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Trigonometry Review
(I)
Introduction
By convention, angles are measured from the initial line
or the x-axis with respect to the origin.
P
If OP is rotated counter-clockwise
positive angle
from the x-axis, the angle so formed
x
O
is positive.
But if OP is rotated clockwise
from the x-axis, the angle so
formed is negative.
O
x
negative angle
P
1
(II)
Degrees & Radians
Angles are measured in degrees or radians.
Given a circle with radius r, the
angle subtended by an arc of length r
measures 1 radian.
r
1
c
r
 rad  180
Care with calculator! Make sure your
calculator is set to radians when you are making
radian calculations.
2
r
(III) Definition of trigonometric ratios
y
P(x, y)
r
y
sin 

sin  
cos  
tan  
hyp
adj
hyp
opp
adj

1
 sin 
x
x
opp
1
Note:



y
cosec  
r
x
r
y
x
sec  

sin 
cos 
1
sin 
1 Do not write
1
1
cos , tan  .
cos 
1
cos 
cot  

tan  sin  3
Graph of y=sin x
y  sin x
1
0
90
180
270
360
1
4
Graph of y=cos x
y  cos x
1
0
90
180
270
360
1
5
Graph of y=tan x
y  tan x
0
90
180
270
360
6
From the above definitions, the signs of sin , cos 
& tan  in different quadrants can be obtained.
These are represented in the following diagram:
sin +ve
2nd
3rd
tan +ve
All +ve
1st
4th
cos +ve
7
(IV) Trigonometrical ratios of special angles
What are special angles?




90
,
0,
180 , 270 , 360



30 , 45 , 60

Trigonometrical ratios of these angles are
worth exploring
8
y  sin x
1
0

2
1
sin 0  0

3
2
sin 2  0
sin   0

sin 0°  0 sin  1
2
sin 180°  0
sin 90°  1
2
sin 360°  0
3
sin  1
2
sin 270°  1
9
1
y  cos x
0

2

1
cos 0°  1
cos 2  1
cos 360°  1
cos   1
cos 0  1
2
3
2
cos 180°  1

cos  0
2
cos 90°  0
3
cos  0
2
cos 270°  0
10
y  tan x
0
tan 0  0
tan 0°  0

2

3
2
tan   0
tan 180°  0
2
tan 2  0
tan 360°  0

tan is undefined.
2
3
tan is undefined.
2
tan 90° is undefined
tan 270° is undefined
11
Using the equilateral triangle
(of side length 2 units) shown
on the right, the following exact
values can be found.
 1
sin 30  sin 
6 2

3

sin 60  sin 
3 2
 1

cos 60  cos 
3 2

3

cos 30  cos 
6 2

 1
tan 30  tan 
6
3


tan 60  tan  3
3

12

1
2
sin 45  sin


4
2
2

cos 45  cos

4

1
2

2
2

tan 45  tan  1
4

Complete the table. What do you observe?
13
14
Important properties:
2nd quadrant
sin(  )  sin 
1st quadrant
sin(2  )  sin 
cos(   )   cos 
cos( 2  )  cos 
tan(  )   tan 
tan(2  )  tan 
3rd quadrant
sin(  )   sin 
cos(   )   cos 
tan(  )  tan 
15
Important properties:
4th quadrant
sin(2  )   sin 
cos( 2  )  cos 
tan(2  )   tan 
sin()   sin 
cos( )  cos 
tan()   tan 
In the diagram,  is acute.
However, these
relationships are true for
all sizes of .
16
Complementary angles
Two angles that sum up to 90° or  radians are called
2
complementary angles.
E.g.: 30° & 60° are complementary angles.
 
 and     are complementary angles.
2

Recall:
1
sin 30  cos 60 
2


1
tan 30  cot 60 
3




3
sin  cos 
3
6 2


tan 60  cot 30  3
17
We say that sine & cosine are complementary
functions.
Also, tangent & cotangent are complementary
functions.


