Download Section 7.1 – Inverse Trigonometric Functions

63
Section 7.1 – Inverse Trigonometric Functions
Recall that if we examine function like p(x) = x2 + 4, it was not one-to-one if
there was a horizontal line that passed through more than one point on the
graph:
Since it is not one-to-one, it does not have an inverse function. But, if we
restricted the domain of f so that you could not find a horizontal line that
passes through more than one point on the graph, then the function with
that restricted domain would be one-to-one and have an inverse function.
Consider p(x) = x2 + 4, x ≤ 0
This function has a restricted
domain, giving only the left
half of the parabola. By the
horizontal test, the function
with this restricted domain
is one-to-one. The domain
is {x | x ≤ 0} and the range
is {y | y ≥ 4}. The inverse
function is p – 1(x) = – x −4 .
We have also seen that the
graph of a function and its
inverse are symmetrical to
the line y = x.
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p
p–1
If we examine the function f(x) = sin(x), we can see that it is not one-to-one.
–π
–
π
2
π
2
π
3π
2
[
However, suppose we restrict the domain to –
–π
–
π
2
π
2
π
3π
2
2π
π
2
,
π
2
5π
2
]:
2π
5π
2
Thus, the sine function with this domain will have an inverse function. In our
definition of the inverse sine function, we reverse the roles of x and y.
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Defintion
[
Let x = sin(y) be a function on the restricted domain of –
π
2
,
π
2
]. Then, the
inverse sine function (arcsine) is defined as:
π
2
y = sin – 1(x) or arcsin(x) where – 1 ≤ x ≤ 1 and –
The value y is the angle between –
π
2
and
π
2
π
2
≤y≤
.
inclusively whose sine is x.
To obtain the graph of sin – 1(x), we can reflect the graph of sin(x) defined
[
on the interval –
π
2
π
2
,
] across the y = x line.
sin – 1(x)
sin(x)
Objective 1:
Find the Exact Value of an Inverse Sine Function.
Find the Exact Value of the following:
Ex. 1a
sin – 1(– 1)
sin – 1(
Ex. 1b
3
2
)
Solution:
a)
[
We are looking for a value of y in –
π
2
,
π
2
] such that
sin(y) = – 1. The sine is equal to – 1 when the terminal side lies
on the negative y-axis. Thus, the angle is –
b)
3
2
. The sine is equal to
in quadrant I. Thus, the angle is
π
3
.
π
π
,
such that
2
2
3
when the terminal
2
[
We are looking for a value of y in –
sin(y) =
π
2
.
]
side is
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Objective 2:
Find an Approximate Value of an Inverse Sine Function.
On most calculators, you will need to hit the "Shift" or "2nd" key followed by
the "SIN" key to obtain the answer.
Find the following. Round to two decimal places:
Ex. 2a
2
5
sin – 1( )
sin – 1(– 0.75)
Ex. 2b
Solution:
Be sure you calculator is in radian mode, before you begin.
2
5
a)
sin – 1( ) = 0.4115168… ≈ 0.41 radians.
b)
sin – 1(– 0.75) = – 0.848062… ≈ – 0.85 radians.
Objective 3:
Using Properties of Inverse Functions to Find the Exact
Value of Certain Composite Functions.
Recall that in our discussion of inverse functions, if we were finding
(f o g)(x) where f and g are inverse functions, the value of x had to be in the
domain of the inside function. Applying this to the sine and inverse sine
functions, we get:
π
2
π
2
1)
sin – 1(sin(x)) = x where –
2)
sin(sin – 1 (x)) = x where – 1 ≤ x ≤ 1 .
Note, if x is not between –
π
2
and
π
2
≤x≤
.
inclusively, the composition
sin – 1(sin(x)) = x will produce an angle that is between –
π
2
π
2
and
inclusively. So, we get the reference angle if sin(x) > 0, or the opposite of
the reference angle if sin(x) < 0.
Find the Exact Value of the Following:
Ex. 3a
sin – 1(sin(
π
7
))
Ex. 3c
sin(sin – 1 (0.62))
Solution:
a)
b)
Since
9π
5
π
7
[
9π
5
Ex. 3b
sin – 1(sin(
Ex. 3d
sin(sin – 1 (– 3))
is in the interval in –
π
2
,
π
2
], then sin
–1
9π
5
) is negative, then sin – 1(sin(
9π
5
π
7
9π
5
(sin(
is in quadrant IV, so the reference angle is 2π –
Since sin(
))
)) = –
π
5
)) =
=
π
5
π
7
.
.
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c)
d)
Since 0.62 is in the interval [– 1, 1], then
sin(sin – 1 (0.62)) = 0.62.
Since – 3 is not in the domain of the inverse sine, then
sin(sin – 1 (– 3)) is undefined.
If we examine the graph of the cosine function, we see that we will have the
same problem that we had with the sine function:
–π
–
π
2
π
2
π
3π
2
2π
5π
2
Clearly, it is not one-to-one. So, we will need to restrict the domain. This
time, we will choose the interval [0, π]:
–π
–
π
2
π
2
π
3π
2
2π
5π
2
Thus, the cosine function with this domain will have an inverse function. In
our definition of the inverse cosine function, we reverse the roles of x and y.
