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Asia Pacific University of Technology and Innovation
Trigonometry
Definition
Trigonometry deals with the relationship between the sides and angles of the triangle.
Examples of things that we can measure in a triangle are the lengths of the sides, the angles and
the area of the triangle and so forth.
Triangles are used in the construction of houses, skyscrapers, bridges or any structure where
stability is desired. This branch of Mathematics is used to solve problems in Surveying,
Geography, Astronomy and Navigation etc.,
Definition on angle
An angle is when two rays (think of a ray as “half” of a line) have their end point in common.
The two rays make up the “sides” of the angle, called the initial and terminal side. A picture of
an angle is shown below.
Terminal
Initial
Pythagorean Theorem
Given a right angled triangle with sides of length a, b and c as shown below (c being the longest
side, which is also called the hypotenuse) then a2 + b2 = c2 .
c
a
b
Example:
Use the Pythagorean Theorem to find the missing side of the right angled triangle shown below.
3
c
7
Solution:
In this triangle we are given the lengths of the “legs” (i.e. the sides joining the right angle) and
we are missing the hypotenuse, or c. So
32 + 72= c2 or c2 = 58 or c = √58 = 7.616
Relation between radians and degrees
We have two ways to measure angles. In degrees a full revolution corresponds to 360° while in
radians a full revolution corresponds to 2π rads.
we have that 360° = 2π rads.
This can be rearranged to give the following useful (if not quite correct) relationship,
180
rad
1
rad
180
This gives a way to convert from radians to degrees or from degrees to radians.
Examples
1.
2.
Convert 240° to radians
Convert 3π/8 radians to degrees.
Solution:
1.
2.
240° = 240°.


=
4
rad
3
180
3 3 180
.
 67.5

8
8 
Review questions
1.
Convert 150° into radian
1
radian int o deg ree
4
Answers
2.
1.
2.
150° = 150°.

180

=
5
rad
6
1 180 1
7

  180 
 14195
4

4
22
Quadrants
Let X OX and YOY  be two lines at right angles to each other in the figure. We call
X OX and YOY  as x-axis and y-axis respectively.
Clearly these axes divide the entire plane into FOUR equal parts, called Quadrants.
Note: Angle is measured anti-clockwise from positive x-axis.
Memory Aid – “ALL Silver Tea Cups”
I
II
III IV
- Positive
The trigonometric functions:\
The six trigonometric functions are sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant
(sec) and cosecant (csc). They are defined in terms of ratios in the following way.
Note: Sine is the Opposite over the Hypotenuse, Cosine is the Adjacent over the Hypotenuse and
Tangent is the Opposite over the Adjacent.
Opposite side b

Hypotenuse
a
Adjacent side c

Cos B =
Hypotenuse
a
Opposite side b

Tan B =
Adjacent side c
Sin B =
Hypotenuse
a

Opposite side b
Hypotenuse
a

Sec B =
Adjacent side c
Adjacent side c

Cot B =
Opposite side b
Cosec B =
Note: Sin = Sine
Cos = Cosine
Tan = Tangent
Cosec [or] csc = cosecant
Sec = secant
Cot = cotangent
When θ is an acute angle, the definition of these ratios are given as above.
Example
If (2, 3) is a point on the terminal side of θ, find all the six trigonometric ratios.
Solution:
x = 2, y = 3, r = x 2  y 2  4  9  13
y
3
sin   
r
13
x
2
cos   
r
13
y 3
tan   
x 2
r
13
cos ec  
y
3
r

x
x
cot   
y
sec  
13
2
2
3
Review question
7
1.
If tan   , find cos  and cos ec .
25
8
2.
If cos   , find tan 
10
Answers
x 8
1.
cos   
 x  8, r  10, r = 252  72  625  49  674  25.96  26
r 10
x 25
cos   
r 26
r 26
cos ec  
y 7
y 6
2.
tan   
x 8
Sine Rule
Sine Rule can be used with any triangle as long as we have been given enough information.
a
b
c
=
=
SinA SinB SinC
Example 1
Find the length of a in the following triangle.
Solution: Given information in the following table.
Side
a=?
b=?
c = 10
Angle
A = 41°
B=?
C = 34°
Matching up the sides and angles, we get
a
c

