Download Answers For Exercise 5

WAJA 2009
ADDITIONAL MATHEMATICS
FORM FIVE
( Teacher’s Copy )
Name: ___________________________
Class : ___________________________
WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
Learning Objective:
5.1 Understand the concept of positive and negative angles measured in degrees and
radians
Learning Outcome:
5.1.1 Represent in a Cartesian plane angles greater than 360o or 2π radians for
a) positive angles
b) negative angles.
1. Fill in the blanks with the symbol ‘+’ or ‘-‘ . Then determine in which quadrant
(i).
y
(a)
lies.
y
+

+


x
x
Quadrant I
Quadrant
(ii).
y
(b)
III
y
x
−
Quadrant IV
x
−
Ө
Ө
Quadrant II
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
2. Fill in the blanks with the symbol ‘+’ or ‘-‘ . Then determine in which quadrant Ө lies.
(a)
y
(b)
y

−
x
−
x

Quadrant III_
(c)
Quadrant
y
(d).
I
y
x
+
Quadrant
IV
x

+
Quadrant

II
Conclusion:
The angle is a positive angle if it is created by turning the rotating ray in an anticlockwise
direction about the origin. The angle is a negative angle if the rotation is in the clockwise
direction.
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
3. Fill in the boxes with correct angle.
(i)
240o
600o = 360o +
o
o
120o
(a)
480 = 360 +
(c)
11
3
 rad = 2 
4
4
(ii)
(b)
(d)
= 2 
o
-500o = -360o + -140
-420o = -360o +

13
3
 rad =  2 
5
5
=  2 
3

4
(e)
405o = 360o +
45o
(g)
-432o = -360o +
-72o
-60o
(f)
(h)
800o = 360o +

440o
15
3
 rad =  3 
4
4
=  3 
(i)
8
2
 rad = 2 
3
3
= 2 
(j)
2

3
3
 
5
3
 
4
14
4
 rad = 2 
5
5
= 2 
4

5
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
4. Match the correct answers.
550o
3
 
4
-417o
Quadrant 1
Quadrant 2
o
-57
5

2

23

4
Quadrant 3
1

2
190o
Quadrant 4
Beginning of
quadrant 2
(End of quadrant
1)
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
Learning Objective:
5.2 Understand and use the six trigonometric functions of any angle
Learning Outcome:
5.2.1 Define sine, cosine and tangent of any angle in a Cartesian plane.
1. Fill in the blanks with the words from the table below:Opposite Ө
Adjacent to Ө
Opposite Ө
Hypotenuse
Hypotenuse
Ө
Adjacent to Ө
2.
r
Sin 
y
r
Cos 
x
r
Tan 
y
x
y
Ө
x
Conclusion:
Sin 
Opposite 
Hypotenuse
Cos 
Adjacent
Hypotenuse
Tan 
Opposite 
Adjacent
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
3.
x 2  32  4 2
x
4
x 
32  4 2
Ө
3
5
=
Conclusion:
a
c
Ө
b
Pythagoras’ Theorem:c2 =
b2
a2
+
a2
c2
=
=
c2
b2
−
−
a2
b2
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
4. Find the length of OA and the values of sine, cosine and tangent of Ө.
(a)
(b)
y
y
A(12, 5)
A(-8, 6)
Ө
0
Ө
x
0
(8) 2  6 2
OA =
12 2  5 2
OA =
=
=
Sin 
x
13
5
13
Cos 
12
13
5
Tan 
12
10
Sin  
6
10
=
3
5
Cos  
=
4
5
Tan  
=
8
10
6
8
3
4
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
(c)
y
Ө
Chapter 5 Trigonometric Functions
(d)
0
y
x
0
Ө
x
A(-3, -4)
A(5, -12)
(3) 2  (4) 2
OA =
=
5
5 2  (12) 2
OA =
=
13
Sin  
4
5
Sin  
Cos 
3
5
Cos 
Tan 
4
3
Tan  
=
12
13
5
13
12
5
4
3
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
(e) Conclusion:Fill in the blanks with the symbol ‘+’ or ‘-‘.
Quadrant II
y
Quadrant I
+
Sin Ө
+
Sin Ө
−
Cos Ө
+
Cos Ө
−
Tan Ө
+
Tan Ө
x
Quadrant III
Quadrant IV
−
Sin Ө
−
Sin Ө
−
Cos Ө
+
Cos Ө
+
Tan Ө
−
Tan Ө
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
5. Find the corresponding reference angle of Ө
(a)
(b)
y
y
120o
50o
x
Reference angle =
180o
=
60o
x
- 120o
Reference angle =
(c)
50o
(d)
y
y
210o
280o
x
Reference angle = 210o -
=
x
180o
Reference angle =
360o
=
80o
30o
- 280o
(e) Conclusion:
i) Reference angle(R.A) is the acute angle formed between the rotating ray of the angle and
the x-axis.
ii)
y
R.A=
180o
-Ө
R.A= Ө
x
R.A= Ө -
180o
R.A=
360o
-Ө
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
6. Given that Cos 51o = 0.6293, find the trigonometric ratios of Cos 231o without using a
calculator or mathematical tables.
Reference angle of 231o
y
231o
180o
= 231o x
=
51o
Cos 231o =
=
- Cos 51o
-0.6293
7. Given that Sin 70o = 0.9397, find the trigonometric ratios of Sin 610o without using a
calculator or mathematical tables.
Reference angle of 610o
y
250o
360o
= 610o x
=
-
180o
70o
Sin 610o =
=
- Sin 70o
- 0.9397
8. Given that Tan 25o = 0.4663, find the trigonometric ratios of Tan 335o without using a
calculator or mathematical tables.
Reference angle of 335o
y
335o
335o
= 360o x
=
25o
Tan 335o =
=
- Tan 25o
- 0.4663
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
Learning Outcome:
5.2.2 Define cotangent, secant and cosecant of any angle in a Cartesian plane.
1.
2.
y
r
y
r
Ө
Ө
x
x
Sin 
y
r
1

