Engineering Mathematics 1, Self Learning

2ND TOPIC : TRIGONOMETRY

Example 2.1

1. If tan = 2138, determine the value of the other five trigonometry ratio.

Solution Method
2

1st : Used calculator to get the

value of angle A.

tan = 18
23
18
= tan−1 23

≈ 38.05

Solution :- continue

2nd : calculate other 5 trigonometry
ratio.
Method 2 – by using calculator, use
value of angle A to calculate all
trigonometry ratio.

Tips:- According to above triangle :-
1. Make sure calculator is in 1. sin 38.05 ≈ 0.6163
2. cos 38.05 = 0.7875
Degree Mode. 1
2. The Value must as follow: 3. cot 38.05 = tan 38.05
• sine and cosine
≈ 1.2776
(−1 ≤ ≤ 1) csc 38.05 1
• cosecant and secant can be 4. = sin 38.05

either ≈ 1.6225
• −1 < 1
• > 1 5. sec 38.05 = cos 38.05

CLICK ME….. ≈ 1.2699
https://youtu.be/5CC4xd5A10A

49

2ND TOPIC : TRIGONOMETRY

Example 2.2

2. If sec = 2253, determine the value of the other five trigonometry ratio.

Solution Method
1

1st : Draw your right-angled 25 23
triangle.
- Sketch and determine a right-
sec = 25 Value of Hypotenuse side
angle then hypotenuse side 23
- Then, label angle follow by
Value of Adjacent side
both adjacent and opposite side.
Solution :- continue
2nd : Used Theorem of Pythagoras
to get value of opposite side:- 3rd : calculate other 5 trigonometry
ratio.
ℎ2 = 2 + 2 Method 1 – without using calculator,
o = ℎ2 − 2 use right-angled triangle only.
= 252 − 232
= 96 ≈ 9.80

Tips:- According to above triangle :-
1. Always use at least 2
1. sin = 96
decimal places (2 d.p) 2. cos = 2235
2. Instead of decimal 25
96
values, root or 3. tan = 2253
fraction value is more 4. c = 96
accurate.

5. c = 23
96

CLICK ME…..
https://youtu.be/NeUQMWEFm2

8

50

2ND TOPIC : TRIGONOMETRY

Example 2.2
2. If sec = 2235, determine the value of the other five trigonometry ratio.

Solution Method
2

1st : Used trigonometry ratio to get
the value of angle .
25
sec = 23

1 25
c = 23
23 = 25 cos
23 CLICK ME…..
cos = 25 https://youtu.be/noQcaLwVVqM

= cos−1 23 Solution :- continue
25
≈ 23.07

Tips:- 2nd : calculate other 5 trigonometry
Value of tan 90 = ∞, not ratio.
math error. Method 2 – by using calculator, use
value of angle, to calculate all
Figure 2.1: Graph of tangent trigonometry ratio.

According to above triangle :-
1. sin 23.07 ≈ 0.3919
2. cos 23.07 ≈ 0.9200
3. tan 23.07 ≈ 0.4259
1
4. csc 23.07 = sin 23.07

≈ 2.5520
1
5. cot 23.07 = tan 23.07

≈ 2.3479

51

2ND TOPIC : TRIGONOMETRY

Test Your Knowledge 2.1

1. If sin = 1238, determine the value of tan and sec
without using calculator. (State your answer in Fraction
Form.)

2. If cot = 4209, determine the value of sin and cos .
(State your answer in Fraction Form.)

3. Find the acute angle of csc = 35 in degrees and
30
minutes.

4. A man drive 10km due north and then 15km due west.
Another woman, starting at the same time with the same
speed with the man, drive 20km due west and then 5km
due north. Find the distance between two of them.

5. A straight bamboo stick 10 m long is placed against a
perpendicular tree with its foot 4.5 m from the tree. How
far up to the tree does the bamboo stick can reach? (state
your answer in 3 decimal places.)

Answer :
1. tan = 18 sec = 23

205 205

2. sin = 40 cos = 29
2441 2441

3. = 28 59′50′′

4. Distance = 7.07

5. Only 8.930 metre

52

2ND TOPIC : TRIGONOMETRY

21.1.4FUPANRDTAIMALENFRTAACLTOIFONTRIGONOMETRIC FUNCTIONS

QUADRANTS FOR TRIGONOMETRY RATIO

Figure 2.2: Trigonometric Function’s Quadrants.

Read the following explanation carefully:-

indicate positive angle movement (anti-clockwise) and indicate
negative angle movement (clockwise).

Each trigonometry ratio has 2 positive quadrants and 2 negative
quadrants.

Each quadrant’s value is 90 representing value of acute angle.