sin A sin C
a
10


sin 41 sin 34
Rearranging to solve for a,
10  0.656
10  sin 41
a

 11.74m
a
0.559
sin 34
Note: b and B are unknown in the above example, so don’t choose the ratio.
Example 2
In triangle ABC, A = 35 017 , C = 45 013 and b = 42.1 cm, Find B, a and c using Sine Law.
Solution: Given information in the following table.
Side
Angle
A = 35 017
a=?
B  ? [1800  35017  45013]  9930
b = 42.1
C = 45 013
c=?
a
42.1


0
sin 35 17 sin 99 030
42.1
a
 sin 35 017 = 24.65
o
sin 99 30
42.1
c


0
sin 99 30 sin 45013
42.1
c
 sin 45 013 = 30.3
sin 99 o 30
Example 3
Triangle PQR has PQ = 7.73 cm, QR = 13 cm, and PRQ  35 . Calculate the two possible
values of RPQ .
Solution:
7.73
13

sin 35 sin P
13 sin 35
sin P 
 0.9646
7.73
P = sin 1 0.9646  74.71or105.29
Review questions
1.
In a triangle ABC, A = 40°, B = 52°, a = 9 cm, find C and b. (Note: c is the side opposite
to the angle C)
2.
In triangle ABC, A = 56 0 , C = 73 0 and a = 11 cm, find B, b and c.
Answers
1.
C = 180° - 52° - 40° = 88°
b
9

0
sin 52
sin 40 0
9
b
 sin 52 0 = 11.03 cm
0
sin 40
2.
B = 180  – 56  – 73  = 51
11
b


sin 56
sin 51
11
 sin 51  10.31 cm

sin 56
11
c


sin 56
sin 73
11
c
 sin 73  12.69 cm

sin 56
b
Cosine Rule
Cosine Rule can be used with any triangle as long as we have been given enough information.
The Cosine rule can be used to find:
1.
2.
An unknown side when two sides of the triangle and the included angle are given (SAS)
An unknown angle when 3 sides are given (SSS).
Finding an unknown side
2
2
2
2
2
2
2
2
2
a = b + c – 2bcCosA
b = a + c – 2acCosB
c = a + b – 2abCosC
Alternate version
Finding an unknown angle
b2  c 2  a 2
2bc
2
a  c2  b2
CosB 
2ac
2
a  b2  c2
CosC 
2ab
CosA 
Example
Angle A in the following triangle is 45o, length b is 2 units and length c is 3 units. Find angles B
and C and the length a using cosine rule.
Solution: Given information in the following table.
Side
Angle
a=?
A = 45°
b=2
B=?
c=3
C=?
Note: If an unknown side to find when two sides of the triangle and the included angle are given
(SAS), the following rule is to be used.
2
2
2
a = b + c – 2bcCosA
a2 = 22 + 32 -2 (2)(3) cos (45o)
a2=4 + 9 -12 cos (45o)
a = 2.1
To work out the angle B, we can rewrite the cosine rule.
b2=a2 + c2- 2ac cos B
4=4.5+9-12.6*cosB
4 – 4.5 – 9 = - 12.6 cos B
 9.5
 cos B
 12.6
 Cos B = 0.75
B = 410 24  41
a2  c2  b2
Note : CosB 
can also be used .
2ac
Note that A + B + C = 180°
So C = 180° - 45° - 41°= 94°
Review questions
1.
In triangle ABC, AB = 7 cm, AC = 13 cm, If its area = 30 cm2, calculate length of BC
2.
For the diagram below, find the length of a.
Answers
1.
BC 2  AC 2  AB 2  2 AB( AC) cos A
 7 2  132  2(13)(7) cos 41.250
= 218 – 136.8422
= 81.1578
BC = 81.1578
= 9.009
1
1
(b)(c) sin A  30  (7)(13)(sin A)
2
2
0
A  41.25
Note: Area =
2.
a  212  32 2  2(21)(32) cos 40 0  20.87
Graph of trigonometric functions
Formula for drawing the trig functions
Amplitude = A = A
Period =
360
B
If we sketch the graph of y = a sin (x),
a > 1 stretches the graph in the y-axis direction
a < 1 compresses the graph in the y-axis direction
Negative value of a flips graph in the x-axis
Here a = A
B defines how many times it repeats itself in 360°
Example 1
Draw the graph of y = 3 sin x
A
B
 60  (91)(sin A)
 sin A 
60
91
Solution:
Amplitude = A = 3  3
Period =
x
y
360
= 360°
1
-2π
0

