1.
P(x; y) is a point on the Cartesian plane. The length of the line drawn from the origin
0 to point P is r. The angle between OP and the x-axis is  .
P (x; y)
r
θ
R
1.1
1.1.1
1.1.2
1.1.3
Complete the terms of x, y and r:
sin  =
cos  =
tan  =
1.2
If OP is called the hypotenuse, what are OR and PR called respectively?
1.3
Give the definition of sin  , cos  and tan  in terms of the three sides mentioned.
Special Angles:
0o ; 30o ; 45o ; 60o ; 90o
1.
1.1
1.2
2.
2.1
2.2
Calculate the value of the following expressions without using a calculator:
3 cos 60°. sin 30°.tan 2 60°
cos 30. sin 30
tan 45. sin 60
Prove the following without using a calculator:
sin 45
tan 45° =
cos45
1 – 2 sin 2  = cos2  if  = 30°.
Solution:
1.1
3 cos 60°.sin 30°.tan 2 60°
1.2
2
3  1  1  3 
9
 
=   
1  2  2  1 
4



= 




1
=2
4
2.1
LHS = tan45°
2
=
2
=1
sin 45
RHS =
cos45
cos 30. sin 30
tan 45. sin 60
2.2
2
= 2
2
2
=1
 LHS = RHS
sin 45
 tan45°=
cos45
3 1
. 
2   2 
2  3 
.

2   2 
=
1
2
LHS = 1 – 2 sin 2  of  = 30°
= 1 – 2sin 2 30°
1
= 1 – 2 ( )2
2
1
=1–2( )
4
1
=1–
2
1
=
2
RHS = cos2  of  = 30°
= cos (2  30°)
= cos60°
1
=
2
 LHS = RHS
 1 – 2 sin 2  = cos if  = 30°
Exercise:
1.
Calculate the value of the following expressions without the use of a calculator:
1.1
sin 2 60° + cos 2 60°
1.2
cos 90° + 8 sin 30° – cos180° – 5tan45°
1.3
tan60. sin 60
cos 45 . sin 45
1.4
tan60°.sin60° – 2sin45°.cos45° – cos 2 45°
1.5
1.6
cos 2 30 – sin 2 45
tan 2 60° – (cos45° sin30°)
Reduction Formulae:
90o
2
1
Sin
All
180o -
90o -
90o +
0o
180o
180o +
360o -
Tan
n
3
Cos
4
270o
2.
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.14
2.15
Complete the table with the help of reduction formulae:
sin 180   
tan360 –  
cos90   
tan180   
sin 90 –  
cos360 –  
sin –  
tan360 –  
cos –  
sin 90   
cos180 –  
cos90 –  
sin 360   
tan360   
cos360   
360o
3.
3.1
3.2
Simplify by means of reduction formulae:
cos 360   . cos 180 –  . cos 90 –  . sin –  
cos 90 –  . tan 360 –  
sin 180 –  . tan –  
4.
Prove using reduction formulae:
cos –  . sin 360 –  
 sin 90 –   .
sin 90   . tan 180 –  
5.
Determine the value of the following expression without using a calculator (the
answer should be in surd form):
sin 390 º. cos 225 º . sin 210 º . sin 315 º . sin 260 º
cos 350 º
1.
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
1.14
Complete the following table:
Trigonometric
function
tan180º  
sin 360º– 
cos 90º  
sin 180º– 
sin 90º– 
cos –  
sin 180º 
tan360º– 
cos360   
sin 90º  
cos360   
tan  
cos180   
sin360   
Quadrant
Sign
Reduced
function
2.
2.1
2.2
2.3
2.4
2.5
Complete the following by means of coordinate and reduction formulae:
cos 17º  sin______
cos 248º  – cos_____
tan 196º  tan______
sin 144º  sin______
sin 144º  cos_______
3.
Simplify using reduction formulae:
3.1
sin 360º –  . tan180º   . cos–  
cos90º   sin 180º –  
3.2
sin – A. sin 180º  A. tan360º – A
tan180º – A. cos90º – A. cos180º – A
3.3
cos90º –  . sin180º –   – sin90º   cos180º   
3.4
tan180º   . cos360º   
– cos90º –  . cos90º    – sin –  . sin 180º –  
3.5
tan180º  A. sin 90º – A. tan360º – A. cos360º  A
cos360º – A. cos180º – A
3.6
cos360º – A. cos90º – A. tan– A
tan180º  A. sin 90º – A. cos180º  A
4.
Use reduction formulae to prove the following:
4.1
cos90º –  
 tan360º –  
cos180º   
4.2
sin 2 360º –   
4.3
cos–  . cos90º   . cos90º –   cos360º –  . sin –  

sin 360º –  . sin 180º –  
sin 180º   
4.4
sin 180º – A  tan180º  A  tan360º – A
 tan180º – A. sin 270º
cos180º  A. cos180º
5.
sin 2 90º   . tan360º –  . cos180º   
1
sin180º –  
Determine the values of the expressions without using a calculator. (Answer should
be in surd form.)
5.1
cos 150 º. sin 330 º
tan 225 º. sin 300 º
5.2
sin 2 – 130º   sin 2 220º
5.3
tan120º sin 300º–2 sin135º tan315º– cos2 135º
5.4
tan 216 º. sin 126 º
sin 144 º
6.
Determine the value of
7.
7.1
7.2
7.3
7.4
7.5
If sin 53º  p , express each of the following in terms of p:
sin 233 º
cos 307 º
tan 233 º
cos 127 º
cos 143 º
cos 