Reference angle (RA) (always start from x-axis or axis of 0 /180 ) is an
acute angle. Therefore, to calculate reference angle refer formula below :-

Quadrant I; Quadrant II; Quadrant III; Quadrant IV;
RA = RA = 180 − RA = − 180 RA = 360 −

53

2ND TOPIC : TRIGONOMETRY

Example 2.3

State the following trigonometric functions either positive or negative and in

which quadrant.
i. tan 45 ii. cos 120 iii. sin 270 iv. sec 410 v. csc −210

Solution Tips:-
1. For any angle which is greater
i. tan 45
than 360 ( > 360 ) or
- Quadrant I – (0 ≤ ≤ 90 ) less than −360
- tan 45 is positive – All ( < −360 ), you can add
or subtract with 360 until
trigonometry ratio are positive in you can get a value less that
Quadrant I. ± 360 .
2. For any negatives angle, you
ii. cos 120 can find its positive angle by
- Quadrant II – (90 ≤ ≤ 180 ) adding with 360 .
- cos 120 is negative – ONLY sine
Solution
and cosecant are positive in iv. sec 410
Quadrant II. - Quadrant I

iii. sin 270 (360 ≤ ≤ 450 )
- Quadrant III and Quadrant IV - 270 - sec 410 is positive

is border value for Quadrant III and All trigonometry ratio are
Quadrant IV. (you may refer to positive in Quadrant I.
Figure 2.2)
- sin 270 is negative – sine is v. csc( −210 )
negative in BOTH Quadrant III and - Quadrant II
Quadrant IV.
(−180 ≤ ≤ −270 )
CLICK ME….. - csc( −210 ) is positive
https://youtu.be/tEAwM0tfGL
Sine and cosecant positive in
M Quadrant II.

54

2ND TOPIC : TRIGONOMETRY

21.1.4FUPANRDTAIMALENFRTAACLTOIFONTRIGONOMETRIC FUNCTIONS

Special Angle
1. Only 3 angle are special angle :- 30 , 45 and 60 .

2. All special angles are reference angle.

3. Most of the question for special angle frequently need to be solved
without using calculator.

Special Angle 30 45 60
( )
sin 1 1 3
cos 2 2 2
tan 1 1
csc 3 2 2
sec 2 1
cot 1 1=1 3
3 1= 3
2 2
1=2 1= 2 2
2 2 3
3 1= 2 2
3 1 1=2
1= 3 1=1 1
3

Table 2.1 : Value of Special Angle for Trigonometric Function.

4. According to Table 2.1, values circled show exact value of the
trigonometric functions, while value in fractions show the value of
right-angled side that being used accordingly.

5. Final answer for trigonometric with special angle must in fraction
and/or with roots included values of special angle use in solutions.
Avoid decimal values.[Refer to Table 2.1]

55

2ND TOPIC : TRIGONOMETRY

Example 2.4

Find the value for each of the following without using calculator :-
i. tan 45 ii. cos 120 iii. sin 210 iv. sec 405 v. csc −210

Solution Tips:-
To solve Special Angle, always
i. tan 45 determine its Reference Angle
- 45 is reference angle according to the Quadrants.
- Quadrant I : tangent is positive
Solution
tan 45 = 1
iv. sec 405
ii. cos 120 - Reference angle for 405 :-
- Reference angle for 120 :-
Quadrant I;
Quadrant II; = 405 − 360
= 180 − 120 = 45
= 60
- Quadrant I : sec is positive.
- Quadrant II : cosine is negative. sec 405 = sec 45
cos 120 = − cos 60 =2
1
= − 2 v. csc −210
- positive angle for −210 :-
iii. sin 210
- Reference angle for 210 :- −210 = 360 −210
= 150
Quadrant III;
= 210 − 180 Quadrant II;
= 30 = 180 − 150
= 30
- Quadrant III : sine is negative.
sin 210 = − sin 30 - Quadrant II : sine is positive.
1 csc −210 = csc 30
= − 2 =2

CLICK ME….. 56
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2ND TOPIC : TRIGONOMETRY

Example 2.5

2. Find the value for each of the following without using calculator:-
i. tan 45 + cos 120 ii. sin 210 − cot 240 iii. sec 405 − csc 240

Solution Tips:-

i. tan 45 + cos 120 For operations of Special
Angle, make sure :-

= tan 45 + [−cos 180 − 120 ] 1. Determine the quadrant
= tan 45 + − cos 60 for each of the
1 trigonometry function.
= 1 − 2
2. Determine either
1 trigonometric function
=2 is positive or negative in
the quadrant.

ii. sin 210 − cot 240 3. Find its Reference
Angle.

= − sin 210 − 180 − cot 240 − 180 4. Substitute the value and
= − sin 30 − cot 60 complete the solution.

= − 1 − 1 Solution
2 3
iii. sec 405 − csc 240
=− 3 − 2
23 23 = sec 405 − 360 − − csc 240 − 180
= sec 45 + csc 60
− 3−2
= 2
= 2+
23
3
=− 3+2 23 2
23 =+

33

62
=+

33

6+3
=

3

CLICK ME….. 57
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2ND TOPIC : TRIGONOMETRY

Test Your Knowledge 2.2

Find the value for each of the following without using
calculator:-

1. tan 30o + cos 60o

2. sin 330o − cot 45o
3. csc 405o − sec 240o
4. tan −45o + sin −60o
5. cos 870o − cot 570o

Answer :

