4 7 INVERSE TRIG FNS

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4.7
INVERSE TRIGONOMETRIC FUNCTIONS
Copyright © Cengage Learning. All rights reserved.
What You Should Learn
• Evaluate and graph the inverse sine function.
• Evaluate and graph the other inverse
trigonometric functions.
• Evaluate and graph the compositions of
trigonometric functions.
2
Inverse Sine Function
3
Inverse Sine Function
For a function to have an inverse function, it must be one-to-one—that
is, it must pass the Horizontal Line Test.
Restrict the domain to the interval – / 2  x
  / 2, the following properties hold.
1. On the interval [– / 2,  / 2], the function
y = sin x is
increasing.
sin x has an inverse function
on this interval.
By definition, the values of inverse
trigonometric functions are always
in radians.
2. On the interval [– / 2,  / 2], y = sin x
takes on its full
range of values, –1  sin x  1.
3. On the interval [– / 2,  / 2], y = sin x is
one-to-one.
4
Inverse Sine Function
On the restricted domain – / 2  x   / 2, y = sin x has a
unique inverse function called the inverse sine function.
y = arcsin x
or
y = sin –1 x.
means the angle (or arc) whose sine is x.
5
Inverse Sine Function
6
Example 1 – Evaluating the Inverse Sine Function
If possible, find the exact value.
a.
Solution:
a. Because
b.
c.
for
.
,
Angle whose sine is
7
Example 1 – Solution
b. Because
for
.
cont’d
,
Angle whose sine is
c. It is not possible to evaluate y = sin –1 x when x = 2
because there is no angle whose sine is 2.
Remember that the domain of the inverse sine function
is [–1, 1].
8
Example 2 – Graphing the Arcsine Function
Sketch a graph of y = arcsin x.
Solution:
In the interval [– / 2,  / 2],
9
Example 2 – Solution
cont’d
y = arcsin x
Domain: [–1, 1]
Range: [– / 2,  / 2]
10
Other Inverse Trigonometric
Functions
11
Other Inverse Trigonometric Functions
The cosine function is decreasing and one-to-one on the
interval 0  x  .
cos x has an inverse function on this interval.
On the interval 0  x   the cosine function has an inverse
function—the inverse cosine function:
y = arccos x or y = cos –1 x.
12
Other Inverse Trigonometric Functions
DOMAIN: [–1,1]
RANGE:
DOMAIN: [–1,1]
RANGE: [0,  ]
DOMAIN:
RANGE:
13
Example 3 – Evaluating Inverse Trigonometric Functions
Find the exact value.
a. arccos
b. cos –1(–1)
c. arctan 0
d. tan –1 (–1)
Solution:
a. Because cos ( / 4) =
, and  / 4 lies in [0,  ],
.
Angle whose cosine is
14
Example 3 – Solution
cont’d
b. Because cos  = –1, and  lies in [0,  ],
cos –1(–1) = .
Angle whose cosine is –1
c. Because tan 0 = 0, and 0 lies in (– / 2,  / 2),
arctan 0 = 0.
Angle whose tangent is 0
15
Example 3 – Solution
cont’d
d. Because tan(– / 4) = –1, and – / 4 lies in (– / 2,  / 2),
tan –1 (–1) =
.
Angle whose tangent is –1
16
Compositions of Functions
17
Compositions of Functions
For all x in the domains of f and f –1, inverse functions have the
properties
f (f –1(x)) = x
and
f –1(f (x)) = x.
18
Compositions of Functions
These inverse properties do not apply for arbitrary values
of x and y.
.
The property
arcsin(sin y) = y
is not valid for values of y outside the interval [– / 2,  / 2].
19
Example 5 – Using Inverse Properties
If possible, find the exact value.
a. tan[arctan(–5)]
b.
c. cos(cos –1 )
Solution:
a. Because –5 lies in the domain of the arctan function, the
inverse property applies, and you have
tan[arctan(–5)] = –5.
20
Example 5 – Solution
cont’d
b. In this case, 5 / 3 does not lie within the range of the
arcsine function, – / 2  y   / 2.
However, 5 / 3 is coterminal with
which does lie in the range of the arcsine function,
and you have
21
Example 5 – Solution
cont’d
c. The expression cos(cos –1) is not defined because
cos –1  is not defined.
Remember that the domain of the inverse cosine
function is [–1, 1].
22
Example 6 – Evaluating Compositions of Functions
Find the exact value: tan
2
arccos
3
2
arccos ,
3
2
3
Let 𝑢 =
then cos 𝑢 = > 0, therefore, u is in QI
2
sin 𝑢
tan arccos
= tan 𝑢 =
=
3
cos 𝑢
2
1− 3
2
3
2
5
=
2
23
Example 6 – Evaluating Compositions of Functions
Find the exact value: cos
Let 𝑢 =
Q IV
3
arc𝑠𝑖𝑛(− ),
5
3
cos arc𝑠𝑖𝑛(− ) =
5
3
arc𝑠𝑖𝑛(− )
5
then sin 𝑢 =
3
1− −
5
3
−
5
2
< 0, therefore, u is in
4
=
5
24
Example
Write the expression as an algebraic expression in x:
1
sin arccos 3𝑥 , 0 ≤ 𝑥 ≤
3
sin arccos 3𝑥 = 1 − 9𝑥 2
25
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