sin
40

cos
50
E.g.:
3

cos

sin
8
8

3
tan  cot
8
8


cot 35  tan 55
18
E.g. 1: Simplify
(i) sin 210
5
(ii) cos
3
2
(iii) tan(– )
3
3
(iv) sin(   )
2
Solution:
(a) sin 210  sin(180°+30)  - sin 30 = 
1
2
3rd quadrant
210° = 180°+30°
19
(b) cos
5
4th


1
 cos(2  )  cos 
3
3 2
3
5
3
quadrant
 2  3
 2 
2



  tan( )   tan(   )
(c) tan 
 3 
3
3



 ( tan
3
)
 3
20
E.g. 2:
If sin x = 0.6, cos x = 0.8, find
(a) sin (3  x)
(b) cos (4  x).
Soln :
sin (3 - x)
cos (4 + x)
 sin (2   - x)
 cos (2 + x)
 sin (  - x)
 sin x
 0.6
 cos x
 0.8
21
(V) Basic Angle
The basic angle is defined to be the positive, acute angle
between the line OP & its projection on the x-axis. For
any general angle, there is a basic angle associated with it.
Let  denotes the basic angle.
So 0    90 or 0     .


2
P

O

P


O
  180°  
or     
22
(0    90 or 0     )
2


O
P
   – 180°
or    – 

O

P
  360°   or
  2  
23
E.g.:

  55
basic     55


(1st quadrant)

(1st quadrant)
P

O

4
basic    

4
24
E.g.:
 (2nd quadrant)
  130


P

basic     180  130  50

2
3
basic      
3


O
  180°  
or     
(2nd quadrant)
2


3
25
E.g.:
 (3rd quadrant)
  200




basic     200  180  20

5
4

P
(3rd quadrant)
basic    
5
4

O
   – 180°
or    – 

4
26
E.g.:
 (4th quadrant)
  300



basic     360  300  60

O

11
6
basic     2 
6

P
  360° 
or   2  
(4th quadrant)
11


6
27
Principal Angle & Principal Range
Example: sinθ = 0.5


2

2
Principal range
Restricting y= sinθ inside the principal range makes it a
one-one function, i.e. so that a unique θ= sin-1y exists
28
E.g. 3(a):
Since
sin (
3
1
 ) 
2
2
3
sin (   )
2
is positive, it is in the 1st or 2nd quadrant

Basic angle, α =
4
3




Therefore 2
4

5
(inadmissib le )
4
Hence,  
. Solve for θ if 0    
or
3

   
2
4
or

3
4
3
4
29
E.g. 3(b): cos (2  250 )  0.8 . Solve for θ if 0    180
0
Since cos (2  25 ) is negative, it is in the 2nd or 3rd quadrant
Basic angle, α = 36.870o
Therefore 2  25  180  36.870
or
2  25  180  36.870
  59.1
or
  95.9
Hence,   59.1
or
95.9
30
(VI) 3 Important Identities
P(x, y)
By Pythagoras’ Theorem,
x2  y 2  r 2
2
2
 x  y
     1
r r
x
y
Since sin A 
and cos A  ,
r
r
sin A2  cos A2  1
sin2 A  cos2 A  1
O
r
y
A
x
Note:
sin 2 A  (sin A)2
cos 2 A  (cos A)2
31
(VI) 3 Important Identities
(1)
sin2 A + cos2 A  1
Dividing (1) throughout by cos2 A,
tan 2 x = (tan x)2
(2)
tan2 A +1  sec2 A
1
Dividing (1) throughout by sin2 A,
(3)
1+
cot2 A

