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6.4 Inverse Trigonometric Functions
and Right Triangles
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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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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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