = −4
4 = ( 3)2 − 5 = (−4)2 − 5 = 11
5 = ( 4)2 − 5 = (11)2 − 5 = 116
6 = ( 5)2 − 5 = (116)2 − 5 = 13451
The first six terms of the sequence are 2, -1, -4, 11, 116, 13451.

EXAMPLE 3

Write down the first six terms of each of these sequences.

1 1 = 2
+1 = 1 − ,

Solution:

Given, 1 = 2, 1 1 1
1 2 2
then 2 = 1 − = 1 − =

11
3 = 1 − 2 = 1 − 1 = −1
2
11
4 = 1 − 3 = 1 − (−1) = 2
1 11
5 = 1 − 4 = 1 − 2 = 2
11
6 = 1 − 5 = 1 − 1 = −1
2
The first six terms of the sequence are 2, 1 , −1, 2, 1 , −1.
22

139

WORKSHEET 2.1

Write down the first six terms of each of these sequences.

a) +1 = 1, 1 = 5 b) +1 = 1 − 1, 1 = 3



Solution: Solution:

1 = 5, 1 = 3,
1 2

2 = 5, 2 = 3,
3 = 5, 1

1 3 = − 2,
4 = 5, 4 = 3,
5 = 5,
2
1 5 = 3,
6 = 5.
1
111 6 = − 2.
5, 5 , 5, 5 , 5, 5
2121
3, 3 , − 2 , 3, 3 , − 2

c) +1 = + −1, 1 = 2, 2 = 3 d) +1 = 2 −1 − 3 , 1 = 4, 2 = 3

Solution: Solution:
1 = 2, 1 = 4,
2 = 3, 2 = 3,
3 = 5, 3 = −1,
4 = 8, 4 = 9,
5 = 13, 5 = −29,
6 = 21. 6 = 105.

2, 3, 5, 8, 13, 21 4, 3, −1, 9, −29, 105

e) +1 = 2 + 1, 1 = 4 f) +1 = , 1 = 6, 2 = 3

Solution: −1
1 = 4,
2 = 9, Solution:
3 = 19,
4 = 39, 1 = 6,
5 = 79, 2 = 3,
6 = 159.
1
4, 9, 19, 39, 79, 159 3 = 2,

1
4 = 6,

1
5 = 3,
6 = 2.

111
6, 3, 2 , 6 , 3 , 2

140

Learning outcome: Find the limit of a convergent sequence.

The Properties of Limits

Law Example

lim = lim 7 = 7

→ →4

lim = lim 3 = 23

→ →2

=8

lim ( ) = lim ( ) lim 4(3 2 − 50) = 4 lim(3 2 − 50)
→ → →5 →5
= 4[3(5)2 − 50]

= 100

lim[ ( ) ± ( )] = lim ( ) ± lim ( ) lim(5 + 3 2) = lim 5 + lim 3 2
→ → → →2 →2 →2
= 5(2) + 3(2)2

= 22

lim[ ( ) ∙ ( )] = lim ( ) ∙ lim ( ) lim [ 3(1 + 2)] = lim 3 ∙ lim (1 + 2)
→ → → →−1 →−1 →−1
= (−1)3 ∙ [1 + (−1)2]

= −2

( ) lim ( ) 2 + 3 lim( 2 + 3)
( ) ( ), − 1 →3
lim = → lim = − 1) ,
lim(
→ lim →3 →3

→ (3)2 + 3

where lim ( ) ≠ 0 = (3) − 1
→

=6

A sequence, is a convergent sequence if lim = , where is a constant.

→∞

141

Determine whether the following sequence converges or diverges. For those which are convergent,
determine the limiting value to which they are tending.

EXAMPLE 1 ≥ 1
2 + 1

= 3 − 2 ,

Solution:

2 + 1
lim = lim (3 2)
−
→∞ →∞

= lim 2 + 1
(3 − 2 )
→∞


= lim (2 + 1 )
(3 − 2 )
→∞

lim

→∞

2+0
=3−0

2
=3

Since 1 → 0 and 2 → 0 as → ∞. converges to 23.


EXAMPLE 2
5 − 2

= 2 + 1

Solution:

5 − 2
lim = lim (2 1)
+
→∞ →∞

= lim 5 − 2
( + 1 )
→∞
2

= 5−0
2−0

5
=2

Since 2 → 0 and 1 → 0 as → ∞. converges to 52.