1. 2+ 3

23
3
2. − 2

3. 2 + 2

4. − 2+ 3
2

3+2 3
5. 2

58

2ND TOPIC : TRIGONOMETRY

21.2.4TPRAIGROTNIAOLMFERTARCITCIOEQNUATIONS & IDENTITIES

TRIGONOMETRY IDENTITIES AND FORMULAE

Basic Identities Double Angle Formulae
1. sin2α + cos2α = 1
2. 1 + tan2α = sec2α 1. sin 2α = 2 sin cos
3. 1 + cot2α = csc2α
2. tan 2α = 2 tan
1− 2

3. cos 2α = 2 − 2

Quotient Identities = 2 2 − 1

1. tan α = sin α = 1 − 2 2
cos α

2. cot α = cos
sin

Compound Angle Formulae : Addition Angle
1. sin + = sin cos + cos sin
2. cos + = cos α cos − sin sin

3. tan + = tan +tan
1−tan tan

Compound Angle Formulae : Subtraction Angle
1. sin − = sin cos − cos sin
2. cos − = cos α cos + sin sin

3. tan − = tan −tan
1+tan tan

Do not expand compound angle, this is not algebraic expression, this is
trigonometric function.
Example : sin − ≠ sin − sin

59

2ND TOPIC : TRIGONOMETRY

21.2.4TPRAIGROTNIAOLMFERTARCITCIOEQNUATIONS & IDENTITIES

Why we need trigonometric identities and formulae?
1. To solve trigonometric equations, the equations should meet 2 conditions

i.e. have SAME trigonometry function and SAME angle.
Quadratic and Linear of Trigonometric Equations with the
same trigonometric ratio are considered as the same
Trigonometric Function.
Example : and , there are still function of sine
but with different index.

2. Therefore, we need trigonometry Identities and Formulae just to convert
existing/given equation to meet point number 1.

Extra Note:-
a) Basic Identities also known as Pythagoras Identity.
b) Quotient Identities is used either to convert tangent into combination of

sine and cosine functions or vice versa.

60

2ND TOPIC : TRIGONOMETRY

Example 2.6

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. sin α = 0.5 ii. cos α = −0.1234 iii. tan 2α = 3.1203

iv. csc 2 = 2.431

Solution Check the value either positive or
negative to determine the angle’s
i. sin α = +0.5 quadrant.
= sin−1 0.5
= 30

Reference Sine positive in Quadrant I and Quadrant II
Angle Quadrant I :- = 30
Quadrant II :- = 180 − 30

= 150
= 30 150 .

Quadrant II = Quadrant I

= 30 =

= 30 0

Figure 2.3 : Diagram to show Reference Angle and Angle in Quadrants.

Tips:-
1. Reference Angle is an acute angle and exist in quadrant.
2. Angle for trigonometric equation should be calculate start from

0 .
3. You may check your answer by using calculator.

Example: sin 30 and sin 150 is equal to 0.5.

61

2ND TOPIC : TRIGONOMETRY

Example 2.6 :- continue

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. sin α = 0.5 ii. cos α = −0.1234 iii. tan 2α = 3.1203

iv. csc 2 = 2.431

Solution Tips:-
ii. cos α = −0.1234 - Ignore negative sign (−)

= cos−1 0.1234 when you calculate inverse
= 82.91 trigonometry functions.
- Positive and negative just
Cosine negative in Quadrant II and Quadrant III. to indicate quadrants of
Quadrant II :- angle.
= 180 − 82.91 - Fail to ignore, you may get
= 97.09 reference angle > −90 .

Quadrant III :-
= 180 + 82.91
= 262.91
= 97.09 and 262.91 .

Solution
ii. tan 2α = 3.1203

2 = tan−1 3.1203
2 = 72.23

Tangent positive in Quadrant I and Quadrant III.

Quadrant I :- 2 = 360 + 72.23 = 432.23
2 = 72.23 = 216.115

= 36.115

Quadrant III :- 2 = 360 + 252.23 = 612.23
2 = 180 + 72.23 = 252.23 = 306.115

= 126.115

= 36.115 , 126.115 , 216.115 and 306.115 .

CLICK ME….. 62
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2ND TOPIC : TRIGONOMETRY

Example 2.6 :- continue

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. sin α = 0.5 ii. cos α = −0.1234 iii. tan 2α = 3.1203

iv. csc 2 = 2.431

Solution

iv. csc = 2.431 Tips:-
2 0 ≤ ≤ 360
1
= 2.431
sin 2
Always check the range as

1 below:
2.431 = sin 2 - ≤ ≤ means:
022 .4==11s2i44n.−=219s 0 i.n41214 i. is 1 complete rotation (4

quadrants)

ii. is 2 complete rotation

(each quadrant will have 2

values.)

iii. is half rotation (only

Sine positive in Quadrant I and Quadrant II. quadrant I and II)
Q2 u==ad24r84a..n25t89I :-
- ≤ ≤ means
Q 2 u==ad31r81a10n. t 4 I−2I : 2- 4.29 = 155.71 ONLY for Quadrant I and II.
= 48.28 and 311.42 .
- ≤ ≤ means 2
completed rotation.

- ≤ ≤ means
ONLY Quadrant III and

Quadrant IV.