csc2 A
cos 2 A
 1 


 cos A 
 (sec A)
2
2
2
 sec A
32
(VII) Important Formulae
(1)
Compound Angle Formulae
sin( A  B)  sin A cos B  cos A sin B
sin( A  B)  sin A cos B  cos A sin B
cos( A  B)  cos A cos B  sin A sin B
cos( A  B)  cos A cos B  sin A sin B
tan A  tan B
tan( A  B) 
1  tan A tan B
tan A  tan B
tan( A  B) 
1  tan A tan B
33
E.g. 4: It is given that tan A = 3. Find, without using calculator,
(i) the exact value of tan , given that tan ( + A) = 5;
(ii) the exact value of tan  , given that sin ( + A) = 2 cos ( – A)
Solution:
(i)
Given tan ( + A)  5 and tan A  3,
tan   tan A
tan(  A) 
1  tan  tan A
tan   3
5
1  3 tan 
5  15 tan   tan   3
1
 tan  
8
34
Solution:
(ii)
Given sin ( + A) = 2 cos ( – A) & tan A  3,
sin  cos A + cos  sin A = 2[ cos cos A + sin sin A ]
(Divide by cos A on both sides)
sin  + cos  tan A = 2(cos  + sin tan A)
sin  + 3cos  = 2(cos  + 3sin )
5sin  = cos 
1
tan  =
5
35
(2)
Double Angle Formulae
(i) sin 2A = 2 sin A cos A
Proof:
sin 2 A
(ii) cos 2A = cos2 A – sin2 A
 sin( A  A)
 sin A cos A  cos A sin A
= 2 cos2 A – 1
= 1 – 2 sin2 A
(iii) tan 2 A 
2 tan A
2
1  tan A
 2 sin Acos A
cos 2 A  cos( A  A)
 cos 2 A  sin 2 A
2
2
 cos A  (1  cos A)
 2 cos 2 A  1
36
(3)
Triple Angle Formulae:
(i) cos 3A = 4 cos3 A – 3 cos A
Proof:
cos 3A = cos (2A + A)
= cos 2A cos A – sin 2A sin A
= ( 2cos2A 1)cos A – (2sin A cos A)sin A
= 2cos3A  cos A – 2cos A sin2A
= 2cos3A  cos A – 2cos A(1  cos2A)
= 4cos3A  3cos A
37
(ii) sin 3A = 3 sin A – 4 sin3 A
Proof:
sin 3A = sin (2A + A)
= sin 2A cos A + cos 2A sin A
= (2sin A cos A )cos A + (1 – 2sin2A)sin A
= 2sin A(1 – sin2A) + sin A – 2sin3A
= 3sin A – 4sin3A
38
E.g. 5: Given sin2 A  16 & A is obtuse, find,
25
without using calculators, the values of
(i) cos 4A
(ii) sin ½A
Solution:
16
2
Since sin A 
25
4
sin A  
5
4
But A is obtuse, sin A =
5
4
5
A
3
3
cos A  
5
39
2
(i) cos 4 A  1  2 sin 2 A
 1  2(2sin A cos A)
2
24


 1  2  
 25 
527

625
4
2
5
A
3
3
cos A  
5
40
A
(ii) cos A = 1 – 2sin2 (
)
2
3
A
2
 = 1 – 2sin (
)
5
2
2 A 4
sin   
2 5
A
Since 90  A  180, 45   90.
2
A
A
st
i.e.
lies in the 1 quadrant. So sin  0
2
2
A
2
sin (
)=
2
5
41
E.g. 6: Prove the following identities:
4
2
(i)
cos 4 A  8 cos A  8 cos A  1
Recall:
cos 2A = cos2 A – sin2 A
= 2 cos2 A – 1
Solution:
(i) LHS = cos 4 A
 2 cos 2 2 A  1
2
= 1 – 2 sin2 A
2
 2(2 cos A  1)  1
4
2
 2( 4 cos A  4 cos A  1)  1
4
2
 8 cos A  8 cos A  1
= RHS
42
1  cos 2 A
E.g. 6: Prove the following identities: (ii)
 tan A
sin 2 A
Solution:
1 cos 2 A
(ii) LHS =
sin 2 A
1  (1  2 sin 2 A)

