Download Lesson 11

LESSON 11 BASIC IDENTITIES
Examples which are worked in this lesson (Click on the number):
Examples Use Basic Identities to find the exact value of the other five
trigonometric functions for the following.
csc    5 and  is in the III quadrant
1.
2
cos  
2.
and csc   0
3
17
tan   
3.
and sin   0
8
Examples Find all the exact solutions for the following equations.
2 cos 2   1  sin 
7  2 cos   8 sin 2 
1.
2.
sec 2   9  7 tan 
 5 cot 2 x  6 csc x  5
3.
4.
Recall the following Basic Identities from Lesson 2:
1.
tan  
sin 
cos 
3.
cot  
1
tan 
5.
csc  
1
sin 
2.
cot  
cos 
sin 
4.
sec  
1
cos 
Recall the following Pythagorean Identities from Lesson 2:
1.
cos 2   sin 2   1
2.
sec 2   tan 2   1
3.
csc 2   cot 2   1
Examples Use Basic Identities to find the exact value of the other five
trigonometric functions for the following.
1.
csc    5 and  is in the III quadrant
Since csc    5 , then sin   
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1
5
2
2
Since csc    5 and csc   cot   1 by one of the Pythagorean
Identities, then
(  5 ) 2  cot 2   1  25  cot 2   1  cot 2   24 
cot   
24 . Since  is in the III quadrant, then cot  
24 , then tan  
Since cot  
24 .
1
24
.
1
and
5
cos 2   sin 2   1 by one of the Pythagorean Identities, then you could
use this identity to find cos  . However, you only need to use the
Pythagorean Identities once to solve these problems.
Now, we need to find cos  and sec  .
Since cot  
Since sin   
cos 
, then cos   cot  sin  . Since cot  
sin 
1
sin    , then cos  
5
Since cos   
24
5
 1
24     
 5
, then sec   
24
5
5
24
.
.
24 and
24
Answers: cos   
5
and cot  
2.
cos  
, sin   
1
, tan  
5
1
24
, sec   
5
24
,
24
2
and csc   0
3
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First, determine what quadrant the angle  is in. Using Method 1 from
Lesson 6, we have:
cos   0  the x-coordinate of the point of intersection of the terminal
side of the angle  with the Unit Circle is positive. That is, x > 0.
csc   0  sin   0  the y-coordinate of the point of intersection
of the terminal side of the angle  with the Unit Circle is negative. That is,
y < 0.
Thus, we have that x > 0 and y < 0. Thus, the angle  is in the IV quadrant.
You may use Method 2 or Method 3 from Lesson 6 if you wish.
Since cos  
Since cos  
2
3
, then sec  
3
2
2
2
2
and cos   sin   1 by one of the Pythagorean
3
Identities, then
2
 
3
2
 sin 2   1 
sin   
5
3
4
 sin 2   1 
9
sin 2  
5

9
. Since  is in the IV quadrant, then sin   
Since sin   
5
3
, then csc   
3
5
.
5
3
.
Since tan  
Now, we need to find tan  and cot  .
5
2
sin   
and cos   , then tan  
3
3
Since tan   
5
2
Answers: sin   
, then cot   
5
3
and cot   
3.
tan   
17
8
, tan   

sin 
cos 
and
5
3
2
3
 
5
2 .
2
5
5
2
.
3
3
, csc   
,
2
5
, sec  
2
5
and sin   0
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First, determine what quadrant the angle  is in. Using Method 1 from
Lesson 6, we have:
sin   0  the y-coordinate of the point of intersection of the terminal
side of the angle  with the Unit Circle is positive. That is, y > 0.
The tangent of the angle  is the y-coordinate of the point of intersection of
the terminal side of the angle  with the Unit Circle divided by the xy
coordinate of the point of intersection. That is, tan   . Since we have
x
that tan   0 and y > 0, then we have the following:
(  )  tan  
()
y

 x  0
x
?
Thus, we have that x < 0 and y > 0. Thus, the angle  is in the II quadrant.
You may use Method 2 or Method 3 from Lesson 6 if you wish.
Since tan   
17
Since tan   
17
8
8
8
, then cot   
17
2
2
and sec   tan   1 by one of the Pythagorean
Identities, then

sec 2    


sec   
17 

8 
2
1 
17
81
 1  sec 2  

64
64
sec 2  
9
9
. Since  is in the II quadrant, then sec    .
8
8
Since sec   
9
8
, then cos    .
8
9
Since tan  
Now, we need to find sin  and csc  .
sin   tan  cos  .
sin   
Since
17  8 
  
