Transcript Right Triangle Trigonometry

Inverse Trigonometric
Functions
Section 4.7
Objectives
• Evaluate inverse trigonometric
functions at given values.
• State the domain and range of each of
the inverse trigonometric functions.
• Use right triangles to find the
composition of a trigonometric function
and an inverse trigonometric function.
• Solve simple trigonometric equations
requiring inverse trigonometric
functions.
Vocabulary
•
•
•
•
arcsine of a number
arccosine of a number
arctangent of a number
arcsecant of a number
Inverse Functions
You should remember that a function has an
inverse that is a function only if the original
function is one-to-one. A one-to-one function
passes the horizontal line test.
Recall that when we had a function that was not
one-to-one, we were able to restrict the domain
of the function so that the function on the
restricted domain was one-to-one.
continued on next slide
Inverse Functions
This will be necessary for all of our trigonometric
functions. When we restrict the domains, we could
chose any restriction on which our function does not
repeat any y-values. This, however, will not necessarily
have any consistency from one person to the next.
Thus, we will need a consistent method to restrict our
domains. The consistent method will be to always
include all angles from quadrant I that are between 0
and 2π. We will also include a quadrant contiguous with
quadrant I that will include all of the negative values of
the trigonometric function.
This method will produce different restrictions for
different trigonometric functions.
The graph of the function
f(x) = sin(x) is not one-to-one
Using our method for restricting the domain, we will need to include
the interval [0, π/2]. We need to determine what other interval to
take. If we take the interval which is quadrant II, we will get
repeated y-values and thus will not be one-to-one. This means that we
must take the interval that is quadrant IV.
continued on next slide
The graph of the function
f(x) = sin(x) is not one-to-one
We normally think of quadrant IV as being between [(3π)/2, 2π]. This
interval for quadrant IV is not contiguous with the quadrant I that we
must use. Thus we need a different interval that is also quadrant IV.
This interval will be [- π/2, 0]. The restricted graph is shown on the
next screen.
continued on next slide
The restricted graph of the
function f(x) = sin(x) is one-toone
Now we need to know what the inverse of this restricted sine function
is. Recall that to find the graph of the inverse function, we reflect the
original function over the line y=x. This reflection will interchange the
x-values and y-values.
continued on next slide
and thus has in inverse function
g (x )  sin (x )
1
What is the domain?
Recall that the domain of the inverse function is the same as the range of
the original function. The range of the sin(x) function is [-1, 1]. Thus the
domain of the sin-1(x) is [-1, 1].
continued on next slide
and thus has in inverse function
g (x )  sin (x )
1
What is the range?
Recall that the range of the inverse function is the same as the domain of
the original function. The restricted domain that we used for the sin(x)
function is [-π/2, π/2]. Thus the range of the sin-1(x) is [-π/2, π/2].
The graph of the function
f(x) = cos(x) is not one-to-one
Using our method for restricting the domain, we will need to include
the interval [0, π/2]. We need to determine what other interval to
take. If we take the interval [π/2, π] which is quadrant II, we will get
not get any repeated y-values. The restricted graph is shown on the
next screen.
continued on next slide
The restricted graph of the
function f(x) = cos(x) is one-toone
Now we need to know what the inverse of this restricted sine function
is. Recall that to find the graph of the inverse function, we reflect the
original function over the line y=x. This reflection will interchange the
x-values and y-values.
continued on next slide
and thus has in inverse function
g (x )  cos (x )
1
What is the domain?
Recall that the domain of the inverse function is the same as the range of
the original function. The range of the cos(x) function is [-1, 1]. Thus
the domain of the cos-1(x) is [-1, 1].
continued on next slide
and thus has in inverse function
g (x )  cos (x )
1
What is the range?
Recall that the range of the inverse function is the same as the domain of
the original function. The restricted domain that we used for the cos(x)
function is [0, π]. Thus the range of the cos-1(x) is [0, π].
The graph of the function
f(x) = tan(x) is not one-to-one
Using our method for restricting the domain, we will need to include
the interval [0, π/2). We need to determine what other interval to
take. If we take the interval which is quadrant II, we will get
repeated y-values and thus will not be one-to-one. This means that we
must take the interval that is quadrant IV.
continued on next slide
The graph of the function
f(x) = tan(x) is not one-to-one
We normally think of quadrant IV as being between [(3π)/2, 2π]. This
interval for quadrant IV is not contiguous with the quadrant I that we
must use. Thus we need a different interval that is also quadrant IV.
This interval will be (- π/2, 0]. The restricted graph is shown on the
next screen.
continued on next slide
The restricted graph of the
function f(x) = tan(x) is oneto-one
Now we need to know what the inverse of this restricted sine function
is. Recall that to find the graph of the inverse function, we reflect the
original function over the line y=x. This reflection will interchange the
x-values and y-values.
continued on next slide
and thus has in inverse function
g (x )  tan (x )
1
What is the domain?
Recall that the domain of the inverse function is the same as the range of
the original function. The range of the tan(x) function is (-, ). Thus
the domain of the tan-1(x) is (-, ).
continued on next slide
and thus has in inverse function
g (x )  tan (x )
1
What is the range?
Recall that the range of the inverse function is the same as the domain of
the original function. The restricted domain that we used for the tan(x)
function is (- π/2, π/2). Thus the range of the tan-1(x) is (- π/2, π/2).
Notice that the endpoints are not included since the tangent function is
not defined on - π/2 or π/2.
Evaluate each of the following
(remember that the output of an
inverse trigonometric function is an
angle)
When we are finding the value of an inverse function, our input is
a number and the output we are looking for an angle.
In order to find our answer, we can ask the question below. The
angles are determined by the range of the inverse function that
we are calculating
sin (1)
1


