# Trigonometric Functions of Special Angles by Z0Kgs0P

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```									By bithun jith
Done by bithun jith binoy
k.v.pattom
You must know and memorize the following.

Pythagorean Identities:                              Tangent/Cotangent Identities:
sin x
sin2   x+     cos2   x=1                              tan x 
cos x
1 + tan2 x = sec2 x
cos x
1+     cot2   x=   csc2   x                           cot x 
sin x
Reciprocal Identities:                               Cofunction Identities:
1                              1                                       
sin x                        csc x                  sin   x   cos x   csc  x   sec x
csc x                         sin x               2                  2    
1                             1                                       
cos x                        sec x                  cos  x   sin x    sec  x   csc x
sec x                         cos x               2                  2    
                      
1                             1             tan   x   cot x   cot  x   tan x
tan x                        cot x                      2                  2    
cot x                         tan x

sin2 x = (sin x)2
11.3 Sum and Difference
Formulas
Objective: To use the sum and difference formulas for sine and cosine.
sin ( + ) = sin  cos  + sin  cos 
sin ( - ) = sin  cos  - sin  cos 

45                  60

45                  30

1. This can be used to find the sin 105.                 HOW?

2. Calculate the exact value of sin 375.
cos ( + ) = cos  cos  - sin  sin 
cos ( - ) = cos  cos  + sin  sin 

Note the similarities and differences to the sine properties.
3. This can be used to find the cos 285.       HOW?

4. Calculate the exact value of cos 345.

5. Pr ove : sin(α  β)  sinα  β  2 sinβ cos α
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.          
If we look at half of
that angle, we have
180  or  radians  .

Looking at the angle                 180                                  0
half-way between 0

and 180 or , we have


   
90 or .
2
3
Looking at the angle half-way                          270   
2
between 180 and 360 , we have
Moving all the way around from 0  to 360
270or 3 radians which is 3 of
2                      4           completes the circle and and gives the 360
the total (360  or 2 radians).
angle which is equal to 2 radians.

.
2
Notice that after counting these angles
based on portions of the full circle,             90

two of these angles reduce to radians                         
with which we are familiar,  and 2.
2

measure to each of these
2                                   0
180     2
4
We can approximate the
2
angle to two decimal places.                                            6.28 radians
know,   3.14 radians. It
will probably be a good idea                      3

to memorize the others.                            2
Knowing all of these                              270
numbers allows you to
quickly identify the location
of any angle.
We can find the trigonometric functions of the
Remember the six
quadrantal angles using this definition. We will
trigonometric functions
defined using a point (x, y)       begin with the point (1, 0) on the x axis.
on the terminal side of an                                   90
angle,  .                                                   
2

y                  r
sin               csc 
r                  y
(1, 0)   0
cos  
x
sec  
r                   x                 radians                         or
y                    x                                                360
tan              cot                                                         2 radians
x                    y
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       csc0  is undefined                  90

cos0  1

sec 0  1
                                2

0
tan 0        0 cot 0  is undefined
1
(1, 0)   0
360
It will be just as easy to find the
270
trig functions of the remaining
3
2
(x, y) and the r value of 1.
                              
sin      1                     csc     1
2                              2
                                                                  90
cos  0                       sec is undefined                         
2
                              
tan is undefined               cot  0
2                              2

(0, 1)
sin   0                csc is undefined                                                0
cos   1               sec  1                  180                                  0 radians
tan   0                cot is undefined            radians (-1, 0)                   or
(0, -1)
360
3                       3
sin        1            csc       1
2                        2                                        270
cos
3
0              sec
3
 is undefined                         3
2                        2                                         2
3                       3
tan      is undefined    cot      0
2                        2
Now let’s cut each quadrant in                        The first angle, half way between 0
half, which basically gives us 8                                        1  
and     would be   .
equal sections.                                              2         2     2   4

We can again count around the
circle, but this time we will count                                2   
                                                          90 
in terms of 4 radians. Counting                                     4   2

we say:                                                               2                 
3  135 
4  45 
1 2 3 4 5 6 7       8                           4
,  ,  ,  ,  ,  ,   , and    .
4 4 4 4 4 4 4              4
Then reduce appropriately.
4                                                     8  2  360 
                                                      0
4
                                    180 
4
Since 0 to 2 radians is 90, we know

that is half of 90or 45. Each

4                 
successive angle is 45 more than the                      5                           7
previous angle. Now we can name all                        4                            4  315 
 225 
of these special angles in degrees.
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,).
                   
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 – 45 – 90 triangle are equal                                     2
to each other because their                    1

corresponding angles are                                                 
45
equal.                                                      1

If we let each leg have a length
of 1, then we find the hypotenuse        You should memorize this
to be 2 using the Pythagorean            triangle or at least be able
theorem.                                 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.

1      2                                    
45
sin 45 

        csc45  2

2
2   2                            1

2                                                     45
cos 45 

sec 45  2                         1
2
tan 45  1             cot 45  1
WARM-UP B) occur frequently
The expressions sin (A + B) and cos (A +
enough in math that it is necessary to find expressions
equivalent to them that involve sines and cosines of single
angles. So….