8  9
Since sin  
17
9
tan   
8
and cos   
8
, then
9
17
9
.
, then csc  
9
8
, sin  
9
17
Answers: cos   
17
sin 
, then
cos 
17
9
.
, sec   
9
, csc  
8
9
17
,
and cot   
8
17
Examples Find all the exact solutions for the following equations.
1.
2 cos 2   1  sin 
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This equation is neither quadratic in cos  nor sin  . We will use the
2
2
2
2
Pythagorean Identity cos   sin   1 to replace cos  by 1  sin  .
2
2
The resulting equation will be quadratic in sin  . Since cos   sin   1,
2
2
then cos   1  sin  . Thus,
2 cos 2   1  sin   2 (1  sin 2  )  1  sin  
2  2 sin 2   1  sin   0  2 sin 2   sin   1 
( sin   1) ( 2 sin   1)  0
Thus, either sin   1  0 or 2 sin   1  0 . Thus, there are two
equations to be solved. NOTE: These types of equations were solved in
Lesson 10. Please read that lesson in order to see how to solve these
equations.
Since sin   1  0  sin    1, then the solutions of the first equation
3
 2 n  , where n is an integer.
are  
2
1
, then the solutions of the second
2
5

 2 n  and  
 2 n  , where n is an integer.
equation are  
6
6
Since 2 sin   1  0  sin  
Answers:  
3
5

 2 n ,  
 2 n  , and  
 2 n  , where n
6
6
2
is an integer
2.
7  2 cos   8 sin 2 
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This equation is neither quadratic in cos  nor sin  . We will use the
2
2
2
2
Pythagorean Identity cos   sin   1 to replace sin  by 1  cos  .
The resulting equation will be quadratic in cos  . Since cos 2   sin 2   1 ,
2
2
then sin   1  cos  . Thus,
2 cos   7  8 sin 2   2 cos   7  8 (1  cos 2  ) 
2 cos   7  8  8 cos 2   8 cos 2   2 cos   1  0 
( 2 cos   1) ( 4 cos   1)  0
Thus, either 2 cos   1  0 or 4 cos   1  0 . Thus, there are two
equations to be solved. NOTE: These types of equations were solved in
Lesson 10. Please read that lesson in order to see how to solve these
equations.
1
, then the solutions of the second
2
4
2



2
n



 2 n  , where n is an integer.
equation are
and
3
3
Since 2 cos   1  0  cos   
To solve the second equation 4 cos   1  0 :
4 cos   1  0 
4 cos   1 
cos  
1
4
1
is not the maximum
4
positive number for the cosine function, then the solutions do not occur at the
coordinate axes. Thus, the solutions occur in two of the four quadrants.
Since cosine is positive in the I and IV quadrants, the solutions for this
equation occur in those quadrants.
Determine where the solutions  will occur. Since
Find the reference angle  ' for the solutions  :
cos  
1

4
cos  ' 
1

4
 '  cos  1
1
4
The solutions in the I quadrant: The one solution in the I quadrant, that is
1 1
between 0 and 2  , is   cos
. Now, all the other solutions in the I
4
quadrant are coterminal to this one solution. Thus, all the solutions in the I
1 1


cos
 2 n  , where n is an integer.
quadrant are given by
4
The solutions in the IV quadrant: The one solution in the IV quadrant, that is
1 1
between  2  and 0, is    cos
. Now, all the other solutions in the
4
IV quadrant are coterminal to this one solution. Thus, all the solutions in the
1 1
 2 n  , where n is an integer.
IV quadrant are given by    cos
4
the solutions of the second equation 4 cos   1  0 are
1
1
  cos  1  2 n  and    cos  1  2 n  , where n is an integer.
4
4
Thus,
Answers:
  c o s 1
   cos  1
4
2
1
 2 n ,  
 2 n ,  
 2 n  and
3
4
3
1
 2 n  , where n is an integer
4
3.
sec 2   9  7 tan 
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This equation is neither quadratic in sec  nor tan  . We will use the
2
2
2
2
Pythagorean Identity sec   tan   1 to replace sec  by 1  tan  .
The resulting equation will be quadratic in tan  . Since sec 2   tan 2   1,
2
2
then sec   1  tan  . Thus,
sec 2   9  7 tan   1  tan 2   9  7 tan  
tan 2   7 tan   8  0  ( tan   1) ( tan   8 )  0
Thus, either tan   1  0 or tan   8  0 . Thus, there are two
equations to be solved. NOTE: These types of equations were solved in
Lesson 10. Please read that lesson in order to see how to solve these
equations.
Since tan   1  0 
ta n   1 , then the solutions of the first equation
5