63

2ND TOPIC : TRIGONOMETRY

Test Your Knowledge 2.3

Find value of for each of the following equations which
satisfy 0 ≤ ≤ 360 ;

1. sin = 0.4756
2. tan = −3.0204
3. cos 2 = 0.1203

4. sec = −2.431
2

Answer :
1. = 28.40 151.60
2. = 108.32 288.32
3. = 41.545 , 138.455 , 221.545 318.455
4. = 228.52

64

2ND TOPIC : TRIGONOMETRY

Example 2.7

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. tan2 + 3 tan − 5 = 0 ii. 6sin2 − sin = 2

Solution

i. tan2 + 3 tan − 5 = 0 Tips:-

tan = 1.1926 and tan = −4.1926 tan2 = tan 2

(Note :- I used calculator to solve this Quadratic  If trigonometric
Trigonometric Equation) function with index of 2,
please apply

For tan = 1.1926 Quadratics Equation
= tan−1 1.1926 solution.
= 50.02 (value of reference angle)
• Linear trigonometric
equation, only have 1

Tangent Positive in Quadrant I and Quadrant III. reference angle.
• Quadratic
Quadrant I :- Quadrant III :-
= 50.02 = 180 + 50.02 trigonometric equation

= 230.02 have 2 reference angle.

For tan = −4.1926 (sign to show
quadrants of reference angle exist)

= tan−1 4.1926
= 76.58

Tangent Negative in Quadrant II and Quadrant

IV.

Quadrant II :- Quadrant IV :-
= 180 − 76.58 = 360 − 76.58
= 103.42
= 283.42

= 50.02 , 103.42 , 230.02 and 283.42 .

CLICK ME….. 65
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2ND TOPIC : TRIGONOMETRY

Example 2.7 :- continue

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. tan2 + 3 tan − 5 = 0 ii. 6sin2 − sin = 2

Solution

ii. 6sin2 − sin = 2

6sin2 − sin − 2 = 0
3 sin − 2 2 sin + 1 = 0
2 1
sin = 3 and sin = − 2

For sin = 2
2 3
= sin−1 3

= 41.81 , 138.19 , 210 and 330 . 66

CLICK ME…..
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2ND TOPIC : TRIGONOMETRY

Example 2.8

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. 3sin2 + 4 cos − 2 = 0 ii. 2csc2 − 5 tan = 3

Solution

i. 3sin2 + 4 cos − 2 = 0 Tips:-

(Note :- Use Basic Identities to convert sine to To SOLVE trigonometric
cosine. Equation, the equation
WHY? must meet 2 conditions:-
It is much easier to convert quadratic trigonometric • Have ONLY 1
function compare to convert linear trigonometric
function.) trigonometric
function.
sin2 + cos2 = 1 • Have THE SAME
sin2 = 1 − cos2 angle

∴ 3 1 − cos2 + 4 cos − 2 = 0
3 − 3cos2 + 4 cos − 2 = 0
−3cos2 + 4 cos + 1 = 0

cos = −0.2153 and cos = 1.5486

For cos = −0.2153
= cos−1 0.2153
= 77.57

Cosine Negative in Quadrant II and Quadrant III.

Quadrant II :- Quadrant III :-
= 180 − 77.57 = 180 + 77.57
= 102.43
= 257.57

For cos = 1.5486 NEVER state your
= cos−1 1.5486 answer as MATH
= ERROR. It is
= 102.43 and 257.57 . unacceptable
answer.

CLICK ME….. 67
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2ND TOPIC : TRIGONOMETRY

Example 2.8 :- continue

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. 3sin2 + 4 cos − 2 = 0 ii. 2csc2 − 5 cot = 3

Solution
ii. 2csc2 − 5 cot = 3

2csc2 − 5 cot − 3 = 0
(Note :- Use Basic Identities to convert cosecant to cotangent. )

1 + cot2 = csc2
csc2 = 1 + cot2

∴ 2 1 + cot2 − 5 cot − 3 = 0
2 + 2 cot2 − 5 cot − 3 = 0
2cot2 − 5 cot − 1 = 0
2ttaa−nn22225 ta−−n5tt a a t5n−na2n t a − n 2−1 tt =aann022
= 0


− tan2t a n−25 tan + = 0 0
2 =
tan = −5.3723 and tan = 0.3723

For tan = −5.3723
= tan−1 5.3723
= 79.46

Tangent Negative in Quadrant II and Quadrant IV.

Quadrant II :- Quadrant IV :-
= 180 − 79.46 = 360 − 79.46
= 100.54
= 280.54

Solution continue on the next page.

CLICK ME….. 68
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2ND TOPIC : TRIGONOMETRY

Example 2.8 :- continue

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. 3sin2 + 4 cos − 2 = 0 ii. 2csc2 − 5 cot = 3

Solution ii continue.

tan = −5.3723 and tan = 0.3723
For tan = 0.3723
= tan−1 0.3723
= 20.42

Tangent positive in Quadrant I and Quadrant III.

Quadrant I :- Quadrant III :-
= 20.42 = 180 + 20.42
= 200.42

= 20.42 , 100.54 , 200.42 and 280.54 .