2 sin A cos A
2
2 sin A

2 sin A cos A
sin A

cos A
 tan A = RHS
43
E.g. 6: Prove the following identities:
(iii) 1  cos   cosec   cot  , where 0    
1  cos 
2
Solution:
1  cos 
LHS 
1  cos 
(1  cos  )(1  cos  )

(1  cos  )(1  cos  )


(1  cos  )2
1  cos 
2
(1  cos  )2
sin 
2
44

(1  cos  )2
sin 
2
1  cos 

sin 
1  cos 

sin 
1
cos 


sin  sin 
( Given 0   

,
2
0  sin   1 and 0  cos   1.)
 cos ec  cot 
 RHS
45
E.g. 6: Prove the following identities:
3
3
cos


cos
3

sin
  sin 3
(iv)

3
cos 
sin 
Solution:
cos3   cos 3 sin 3   sin 3
LHS =

cos 
sin 
cos 3
sin 3
2
2
 cos  
 sin  
cos 
sin 
sin 3 cos   sin  cos 3
 1
sin  cos 
 1
sin( 3   )
1 sin 2
2
 1  2  3  RHS
46
(5)
The Factor Formulae (Sum or difference of
similar trigo. functions)
Recall compound angles formulae:
sin( A  B)  sin A cos B  cos A sin B …. 
sin( A  B)  sin A cos B  cos A sin B …. 
cos( A  B)  cos A cos B  sin A sin B …. 
cos( A  B)  cos A cos B  sin A sin B …. 
 +  : sin( A  B)  sin( A  B)  2 sin A cos B
   : sin( A  B)  sin( A  B)  2 cos A sin B
 +  : cos( A  B)  cos( A  B)  2 cos A cos B
   : cos( A  B)  cos( A  B)  2 sin A sin B47
By letting X = A + B and
obtain the factor formulae:
Y = A – B, we
 X Y   X Y 
(1) sin X  sin Y  2 sin
 cos

 2   2 
 X Y   X Y 
(2) sin X  sin Y  2 cos
 sin

 2   2 
 X Y   X Y 
(3) cos X  cos Y  2 cos
 cos

 2   2 
 X Y   X Y 
(4) cos X  cos Y  2 sin
 sin

 2   2 48
E.g. 8: Show that
2
(i)cos   cos 3  cos 5  cos 3 (4 cos   1)
Solution:
(i) LHS
Using cos X  cos Y
= cos  + cos 3 + cos 5
= (cos 5 + cos ) + cos 3
= 2cos 3 cos 2 + cos 3
X Y   X Y 

 2 cos
 cos

 2   2 
= cos 3 [2cos2 + 1]
= cos 3 [ 2(2 cos2 – 1) + 1 ]
= cos 3 (4 cos2 – 1) = RHS
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sin A  sin B
A B

E.g. 8: Show that (ii)
  cot 

cos A  cos B
 2 
Soln:
sin A  sin B
(ii) LHS =
cos A  cos B
 A B  A B
2 sin
 cos

2   2 


 A B  A B
 2 sin
 sin

 2   2 
A B

cos

A B
2 



  cot 
= RHS

A B

2


sin

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 2 
E.g. 8: Show that
(iii) sin  + sin 3 + sin 5 + sin 7 = 16 sin cos2 cos2 2
Soln:
(iii) LHS = sin  + sin 3 + sin 5 + sin 7
= (sin 3 + sin  ) + (sin 7 + sin 5 )
4
2
12
2
 2 sin cos  2 sin
cos
2
2
2
2
= 2sin 2 cos + 2sin 6 cos
= 2cos [ sin 6 + sin 2 ]

8
4 
 2 cos   2 sin cos 
2
2

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
8
4 
 2 cos   2 sin cos 
2
2

= 4 cos cos 2 sin 4
= 4 cos cos 2 [ 2 sin 2 cos 2 ]
= 8 cos cos2 2 sin 2
= 8 cos cos2 2 ( 2 sin cos )
= 16 sin cos2 cos2 2
= RHS
52