Mathematical Investigations III
Name
Mathematical Investigations III
Trigonometry - Modeling the Seas
ANOTHER WAY TO VIEW THE TRIGONOMETRIC FUNCTIONS
Consider APO .
P
P(x, y)
1
1
y

x
O
O
y

A
x
From the right triangle trigonometry you have
studied previously, you know that
OA OA
cos( ) 

 OA = x-coordinate
OP
1
and
AP AP
sin( ) 

 AP = y-coordinate
OP
1
That is, in the unit circle,
x  cos( ) and y  sin( ) .
A
There is another idea that is clear from this diagram, ( AO)2  (PA)2  1 .
This can also be stated as
cos 2 ( )  sin 2 ( )  1
(Notice that the function is squared, NOT the argument of the function.)
Use these concepts to complete the following pages.
1.
(
,
Fill in the coordinates around the circle. Then find the trigonometric values as requested.
)
(4 / 5, 3 / 5)

1
(
,
)
(
,
Find:
cos   
Find:
sin   
cos   
sin   
cos     
sin     
cos     
sin     
) cos 2    
sin 2    


cos     
2



sin     
2

Trig. 6.1
Rev. F07
Mathematical Investigations III
Name
2.
3.
5
and sin() > 0, find each of the following.
13
[Hint: Draw a unit circle and place  on it.]
cos(–) =
cos( – ) =
If cos( ) 


cos     
2

sin() =
sin(–) =
sin( – ) =
sin( + ) =


sin     
2

If cos(  ) 
cos(–) =
4.
cos( + ) =
7
and sin() < 0, find the following.
25
cos( – ) =
cos( + ) =


cos     
2

sin() =
sin(–) =
sin( – ) =
sin( + ) =


sin     
2

If sin( ) 
cos() =
2
and cos() < 0, find the following.
3
cos(–) =
cos( – ) =
cos( + ) =


cos     =
2

sin(–) =
sin( – ) =
sin( + ) =


sin     
2

Trig. 6.2
Rev. F07
Mathematical Investigations III
Name
5. Find each of the following. (Be careful. This is not a unit circle.)
A
(8, 15)
6.
Find each of the following.
B
(4, 2)
cos(A) =
sin(A) =
cos( –) =
sin( –) =
cos(A + ) =
sin(A + ) =
cos(A + 5) =
(Think about what adding
5 does to the angle.)
sin(A + 5) =


cos   A =
2



sin   A 
2



cos  A   
2



sin  A   
2

cos B 
sin B 
cos   B  
sin   B 
cos B    
sin B    
cos B  3  
sin B  2  


cos   B  
2



sin   B 
2



cos  B   

2


sin  B   

2
Trig. 6.3
Rev. F07