What angle between 
and
has a
2
2
sine value of 1?
You should know what angle has a sine value of 1. Thus
sin (1) 
1

2
continued on next slide
Evaluate each of the following
(remember that the output of an
inverse trigonometric function is an
angle)
1
1 
sin   
 2
What angle between 
sine value of -1/2?


and
has a
2
2
You should know what angle has a sine value of -1/2. Thus

 1
sin     
6
 2
1
continued on next slide
Evaluate each of the following
(remember that the output of an
inverse trigonometric function is an
angle)
cos (0)
1
What angle between 0 and π has a cosine
value of 0?
You should know what angle has a cosine value of 0. Thus
cos (0) 
1

2
continued on next slide
Evaluate each of the following
(remember that the output of an
inverse trigonometric function is an
angle)

tan 1 ( 3 )

What angle between 
and 2 has a
tangent value of 3 ? 2
Unlike previous problems where I said “you should know this angle”,
I don’t expect you to have any of the tangent values memorized.
Instead, you need to remember that the tangent function is
defined as the sine function divided by the cosine function. Now
what we need is an angle whose sine value or cosine value have the 3
in them.
The angles that fit this bill and also have positive tangent values
are π/6 and π/3. Basically, now we will check each of these
angles to see which one has a tangent value of 3 .
continued on next slide
Evaluate each of the following
(remember that the output of an
inverse trigonometric function is an
angle)


tan ( 3 )
1

1
sin 
 

1
2
6


tan   
   
3
 6  cos     3 
  

6  2 


Check π/6:
Check π/3:
Thus
What angle between 
2 and 2 has a
tangent value of 3 ?
This is not what we want for
a tangent value.
   3 

sin  


3  2 



tan   

 3
1
 3  cos   
 
 
3
2
tan ( 3) 
1

3
continued on next slide
Evaluate each of the following
(remember that the output of an
inverse trigonometric function is an
angle)
 2
1

sin 

2




What angle between 
and
has a
2
2
2
sine value of
2 ?
You should know what angle has a sine value of
2
. Thus
2
 2 

sin 
 4
2


1
continued on next slide
Evaluate each of the following
(remember that the output of an
inverse trigonometric function is an
angle)
cos ( 1)
1
What angle between 0 and π has a cosine
value of -1?
You should know what angle has a cosine value of -1. Thus
cos 1 ( 1)  
Composition of Functions
The following slides will have us explore the
composition of trigonometric function and inverse
trigonometric function.
You should note in these exercises that when a
function is composed with its inverse, you do not
always get back the original input because of the
restriction that had to be placed on the domain of
the original function in order to get a part of the
function that was one-to-one.
Evaluate each of the following
sinsin (1) 
1
The first step in evaluating this expression is to find the value
of the inside parentheses.
To evaluate sin 1 (1) we ask the question “What angle between
–π/2 and π/2 has a sine value of 1?” This is one that we should
know.
sin 1 (1) 

2
Now we plug this into the original expression replace the sin 1 (1)


sin 
2


This we can evaluation to get 1. Thus sin sin 1 (1)  1
continued on next slide
Evaluate each of the following

 3 

1

sin cos 


2



The first step in evaluating this expression is to find the value
of the inside parentheses.


