Transcript Verifying-Trigonomet..

Verifying Trigonometric
Identities
Section 5.1
Objectives
• Apply a Pythagorean identity to solve a
trigonometric equation.
• Rewrite an expression in terms of sines
and cosines using definitions and
identities.
• Verify a trigonometric identity.
Pythagorean Identities
sin (x )  cos (x )  1
2
2
1  cot (x )  csc (x )
2
2
tan (x )  1  sec (x )
2
2
Reciprocal Identities
1
sin(t ) 
csc(t )
cos(t ) 
1
sec(t )
1
tan(t ) 
cot(t )
1
csc(t ) 
sin(t )
sec(t ) 
1
cos(t )
1
cot(t ) 
tan(t )
Quotient Identities
sin(t )
tan(t ) 
cos(t )
cos(t )
cot(t ) 
sin(t )
Even-Odd Identites
sin( t )   sin(t )
csc( t )   csc(t )
cos( t )  cos(t )
sec( t )  sec(t )
tan( t )   tan(t )
cot( t )   cot(t )
Verifying Trigonometric
Identities
When verifying trigonometric identities, a
main way to start is to change everything
to be in terms of sine and cosine function
before trying to simplify. In the next
two problems, we can see the process of
changing everything to be in terms of
sines and cosines. In the first problem,
we use the Pythagorean identity to find
trigonometric function values. In the
second, we verify a trigonometric
identity.
Evaluate the following
expressions using identities if
8
tan( x )  
and sin(x )  0
7
sin(x )
cot(x )
cos(x )
sec(x )
tan(x )
csc(x )
To solve this problem, we are going to rewrite the tangent function in
terms of the sine and the cosine function. Once we have done that,
we will use the Pythagorean identity to find the values of the sine
and cosine functions. Once we have these, we can find the values
of all of the other functions.
continued on next slide
Evaluate the following
expressions using identities if
8
tan( x )  
and sin(x )  0
7
Based on the definition of the tangent function:
sin( x )
tan( x ) 
cos( x )
Thus
sin( x )
8

cos( x )
7
continued on next slide
Evaluate the following
expressions using identities if
8
tan( x )  
and sin(x )  0
7
We now have two equations with two unknowns. Such equations we can
solve using substitution. We will solve the first equation for sin(x) and then
substitute that into the second equation.
sin( x )
8

cos( x )
7
sin (x )  cos (x )  1
2
2
continued on next slide
Evaluate the following
expressions using identities if
8
tan( x )  
and sin(x )  0
7
sin( x )  
8
cos( x )
7
Plug into second equation and solve for cos(x)
2
 8

2

cos(
x
)

cos
(x )  1


 7

64
cos 2 (x )  cos 2 (x )  1
49
2
2
64 cos (x )  49 cos (x )  49
continued on next slide
Evaluate the following
expressions using identities if
8
tan( x )  
and sin(x )  0
7
113 cos 2 (x )  49
49
cos (x ) 
113
49
cos( x )  
113
2
Since the tangent value is negative and the sine value is negative, the
angle x must be in quadrant IV. Thus the cosine value is positive.
cos( x ) 
7
113
continued on next slide
Evaluate the following
expressions using identities if
8
tan( x )  
and sin(x )  0
7
Now we plug this value in for cos(x) to find the sin(x)
8
sin( x )   cos( x )
7
8 7 
sin( x )   

7  113 
8
sin( x )  
113
continued on next slide
Evaluate the following
expressions using identities if
8
tan( x )  
and sin(x )  0
7
Now we are ready to use the sine and cosine values to find the
other trigonometric function values.
8
sin( x )  
113
7
cos( x ) 
113
8 



sin( x ) 
8
113 
tan( x ) 


7 
cos( x )
7

cot(x )


 113 
continued on next slide
Evaluate the following
expressions using identities if
8
tan( x )  
and sin(x )  0
7
Now we are ready to use the sine and cosine values to find the
other trigonometric function values.
8
sin( x )  
113
7
cos( x ) 
113
 7 


cos( x )
7
113 

cot( x ) 


8 
sin( x ) 
8


113 

continued on next slide
Evaluate the following
expressions using identities if
8
tan( x )  
and sin(x )  0
7
Continuing with the final two trigonometric functions.
8
sin( x )  
113
7
cos( x ) 
113
1
1
113
sec( x ) 


cos( x )  7 
7


 113 
1
1
113
csc( x ) 


8 
sin( x ) 
8


113 

Verify the trigonometric
identity.
2  tan 2 (x )
2

1

cos
(x )
2
sec (x )
When verifying a trigonometric identity, we work with one side of the equation
until we make it look exactly like the other side of the equation. When working
with the one side of the equation, we can multiply that side by 1 (or any
fraction that is equal to 1), rearrange terms within the side we are working on,
combine or take apart fractions that are on the side we are working on, and
replace terms with equal terms. The one thing that we cannot do is move
things from one side of the equation to the other.
For this example, since the left side is more complicated, we will work with that
side trying to simplify it.
continued on next slide
Verify the trigonometric
identity.
2  tan 2 (x )
2

1

cos
(x )
2
sec (x )
We will start by changing all the functions on the left side to be in terms of
sine and cosine functions.

sin 2 (x ) 
 2 

2
cos (x ) 

1


1


2
 cos (x ) 
Notice that since we are only working with the left side, we do not even
write the right side. This will keep us from trying to do things to both
sides of the equation.
continued on next slide
Verify the trigonometric identity.
2  tan 2 (x )
2

1

cos
(x )
2
sec (x )
Our next step will be to get a common denominator for the two parts of the
numerator. This will allow us to write the numerator as a singe fraction.
 2 cos 2 (x ) sin 2 (x ) 



2
2
 cos (x ) cos (x )   1


1


2
cos
(
x
)


 2 cos 2 (x )  sin 2 (x ) 


2
cos (x )

 1


1


2
 cos (x ) 
In our next step we will simplify the complex fraction. Since one
fraction is being divided by another fraction, we can multiply the
fraction in the numerator of the large fraction by the multiplicative
inverse of the fraction in the denominator of the large fraction.
continued on next slide
Verify the trigonometric identity.
2  tan 2 (x )
2

1

cos
(x )
2
sec (x )
 2 cos 2 (x )  sin 2 (x ) 


2
cos (x )

 1


1


2
 cos (x ) 
 2 cos 2 (x )  sin 2 (x )  

1



  1


2
2


cos (x )
  cos (x ) 

 2 cos 2 (x )  sin 2 (x )   cos 2 (x ) 
  
  1

2
cos (x )
1
 


2 cos 2 (x )  sin 2 (x )  1
Next we should be reminded that what is left looks similar to the
Pythagorean identity. The difference between the Pythagorean identity
and what we have is the 2 in front of the cos2(x). We are going to
rewrite this expression to change it from 2 cos2(x) to cos2(x). + cos2(x).
continued on next slide
Verify the trigonometric identity.
2  tan 2 (x )
2

1

cos
(x )
2
sec (x )
2 cos 2 (x )  sin 2 (x )  1
cos 2 (x )  cos 2 (x )  sin 2 (x )  1
Now we can see that the encircled part matches one side of the
Pythagorean identity. We can replace this part with what the
Pythagorean identity says that it is equal to.
Pythagorean identity:
sin2 (x )  cos 2 (x )  1
Thus we can replace the encircled part with 1.
cos 2 (x )  1  1
cos 2 (x )
Notice that we have now made what was on the left side of
the equation look exactly like what was on the right side. This
means that we have verified the identity.