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Defintion
Let x = cosine(y) be a function on the restricted domain of [0, π]. Then, the
inverse cosine function (arccosine) is defined as:
y = cos– 1(x) or arccos(x) where – 1 ≤ x ≤ 1 and 0 ≤ y ≤ π.
The value y is the angle between 0 and π inclusively whose cosine is x. To
obtain the graph of cos – 1(x), we can reflect the graph of cos(x) defined on
the interval [0, π] across the y = x line.
cos – 1(x)
cos(x)
Find the Exact Value of the following:
Ex. 4a
cos – 1(– 1)
Ex. 4b
cos – 1(
3
2
)
Solution:
a)
We are looking for a value of y in [0, π] such that
cos(y) = – 1. The cosine is equal to – 1 when the terminal side
lies on the negative x-axis. Thus, the angle is π.
b)
We are looking for a value of y in [0, π]. such that
3
2
cos(y) =
. The cosine is equal to
side is in quadrant I. Thus, the angle
3
when
2
π
is .
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the terminal
On most calculators, you will need to hit the "Shift" or "2nd" key followed by
the "COS" key to obtain the answer.
Find the following. Round to two decimal places:
Ex. 5a
cos – 1(
2
5
)
Ex. 5b
cos – 1(– 0.75)
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Solution:
Be sure you calculator is in radian mode, before you begin.
2
5
a)
cos – 1(
b)
cos – 1(– 0.75) = 2.418858… ≈ 2.42 radians.
) = 1.159279… ≈ 1.16 radians.
Applying the inverse properties with composition of the cosine and inverse
cosine functions, we get:
1)
cos – 1(cos(x)) = x where 0 ≤ x ≤ π.
2)
cos(cos – 1 (x)) = x where – 1 ≤ x ≤ 1 .
Note, if x is not between 0 and π inclusively, the composition
cos – 1(cos(x)) = x will produce an angle that is between 0 and π inclusively.
So, we get the reference angle if cos(x) > 0, or π minus the reference angle
if cos(x) < 0.
Find the Exact Value of the Following:
π
8
Ex. 6a
cos – 1(cos(
Ex. 6c
cos(cos – 1 (– 0.78))
))
8π
))
7
3π
cos(cos – 1 ( ))
2
cos – 1(cos(
Ex. 6b
Ex. 6d
Solution:
a)
b)
Since
8π
7
π
8
is in the interval in [0, π] then cos – 1(cos(
is in quadrant III, so the reference angle is
Since cos(
8π
7
) is negative, then cos – 1(cos(
angle in quadrant II which is π –
cos – 1(cos(
8π
7
)) =
π
7
=
6π
7
)) =
–π=
6π
7
Since – 0.78 is in the interval [– 1, 1], then
cos(cos – 1 (– 0.78)) = – 0.78.
d)
Since
is not in the domain of the inverse cosine, then
cos(cos – 1(
3π
2
)) is undefined.
π
.
8
π
.
7
)) will yield an
. Hence,
c)
3π
2
8π
7
8π
7
π
8
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If we examine the function f(x) = tan(x), we can see that it is not one-toone.
– 2π
–π
–
π
2
π
2
π
(
However, suppose we restrict the domain to –
– 2π
–π
–
π
2
π
2
π
2
,
2π
π
2
):
π
2π
Thus, the tangent function with this domain will have an inverse function. In
our definition of the inverse tangent function, we reverse the roles of x & y.
Defintion
(
Let x = tan(y) be a function on the restricted domain of –
π
2
,
π
2
). Then,
the inverse tangent function (arctangent) is defined as:
y = tan – 1(x) or arctan(x) where – ∞ < x < ∞ and –
π
2
<y<
π
2
.
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The value y is the angle between –
π
2
and
π
2
inclusively whose tangent is
x. To obtain the graph of tan – 1(x), we can reflect the graph of tan(x)
(
defined on the interval –
π
2
,
π
2
) across the y = x line.
tan(x)
tan – 1(x)
Find the Exact Value of the following:
Ex. 7a
tan – 1(1)
Ex. 7b
Solution:
a)
(
We are looking for a value of y in –
tan – 1(0)
π
2
π
2
,
) such that
tan(y) = 1. The tangent is equal to 1 when the terminal side lies
π
4
in the first quadrant. Thus, the angle is
b)
(
We are looking for a value of y in –
π
2
.
π
2
,
) such that
tan(y) = 0. The tangent is equal to 0 when the terminal side lies
on the x-axis. Thus, the angle is 0.
On most calculators, you will need to hit the "Shift" or "2nd" key followed by
the "TAN" key to obtain the answer.
Find the following. Round to two decimal places:
Ex. 8a
5
8
tan – 1( )
Ex. 8b
tan – 1(– 3.2)
Solution:
Be sure you calculator is in radian mode, before you begin.
