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TRIGONOMETRIC FUNCTION
ADDITIONAL MATHEMATICSFORM 5
MODULE 9
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CONTENTS
CONTENTS PAGES
9.0 CONCEPT MAP 2
9.1 EXERCISE 1 3
9.2 PAST YEAR SPM QUESTION 5
9.3 ASSESMENT 6
ANSWER 7
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9.0 CONCEPT MAP
TRIGONOMETRICFUNCTION
POSITIVE ANGLES AND NEGATIVE ANGLES
THE SIXTRIGONOMETRICFUNCTION
Special angles 30o,___ and 60o
Tan Ө = sin Ө
sec Ө = 1cos Ө
cosec Ө= 1sin Ө
cot Ө = cosec Ө
Basic Identities sin2 Ө + cos2 Ө 1
____+ tan2 Ө sec2 Ө
1 + cot2 Ө______
Complementary Angle
sin cos(90 )o
cos sin(90 )o
tan cot(90 )o
sec cos (90 )oec
cos sec(90 )oec
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9.1 Exercise 1
1. Given that sin Ө = 8
17and cos Ө = -
15
17. Find the value of :-
a) sec Ө = 1
cos
=1
1517
= -17
15
b) cot Ө
2. Given that sin 36o = 0.5878, cos 36o = 0.8090 and tan36o = 0.7265 . Find the value of
a) cos 54 = sin (90 o – 54 o)= sin 36= 0.5878
b) cosec 54 o
3. Solve the trigonometric equation for 0 o x 360 o
a) cos 2x = - 0.7660
basic angle = 40 o
Given 0 o x 360 o
So 0 o 2x 720 o
Cos is negative at quarter II and III2x = 180 o - 40 o, 180 o + 40 o,
(180 o - 40 o ) + 360 o,(180 o + 40 o ) + 360 o,
= 140 , 220, 500, 580x = 70 o, 110 o, 250 o, 290 o
e) 2 sin x = tan x
b) tan x = cot 60 o f) 3 sin x cos x = 2 sin2 x
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c) sin ( x – 30 o) = -1
2
g) 2 sin2 x + cos x – 2 = 0
d) sin x + 2 sin x cos x = 0sin x ( 1 + 2cos x) = 0sin x = 0
x = 0 o, 180 o, 360 o
or1 + 2cos x = 0
2cos x = -1
cos x = -1
2x = 120 o, 240 o
so x = 0 o, 120 o, 180 o, 240 o
h) tan2 x + sec x – 5 = 0
4. Prove that identity
a) 2 2 2 2cot tan cos secx x ec x x b)
21 2sincos sin
sin cos
xx x
x x
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9.2 PAST YEAR SPM QUESTION
2003, Paper 1
1. Given that tan ,0 , 90o ot , express in terms of t :
) cot
)sin(90 )o
a
b
[3 marks]
2. Solve the equation 26sec 13tan 0,0 360o oA A A [ 4 marks]
2004, Paper 1
3. Solve the equation 2 2cos sin sinx x for 0 360o ox [ 4 marks]
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9.3 ASSESSMENT
1. Given that3
cos5
x and 0 180o ox , find sec x + cosec x .
Solution:
2. Given that sin Ө = k and Ө is acute angle, express in term of k:a. tan Өb. cosec Ө
Solution:
3. Solve the equation 2 25sin 2 3 0,0 360o oA cos A A
Solution:
4. Prove that 2 2 2 2tan sin tan sinA A A A
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ANSWER
Exercise 1
1. b) -15
82. b) 1.236
3. b) 30 ,210o ox c) 30 ,90 , 210 , 270o o o ox
e) 0 ,60 ,180 ,300 ,360o o o o ox f) 0 ,56 19 ,180 , 236 19 ,360o o o o ox
g) 60 ,90 , 270 ,300o o o ox h) 60 ,109 28 , 250 32 ,300o o o ox
Past Year SPM Question
1. a)1
tb)
2
1
1t
2. 33.69 ,56.31 , 213.69 ,236.31o o o oA
3. 30 ,150 , 270o o ox
Assesment
1.35
12
2. a)2
1
1 kb)
1
k
3. 35.26 ,144.74 ,180 ,215.26 ,324.74o o o o oA
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TRIGONOMETRICFUNCTIONS
ADDITIONAL MATHEMATICSFORM 5
MODULE 10
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Contents
10. 1 ADDITION FORMULAE AND DOUBLE-ANGLE FORMULAE
10.1.a Understand and use addition formulae and double-angle formulae
10.1.b Solving Trigonometric Equations by using Addition Formulae and
Double- angle Formulae
10.2 GRAPHS OF FUNCTIONS OF SINE, COSINE AND TANGENT
10.2.1 Sketch the graph of functions sine
10.2.2 Sketch the graph of functions cosine
10.2.3 Solution of equations involving the graphs
10.3 SPM QUESTIONS
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10. TRIGONOMETRIC FUNCTIONS
10. 1 ADDITION FORMULAE AND DOUBLE-ANGLE FORMULAE
10.1.a Understand and use addition formulae and double-angle formulae
List down the Addition Formulae and Double-Angle FormulaeADDITION FORMULAE DOUBLE-ANGLE FORMULAE
Example
Given that sin A =5
4and cos B =
13
12where A and B are angles in the second
and fourth quadrants respectively. Calculate the value of tan ( B – A )
Solution
A in the second quadrant B in the fourth quadrant
( 12 , -5 )
-5
12B
13
( -3 , 4 )
45
A
- 3
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sin A =5
4
tan A =3
4
http://mathsm
cos B =13
12
10
tan B =12
5
ozac.blogspot.com
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tan ( B – A ) = BA
AB
tantan1
tantan
=
3
4
12
51
3
4
12
5
=
9
1412
11
=56
33
EXERCISE 1
1. Given that sin α =17
8, 90o < α < 270o and sin β =
13
12 , 90o < β < 270o.