Sin 
1
y
r
Cos 
x
r
1

Cos
1
x
r
Tan 
y
x
1

Tan
3. Fill in the blanks with Cot Ө, Sec Ө or
Cosec Ө.
1

Sin 
Cosec Ө
1
=
Cos
Sec Ө
1

Tan
Cot Ө
1
y
x
4. Since tan  
=
r
y
=
r
x
=
x
y
Sin 
, then
Cos
Cot  
Cos
.
Sin 
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
5.
α
90o
α = 180o – (
y
+
Ө
)
r
Ө
Ө
= 90o -
x
Sin 
y
r
Sin  
x
r
Cos 
x
r
Cos  
y
r
Tan 
y
x
Tan  
x
y
6. Conclusion:
Fill in the blanks with Sin, Cos, Tan, Cot, Sec or Cosec.
Sin Ө =
Cos
s
(90o - Ө)
Cos Ө =
Sin
(90o - Ө)
Tan Ө =
Cot
(90o - Ө)
Cot Ө =
o
Tan (90 - Ө)
Sec Ө =
Cosec Ө =
Cosec
(90o - Ө)
Sec
(90o - Ө)
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
7. Given that Sin 48o = 0.7431, Cos 48o = 0.6691 and Tan 48o = 1.1106, evaluate the value of
Cos 42o.
90o – 48o
=
Cos 42o =
Sin 48o
42o
o
48
=
0.7431
8. Given that Sin 67o = 0.9205, Cos 67o = 0.3907 and Tan 67o = 2.3559, evaluate the value of
Cot 23o.
90o – 67o
=
Cot 23o =
Tan 67o
23o
o
67
=
2.3559
9. Given that Sin 42o = 0.6691, Cos 42o = 0.7431 and Tan 42o = 0.9004, evaluate the value of
Sec 48o.
90o – 42o
=
Sec 48o =
Cosec 42o
48o
o
42
=
1.4945
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
Learning Outcome:
5.2.3 Find values of six trigonometric functions of any angle
1.
A
2.
A
1
B
60o
2
2
1
60o
B
60o
1
C
2
D
1
C
A
A
30o
2
45o
60o
M
1
45o
C
1
AM =
=
Sin 60o =
Cos 60o =
D
2 2  12
3
AC =
3
2
=
1
2
Sin 45o =
Tan 60o =
Sin 30o =
Cos 30o =
Tan 30o =
C
1
3
2
2
1
2
3
1
2
12  12
Cos 45o =
1
Tan 45o =
1
2
1
3
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
60o
45o
30o
Sin Ө
3
2
1
1
2
Cos Ө
1
2
1
3
Tan Ө
2
2
3
2
1
1
3
3 Complete the table below.
4. Find the value of the trigonometric functions below :
Example : (i) Evaluate Sin 210o
Draw diagram to determine positive or negative.
y
210o
x
Find reference angle.
Reference angle of 210o
= 210o - 180o
=
30o
Solve.
Sin 210o = - Sin 30o
=

