ADDITIONAL MATHEMATICS FORM 5 MODULE 9

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ADDITIONAL MATHEMATICS FORM 5 MODULE 9 TRIGONOMETRIC FUNCTION

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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

6

ASSESMENT

ANSWER

7

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9.0

CONCEPT MAP

TRIGONOMETRIC FUNCTION

POSITIVE ANGLES AND NEGATIVE ANGLES

Tan Ө = sin Ө Special angles 30o, ___ and 60o

THE SIX TRIGONOMETRIC FUNCTION sec Ө =

1 cos Ө

Complementary Angle  sin   cos(90o   )  cos   sin(90o   )  tan   cot(90o   )  sec   cos ec(90o   )  cos ec  sec(90o   )

cosec Ө= 1 sin Ө cot Ө = cosec Ө

Basic Identities  sin2 Ө + cos2 Ө  1 

____+ tan2 Ө  sec2 Ө



1 + cot2 Ө  ______



9.1 Exercise 1 1. Given that sin Ө =

8 and cos Ө = 17

1 cos  1 = 15 17 17 =15

a) sec Ө =

-

15 . Find the value of :17 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

e) 2 sin x = tan x

basic angle = 40 o Given 0 o  x  360 o So 0 o  2x  720 o Cos is negative at quarter II and III 2x = 180 o - 40 o, 180 o + 40 o, (180 o - 40 o ) + 360 o, (180 o + 40 o ) + 360 o, = 140 , 220, 500, 580 x = 70 o, 110 o, 250 o, 290 o b) tan x = cot 60 o

f) 3 sin x cos x = 2 sin2 x



c) sin ( x – 30 o) = -

g) 2 sin2 x + cos x – 2 = 0

1 2

d) sin x + 2 sin x cos x = 0 sin x ( 1 + 2cos x) = 0 sin x = 0 x = 0 o, 180 o, 360 o or 1 + 2cos x = 0 2cos x = -1 1 cos x = 2 x = 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) cot 2 x  tan 2 x  cos ec 2 x  sec 2 x

b)

1  2sin 2 x  cos x  sin x sin x  cos x



9.2 PAST YEAR SPM QUESTION 2003, Paper 1 1. Given that tan   t , 0o ,    90o , express in terms of t : a ) cot  b)sin(90o   )

[3 marks]

2. Solve the equation 6sec2 A  13 tan A  0, 0o  A  360o

[ 4 marks]

2004, Paper 1 3. Solve the equation cos 2 x  sin 2  sin x for 0o  x  360o


[ 4 marks]


9.3 ASSESSMENT 1. Given that cos x 

3 and 0o  x  180o , find sec x + cosec x . 5

Solution:

2. Given that sin Ө = k and Ө is acute angle, express in term of k: a. tan Ө b. cosec Ө Solution:

3. Solve the equation 5sin 2 A  2cos 2 A  3  0, 0o  A  360o Solution:

4. Prove that tan 2 A sin 2 A  tan 2 A  sin 2 A



ANSWER Exercise 1

2.

15 8 b) 1.236

3.

b) x  30o , 210o

c)

x  30o ,90o , 210o , 270o

e) x  0o , 60o ,180o ,300o ,360o

f)

x  0o ,56o19,180o , 236o19,360o

g) x  60o ,90o , 270o ,300o

h)

x  60o ,109o 28, 250o 32,300o

1.

b) -

Past Year SPM Question 1.

a)

1 t

1

b)

t2 1

A  33.69o ,56.31o , 213.69o , 236.31o

2.

x  30o ,150o , 270o

3.

Assesment 1.

2.

35 12

a)

1 1 k 2

b)

1 k

3. A  35.26o ,144.74o ,180o , 215.26o ,324.74o



ADDITIONAL MATHEMATICS FORM 5

MODULE 10 TRIGONOMETRIC FUNCTIONS



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



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 Formulae ADDITION FORMULAE DOUBLE-ANGLE FORMULAE

Example 4 12 and cos B = where A and B are angles in the second 5 13 and fourth quadrants respectively. Calculate the value of tan ( B – A )

Given that sin A =

Solution A in the second quadrant

B in the fourth quadrant

( -3 , 4 ) B 5

12

4 A

-5

13 -3

( 12 , -5 )

4 5 4 tan A = 3

sin A =

12 13 5 tan B = 12

cos B =



tan B  tan A tan ( B – A ) = 1  tan A tan B

5  4    12   3  =   5  4  1     12   3  11 33 = 12 = 14 56 9

EXERCISE 1 8 12 , 90o < α < 270o and sin β =  , 90o < β < 270o. 17 13 Calculate the value of ( a ) sin (α + β ) ( b ) cos ( β – α )

1. Given that sin α =

……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… 4 4 , sin y =  where x and y are angles in the same quadrant. 3 5 Without using the calculator, calculate the value of ( a ) sin ( x – y ) ( b ) cos ( x + y ) ( c ) tan ( y + 45o ) ( d ) tan ( x + y )

2. Given that tan x =

……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………



……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………

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 Doubleangle 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



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

θ = 0 o , 180 o , 360 o

4cos θ-1=0 1 cosθ = 4 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 1 sin θ = 2 θ = 30 o , 150 o So that θ = 30 o , 150 o