Calculate the value of( a ) sin (α + β ) ( b ) cos ( β – α )
………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
2. Given that tan x =3
4, sin y =
5
4 where x and y are angles in the same quadrant.
Without using the calculator, calculate the value of( a ) sin ( x – y ) ( b ) cos ( x + y ) ( c ) tan ( y + 45o ) ( d ) tan ( x + y )
………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
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………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
3. Given that tan θ = p and θ is an acute angle. Express each of the following in terms of p
( a ) sin 2 θ ( b ) cos 2 θ ( c ) tan 2 θ
………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
10.1.b Solving Trigonometric Equations by using Addition Formulae and Double-angle Formulae
Example
Find all the angles between 0o and 360o which satisfy( a ) 2 sin 2 θ = sin θ( b ) 2cos2 θ + 3 = 8 sin θ
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Solution
( a ) 2 sin 2 θ = sin θ 2 ( 2sin θ cos θ ) = sin θ 4sin θ cos θ - sin θ = 0 sin θ (4cos θ – 1 ) = 0 sin θ = 0 or 4cos θ-1=0
θ = 0 o , 180 o , 360 o cosθ = 4
1
θ = 75 o 31’ , 284o29’
So that , θ = 0 o , 75 o 31’ , 180 o , 284o29’, 360 o
( b ) 2cos2 θ + 3 = 8 sin θ 2 ( 1- 2sin2θ ) + 3 = 8 sin θ 2 - 4sin2θ + 3 = 8 sin θ 4sin2θ + 8 sin θ – 5 = 0 ( 2 sin θ – 1 )( 2 sin θ + 5 ) = 0 2 sin θ – 1 = 0 or 2 sin θ + 5 = 0
sin θ =2
1sin θ =
2
5
θ = 30 o , 150 o θ is undefinedSo that θ = 30 o , 150 o
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EXERCISE 2
1. Find all the angles between 0o and 360o which satisfy( a ) 3sin 2A = 4sin A( b ) 5sin2A = 5 – sin 2A
………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
2. Find all the angles between 0o and 360o which satisfy
( a ) 3cos2α – 5 = 8 cos α( b ) tan 2α tan α = 1
………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
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10.2 GRAPHS OF FUNCTIONS OF SINE, COSINE AND TANGENT
10.2.1 Sketch the graph of each of the following functions
1 ) y = 2sin θ
2 ) y =2
1sin θ
3 ) y = - sin θ
4 ) y = sin 2θ5 ) y = sin
2
1θ
θO
y
180o
360o
270o
90o
y = sin θ
1
-1
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6 ) y = sin θ + 1 7 ) y = sin θ – 1
8 ) y = | sin θ | 9) y = 2sin 2θ
10 ) y = | sin 2θ | 11 ) y = | sin θ | + 1
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10.2.2 Sketch the graph of each of the following functions
1 ) y = 2cos θ
2 ) y =2
1cos θ
3 ) y = - cos θ
4 ) y = cos 2θ5 ) y = cos
2
1θ
θO
y
180o
360o270
o
90o
y = cos θ
1
-1
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6 ) y = cos θ + 1 7 ) y = cos θ – 1
8 ) y = | cos θ| 9) y = 2cos 2θ
10 ) y = | cos 2θ | 11 ) y = | cos θ | + 1
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10.2.3 SOLUTION OF EQUATIONS INVOLVING THE GRAPHS
Determining number of solutions by sketching the graphs
Example
Sketch the graph of the function y = cos 2θ for 0 < θ < 2π. Hence find the number ofsolutions of each of the equation
cos 2θ + 0.5 = 0
Solution
cos 2θ = - 0.5
The number of solutions = 4
y
θO π 2π
2
2
3
1
- 1
y = cos 2θ
y = - 0.5
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Exercise 3
1. Sketch the graphs of the functions y = - 2sin 2x and y =2
xfor 0 < θ < 2π on the
same axes. Hence, state the number of solutions of the equation 4πsin 2x + x = 0.
2. Sketch the graphs of the functions y = 3cos2x and 2y + x = 0 for 0 < θ < 2π on the same axes. Hence, state the number of solutions of the equation 6 cos2x + x = 0.