3
1
To evaluatecos  2  we ask the question “What angle between


0 and π has a cosine value of 3 ?” This is one that we should
2
know.
 3 

cos 1 

 2  6
 3


2


Now we plug this into the original expression replace the cos 1 

sin 
6

 3  1
1

 
This we can evaluation to get 1/2. Thus sin cos 


 2  2

continued on next slide
Evaluate each of the following
cos tan (1) 
1
The first step in evaluating this expression is to find the value
of the inside parentheses.
To evaluate tan-1(1) we ask the question “What angle between
-π/2 and π/2 has a tangent value of 1?” This is one that should
not be too difficult to determine since to get a tangent of 1,
the sine and the cosine values must be the same. This happens
only one place in the interval.
tan 1 (1) 

4
Now we plug this into the original expression replace the tan-1(1)


cos 
4
This we can evaluation to get
2
2
. Thus cos tan 1 (1) 
2
2


Evaluate each of the following
 5  
1 
sin  sin

  4 
The first step in evaluating this expression is to find the value
of the inside parentheses. This is a trigonometric function
value that we should be able to get quickly.
2
 5 
sin

2
 4 
5
Now we plug this into the original expression replace the sin  
 4 


2

sin 1  

 2 

To evaluate sin 1  

2

2 
we ask the question “What angle between
–π/2 and π/2 has a sine value of
should know.

2


sin 1  

4
 2 

2
2
?” This is one that we

5
 4
Thus sin 1  sin



  
4

continued on next slide
Evaluate each of the following
 5  
1 
cos  cos 

 3 

The first step in evaluating this expression is to find the value
of the inside parentheses. This is a trigonometric function
value that we should be able to get quickly.
 5
cos 
 3
 1

 2
 5 
Now we plug this into the original expression replace the cos 

 3 
1
cos 1  
2
1
2
To evaluate cos 1   we ask the question “What angle between
0 and π has a sine value of 1/2?” This is one that we should
know.
1 

 5   
cos 1   
Thus cos 1  cos 
 
2 3
 3  3

continued on next slide
Evaluate each of the following

 3
cos  cos 
 4

1



The first step in evaluating this expression is to find the value
of the inside parentheses. This is a trigonometric function
value that we should be able to get quickly.
2
 3 
cos 

2
 4 
 3 
Now we plug this into the original expression replace the cos  
 4 


2

cos 1  

 2 
To

2
1


cos

evaluate
 2 


we ask the question “What angle between
2
0 and π has a sine value of 
?” This is one that we should
2
know.

 3   3
2  3
1 
1
cos
cos

Thus



 
cos  


 4  4

4
 2 
Evaluate each of the following
 3  
1 
tan  tan 

 4 

The first step in evaluating this expression is to find the value
of the inside parentheses. This is a trigonometric function
value that we should be able to get quickly.
 2
 3 


sin



2
3



 4   
  1
tan 

 4  cos  3  
2

 


 4   2 
 3 
Now we plug this into the original expression replace the tan 

 4 
tan 1  1
continued on next slide
Evaluate each of the following
 3  
1 
tan  tan 

 4 

To evaluate tan-1(-1) we ask the question “What angle between
- π/2 and π/2 has a tangent value of -1?” This is one that
should not be too difficult to determine since to get a tangent
of -1, the sine and the cosine values must be negatives of each
other. This happens only one place in the interval.
tan 1  1  

4


 3  
Thus tan 1  tan     
4
 4 

Rewrite the expression as an
algebraic expression in x
x 


tan  arcsin   
 2 

For this problem, we must start inside the parentheses with arcsin(x/2).
This is the angle θ between –π/2 and π/2 whose sine value is x/2.
Although we don’t know the measure of the angle θ, we can draw a
triangle with the angle θ whose sine value is x/2.
x
2
θ
This triangle has an angle θ whose sine value is x/2.
continued on next slide
Rewrite the expression as an
algebraic expression in x
x 


tan  arcsin   
 2 

This means that in this problem, we can replace the arcsin(x/2) in the
problem with θ.
tan  
x
2
θ
Using the triangle that we drew, we can find the tan(θ) if we
have the lengths of the adjacent side and the opposite side.
continued on next slide
Rewrite the expression as an
algebraic expression in x
x 

tan  arcsin   
 2 

tan  
2
x
θ
adj
Since all we need for our answer is an expression in terms of x, we do not
need a numerical value of the length of the adjacent side or the length of
the opposite side. We will use the Pythagorean theorem to find an
expression for the length of the adjacent side.
x 2  ( adj)2  22
x 2  ( adj)2  4
( adj)2  4  x 2
( adj)   4  x 2
continued on next slide
Rewrite the expression as an
algebraic expression in x
x 

tan  arcsin   
 2 

tan  
2
x
θ
adj
Since length must be positive, we will keep the positive square root.
( adj)   4  x 2
Thus we can write the tan(θ) as the ratio of the opposite side to
the adjacent side. This will give us the following algebraic
expression in x:
x 

tan  arcsin    tan( ) 
 2 

x
4x2