Transcript Trig functions of Special Angles

Special Angles and their Trig Functions
By Jeannie Taylor
Through Funding Provided by a
VCCS LearningWare Grant
We will first look at the special angles called the quadrantal
angles.
The quadrantal angles are those angles that lie on the axis of
the Cartesian coordinate system: 0, 90, 180, and 270.
90
180
0
270
We also need to be able to recognize these angles when they
are given to us in radian measure. Look at the smallest

possible positive angle in standard position, other than 0 , yet


having the same terminal side as 0 . This is a 360 angle
 
which is equivalent to 2 radians.
90 2 radians
If we look at half of
that angle, we have
180 or radians
.
Looking at the angle

half-way between 0

and 180 or , we have


90 or .
2
Looking at the angle half-way
between 180 and 360 , we have
270or 32 radians which is 34 of
the total (360  or 2 radians).
180
  radians
0
360  2 radians
270

3
radians
2
Moving all the way around from 0  to 360
completes the circle and and gives the 360
angle which is equal to 2 radians.
We can count the quadrantal angles in terms of
Notice that after counting these angles
based on portions of the full circle,
two of these angles reduce to radians
with which we are familiar,  and 2.
Add the equivalent degree
measure to each of these
quadrantal angles.
180
We can approximate the
radian measure of each
angle to two decimal places.
One of them, you already
know,   3.14radians. It
will probably be a good idea
to memorize the others.
Knowing all of these
numbers allows you to
quickly identify the location
of any angle.

2
radians.
90

2
radians 1.57 radians
0
2
radians
2
  radians
 3.14 radians
0 radians
4
2
radians
360
 2 radians
 6.28radians
3
radians 4.71radians
2
270
We can find the trigonometric functions of the
quadrantal angles using this definition. We will
begin with the point (1, 0) on the x axis.
Remember the six
trigonometric functions
defined using a point (x, y)
on the terminal side of an
angle,  .
90

2
y
sin  
r
r
csc  
y
x
cos  
r
r
sec  
x
tan  
y
x
cot  
radians
(1, 0)
180
0
0 radians
  radians
or
360
x
y
 2 radians
270

3
For the angle 0 , we can see that x = 1 and
radians
2
y = 0. To visualize the length of r, think

about the line of a 1 angle getting closer As this line falls on top of the x axis,

and closer to 0 at the point (1, 0).
we can see that the length of r is 1.

Using the values, x = 1, y = 0, and r = 1, we list the six trig functions of 0.

And of course, these values also apply to 0 radians, 360 , 2 radians, etc.
sin 0   0
csc 0  is undefined
90
cos0  1
sec 0  1
2

tan0  


radians
0
 0 cot 0  is undefined
1
(1, 0)
180
0
0 radians
  radians
or
360
 2 radians
It will be just as easy to find the
trig functions of the remaining
quadrantal angles using the point
(x, y) and the r value of 1.
270
3
radians
2
sin

1
2
csc

cos  0
2
tan

1
2
90

sec is undefined
2

is undefined
2
cot

2

0
2
radians
(0, 1)
sin   0
cos   1
tan  0
csc  is undefined
sec   1
cot is undefined
0
180
  radians (-1, 0)
0 radians
(0, -1)
csc
3
 1
2
270
cos
3
0
2
sec
3
 is undefined
2
3
radians
2
tan
3
is undefined
2
cot
3
0
2
360
 2 radians
3
 1
2
sin
or
Now let’s cut each quadrant in
half, which basically gives us 8
equal sections.
The first angle, half way between 0

1  
and 2 would be 2  2  4 .
We can again count around the
circle, but this time we will count

in terms of 4 radians. Counting
we say:
2

  90 
4
2
3  135 
4
1 2 3 4 5 6 7
8
,
,
,
,
,
,
, and
.
4 4 4 4 4 4 4
4
Then reduce appropriately.

Since 0 to 2 radians is 90, we know


that 4 is half of 90or 45. Each

successive angle is 45 more than the
previous angle. Now we can name all
of these special angles in degrees.

2

4  45 
4
4
 180 

0
5
4
 225 
7
4  315 
6 3

4
2
 270 
It is much easier to construct this picture of angles in
both degrees and radians than it is to memorize a table


involving these angles (45 or 4 reference angles,).
8  2  360 
4



Next we will look at two special triangles: the 45 – 45 – 90



triangle and the 30 – 60 – 90 triangle. These triangles will allow

us to easily find the trig functions of the special angles, 45 , 30 ,
and 60 .
The lengths of the legs of the



45 – 45 – 90 triangle are equal
to each other because their
corresponding angles are
equal.
If we let each leg have a length
of 1, then we find the hypotenuse
to be 2 using the Pythagorean
theorem.
45

2
1
45

1
You should memorize this
triangle or at least be able
to construct it. These
angles will be used
frequently.
Using the definition of the trigonometric functions as the
ratios of the sides of a right triangle, we can now list all six
trig functions for a 45 angle.
sin 45 

1
2

2
2
45
csc 45  2


2
1
2

cos 45 
2
sec 45  2
tan 45  1
cot 45  1
45
1

For the 30– 60– 90triangle, we will construct an equilateral
triangle (a triangle with 3 equal angles of 60 each, which
guarantees 3 equal sides).
If we let each side be a length
of 2, then cutting the triangle
in half will give us a right
triangle with a base of 1 and a
hypotenuse of 2. This smaller
triangle now has angles of 30,