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10.3 SPM QUESTIONS
SPM 2003 PAPER 2 NO 8
( a ) Prove that tan θ + cot θ = 2 cosec 2θ [ 4 marks ]
( b ) ( i ) Sketch the graph y= 2 cos x2
3for 0 ≤ x ≤ 2π
( ii ) Find the equation of a suitable straight line for solving the equation
cos x2
3= 1
4
3x
Hence, using the same axes, sketch the straight line and state the number of
solutions to the equation cos x2
3= 1
4
3x
for 0 ≤ x ≤ 2π [ 6 marks ]
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SPM 2004 PAPER 2 NO 3
( a ) Sketch the graph of y = cos2x for 0o < x < 180o. [ 3 marks ]
( b ) Hence, by drawing a suitable straight line on the same axes, find the number of
solutions satisfying the equation 2 sin2 x = 2 -180
xfor 0o < x < 180o [ 3 marks ]
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SPM 2005 PAPER 2 NO 5
( a ) Prove that cosec2x – 2 sin2x – cot2x = cos 2x [ 2 marks ]
( b ) ( i ) Sketch the graph of y = cos 2x for 0 ≤ x ≤ 2π.( ii ) Hence, using the same axes, draw a suitable straight line to find the number of
solutions to the equation 3( cosec2x – 2sin2x – cot2x ) = 1
xfor 0 ≤ x ≤ 2π.
State the number of solutions [ 6 marks ]
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10.4 ASSESSMENT
1. Given that sin θ = k such that θ is an acute angle, express cos 2θ in terms of k[ 2 marks ]
2. Find all values of x between 0o and 360o which satisfy the equation3 tan2y = cot y [ 3 marks ]
3. Sketch the graph of y = | sin 2x | for 0 ≤ x ≤ 2π. Determine the equation of a suitablestraight line to solve the equation x - 2π| sin 2x | = 0.Sketch the straight line and hence, state the number of solutions to the equationx - 2π| sin 2x | = 0 for 0 ≤ x ≤ 2π.
[ 5 marks ]
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ANSWER:Exercise 1.
1. a)221
140b) -
221
21
2. a) 6 b ) -25
7d ) -
7
24
3. a)1
22 c
cb )
1
12
2
c
cc)
21
2
c
c
Exercise 2.
1) a) 0 0 , 48011’, 1800, 3110 480 , 3600
b ) 680 12’,900 , 248012’, 2700
2) a) 1310 49’ , 228011’b ) 300 , 1500 , 2100 , 3300
10.2
θO
y
180o
360o
270o
90o
y = sin2 θ
1
-1
θO
y
180o
360o
270o
90o
y = -sin θ
1
-1
θO
y
180o
360o
270o
90o
y =2
1sin θ
2
1
2
1
θO
y
180o
360o
270o
90o
y = 2sin θ
2
-2
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O
y
180o
360o270
o90
o
y =| sin2 θ|
1
-1
θθO
y
180o
360o
270o
90o
y = 2sin2 θ
2
-2
θO
y
180o
360o
270o
90o
y = |sin θ|
1θ
y180
o360
o270o
90o
y = sin θ-1
-2
-1
y
180o
360o
270o
90o
y = sin θ+1
2
1θO
y
180o
360o
270o
90o
y = sin2
1θ
1
-1
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θO
y
180o
360o270
o
90o
y = cos2θ
1
-1
θO
y
180o
360o270
o
90o
y =- cos θ
1
-1
θO
y
180o
360o270
o
90o
y =2
1cos θ
2
1
θO
y
180o
360o
270o
90o
y = 2cos θ
2
-2
y = |sin θ|+1
90o
180o
270o
360o
2
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O
1
y =|cos2 θ|
180o
270o
360o
90o
θO
2
y =| 2cos2 θ|
180o
270o
360o
90o
θ
O
1
2y =| cos θ|
180o
270o
360o
90o
θ
θO 180
o360
o270o
90o
y = cos θ+1
1
2
θO
y
180o 360
o270o
90o
y = cos2
1θ
1
-1
θO 180
o360
o270o
90o
y = cos θ -1
-1
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Exercise 31.X = 0 , 0.53 , 0.95 , 1.58 , 1.902.Number of intersection = 4
SPM 2003
Number of intersection = 3
1
y =| cos θ|+ 1
90o
180o
270o
360o
y
θO π 2π
1
- 1
y = - 0.5
y = 3cos x2
3
y = 22
3x
3
3
5
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SPM 2004
(a ) (b )
Number of intersection = 2
SPM 2005
Number of intersection = 4
y
θO
4
2π
1
- 1
y = cos2x
3
5
θO
y
180o
90o
1
-1
y =180
x-1
y =3
1
3
x
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10.4 ASSESSMENT1. 1 -2k2
2. y = 300, 1500,2100,3300
3.
,Number of intersection
O
y
90o
y = |2sin2 θ|
2
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= 4
θ180
o
)(2 xy
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