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5
8
a)
tan – 1( ) = 0.558599… ≈ 0.56 radians.
b)
tan – 1(– 3.2) = – 1.267911… ≈ – 1.27 radians.
Applying the inverse properties with composition of the tangent and inverse
tangent function, we get:
π
2
π
2
1)
tan – 1(tan(x)) = x where –
2)
tan(tan – 1 (x)) = x where – ∞ < x < ∞ .
Note, if x is not between –
π
2
and
π
2
<x<
.
inclusively, the composition
tan – 1(tan(x)) = x will produce an angle that is between –
π
2
and
π
2
inclusively. So, we get the reference angle if tan(x) > 0, or the opposite of
the reference angle if tan(x) < 0.
Find the Exact Value of the Following:
11π
5
Ex. 9b
tan – 1(tan(
Ex. 9c
tan(tan – 1 (4.67))
Solution:
Ex. 9d
tan(tan – 1 (– 3))
a)
tan – 1(tan(–
2π
5
))
Ex. 9a
Since –
2π
5
tan – 1(tan(–
11π
5
b)
d)
2π
5
))= –
2π
5
π
2
,
π
2
), then
.
is in quadrant III, so the reference angle is
Since tan(
c)
(
is in the interval in –
11π
5
))
) is positive, then tan – 1(tan(
11π
5
11π
5
)) =
–π=
π
5
π
5
.
Since 4.67 is in the interval (– ∞, ∞), then
tan(tan – 1 (4.67)) = 4.67
Since – 3 is in the interval (– ∞, ∞), then tan(tan – 1 (– 3)) = – 3.
Objective 4:
Finding the Inverse Function of a Trigonometric Function.
To find the inverse function algebraically, we interchange x and y and solve
for y. We can do the same thing for trigonometric functions.
Find the inverse function of the following. State the domain of the
function and the inverse function:
Ex. 10a
f(x) = 2cos(3x + 1) + 4
Ex. 10b
g(x) = – tan(x + 2) – π
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Solution:
a)
Domain of f is (– ∞, ∞).
Replace f(x) by y:
y = 2cos(3x + 1) + 4
x = 2cos(3y + 1) + 4
x – 4 = 2cos(3y + 1)
x−4
2
= cos(3y + 1)
(interchange x and y)
(solve for the cosine function)
(use the defintion of the inverse cosine)
( x−4
2 )
–1
( x−4
2 )
3y + 1 = cos – 1
(solve for y)
3y = cos – 1
y=
 x− 4 
−1
cos −1
 2 
3
So, f – 1(x) =
 x− 4
cos −1
 2
3

−1

To find the domain of f – 1, the argument for inverse cosine has to be
between – 1 and 1 inclusively:
–1≤
x−4
2
≤1
(now, solve for x)
–2≤x–4≤2
2≤x≤6
So, the domain of f – 1 is [2, 6].
b)
The argument of the tangent function cannot be equal to odd
multiples of
x≠
(2k+1)π
2
π
2
. Thus, x + 2 ≠
(2k+1)π
2
where k is an integer or
– 2 where k is an integer.
Thus, the domain of f is {x| x ≠
(2k+1)π
2
– 2, k is an integer}
Replace g(x) by y:
y = – tan(x + 2) – π
(interchange x and y)
x = – tan(y + 2) – π
(solve for the tangent function)
x + π = – tan(y + 2)
– x – π = tan(y + 2)
(use the defintion of the inverse tangent)
–1
(solve for y)
y + 2 = tan (– x – π)
–1
y = tan (– x – π) – 2
The domain of f – 1 is all real numbers.
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Objective 5:
Solve Equations Involving Inverse Trigonometric
Functions.
If an equation has an inverse trigonometric function in it, we will need to
first isolate the inverse trigonometric function and then apply the definition
of the appropriate inverse trigonometric function.
Solve the following:
Ex. 11a
6cos – 1(3x) = 11π
Ex. 11b
5π + 2tan – 1(x) = 4π – 2tan – 1(x)
Solution:
a)
First, solve for the inverse cosine:
6cos – 1(3x) = 11π
11π
6
cos – 1(3x) =
3x = cos(
11π
6
(now, apply the defintion)
)
The angle is in quadrant IV, hence the cosine is positive and the
reference angle is 2π –
3x = cos(
3x =
x=
b)
11π
6
11π
6
=
π
6
. So, cos(
11π
6
) = cos(
π
6
)=
3
2
.
)
3
2
3
6
(solve for x)
The solution is
3
6
{
}.
First, solve for the inverse tangent:
5π + 2tan – 1(x) = 4π – 2tan – 1(x)
5π + 4tan – 1(x) = 4π
4tan – 1(x) = – π
tan – 1(x) = –
x = tan(–
π
4
π
4
(now, apply the definition)
)
The angle is in quadrant IV and so, the tangent function is negative.
The reference angle is
x = tan(–
x=–1
π
4
π
4
. Thus, tan(–
π
4
) = – tan(
)
The solution is {– 1}.
π
4
) = – 1.