Download 6.3 Properties of Trigonometric Functions

Section 6.3 Notes Page 1
6.3 Properties of Trigonometric Functions
The following Even – Odd Properties will allow you to not deal with negative angles.
Even – Odd Properties
cos(t )  cos t
sec(t )  sec t
sin(t )   sin t
csc(t )   csc t
tan(t )   tan t
cot(t )   cot t
Periodic Properties
If we start at an angle and go around one revolution ( 360  or 2 radians) we will end up at the same angle we
started with. The k value is any integer, and represents how many revolutions are going around. If you want to
use degrees, replace the 2k in the equations below with 360k.
sin(t  2k )  sin t
csc(t  2k )  csc t
cos(t  2k )  cos t
sec(t  2k )  sec t
tan(t  2k )  tan t
cot(t  2k )  cot t
EXAMPLE: Find the EXACT value of sin 1485 .
We need to divide 1485 by 360 to see how many revolutions we have. If you divide 1485 by 360 you will get 4
with a remainder of 45. So we can rewrite our problem as: sin(45  360(4)) . From our definitions above we
know that sin(45  360(4))  sin 45 . From our previous values we know that sin 45 
that sin 1485 
2
. So we know
2
2
.
2
 17 
EXAMPLE: Find the EXACT value of tan 
.
 4 
First we will use our even-odd property to change the negative angle into a positive angle:
17 180
 17 
 17 

 765 . Dividing this by
tan 
   tan
 . It would be easier to change this into degrees:
4

4
4




360 we will get 2 with a remainder of 45. So our problem becomes tan(45  360(2)) . From our periodic
properties, tan(45  360(2))  tan 45 . From our previous values we know that tan 45  1 . So we know that
 17 
 tan
  1 .
 4 
Section 6.3 Notes Page 2
EXAMPLE: Find the exact value of tan
14
using reference angles. Draw the angle in standard position and
3
indicate the reference angle.
14 180

 840  . We can subtract two revolutions from this:
3





840  360  360  120 . From our rules, we can rewrite this problem as: tan(120  360(2))  tan120
2
In radian form it would look like this: tan
.
3
We can change this into degrees:
1.) First we will draw this angle in standard position. The
reference angle is indicated by the double curved lines.
2 
We found the reference angle by taking  
 . (180 – 120)
3
3
2.) We need to apply the trig function to our reference angle. We
have tan

3
 3.
3.) We need to apply the appropriate sign. This angle is in the
second quadrant, so tangent needs to be negative here. So now we
14
can write our answer: tan
  3.
3
EXAMPLE: Find the reference angle of tan(225 ) and find its EXACT value.
Using the even and odd properties, tan(225 )   tan 225 . We have changed the problem, so now we will
look at positive 225 degrees. This is in the third quadrant, so our reference angle is 225  180  45 . We will
now calculate  tan 45 . The value for tan 45 is 1. So now we have  tan(45 )  1 In the third quadrant,
tangent is positive, so we don’t need to change the sign of our answer, so tan(225 )  1 .
EXAMPLE: Find the reference angle of sec(210  ) and find its EXACT value.
Using the even and odd properties, sec(210 )  sec 210 . We have changed the problem, so now we will look
at positive 210 degrees. This is in the third quadrant, so our reference angle is 210  180  30 . We will do
1
. Now we can put in the value off our table, and we will get
sec 30  . This is the same as
cos 30 
2 3
1
2
2 3


. Secant is negative in the 3rd quadrant, so we need to change the sign: sec( 210  )  
.
3
3
3
3
2
Section 6.3 Notes Page 3
 11 
EXAMPLE: Find the reference angle of sin  
 and find its EXACT value.
 3 
11 180

 660 . So now our problem becomes: sin(660  ) .
3 

Using the even and odd properties, sin(660 )   sin 660  . We have changed the problem, so now we will
look at positive 660 degrees. We can rewrite the problem as  sin(300  360 ) which becomes  sin(300  ) .
We want to change our angle into degrees: 
This is in the fourth quadrant, so our reference angle is 360  300  60 , or

. We will now look at
3
3
. In the fourth quadrant, secant is
2
3
 11 
negative, so we need to change the sign of our answer, so sin  
.

 3  2
 sin 60 . We can put in the value off our table, and we will get 
 19 
EXAMPLE: Find the reference angle of csc 
 and find its EXACT value.
 4 
19 180

 855 . Using the even and odd properties,
4

csc(855 )   csc 855 . This can be rewritten as  csc(135  360 (2)) . This reduces to  csc(135 ) . This is
in the second quadrant, so the reference angle is 180  135  45 . So now we have  csc 45 . This is the
1
2
2 2
1
same as 
. From our table we have: 


  2 . Since we are in the second

2
sin 45
2
2
Changing this into degrees we get: 
2
 19
quadrant, sine is positive, so we don’t need to change our sign. Our answer is csc 
 4

 2.

Fundamental Identities
sin t 
1
csc t
csc t 
1
sin t
sin 2 t  cos 2 t  1
cos t 
1
sec t
sec t 
1
cos t
tan t 
tan t 
1
cot t
cot t 
1
tan t
sin t
cos t
1  tan 2 t  sec 2 t
cot t 
cos t
sin t
1  cot 2 t  csc 2 t
Section 6.3 Notes Page 4
EXAMPLE: Given sin t 
5
2
and cos t 
, find the other 4 trig values using identities.
3
3
We just use the identities and plug in our values for sine and cosine where necessary:
1
1
3
3 5



cos t
5
5 3
5
csc t 
1
1
3


sin t 2 / 3 2
cot t 
1
1
5
5



tan t 2 5 5 2 5
2
EXAMPLE: Given sin t 
sec t 
tan t 
23
sin t
2 3
2
2 5
,

 


cos t
5
5 3 3 5
5
7

and 0  t  , use the Pythagorean Identity sin 2 t  cos 2 t  1 to find cos t .
8
2
2
7
7
We first start with the identity sin t  cos t  1 and plug in
for sin t . You will get    cos 2 t  1 .
8
8
49
49
Simplifying will give you
from both sides. You will
 cos 2 t  1 . Now isolate the cosine by subtracting
64
64
15
15

get cos 2 t 
. Square root both sides to get cos t  
. Since the angle for t needs to be 0  t  , we
64
2
8
know that this angle must be in the first quadrant. What we know from this is that the x-coordinate of the unit
15
circle must be positive in the first quadrant, so cos t 
.
8
2
2