 2 n  and  
 2 n  , where n is an integer.
are  
4
4
To solve the second equation tan   8  0 :
tan   8  0 
tan    8
Determine where the solutions  will occur. Since tangent is negative in the
II and IV quadrants, the solutions for this equation occur in those quadrants.
Find the reference angle  ' for the solutions  :
tan    8 
tan  '  8 
 '  tan
1
8
The solutions in the II quadrant: The one solution in the II quadrant, that is
1
between 0 and 2  , is      '    tan 8 . Now, all the other
solutions in the II quadrant are coterminal to this one solution. Thus, all the
1
solutions in the II quadrant are given by     tan 8  2 n  =
 tan  1 8  ( 2 n  1)  , where n is an integer.
The solutions in the IV quadrant: The one solution in the IV quadrant, that is
1
between  2  and 0, is    tan 8 . Now, all the other solutions in the
IV quadrant are coterminal to this one solution. Thus, all the solutions in the
1
IV quadrant are given by    tan 8  2 n  , where n is an integer.
Thus, the solutions of the second equation tan   8  0 are
   tan  1 8  ( 2 n  1)  and    tan  1 8  2 n  , where n is an
integer. NOTE: Since the period of the tangent function is  , then this
1
answer may also be written as    tan 8  n  , where n is an integer.
Answers:
 
5

 2 n ,  
 2 n  , and    tan  1 8  n  ,
4
4
where n is an integer
4.
 5 cot 2 x  6 csc x  5
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This equation is neither quadratic in cot x nor csc x . We will use the
2
2
2
2
Pythagorean Identity csc x  cot x  1 to replace cot x by csc x  1 .
2
2
The resulting equation will be quadratic in csc x . Since csc x  cot x  1 ,
2
2
then cot x  csc x  1 . Thus,
 5 cot 2 x  6 csc x  5   5 ( csc 2 x  1)  6 csc x  5 
 5 csc 2 x  5  6 csc x  5 
 5 csc 2 x  6 csc x 
0  5 csc 2 x  6 csc x  csc x ( 5 csc x  6 )  0
Thus, either csc x  0 or 5 csc x  6  0 . Thus, there are two equations to
be solved.
To solve the first equation csc x  0 : Since csc x  0 and csc x 
1
,
sin x
1
 0 . A fraction is zero if and only if its numerator is zero. The
sin x
1
numerator of the fraction
will always be 1. Thus, the equation
sin x
csc x  0 has no solutions.
then
To solve the second equation 5 csc x  6  0 :
5 csc x  6  0 
5 csc x   6 
csc x  
6
5
 sin x  
5
6
5
is not the minimum
6
negative number for the sine function, then the solutions do not occur at the
coordinate axes. Thus, the solutions occur in two of the four quadrants.
Since sine is negative in the III and IV quadrants, the solutions for this
equation occur in those quadrants.
Determine where the solutions x will occur. Since 
Find the reference angle x ' for the solutions x :
sin x  
5

6
sin x ' 
5

6
x '  sin
1
5
6
The solutions in the III quadrant: The one solution in the III quadrant, that is
1 5
between 0 and 2  , is x    x '    sin
. Now, all the other
6
solutions in the III quadrant are coterminal to this one solution. Thus, all the
1 5
 2 n =
solutions in the III quadrant are given by x    sin
6
5
x  sin  1  ( 2 n  1 )  , where n is an integer.
6
The solutions in the IV quadrant: The one solution in the IV quadrant, that is
5
. Now, all the other solutions in the
6
IV quadrant are coterminal to this one solution. Thus, all the solutions in the
1 5
 2 n  , where n is an integer.
IV quadrant are given by x   sin
6
between  2  and 0, is x   sin
1
the solutions of the second equation 5 csc x  6  0 are
5
5
x  sin  1  ( 2 n  1 )  and x   sin  1  2 n  , where n is an
6
6
integer.
Thus,
1
Answers: x  sin
is an integer
5
 ( 2 n  1 )  and x   sin
6
1
5
 2 n  , where n
6