Trigonometric Ratios

30 
45 
60 
90 
c
6
c
4

c
3
c
2
sin 
1
2
1
2
3
2
1
cos
3
2
1
2
1
2
0
tan 
1
3
1
3
-
Remember:
y
x
y
sin   ,cos  , tan  
r
r
x
II
S
p(x,y)
T
III
Show that
I
A
C
IV
:
y
x
,r 
sin 
cos
y
x

sin  cos
sin  y
  tan 
cos x
sin 
so tan 
cos
r
To find the value of cos we follow these steps:
1. Draw the unit circle with the angle
2. Find the reference angle in the first quadrant.
3. Use symmetry to find the relationship between the angles (CAST)
4. State the relationship and give the answer.
Remember:
Example: Find the following: sin 210, cos
7
5
, tan
.
4
4
sin 210
reference angle is 30, it is the third quadrant so sin 210 = - ½
cos
7
4
reference angle is
tan
5
4
reference angle is

7
1
, it is the fourth quadrant so cos

4
4
2

5
, it is the second quadrant so tan
 1
4
4
Example: Given that sin  
A
3
B 
32  12  a 2
9  1  a2
a 82 2
cos 
2 2
3
1
C
1

and 0    , find cos
3
2
Given that sin  0.3, where 0   
a) sin    
a)
b)
c)
b) sin  2   
sin       sin  
sin      0.3
sin  2      sin  
sin  2     0.3


cos      sin 
2



cos      0.3
2

Example:

2
, find:


c) cos    
2

Solving trigonometric equations: remember to use the unit circle
or a graph to help.
1. Solve tan   1, where0    2
 
tan    1 Are there any other answers? Remember CAST
4
 5 
tan    1
 4 
1
2. Solve sin   , where  2    2
2
 1
sin 
6 2
Using CAST it is also positive in the second quadrant so
 5
 
6
6
We also need to consider the negative part of the graph so:

11
5
7
 2  
 2  
and
6
6
6
6
3. Solve 5cos2  2.5, where0    2
If you look at your graph there is an additional wave because it’s
2θ so we need to look at
1
5cos 2  2.5
cos 2 
2
2 



6
3


5
,
6
2 

6

11
6

6
and  

6

7
6
4. Use your calculator to approximate 5tan 2  3, where0    
3
tan 2 
5
3
2  tan 1
5
  0.2702

since each tan
2
has a period of π but it’s 2θ so there will be 2 waves in that span of

the graph so 0.2702   1.8410
2
Since there is 2 possible solutions we need to add
5. Find the exact solutions to sin(3x)  cos(3x) for 0  x  2
sin(3 x)
1
cos(3x)
tan(3 x)  1
3x 
x

4
,3 x 
 5
5
4
,
12 12
2
 2 9 9 2 17
period 
, so


,


3
12 3 12 12
3
12
5 2 13 13 2 21


,


12
3
12 12
3
12
Example: Show that the following relationship is true
cos 2  2cos 2   1, where  
 
 
cos 2    2cos 2    1
6
6
2
 3
 
cos    2 
 1
3
2
 


1 3
 1
2 2

6
Extra material:
Given that cos  k and 0   


find cos    
2
2




cos       sin 
2

cos  
B
1
k
1
A

C
k
12  k 2  BC 2
To find the opposite side use Pythagoras: 1  k 2  BC 2
 1  k 2  BC
positive since it is in the first quadrant.
sin  
1 k2
then  sin   
1 k2
must be