60, and 90 .
30
2
3
60
1
You should memorize this
triangle or at least be able to
construct it. These angles,
also, will be used frequently.
We find the length of the other
leg to be 3 , using the
Pythagorean theorem.
Again, using the definition of the trigonometric functions as the
ratios of the sides of a right triangle, we can now list all the trig


functions for a 30 angle and a 60 angle.
sin 30 
1
2
csc 30  2
cos30 
3
2
t an30 
1
30
2
3
2
sec 30 

3
3
3
cot 30 

2 3
3

2 3
3

3
3
3
3
60
1
sin 60 
3
2
csc 60 
cos 60 
1
2
sec 60  2
t an60 
3
cot 60 
2
3
1
3
45
30

2
2
1
3
45

60
1
1
Either memorizing or learning how to construct these
triangles is much easier than memorizing tables for the



45 , 30 , and 60 angles. These angles are used frequently
and often you need exact function values rather than
rounded values. You cannot get exact values on your
calculator.
45

30
2
1
2
45

3
1
60
1
Knowing these triangles, understanding the use of reference
angles, and remembering how to get the proper sign of a
function enables us to find exact values of these special
angles.
Sine
All
A good way to
II
I
remember this chart is
that ASTC stands for
III
IV
All Students Take
Tangent
Cosine
Calculus.
Example 1: Find the six trig functions of 330.
First draw the 330 degree angle.
Second, find the reference angle, 360 - 330 = 30
To compute the trig functions

of the 30 angle, draw the
“special” triangle.
y
30
S
2
A
3


60
1
Determine the correct sign for the trig functions
of 330 . Only the cosine and the secant are “+”.
330
T
C
30
x
Example 1 Continued: The six trig functions of 330 are:
1
2
3
cos 330 
2
sin 330  
csc 330   2
sec 330 
1
3

3
3
tan 330  
2 2 3

3
3
y
cot 330   3
30
S
2
A
3

30

60
1
330
T
C
x
Example 2: Find the six trig functions of
First determine the location of
4
3
4
3.
(Slide 1)
.
With a denominator of 3, the distance from 0 to  radians is cut into
thirds. Count around the Cartesian coordinate system beginning at 0 until
4
we get to .
3
We can see that the reference
angle is  , which is the same as
3
60. Therefore, we will
compute

the trig functions of 3 using the
60 angle of the special triangle.
3
3
3
30

3
3
60
1

3
3
2

y
2
3
4
3
x
4
3.
Example 2: Find the six trig functions of
(Slide 2)
Before we write the functions, we need to determine the signs for each
function. Remember “All Students Take Calculus”. Since the angle, 43 , is
located in the 3rd quadrant, only the tangent and cotangent are positive. All
the other functions are negative..
sin
4
3

3
2
csc
4
2
2 3


3
3
3
cos
4
1

3
2
sec
4
 2
3
tan
4
 3
3
cot
4
1
3


3
3 3

3
y
2
3
S
A

30
x

3
2
3
T
60
1

4
3
C
Example 3: Find the exact value of cos   5  .
 4
We will first draw the angle to determine the quadrant.
5
We see that the angle    is
 4
located in the 2nd quadrant
and the cos is negative in the
2nd quadrant.
Note that the reference angle is
 5 


 4 

4
 4 


 4 

4
0 radians
.
T
 3 


 4 

4
We know that is the
same as 45 , so the

reference angle is 45 .
Using the special triangle
we can see that the cos of

45 or 4 is 12 .
A
S
45
C
 
 
 4
 2 


 4 

2
1
45
1

cos   54  = 
1
2

2
2
Practice Exercises

1. Find the value of the sec 360 without using a calculator.

2. Find the exact value of the tan 420 .
5 
.
 6
3. Find the exact value of sin 

4. Find the tan 270 without using a calculator.
5. Find the exact value of the csc   73  .

6. Find the exact value of the cot (-225 ).
 13 

4 
 11 


 6 
7. Find the exact value of the sin 
.
8. Find the exact value of the cos
.
9. Find the value of the cos(- ) without using a calculator.

10. Find the exact value of the sec 315 .
Key For The Practice Exercises

1. sec 360 = 1

3
2. tan 420 =
3. sin
 5 


 6 
=

1
2
4. tan 270  is undefined
5. csc
 7 


 3 
2
=

3

2 3
3
6. cot (-225 ) = -1
7. sin
8. cos
 13 


 4 
 11 


 6 
=

=
3
2
1
2

2
2
9. cos(-  ) = -1

10. sec 315 = 2
Problems 3 and 7 have solution explanations following this key.
Problem 3: Find the sin   5  .

6 
We will first draw the
angle by counting in a
negative direction in
units of  .
6

A
S
0 radians

6


5
We can see that 6 is the
6

6
T
C
reference angle and we know
2


4


that 6 is the same as 30 . So
6
3
6


we will draw our 30 triangle
6
1
and see that the sin 30 is 2 .
All that’s left is to find the correct sign.
30
And we can see that the correct sign is “-”, since
the sin is always “-” in the 3rd quadrant.
2
3
60
1

Answer:
 5 
sin  6 
=

1
2
 13 

.
 4 
Problem 7: Find the exact value of cos
We will first draw the angle to determine the quadrant.
 13 


4


We see that the angle
is
located in the 3rd quadrant
and the cos is negative in the
3rd quadrant.
Note that the reference angle is
We know that 4 is the
same as 45 , so the

reference angle is 45 .
Using the special triangle
we can see that the cos of

45 or 4 is 12 .
2
4 10
4
3
4
4
4

4
45
A
S
11
4
9
4 
4
8
4
0 radians

4
 12 


 4 
.
5
13
4
4
T
C
7
4
6
4

2
1
cos 13  = 
45
1

 4 
1
2

2
2