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Trigonometric Functions:
Right Triangle Approach
Copyright © Cengage Learning. All rights reserved.
6.4 Inverse Trigonometric Functions
and Right Triangles
Copyright © Cengage Learning. All rights reserved.
Objectives
► The Inverse Sine, Inverse Cosine, and Inverse
Tangent Functions
► Solving for Angles in Right Triangles
► Evaluating Expressions Involving Inverse
Trigonometric Functions
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The Inverse Sine, Inverse Cosine,
and Inverse Tangent Functions
4
Real Numbers
Let's first consider the sine function. We restrict the domain
of the sine function to angles  with – /2     /2.
From Figure 1 we see that on this domain the sine function
attains each of the values in the [–1, 1] interval exactly
once and so is one-to-one.
Restricted domains of the sine, cosine, and tangent functions
Figure 1
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Real Numbers
Similarly, we restrict the domains of cosine and tangent as
shown in Figure 1. On these restricted domains we can
define an inverse for each of these functions.
By the definition of inverse function we have
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Real Numbers
We summarize the domains and ranges of the inverse
trigonometric functions in the following box.
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Example 1 – Evaluating Inverse Trigonometric Functions
Find the exact value.
(a) sin–1
(b) cos–1
(c) tan–1 1
Solution:
(a) The angle in the interval [– /2,  /2] whose sine is
is  /3. Thus sin–1(
) =  /3.
(b) The angle in the interval [0, ] whose cosine is
2 /3. Thus cos–1
= 2 /3.
is
(c) The angle in the interval [– /2,  /2] whose tangent is 1
is  /4. Thus tan–1 1 =  /4.
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Solving for Angles in Right Triangles
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Example 3 – Finding an Angle in a Right Triangle
Find the angle  in the triangle shown in Figure 2.
Figure 2
Solution:
Since  is the angle opposite the side of length 10 and the
hypotenuse has length 50, we have
sin 
sin
Now we can use sin–1 to find  :
=
sin–1
  11.5
Definition of sin–1
Calculator (in degree mode)
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Evaluating Expressions Involving Inverse
Trigonometric Functions
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Example 7 – Composing Trigonometric Functions and Their Inverses
Find cos(sin–1 ).
Solution 1:
Let  = sin–1 . Then is the number in the interval [– /2,  /2]
whose sine is .
Let's interpret  as an angle
and draw a right triangle with
 as one of its acute angles,
with opposite side 3 and
hypotenuse 5 (see Figure 6).
cos  =
Figure 6
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Example 7 – Solution 1
cont’d
The remaining leg of the triangle is found by the
Pythagorean Theorem to be 4.
From the figure we get
cos(sin–1 ) = cos 
=
So cos(sin–1 ) = .
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Example 7 – Solution 2
cont’d
It's easy to find sin(sin–1 ). In fact, by the cancellation
properties of inverse functions, this value is exactly .
To find cos(sin–1 ), we first write the cosine function in
terms of the sine function.
Let u = sin–1 . Since – /2  u  /2, cos u is positive, and
we can write the following:
cos u =
Cos2 u + sin2 u = 1
u=
Property of inverse functions:
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Example 7 – Solution 2
cont’d
Calculate
So cos(sin–1 ) = .
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