3
2
3
-π
0


0
2
-3
0

2
3
π
0
3
2
-3
2π
0
Note: As this curve has a period 2π, we can extend this curve in both directions along the x-axis
for as many cycles as we please.
Example 2
Draw the graph of y = - cos 3x
Solution:
Amplitude = A =  1  1
Period =
x
y
360
= 120°
3
0
-1
30°
0
60°
1
90°
0
120°
-1
Note: The "minus" sign tells us that the graph is upside down
Review question
Draw the graph of y = 5 cos
x
2
Answer
Amplitude = A = 5  5
Period =
x
y
360
= 720°
1
2
0
180°
5
0
360°
-5
540°
0
720°
5
Note: you are reminded to draw using the table of values.
Area for any triangle
1
Area =  ab sin C
2
1
Area =  bc sin A
2
1
Area =  ac sin B
2
Example
Find the area of triangle ABC which has AB = 12 cm, BC = 14 cm and ABC = 35o.
Solution:
12  14  sin 35 0
Area =
= 48.18 cm2
2
Review questions
1.
In ABC , A  57 0 , AB  8.5cm and AC  6.3cm. Calculate the area of ABC.
2.
ABC is a triangle such that AB = 5 cm and AC = 10 cm and A  120 o . Find the area of
the triangle ABC.
Answers
1
1.
Area of ABC  bc sin A
2
1
 (6.3)(8.5) sin 57 0  22.46cm 2
2
1
2.
Area of the triangle ABC = (5)(10) sin 120 o
2
 21.65cm 2`
Table of Fundamental trigonometric Identities
Reciprocal identities
Pythogorean Identities
Quotient Identities
sin u
1
Sin u =
Sin2u + Cos2u = 1
Tan u 
csc u
cos u
1
cos u
Cos u =
Sin2u = 1 - Cos2u
Cot u 
sin u
sec u
1
Tan u =
Cos2u = 1- Sin2u
cot u
1
Csc u =
1 + tan2u = Sec2u
sin u
1
Sec u =
1 + cot2u = Csc2u
cos u
1
Cot u =
tan u
Co-Function Identities
Even Identities


Sin   u   cos u
Sin(-u) = - sin u
2



Cos   u   sin u
Cos(-u) = cos u
2



Tan   u   cot u
Tan(-u) = – tan u
2



Csc   u   sec u
Csc(-u) = - csc u
2



Sec   u   Cscu
Sec(-u) = sec u
2



Cot   u   tan u
Cot(-u) = - cot u
2

Example 1
Show that sec2 x + csc2 x = sec2x csc2 x
Solution:
The problem means that we are to write the left-hand side, and then show, through substitutions
and algebra, that we can transform it to look like the right hand side. We begin:
1
1
=

Sec2x + csc2x
Using Reciprocal identities
2
cos x sin 2 x
sin 2 x  cos 2 x
cos 2 x sin 2 x
1

2
cos x sin 2 x
1
1

2
cos x sin 2 x
= Sec2x + csc2x =RHS

on adding the fractions
Using Pythagorean identities
Using Reciprocal identities
Example 2
Show that
1
1