1
2
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
(a) Evaluate Cos 300o
Draw diagram to determine positive or negative.
y
300o
x
Find reference angle.
Reference angle of 300o
= 360o - 300o
= 60o
Solve.
Cos 300o = Cos 60o
=
1
2
(b) Evaluate Cot 150o
Draw diagram to determine positive or negative.
y
150o
x
Find reference angle.
Reference angle of 150o
= 180o - 150o
= 30o
Solve.
Cot 150o = -Cot 30o
=  3
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
(c) Evaluate Tan (- 225o)
Draw diagram to determine positive or negative.
y
x
- 225o
Find reference angle
Reference angle of 225o
= 225o - 180o
= 45o
Solve.
Tan (- 225o) = - Tan 45o
= -1
(d) Evaluate Cosec 135o
Draw diagram to determine positive or negative.
y
135o
x
Find reference angle.
Reference angle of 150o
= 180o - 135o
= 45o
Solve.
Cosec 135o = Cosec 45o
=
2
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
(e) Evaluate Sec 330o
Draw diagram to determine positive or negative.
y
330o
x
Find reference angle.
Reference angle of 150o
= 360o - 330o
= 30o
Solve.
Sec 330o
= Sec 30o
2
=
3
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
Learning Outcome:
5.2.4 Solve trigonometric equations
1. Given that Sin Ө = 0.9135, find the reference angle of Ө and determine in which quadrant
Ө lies.
Example :
(i) Sin Ө = 0.9135
66o
Reference angle of Ө =
y
y
66o
66o
x
x
Quadrant: I
Quadrant: II
Ө = 66o
Ө = 180o - 66o
= 114o
(a) Given that Cot Ө = -1.4826, find the reference angle of Ө and determine in which
quadrant Ө lies.
Cot Ө = -1.4826
Reference angle of Ө =
34o
y
y
34o
34o
x
Quadrant: II
Quadrant: IV
Ө = 180o - 34o
Ө = 360o - 34o
=
146o
=
x
326o
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
(b) Solve tan Ө = 0.5317 where 0 0    360 0 .
tan Ө = 0.5317
Reference angle of Ө =
28o
y
y
28o
28o
x
Quadrant: I
Ө = 28o
x
Quadrant: III
Ө = 180o + 28o
=
208o
(c) Solve Cos Ө = -0.8988 where 0 0    360 0 .
Cos Ө = 0.8988
Reference angle of Ө =
26o
y
y
26o
x
Quadrant: II
Ө = 180o -
=
154o
26o
x
Quadrant: III
26o
Ө = 180o +
=
26o
206o
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
(d) Solve Cosec Ө = -2.2027 where 0 0    360 0 .
Cosec Ө = 2.2027
Reference angle of Ө =
27o
y
y
27o
x
x
o
27
Quadrant: III
Ө = 180o +
=
Quadrant: IV
27o
Ө = 360o - 27o
=
207o
333o
(e) Solve Sec 2Ө = 2 where 0 0    360 0 .
Sec 2Ө = 2
Reference angle of 2Ө =
60o
y
y
60o
60o
x
Quadrant: I
2Ө =
Ө=
60o
30o
x
Quadrant: IV
2Ө = 360o - 60o
= 300o
Ө = 150o
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
(f) Find all the values of x that satisfy
Chapter 5 Trigonometric Functions
2Sinx  1  0 .
2Sinx  1  0
Sinx = 
1
2
Reference angle of Ө =
45o
y
y
45o
x
x
o
45
Quadrant: III
Ө = 180o + 45o
= 225o
Quadrant: IV
Ө = 360o - 45o
= 315o
(g) Solve the trigonometric equation 3 tan x  1  0 for 0 0    360 0 without using a
calculator or mathematical tables..
3 tan x  1  0

tanx =
1
3
Reference angle of Ө =
30o
y
y
30o
30o
x
Quadrant: II
Quadrant: IV
Ө = 180o - 30o
Ө = 360o - 30o
=
150o
=
x
330o
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WAJA 2009 – ADDITIONAL MATHEMATICS(FORM 5)
Chapter 5 Trigonometric Functions
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