2 sin θ + 5 = 0 5 sin θ =  2 θ is undefined



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 ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………



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

y = sin θ y 1 o

o

90

270 o

180

O

o

360

θ

-1

2) y=

1 sin θ 2

4 ) y = sin 2θ

3 ) y = - sin θ

5 ) y = sin

1 θ 2



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



10.2.2 Sketch the graph of each of the following functions 1 ) y = 2cos θ

y = cos θ y 1

O

o

90

o

180

o

270

o

360

θ

-1

2) y=

1 cos θ 2

4 ) y = cos 2θ

3 ) y = - cos θ

5 ) y = cos

1 θ 2



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



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 of solutions of each of the equation cos 2θ + 0.5 = 0 Solution

y y = cos 2θ

1

O

 2

π

3 2

θ

2π y = - 0.5

-1

cos 2θ = - 0.5 The number of solutions = 4



Exercise 3 x for 0 < θ < 2π on the 2 same axes. Hence, state the number of solutions of the equation 4πsin 2x + x = 0.

1. Sketch the graphs of the functions y = - 2sin 2x and y =

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.



10.3 SPM QUESTIONS SPM 2003 PAPER 2 NO 8 ( a ) Prove that tan θ + cot θ = 2 cosec 2θ

[ 4 marks ]

3 x for 0 ≤ x ≤ 2π 2 ( ii ) Find the equation of a suitable straight line for solving the equation 3 3 cos x = x 1 2 4 Hence, using the same axes, sketch the straight line and state the number of 3 3 solutions to the equation cos x = x  1 for 0 ≤ x ≤ 2π [ 6 marks ] 2 4

( b ) ( i ) Sketch the graph y= 2 cos



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 x solutions satisfying the equation 2 sin2 x = 2 for 0o < x < 180o [ 3 marks ] 180



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 x solutions to the equation 3( cosec2x – 2sin2x – cot2x ) =  1 for 0 ≤ x ≤ 2π.  State the number of solutions [ 6 marks ]



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 equation 3 tan2y = cot y [ 3 marks ]

3. Sketch the graph of y = | sin 2x | for 0 ≤ x ≤ 2π. Determine the equation of a suitable straight line to solve the equation x - 2π| sin 2x | = 0. Sketch the straight line and hence, state the number of solutions to the equation x - 2π| sin 2x | = 0 for 0 ≤ x ≤ 2π. [ 5 marks ]



ANSWER: Exercise 1. 140 21 1. a) b) 221 221 7 24 2. a) 6 b ) d)25 7 2 2c 1 c 2c 3. a) 2 b) 2 c) c 1 c 1 1 c2 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 y = 2sin θ y

y=

y

2

1 2 o

o

90

o

270 o

o

180

O

360

θ

270 o



o

180

O

360

1 2

y = -sin θ

y = sin2 θ

y

y

1

1 o

o

90

-1

o

90

-2

O

1 sin θ 2

o

270 o

180

o

360

o

90

θ O

270 o

180

-1


o

360

θ

θ


y = sin θ+1

1 θ 2

y = sin

y

y

1

2 o

o

90

270 o

o

180

O

360

θ

1

-1

o

o

o

90 180

y = sin θ-1 y

o

o 90 180

o

o

360

y = |sin θ| y

o

270

270

360

θ

1 o

o

90

-1

180

o

o

270

360

θ

O -2

y = 2sin2 θ

y =| sin2 θ|

y

y

2

1 o

o

o

90

O -2

o

270 o

180

o

360

90

θ

180

O -1


o

270

o

360

θ


y = |sin θ|+1

2

o

90

o

180

o

o

270

360

y = 2cos θ y

1 cos θ 2

y=

y 1 2

2

O

o

90

o

180

o

270

o

θ

360

O

o

90

o

180

o

270

o

θ

o

θ

360

-2

y =- cos θ

y = cos2θ

y

y

1

O -1

1

o

90

o

180

o

270

o

360

θ O

o

90

o

180

-1


o

270

360


y = cos

y

y = cos θ+1

1 θ 2

2

1

O

1

o

90

o

180

o

270

o

360

θ O

o

90

o

180

o

270

o

360

θ

-1

y =| cos θ|

2

y = cos θ -1

1 O

o

90

o

180

o

270

o

360

θ O

-1

o

90

y =| 2cos2 θ|

o

270

o

θ

360

y =|cos2 θ|

2

O

o

180

1

o

90

o

180

o

270

o

360

θ O

o

90

o

180


o

270

o

360

θ


y =| cos θ|+ 1 1

o

o

90

180

o

270

o

360

Exercise 3 1.X = 0 , 0.53  , 0.95  , 1.58  , 1.90  2.Number of intersection = 4

SPM 2003 y y = 3cos 1

O

3 x 2

y=

 3

π

3 x2 2

5  3

θ

2π y = - 0.5

-1

Number of intersection = 3



SPM 2004

(a ) (b ) y 1

y=

O

o

180

o

90

θ

x -1 180

-1


SPM 2005 y y = cos2x 1

O

y= 5  3

 4

2π

x 1  3 3

θ

-1




10.4 ASSESSMENT 1. 1 -2k2 2. y = 300, 1500,2100,3300 3. y = |2sin2 θ| y y

2 o

90

2(  x)  o

180

θ

O

, Number of intersection = 4