Does sin (A + B) = Sin A + Sin B

Try letting A = 30 and B = 60
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
30
triangle with a base of 1 and a
2
hypotenuse of 2. This smaller
3                                                
triangle now has angles of 30,
       
60, and 90 .
60

1
We find the length of the other
You should memorize this            leg to be 3 , using the
triangle or at least be able to     Pythagorean theorem.
construct it. These angles,
also, will be used frequently.
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.
1
sin 30                      csc30   2
2
3                             2          2 3
cos30                       sec 30              
2                              3           3
30
1            3
tan 30                      cot 30         3
2                          3       3

3

60
3                         2           2 3
1              sin 60                     csc60               
2                           3           3
1
cos 60                    sec 60   2
2
1            3
tan 60           3        cot 60              
3          3
30
45

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
3

45
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
II                             A good way to
I
remember this chart is
III             IV             that ASTC stands for
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
                                          y
of the 30 angle, draw the
“special” triangle.

30
S        A
2
3
   x
30
60

330
1
T        C
Determine the correct sign for the trig functions
of 330 . Only the cosine and the secant are “+”.
Example 1 Continued: The six trig functions of 330 are:

1
sin 330                       csc 330   2
2
3                                 2 2 3
cos 330                        sec 330       
2                                   3   3
1      3                                  y
tan 330                     cot 330   3
3    3

30
S       A
2
3

30   x
60

330
1
T       C
4
Example 2: Find the six trig functions of                  3
.   (Slide 1)
4
First determine the location of
3
.
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
y           
We can see that the reference                    2
3
angle is  , which is the same as
3

3
60. Therefore, we will compute

the trig functions of 3 using the
60 angle of the special triangle.
3
3
3
30
3
                          x
2                                3
3

4

60
3
1
4
Example 2: Find the six trig functions of                          3
.   (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..
4     3                    4    2     2 3
sin                     csc            
3    2                      3     3     3
y           
2
4    1                     4                                                 3
cos                    sec          2             3
3    2                      3
4                          4   1   3
tan
3
 3             cot
3
                       S              A
3 3

30                           
                              x
2                                    3
3
T              C
4

60
3
1
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
 5 

                4 

 4                                            A
located in the 2nd quadrant                              S
and the cos is negative in the                            
   
4
 4 

Note that the reference angle is  .
4
T            C             
 
 3                                    4
                               
We know that is the
4
 4 
 2 
   
same as 45 , so the                    
 4 
            45
reference angle is 45 .        1
2
cos   54  = 
1         2
Using the special triangle                                                                     
2        2
we can see that the cos of                                
                                                 45
45 or  is 12 .
4                                    1
Key For The Practice Exercises

1. sec 360 = 1

2. tan 420 =                   3
 5             
1
3. sin              =
 6                   2

4. tan 270  is undefined
 7              2            2 3
5. csc          =               
 3                   3         3

6. cot (-225 ) = -1
 13        
1

2
7. sin           =
 4                  2         2
 11         3
8. cos           =
 6          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                                    S             A
negative direction in
units of  .                                                                    0 radians
6                                        
6
                                                 
5                                         
We can see that 6 is the                                            6
reference angle and we know
6        T          C
                                 4                    
2
that 6 is the same as 30 . So         
6         3            6
                           
we will draw our 30 triangle                        6
and see that the sin 30 is 2 .
1

All that’s left is to find the correct sign.
30
And we can see that the correct sign is “-”, since
2
3                        the sin is always “-” in the 3rd quadrant.
 5         
1
60   
     
=       2
1
11.1 - Basic Trigonometry Identities

Objective:                    to be able to verify basic trig identities
You must know and memorize the following.

Pythagorean Identities:                              Tangent/Cotangent Identities:
sin x
sin2   x+     cos2   x=1                              tan x 
cos x
1 + tan2 x = sec2 x
cos x
1+     cot2   x=   csc2   x                           cot x 
sin x
Reciprocal Identities:                               Cofunction Identities:
1                              1                                       
sin x                        csc x                  sin   x   cos x   csc  x   sec x
csc x                         sin x               2                  2    
1                             1                                       
cos x                        sec x                  cos  x   sin x    sec  x   csc x
sec x                         cos x               2                  2    
                      
1                             1             tan   x   cot x   cot  x   tan x
tan x                        cot x                      2                  2    
cot x                         tan x

sin2 x = (sin x)2
 13 
Problem 7: Find the exact value of cos                             .
 4 

We will first draw the angle to determine the quadrant.
2
4 10
 13                                             4        9
We see that the angle        
is                      3                              4 
 4 
4
11       S                 A      4

and the cos is negative in the                       4
8
4
4    12             
     
 4               4

Note that the reference angle is  .
4                                                      7
5
13   T                C        4
4
We know that  is the
4
6
    4                                                       4
same as 45 , so the                        
               45
reference angle is 45 .           1
2
cos 13  = 
1         2
Using the special triangle                                                                          
2        2
 4 
we can see that the cos of                                     
                                                      45
45 or  is 12 .
4                                        1

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