Inverse Trigonometric Functions Section 4.7. Objectives Evaluate inverse trigonometric functions at...

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 FunctionsThis 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.


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.


The restricted graph of the function f(x) = sin(x) is one-to-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.

and thus has in inverse function

What is the domain?

)(sin)( 1 xxg


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].


What is the range?

)(sin)( 1 xxg

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.


The restricted graph of the function f(x) = cos(x) is one-to-one




What is the domain?

)(cos)( 1 xxg


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].


What is the range?

)(cos)( 1 xxg

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.


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.

The restricted graph of the function f(x) = tan(x) is one-to-one




What is the domain?

)(tan)( 1 xxg


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 (-, ).


What is the range?

)(tan)( 1 xxg

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)

)1(sin 12

What angle between and has a sine value of 1? 2


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

You should know what angle has a sine value of 1. Thus

2)1(sin 1


621

sin 1

What angle between and has a sine value of -1/2? 2

2


You should know what angle has a sine value of -1/2. Thus

21

sin 1


)0(cos 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

2)0(cos 1

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 .

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 in them.

What angle between and has a tangent value of ?


)3(tan 1 2

3

2

3


3

What angle between and has a tangent value of ?


)3(tan 1 2

3

1

23

21

6cos

6sin

6tan

2

3


Check π/6:

Check π/3:

3

21

23

3cos

3sin

3tan

This is not what we want for a tangent value.

Thus

3)3(tan 1

You should know what angle has a sine value of . Thus

What angle between and has a

sine value of ?


22

22

sin 12

2

22

422

sin 1



)1(cos 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

)1(cos 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.

Now we plug this into the original expression replace the

To evaluate 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.

Evaluate each of the following

)1(sinsin 1

)1(sin 1

2)1(sin 1

The first step in evaluating this expression is to find the value of the inside parentheses.

)1(sin 1

2

sin

This we can evaluation to get 1. Thus 1)1(sinsin 1


To evaluate we ask the question “What angle between

0 and π has a cosine value of ?” This is one that we should know.


23

23

cossin 1

23

cos 1




This we can evaluation to get 1/2. Thus

623

cos 1

6

sin

21

23

cossin 1

23

cos 1

Now we plug this into the original expression replace the tan-1(1)


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.


)1(tancos 1

4)1(tan 1

4

cos

This we can evaluation to get . Thus 22

22

)1(tancos 1


45

sinsin 1

22

45

sin

45

sin


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.


22

sin 1


–π/2 and π/2 has a sine value of ?” This is one that we should know.

22

sin 1

22

422

sin 1

445

sinsin 1

Thus



35

coscos 1

21

35

cos

35

cos




0 and π has a sine value of 1/2?” This is one that we should know.

21

cos 1

21

cos 1

321

cos 1

335

coscos 1

Thus



43

coscos 1

22

43

cos

43

cos



0 and π has a sine value of ?” This is one that we should know. Thus

43

43

coscos 1

22

cos 1

22

cos 1

22

43

22

cos 1



1

22

22

43

cos

43

sin

43

tan

43

tantan 1

43

tan



1tan 1

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.


4

1tan 1

43

tantan 1

Thus44

3tantan 1

Rewrite the expression as an algebraic expression in x

2

arcsintanx

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.

θ

x 2

This triangle has an angle θ whose sine value is x/2. continued on next slide


2

arcsintanx

This means that in this problem, we can replace the arcsin(x/2) in the problem with θ.

θ

x 2

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.

tan



2

arcsintanx

tan

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.

θ

x 2


adj

2

22

22

222

4)adj(

4)adj(

4)adj(

2)adj(

x

x

x

x


2

arcsintanx

tan

24)adj( x

Since length must be positive, we will keep the positive square root.

θ

x 2

adj

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:

24)tan(

2arcsintan

x

xx

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