69

2ND TOPIC : TRIGONOMETRY

Test Your Knowledge 2.4

Find value of for each of the following equations which
satisfy 0 ≤ ≤ 360 : −

1. 5sin2 − 2 sin − 2 = 0
2. 2tan2 − 3 tan = 5
3. 4sin2 = 5 cos − 1
4. 8 − 2cot2 = 4 2 − 3

Answer :
1. = 59.69 , 120.31 , 207.60 332.40
2. = 68.20 , 135 , 248.20 315
3. = 49.01 310.99
4. = 42.79 , 137.21 , 222.79 317.21

70

2ND TOPIC : TRIGONOMETRY

Example 2.9

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. 2 cos 2 − 5 sin = 3 ii. 5cos2 + 2 sin 2 = 0

Solution Double Angle (2A) versus Single Angle (A)

i. 2 cos 2 − 5 sin = 3

(Note :- Use Double Angle Formulae to convert Tips:-

cosine to sine. ) • Used Double Angle
cos 2 = 1 − 2 2 Formulae for a question
consists of combination
2 1 − 2 2 − 5 sin = 3 Double angle and Single
2 − 4 2 − 5 sin = 3 angle.
−4 2 − 5 sin − 1 = 0
Now, only sine functions in the question. • Convert trigonometry
with double angle to
sin = −1 and sin = −0.25 trigonometry with
(Note :- I used calculator to solve this Quadratic single angle.

Trigonometric Equation) • Choose a formulae that
will give you only 1
For sin = −1 trigonometry function
= sin−1 1 in a question.
= 90
Double Angle Formulae for
Sine Negative in Quadrant III and Quadrant IV.
cosine:-
Quadrant III :- Quadrant IV :-
= 180 + 90 = 360 − 90
= 270
= 270

For sin = −0.25 = 2 − 2
= sin−1 0.25 = 2 2 − 1
= 14.48 = −

Sine Negative in Quadrant III and Quadrant IV. Choose cos 2 = 1 − 2 2

Quadrant III :- Quadrant IV :- to solve the question.
= 180 + 14.48 = 360 − 14.48
= 194.48
= 345.52

= 194.48 , 270 and 345.52 .

CLICK ME….. 71
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2ND TOPIC : TRIGONOMETRY

Example 2.9 : Continue

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. 2 cos 2 − 5 sin = 3 ii. 5cos2 + 2 sin 2 = 0

Solution Tips:-
Refer to the graph of cosine,
ii. 5cos2 + 2 sin 2 = 0 identify the value for cosine = 0
and get the value of the angle.
(Note :- Use Double Angle Formulae to convert (refer to the circle)
Double Angle to Single Angle. )
sin 2 = 2 sin cos

5cos2 + 2 2 sin cos = 0
5cos2 + 4 sin cos = 0
We have no choice for this, but do not worry. I
will use Quotient Law later to solve it.

cos 5 cos + 4 sin = 0

cos = 0 and + =

For cos =0
= cos−1 0
= 90

0 is neutral value. Easier by refer to the graph

of cosine to get the value of cosine = 0. Figure 2.4: Graph of cosine

While for quadrants, you must calculate for all Double Angle Formulae for
sine:-
quadrants.
=
Quadrant I :- Quadrant II :- I have no options to choose,
= 90 = 180 − 90 so just use it.
= 90 = 90

Quadrant III :- Quadrant IV :-
= 180 + 90 = 360 − 90
= 270
= 270

Solution continue on the next page.

72

2ND TOPIC : TRIGONOMETRY

Example 2.9 : Continue

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. 2 cos 2 − 5 sin = 3 ii. 5cos2 + 2 sin 2 = 0

Solution ii continue.

For + = Used Quotient Law to
5 cos = −4 sin simplify the trigonometry

− 5 = sin equation:-
4 cos

5 =
4
− = tan

tan = − 5 Tips:-
4
5ta1n.3−41 54
= If we need to use
=
Quotient Law, please use

Tangent Negative in Quadrant II and Quadrant =

IV.

Quadrant II :- Quadrant IV :- Avoid using
= 180 − 51.34 = 360 − 51.34
= 128.66 =
= 308.66

= 90 , 128.66 , 270 and 308.66 . To avoid longer steps

73

2ND TOPIC : TRIGONOMETRY

Test Your Knowledge 2.5
Find value of for each of the following equations which
satisfy 0 ≤ ≤ 360 : −

1. 3 cos 2 = 4 cos + 5
2. 2 cos 2 − 5 2 = 4 sin − 3
3. 6cos2 = 5 sin 2

Answer :
1. = 150.28 209.72
2. = 10.28 , 169.72 , 218.52 321.48
3. = 30.96 , 90 , 210.96 270

74

2ND TOPIC : TRIGONOMETRY

Example 2.10

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. sin 50o + = 0 ii. tan − 35 = 4

Solution Tips:-
i. sin 50o + = 0 DO NOT expand +

sin 50o + = sin 50o cos + cos 50o sin + ≠ +
Therefore;
sin 50o cos + cos 50o sin = 0 Because sin is the main function
0.7660 cos + 0.6428 sin = 0 with + is the angle of it
0.7660 cos = −0.6428 sin
0.7660 sin Check your answer:
−0.6428 = cos Used your calculator and check
−1.1917 = tan
tan = −1.1917 sin 50o + = 0
sin 180 = 0

Do you get 0 = 0 ?

