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Engineering
MATHEMATICS
MET 3403
1.Trigonometric Functions



Every right-angled triangle contains two acute angles.
With respect to each of these angles, there are six
functions, called trigonometric functions, each involving
the lengths of two of the sides of the triangle.
Consider the following triangle ABC
B
hypotenuse
opposite
a
C


adjacent
A
AC is the side adjacent to angle a, BC is the side opposite to
angle a.
Similarly, BC is the side adjacent to angle b, AC is the side
opposite to angle b.

Six trigonometric functions with respect to angle a:
B
Hypotenuse (r)
opposite (y)
a
C
Adjacent (x)
opposite
BC y


hypotenuse AB r
adjacent
AC x
cos(a ) 


hypotenuse AB r
opposite BC y
tan( a ) 


adjacent AC x
sin( a ) 

Note: sin( a)
cos(a)
 tan( a)
A
hypotenuse AB
1


opposite
BC sin( a)
hypotenuse
AB
1
sec(a) 


adjacent
AC cos(a)
adjacent AC
1
cot( a) 


opposite
AB tan( a)
csc(a) 
cos(a)
 cot(a)
sin( a)

Example: Consider the right-angled triangle, with lengths
of sides indicated, find sin(d), cos(d), tan(d), sin(e), cos(e),
tan(e).
E
e
13
12
d
F
EF 12
sin( d ) 

ED 13
DF 5
cos(d ) 

ED 13
EF 12
tan( d ) 

DF 5
5
D
DF 5
sin( e) 

ED 13
EF 12
cos(e) 

ED 13
DF 5
tan( e) 

EF 12

Pythagorean Theorem (畢氏定理)
B
C
A
BC 2  AC 2  AB 2

Pythagorean Identities

derived from Pythagorean theorem
sin 2 ( a )  cos 2 ( a )  1
tan 2 (a )  1  sec 2 (a )
cot 2 ( a )  1  csc 2 (a )

Example:
 In
right-angled triangle, sin(a)=4/5, find the values of
the other five trigonometric functions of a.
B
5
4
a
A
 Since
C
sin(a)=opposite over hypotenuse=4/5
BC  4, AB  5
AC 
 AB2  BC 2

52  42
3
3
4
3
5
5
cos(a)  , tan( a)  , cot(a)  , sec(a)  , csc(a) 
5
3
4
3
4

Example:
 If,
in right-angled triangle, sin(a)=7/9, find the values
of cos(a) and tan(a).
 Using trigonometric identity,
sin 2 (a)  cos 2 (a)  1
2
7
cos 2 (a)  1     cos(a) 
9
sin( a)
 Since tan( a) 
cos(a)
79
7
7 2
 tan( a) 


8
4 2 9 4 2
32 4 2

81
9

Angle in degree



Each degree is divided into 60 minutes
Each minute is divided into 60 seconds
Example:
 Express the angle 265.46  in Degree-Minute-Second (DMS)
notation
60'
1
60"
 265  27'0.6'
1'
 26527'36"
265.46  265  0.46 

Angle in radian

A unit circle has a circumference of 2
One complete rotation measures 2  radian
Angle of 360 = 2 radian

Example: 30  30 




180 6
  180
 
 45
4 4


Special Angles (1)
 For
a 30-60-90 right-angled triangle
B
60◦
2
1
30◦
A
 From
3
C
the triangle,
sin( 30 ) 
sin( 60 ) 
1
3
1
3
, cos(30 ) 
, tan( 30 ) 

2
2
3
3
3
1
, cos(60 )  , tan( 60 ) 
2
2
3

Special Angles (2)
 For
a 45-45-90 right-angled triangle
B
45◦
2
1
45◦
A
 From
1
C
the triangle,
sin( 45 )  cos( 45 ) 
1
2

, tan( 45 )  1
2
2

Unit circle and sine, cosine functions
 Start
measuring angle from positive x-axis
‘+’ angle = anticlockwise
θ
‘’ angle = clockwise

Angle and quadrants
of a trigonometric function for an angle in 2nd,
3rd or 4th quadrants is equal to plus or minus of the
value of the 1st quadrant reference angle
 Value
y
(x,y)

’
0
(x,y)
’
Quadrant IIQuadrant I
x
Quadrant IIIQuadrant IV
y
y
(x,y)

’0
(x,y)
’
(x,y)
x

’
0 ’
(x,y)
x
 The
sign of the value is dependent upon the quadrant
that the angle is in.
Quadrant II
SINE +ve
Quadrant I
ALL +ve
Quadrant III
TANGENT +ve
Quadrant IV
COSINE +ve

Exercise: find WITHOUT calculator:
sin(30 °)
cos(45 °)
tan(315 °)
sin(60 °)
cos(180 °)
tan(135 °)
sin(240 °)
cos(-45 °)
= _________
= _________
= _________
= _________
= _________
= _________
= _________
= _________
Hint
1
3
1
3
sin( 30 )  , cos(30 ) 
, tan( 30 ) 

2
2
3
3
sin( 60 ) 
3
1
, cos(60 )  , tan( 60 )  3
2
2
•Simple trigonometric equations
Notation :

If sin  = k then  = sin-1k (sin-1 is written as inv sin or
arcsin). Similar scheme is applied to cos and tan.
e.g. Without using a calculator, solve sin  =  0.5, where
0o    360o
e.g. Solve cos 2 =  0.4 ,where 0    2