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Trigonometric Functions:
Right Triangle Approach
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2. 6.4 Inverse Trigonometric Functions and Right Triangles
Copyright © Cengage Learning. All rights reserved.

3. Objectives
The Inverse Sine, Inverse Cosine, and Inverse Tangent Functions
Solving for Angles in Right Triangles
Evaluating Expressions Involving Inverse
Trigonometric Functions

4. The Inverse Sine, Inverse Cosine, and Inverse Tangent Functions

5. 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

6. 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

7. Real Numbers
We summarize the domains and ranges of the inverse trigonometric functions in the following box.

8. Example 1 – Evaluating Inverse Trigonometric Functions
Find the exact value.
(a) sin– (b) cos– (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 is  /3. Thus cos– = 2 /3.
(c) The angle in the interval [– /2,  /2] whose tangent is is  /4. Thus tan–1 1 =  /4.

9. Solving for Angles in Right Triangles

10. Example 3 – Finding an Angle in a Right Triangle
Find the angle  in the triangle shown in Figure 2.
Solution: Since  is the angle opposite the side of length 10 and the hypotenuse has length 50, we have
sin 
Now we can use sin–1 to find  :
 = sin–1
  11.5
Figure 2
sin
Definition of sin–1
Calculator (in degree mode)

11. Evaluating Expressions Involving Inverse Trigonometric Functions

12. 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

13. 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 ) = .

14. 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– 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:

15. Example 7 – Solution 2
cont’d
So cos(sin–1 ) = .
Calculate