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CMV6120
Foundation Mathematics
Unit 11: Reduction Principle and
Simple Trigonometric Equations
Objectives
Students should be able to
 state the reduction principle
 use the reduction principle to find the trigonometric
ratios of angles related to special angles such as 30, 45
and 60
 solve simple trigonometric equations
Unit 11: Trigonometric Equations
page 1 of 5
CMV6120
Foundation Mathematics
Reduction principle and simple trigonometric equations
1.
Reduction Principle
1.1
Reference Angle
The reference angle  for an angle  is the positive acute angle formed by
the terminal side of  and the x-axis.
Given sin = 4/5, cos = 3/5 and tan = 4/3, you can easily fill in the table
below.
Angles
=
Quadrant
I
 = (180-)
II
 = (180+)
III
 = (360-)
IV
sin
4
5
4
5
4

5
4

5
sin
4
5
4
5
4
5
4
5
cos
3
5
3

5
3

5
3
5
cos
3
5
3
5
3
5
3
5
tan
4
3
4

3
4
3
4

3
tan
4
3
4
3
4
3
4
3
The trigonometric ratios of an angle  and its reference angle  have the
same numerical value,
i.e.
sin    sin 
cos   cos
tan    tan 
where the choice of sign (+ or -) depends on the quadrant in which  lies.
1.2
The Reduction Principle
In the first quadrant,
sin   sin 
cos  cos
tan   tan 
Unit 11: Trigonometric Equations
 = , we have
page 2 of 5
CMV6120
Foundation Mathematics
In the second quadrant,  = (180-), we have
sin   sin( 180   )  sin 
cos   cos(180   )   cos 
tan   tan(180   )   tan 
In the third quadrant,
 = (180+), we have
sin   sin( 180   )   sin 
cos   cos(180   )   cos 
tan   tan(180   )  tan 
In the fourth quadrant,
 = (360-), we have
sin   sin( 360   )   sin 
cos   cos(360   )  cos 
tan   tan( 360   )   tan 
N.B. It is also interesting to note that
sin(90+) = cos;
cos(90+) = -sin;
tan(90+) = -cot;
sin2 + cos2 = 1
sin(-) = -sin
cos(-) = cos
tan(-) = -tan
1.3 Trigonometric ratios of some special angles
The trigonometric ratios of some special angles are listed below for easy
reference. These could be found by using Pythagoras’ theorem.
Ratio\θ
0o
sinθ
0
1
2

)
4
2
2
cosθ
1
tanθ
0
3
2
1
2
2
1
30o(

)
6
45o(

)
3
3
2
60o(
1
2
3
90o(

)
2
1
0
∞
3
Unit 11: Trigonometric Equations
page 3 of 5
CMV6120
Foundation Mathematics
Example 1
Find the values of the following without using a calculator.
(i)
(ii)
cos150
tan(-

)
3
Solution
(i)
cos 150 = cos (180o – 30o)
= -cos30o
=-
(ii)
tan(-
3
2

)=tan(-60o)
3
= -tan 60o
= _________
Example 2
sin( 90   )
Simplify
sin( 90   )
Solution
sin( 90  ) cos 

1
sin( 90  ) cos 
Example 3
Simplify
sin( 180   )
cos(180   )
Solution
sin( 180   )
=
cos(180   )
2.
Simple Trigonometric Equations
We shall demonstrate some methods in solving simple equations involving
trigonometric ratios of angle.
Unit 11: Trigonometric Equations
page 4 of 5
CMV6120
Foundation Mathematics
Example 4
Solve the equation
2 sin   1  0 , 0    360
Solution
sin  
1
, θ = 30o, 150o
2
Example 5
4 cos 2   3  0 , 0    2
Solve the equation
Solution
Factorizing, (cos  
when (cos  

3
3
3
)(cos  
)  0 or simply cos   
2
2
2
3
)  0,
2
5 7 
,
6 6
3
)  0,
2
θ= ___________,__________
when (cos  
Example 6
Solve the equation
5 sin   2 cos 2   1  0 , 0    360
Solution
Since sin2θ + cos2θ=1,
the equation becomes
5sinθ –2(1- sin2θ) –1 = 0
(2sinθ– 1)(sinθ +3) = 0
gives sinθ=1/2
whence
θ= _________,________
[Note: (sinθ+3) = 0 has no solution.]
Unit 11: Trigonometric Equations
page 5 of 5