For tan = −1.1917
= tan−1 1.1917
= 50.00

Tangent Negative in Quadrant II and Quadrant IV.

Quadrant II :- Quadrant IV :-
= 180 − 50 = 360 − 50
= 130
= 310

= 130 and 360 .

Compound Angle Formulae for sine:-
+ = +

CLICK ME….. 75
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2ND TOPIC : TRIGONOMETRY

Example 2.10 : Continue

Find value of for each of the following equations which satisfy 0 ≤ ≤ 360 ;
i. sin 50o + = 0 ii. tan − 35 = 4

Solution

ii. tan − 35 = 4 Compound Angle Formulae for

tan − 35 tan − tan 35 tangent:- −
= 1 + tan tan 35 − = +

Therefore;
tan − tan 35
t a tna3n53 5 = =4 4
1 + tan 1 + tan tan 35
tan −
tan − tan 35 = 4 + 4 tan tan 35
tan − 0.7002 = 4 + 2.8008 tan
tan − 2.8008 tan = 4 + 0.7002
−1.8008 tan = 4.7002
4.7002
tan = −1.8008

tan = −2.6101

For tan = −2.6101
= tan−1 2.6101
= 69.04

Tangent Negative in Quadrant II and Quadrant IV.

Quadrant II :- Quadrant IV :-
= 180 − 69.04 = 360 − 69.04
= 110.96
= 290.96

= 110.96 and 290.96 .

76

2ND TOPIC : TRIGONOMETRY

Test Your Knowledge 2.6

Find value of for each of the following equations which
satisfy 0 ≤ ≤ 360 : −

1. 2 sin 30o + = 0
2. tan + 35o = 4
3. 3 cos − 15o = 0
4. 5 tan 20o − = 3

Answer :
1. = 150 330
2. = 40.96 220.96
3. = 75.00 255
4 . = 169.04 349.04

77

2ND TOPIC : TRIGONOMETRY

21.3.4SIPNAERT&IACOLSFIRNAECRTUIOLNES

INTRODUCTION

1. Sine and Cosine rules can be used to all types of triangle.

2. Used formulae of area for triangle, = if the question stated

value of Area.

3. Never use Theorem of Pythagoras for this subtopic unless the question
stated it is right-angled triangle.

4. Frequently, labelling triangle consist of pairing alphabet; capital letter
and small letter. Example; A and a.

• Capital letter – to label angle

• Small letter – to label side

• Angle and side is opposite to one another.

Length of a

Angle of A

Figure 2.5 : Labelling Angle and Length Using Pairing Alphabet.
5. Formulae of Interior Angles for Triangles i.e. = + + always

being used to find the value of third angle once value of first and second
angles are found.
6. For non-right angled triangle, trigonometric ratio cannot be used.

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2ND TOPIC : TRIGONOMETRY

21.3.4SIPNAERT&IACOLSFIRNAECRTUIOLNES

SINE AND COSINE RULES

Sine Rules Cosine Rules
• = + − , or
• = + −
= =
Note :-

Or 1. You must know how to re-
create cosine rules to meet your
need.
= =
Note :- 2. Above formulae can be used to
1. Used only 2 parts on Sine Rule find length of a or angle A (first
formulae not 3 parts. formulae) and length of b or
2. You need to know at least 1 angle B (second formulae).
pair of angle and its side to
used this rule. 3. This law mostly suitable for a
question which:-

 All values of side are given

without value of angle.

 2 sides are given and 1 angle

a BUT there are not opposite to
B each other. And
 To calculate second angle if
= C first angle is an acute angle
since Law of Cosine can trace
c

Ab

Figure 2.6 : Triangle ABC. obtuse angle.

Tips:-
Obtuse angle is angle with range 90 < < 180 .

79

2ND TOPIC : TRIGONOMETRY

Example 2.11

In triangle ABC, AB is 15cm, BC is 30 cm and AC is 20cm. Sketch the triangle,
solve the triangle and find its area.

Solution

First : Sketch your triangle completed with relevant label.

A

c = 15 cm b = 20cm A = 117.28 a = 30cm
B = 36.33 b = 20cm
B C C = _______ c = 15cm
a = 30cm
This table is optional
Second : Calculate all angle for triangle ABC. but I always used it for
fast checking.
1. Used Law of Cosine to calculate angle A. You can get the data
= + − from your triangle.
302 = 202 + 152 − 2 20 15 cos
302 − 202 − 152
cos = −2 20 15

cos = 275
−600
275
= cos−1 − 600

= 117.28

2. Used Law of Sine to calculate angle B.

sin 11=7.28 sin
×=sin21017.28
sin 30 20 30
=

sin = 0.5925
= sin−1 0.5925
= 36.33

Solution continue on the next page.

80

2ND TOPIC : TRIGONOMETRY

Example 2.11

In triangle ABC, AB is 15cm, BC is 30 cm and AC is 20cm. Sketch the triangle,
solve the triangle and find its area

Solution : Continue

3. Used Interior Angle of Triangle Formula to calculate angle C.
180 = 117.28 + 36.33 +
= 180 − 117.28 − 36.33
= 26.39
A = 117.28 a = 30cm
B = 36.33 b = 20cm
Third : Calculate area of triangle ABC. C = 26.39 c = 15cm

= 1 sin
2
1
= 21333.034 2 0 2 sin 26.39 My triangle data is now
= Or completed.