. cos . cos 2 without using a calculator if   60º .
2
2
Identities:
You must be able to deduce and solve the following two fundamental identities:

Place  POR in a standard position with P O R =  and the coordinates of P equal to
(x; y).
1.
tan  =
sin 
cos 
y
x
y
sin 
RHS =
 r
x
cos
r
y r
= 
r x
y
=
x
LHS = RHS
sin 
 tan  =
cos 
LHS = tan  =
P (x; y)
r
y
θ
x
O
This identity is called a quotient identity.
2.
sin 2   cos2   1
LHS = sin 2   cos2 
y
x
= ( )2 + ( )2
r
r
2
2
x
y
= 2 + 2
r
r
2
2
y x
=
r2
r2
= 2 (according to Pythagoras)
r
=1
= RHS
 sin 2   cos2   1
sin2  + cos2 = 1
sin2  = 1 - cos2
cos2 = 1 - sin2 
This identity is called a quadratic / square identity.
R
Examples:
1.
1.1
Use the fundamental identity to simplify the following:
1
– tan 2 
2
cos 
1.2
cos  – cos  . sin 2 
2.
Prove the following with the help of fundamental identity:
tan  1 sin   cos
=
tan – 1 sin  – cos
1
1

(1  sin A) (1 – sin A) cos2 A
2.1
2.2
Solution:
1.
1.1
1.2
1
1
sin 2 
2

–
tan
=
–
cos 2 
cos 2 
cos2 
1 – sin 2 
=
cos2 
cos2 
=
cos2 
=1
cos  – cos  . sin 2  = cos  (1 – sin 2  )
= cos  cos2 
= cos 3 
2.
2.1
LHS
tan   1
tan   1
sin 
1
cos

=
sin 
1
cos
sin   cos
cos
=
sin   cos
cos
sin   cos
cos


cos
sin   cos
sin   cos

sin   cos
= RHS

2.2
1
(1  sin A) (1 – sin A)
1
=
1 – sin 2 A
1
=
cos 2 A
= RHS
LHS =
Exercise :
1.
1.1
1.2
1.3
1.4
1.5
1.6
2.
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
Use the fundamental identity to simplify the following:
(1 – cos A)(1  cos A)
sin A
2
1 – cos  – sin 2 
(1 – cos )(1 cos )
(1 – sin  )(1 sin  )
(sin  + cos  ) 2
sin 
tan 
tan 2  .sin 2  + tan 2  .cos 2 
Prove the following with the help of fundamental identity:
1
sin A.tan A + cos A =
cos A
4
4
2
sin A – cos A  sin A – cos2 A
1
cos x + sin x.tan x =
cos x
tan2 
1 tan2 
 sin  
(sin  + tan  ) 
 = sin  .tan 
 1  cos 
cos A 1 – sin A

1  sin A cos A
cos  sin  1 – 2 sin  cos

cos – sin  cos2  – sin 2 
1
cos A
sin A
–

sin A sin A 1  cos A
1  cos
sin 
2


1  cos
sin 
sin 
sin 2  =
sin  cos (1  tan2  )
= (sin  + cos  ) 2 – 2 sin  cos 
tan
1.
1.1
sin 2 60º  cos2 60º
2
 3   1 2
  
 
 2
2


3 1
 
4 4
4

4
1
1.2
cos90º  8 sin 30º  cos180º  5 tan45º
0 1
  8    1  51
2 2
 0  4 1 5
0
1.3
tan 60 º. sin 60 º
cos 45º. sin 45º
 3  3 



 1 . 2 



 2  2 



 2 . 2 



3
 2
2
4
3 4
 
2 2
3
1.4
tan60º. sin 60º  2 sin 45º. cos45º  cos2 45º
 3   3   2  2   2 
.
 

 

 
  3   2 2  2    2 
1


 

 

3
2
 1
2
4
3
1
 1
2
2
0
cos2 30º sin 2 45º
1.5

32  2 

– 

2
 2 
2
3 2

4 4
1

4
1

2

1.6
tan2 60ºcos45º. sin 30º 
2
 3  2 1
 
 
. 
 
 1   2 2
2
4
12  2

4
 3
2