 2 sec 2 
1  sin  1  sin 
Solution:
LHS =
1
1
1  sin   1  sin 


on adding the fractions
1  sin  1  sin 
1  sin 2 
1  sin   1  sin 
2
2


by Pythogorean Identities
2
2
1  sin 
1  sin  cos 2 
2
 2 sec 2   RHS by reciporcal identities.
2
cos 
Note: LHS = Left hand side, RHS = Right hand side
Review questions
1.
2.
Prove cos x  tan x sin x  sec x
Prove the trigonometric identity sin 2   cos 2   1  2 cos 2  .
Answers
1.
2.
sin x
cos 2 x  sin 2 x
cos x  tan x sin x  cos x 
. sin x 
 sec x
cos x
cos x
sin 2   cos 2   (1  cos 2  )  cos 2   1  2 cos 2 
Solving trigonometric equations
A trigonometric equation is any equation that contains a trigonometric function.
Example 1
Solve for x in the following equation for 0  x  2
2 sin x  1  0
Solution:
2 sin x  1  sin x 
1
2
sin x  0.5
x = sin 1 0.5  30
Since the value is positive, the angle should be in QI and QII, hence
5

x  and x 
6
6
Note: sin 30° = 0.5 and also sin 150° = 0.5 So, the angle should be in QI and QII
The answers can be checked numerically and graphically.
Example 2
Solve for x in the following equation for 0  x  2
x
(i)
cos = 0.866
2
(ii)
tan 2x  3.732
Solution:
x
(i)
cos  0.866
2
Hence, reference angle = 300
x
2
Since 0 0 x360 0 ,0 0  180 0
Thus,
(ii)
x
is in 1st quadrant only.
2
x
 x  60 0
 30 0
2
tan 2 x  3.732, tan 75 0  3.732
Hence, reference angle = 750
0
0
0
0
Since 0 x360 ,0 2 x720
Thus, 2x is in 1st quadrant and 3rd quadrant
2 x  180 0  750 ,360 0  750 ,360 0  (180 0  750 ),360 0  (360 0  750 )
x  52.50 ,142.50 ,232.50 ,322.50
Review questions
1.
2.
Solve the following trigonometric equations cos  0.807 for 0    360 .
Solve the equation 2 cos ec  1  6 sin  for 0 o    360 o .
Answers
cos  0.807
1.
cos 36.20  0.807
2.
Hence the reference angle  36.20
Since cos is a negative value,  is in quadrant II or III.
Therefore,  =180o-36.20o, 180o+36.20o
= 143.80o, 216.20o
2 cos ec  1  6 sin 
2
 1  6 sin 
sin 
6 sin 2   sin   2  0
(3 sin   2)( 2 sin   1)  0
2
1
sin    or sin  
3
2
o
o
  221.81 , 318.9 , 30o, 150o
Points to remember:
2
Pythagorean theorem
Relation between radians and degrees
Six trigonometric functions
Sine Rule
Cosine Rule
Graph of trigonometric functions
a
b
a
a
Sec B =
c
d
a  Cb( x)  Cc( x)
=
dx =
SinA SinB SinC
Sin B =
a2 = b2 + c2 – 2bc sin A
b2 = c2 + a2 – 2ca sin B
c2 = a2 + b2 – 2ab sin C
Amplitude = A
Period =
Area for any triangle
Trigonometric Identities
Solving trigonometric equation
Formula/Concept
+ b = c2
π radian = 180°
1 radian = 57.3° [ approximately]
c
b
a
; Cos B =
;Tan B =
;Cosec B =
a
c
b
c
; Cot B =
b
2
360
B
1
 ab sin C
2
1
Area =  bc sin A
2
1
Area =  ac sin B
2
Refer the table given
Refer the following table. Use Identity when necessary.
Area =
For plotting the graphs, the values of y
 
0,
,
corresponding to x.
6 4
3 5
,
,
2 3
,
  2 3 5
7 5 4
, ,
,
,
, ,
,
,
3 2 3
7 11
,
,
.
4
6
4
6
6
4
3