= 1 sin
2
1
= 12332.032 1 5 2 sin 117.28
= Or

= 1 sin
2
1
= 12333.030 1 5 2sin 36.33
=

As you might have seen, three formula with different data of angles and
sides give different value of area, however, do not worry, all of the answer
are acceptable accordingly to the formulae that being used.

Note :- Just calculate the area once. I did three times just to show you that
even which formulae you used, the answer almost same.

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2ND TOPIC : TRIGONOMETRY

Example 2.12
Solve the triangle PQR and find its area. Given that p = 8.5 m, q = 10 m and
∠ = 85 .

Solution P = 42.43 p = 8.5 m
Q = ________ q = 10 m
Calculate side r by using Law of Cosine. R = 85 r = 12.55 m
= + −
2 = 8.52 + 102 − 2 8.5 10 cos 85 Tips:-
2 = 157.43
= 157.43 The longest side of triangle
= 12.55 will have largest angle
opposite to it.
Used Law of Sine to calculate angle P.

According to the above data,
= I‘m not using Law of Cosine to
sin 85 sin calculate second angle since
= 8.5 the longest side is 12.55 m
12.55 8.5 × sin 85 and its angle is 85 . So there
sin = is no Obtuse angle in this
12.55 triangle.
sin = 0.6747
= sin−1 0.6747
= 42.43
= + −
8.52 = 102 + 12.552 − 2 10 12.55 cos
8.52 − 102 − 12.552
Lets check angle P by cos = −2 10 12.55
using Law of Cosine.
−185.25
cos = −251

cos = 0.7380

= cos−1 − 275
600
= 42.44

Solution continue on the next page.

82

2ND TOPIC : TRIGONOMETRY

Example 2.12 : Continue
Solve the triangle PQR and find its area. Given that p = 8.5 m, q = 10 m and
∠ = 85 .

Solution P = 42.43 p = 8.5 m
Q = 52.57 q = 10 m
Used Interior Angle of Triangle Formula R = 85 r = 12.55 m
to calculate angle Q
Tips:-
180 = 42.43 + + 85
= 180 − 42.43 − 85 It is normal to have some
= 52.57 differences in value due to
numbers of significant figure
Lets check angle Q by and decimal places being
using Law of Sine. used.
More significant
figure/decimal places being
= used, more accurate answer
sin sin 85 will be.
=
10 = 12.55 85
sin 10 × sin

12.55
sin = 0.7938
= sin−1 0.7938
= 52.54

Calculate area of triangle PQR.

= 1 sin
2
1
= 2 8.5 10 sin 85
= 42.34 2

83

2ND TOPIC : TRIGONOMETRY

Test Your Knowledge 2.7

1. Solve the triangle XYZ and find its area. Given ∠ =
36 , XY = 20 cm and XZ = 30 cm.

2. A triangle RST has sides r = 7.8 m, s = 8.3 m and t =
9.5m. Determine its three angles and its area.

3. An area of triangle ABC is 144 2 with side a = 18 cm
and b = 23 cm. Solve the triangle.

Answer :

1. = 1 6 . 1248 , = 111.74 , = 38.26 , Area =
150

2 . 3 0 =.835 1 . 425 , = 56.27 , = 72.28 , Area =

3 . = 51.19 , = 84.73 , = 44.08 , and c =
16.068 cm

84

2ND TOPIC : TRIGONOMETRY

Let’s play games…………..

Fundamental of
Trigonometry

Trigonometric
Equations

85

COMPLEX NUMBER

SUBTOPIC :-
3.1 CONCEPT OF COMPLEX NUMBER
3.2 OPERATION OF COMPLEX NUMBER

Addition and Subtraction of Complex Number
Multiplication of Complex Number
Division of Complex Number
3.3 ARGAND DIAGRAM
Draw argand Diagram
Modulus and argument of complex number
3.4 OTHER FORM OF COMPLEX NUMBER
Other Forms of Complex Number
Multiplication and division of Complex Number in Polar and
Trigonometric form

86

3RD TOPIC : COMPLEX NUMBER

31.1.4CPOANRCTEIAPLT FORFACCOTMIOPNLEX NUMBER

What is Complex Number?
A combination of real part and imaginary parts

= + i here is
imaginary

parts

a is real b is

number imaginary
number

Example of complex number:

Real Imaginary Imaginary
number number part

= 2 + 3 23 3

w= 5 + 1 51 1
4 4 4

= −4 0 −4 −4

Notes: ≠ 0

Why complex number are created?