142

EXAMPLE 3
5 2 + 3

= 2 − 2

Solution: TIPS

lim = lim ( 5 2 + 3 ) If the highest power of n in the
numerator and denominator are the
→∞ →∞ 2 − 2 same, divide each term throughout by
this highest power terms as shown.
(5 2222−+ 32 2 2 )
= lim

→∞

= lim (15−+ 32 2)

→∞

5+0
=1−0

=5

Since 3 → 0 and 2 → 0 as → ∞. converges to 5.
2

EXAMPLE 4
1 + 3 × 2

= 5 − 7 × 2

Solution:

1 + 3 × 2
lim = lim (5 2 )
− 7 ×
→∞ →∞

1 + 3 × 2
(25 − 7
= lim 2 )
2 × 2
→∞

2

= lim 1 + 3
(25 − )
→∞
2 7

0+3
=0−7
3
= −7

Since 1 → 0 and 5 → 0 as → ∞. converges to − 3.
2 2
7

143

WORKSHEET 2.2

Determine whether the following sequence converges or diverges. For those which are convergent,
determine the limiting value to which they are tending.

a) = 2+1 b) = 3 + 1
( +1)
( +2)(2 +3)

Solution: Solution:

2 + 1 lim = lim (3 + 1 1))

lim = lim (( + 2)(2 + 3)) →∞ →∞ ( +

→∞ →∞

2 + 1 =3

= lim (2 2 + 7 + 6)

→∞ 1 → 0 → ∞.
( +1)
1 + 1 Since as
+ + 2 6 2)
= lim ( 7 converges to 3.
2
→∞

1
=2
1 7 6
Since 2 → 0 , → 0 and 2 → 0 as → ∞.

converges to 1
2

c) = (−1) +1 d) = (−1) +1
2 +1

Solution: Solution:

lim = lim ( (−1) +1 ) lim = lim ( (−1) +1 )

→∞ →∞ 2 →∞ →∞ + 1

=0 =0

Since (−1) +1 → −1 for even + 1 Since (−1) +1 → −1 for even + 1
and (−1) +1 → 1 for odd + 1
and (−1) +1 → 1 for odd + 1 but 1 → 0 as → ∞.

but 1 → 0 as → ∞.
2
converges to 0.
converges to 0.

144

e) = 2−1 f) = 4 + (−0.1)
2+1 Solution:

Solution:

= 2 − 1 lim = lim (4 + (−0.1) )
lim lim ( 2 1)
+ →∞ →∞
→∞ →∞

1 = lim (4) + lim (−0.1)
→∞ →∞
1 − 2
= lim ( + 1 ) =4

→∞ 1

2

=1 Since (−0.1) → 0 as → ∞.

Since 1 → 0 as → ∞. converges to 4.
2

converges to 1.

g) = 4 − 3 h) = (−0.1)
Solution: Solution:

lim = lim (4 − 3) lim = lim (−0.1)

→∞ →∞ →∞ →∞

= lim (4 ) − lim (3) =0
→∞ →∞ since (−0.1) → 0 as → ∞.

=∞ converges to 0.
since 4 → 0 as → ∞.

diverges

145

EXERCISE 2.1

STPM PAST YEAR

STPM MT 2016 (Section A)

Q1. A convergent sequence is defined by +1 = 1 + 1 and 1 = 1 .
3
(1)
a) Write down each of the terms 2 , 3 and 4 in the form ∑ −=10 , and show that an
3

explicit formula for is given by = 3 [1 − (1) ]. [5 marks]

2 3

b) Determine the limit of as r tends to infinity. [2 marks]

STPM MT 2017 (Section A)

Q2. The sum of the first n terms of a sequence 1 , 2 , 3, … is given by = 3 2 − . Find an

explicit and a recursive formulae for . [5 marks]

ANSWER Q1(a) 1
Section +1 = 1 + 3
2.1 1 = 1
11
Exercise 2 = 1 + 3 1 = 1 + 3
2.1 1 1 12
3 = 1 + 3 2 = 1 + 3 + (3)
STPM 1 1 12 13
Past Year

4 = 1 + 3 3 = 1 + 3 + (3) + (3)
1 12 13 1 −1
∴ = 1 + 3 + (3) + (3) + ⋯ + (3)
[1 − (31) ]
1

= 1
3
1 −

3 1
= 2 [1 − (3) ] (shown)

b) 3 1
lim = lim [1 − (3) ]
2
→∞ →∞

3
= 2 (1 − 0)

3
=2