To find the square root of negative numbers
−1 = 2
−1 =

It happened because we assume − =

87

3RD TOPIC : COMPLEX NUMBER

Example 3.1

Simplify the following In terms of complex number:
a) −49
b) − −25

Solution 3.1 (a)

−49 separate number
= (−1)(49) with (−1)
= ( 2)(49)
= 2 49 Change −1 to 2
= = 7
Solution 3.1 (b)
− −25
= − −1 25
= − 2 25
= − 2 25
= − = −5

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88

3RD TOPIC : COMPLEX NUMBER

Example 3.1

Simplify the following In terms of complex number:
c) −4 + 3
d) 5 − −8

Solution 3.1 (c)

−4 + 3 Solution 3.1 (d)
= (−1)(4) + 3 5 − −8
= 2(4) + 3 = 5 − −1 8
= 2 + 3 = 5 − −1 8
= 5 − 2 8
= 5 − 8

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89

3RD TOPIC : COMPLEX NUMBER

Example 3.1
Simplify the following In terms of complex number:
e) 5
f) 8

Solution 3.1 (e)

5 If the index/power
= 4 � is odd number, take
= 2 2 � out one i to make it
= (−1)2�
= 1 = even number

Then change i into
power of 2 2 so
that we can get (-1)

Solution 3.1 (f)

8 If the index/power is
= 2 4 even number, we can
= −1 4 directly change i into
=1
power of 2 ( 2)

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90

3RD TOPIC : COMPLEX NUMBER

Example 3.1
Simplify the following In terms of complex number:
g) 5 7 + 6
h) 3 10 − 4

Solution 3.1 (g) (−1)odd number= (−1)
5 7 + 6 (−1)even number= (+1)
= 5 6 � + 6
= 5 2 3 � + 6 Solution 3.1 (h)
= 5(−1)3 � + 6 3 10 − 4
= 5 −1 + 6 = 3 2 5 − 4
= −5 + 6 = 3 −1 5 − 4
= 3(−1) − 4
= −3 − 4

91

3RD TOPIC : COMPLEX NUMBER

Test Your Knowledge 3.1

Simplify the following In terms of complex number:
a) −9 + 8
b) 3 − −41
c) 4 + −5
d) 5 6 + 2 9
e) 3 13 − 7 4
f) 4 + 11
Answer :

a) 3 + 8
b) 3 − 41

c) 2 + 5
d) −5 + 2
e) −4
f) 4 −

92

3RD TOPIC : COMPLEX NUMBER

31.2.4OPPAERRTAITAILONFROAFCCTOIOMNPLEX NUMBER

3.2.1 Addition and Subtraction of Complex Number
Remember, adding or subtracting complex number is like add or
subtract normal algebraic term.
So, what you should do?

1. You have to sort between real number and imaginary number.
2. Just add or subtract real number with real number
3. Then, add or subtract imaginary number with imaginary number

Addition: Expand the
+ + + negative value to
= + + +
c and d
Subtraction:
+ − +
= + − −
= − + −

93

3RD TOPIC : COMPLEX NUMBER

Example 3.2.1
If given = 3 + 4 and = 5 − , find
a) +
b) −

Solution 3.2.1 (a) Solution 3.2.1 (b)

+ −
= 3 + 4 + 5 − = 3 + 4 − 5 −
= 3 + 4 + 5 − = 3 + 4 − 5 +
= 3 + 5 + 4 − = 3 − 5 + 4 +
= 8 + 3 = −2 + 5

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94

3RD TOPIC : COMPLEX NUMBER

Example 3.2.1
If given = −4 + 2 and = 2 − 5 , find
c) 2 +
d) − 3

Solution 3.2.1 (c) Solution 3.2.1 (d)

2 + c) − 3
= 2 −4 + 2 + 2 − 5 = −4 + 2 − 3(2 − 5 )
= −8 + 4 + 2 − 5 = −4 + 2 − 6 + 15
= −8 + 2 + 4 − 5 = −4 − 6 + 2 + 15
= −6 − = −10 + 17

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95

3RD TOPIC : COMPLEX NUMBER

Test Your Knowledge 3.2.1

Solve the operation of the complex number below:
a) 3 + 2 + 3 − 4
b) 2 + − 3 − 5
c) 4 − 3 + 5 − 4
d) 3 − 4 + 2
e) 2 5 − 3 + 5 1 − 2
f) 3 2 − − 2 − 3
g) 4 + 2 2 − 3 − 5
h) −8 − 7 − 2 + (3 + 3 )

Answer :
a) 6 − 2
b) −1 + 6
c) 2
d) −4 +
e) 15 − 16
f) 4
g) 8 − 11
h) −7 − 4

96

3RD TOPIC : COMPLEX NUMBER

31.2.4OPPAERRTAITAILONFROAFCCTOIOMNPLEX NUMBER

3.2.2 Multiplication of Complex Number

Multiplication of complex numbers is like multiplying binomial functions.
If given = + and = + ,
Hence, � = ( + )( + )

We need to expand the function just like below

= ( + )( + )

= + + + 2 Don’t forget
= + + + −1 that = −
= − + +
Combine real
number with real

number,
imaginary
number with
imaginary

97

3RD TOPIC : COMPLEX NUMBER

Example 3.2.2
Multiply the complex numbers below
a) 2 3 + 4
b) 5 (4 − 2 )

Solution 3.2.2 (a)

2 3 + 4 Expand the
= 6 + 8 function

Solution 3.2.2 (b)

5 4 − 2 Expand the
= 20 − 10 2 function
= 20 − 10 −1
= 20 + 10 2 = −1

98