1. Copyright © 2005 Pearson Education, Inc.

2. Chapter 6 Inverse Circular Functions and Trigonometric Equations

3. Copyright © 2005 Pearson Education, Inc. 6.1 Inverse Circular Functions

4. Copyright © 2005 Pearson Education, Inc. Slide 6-4 Horizontal Line Test Any horizontal line will intersect the graph of a one-to-one function in at most one point. The inverse function of the one-to-one function f is defined as

5. Copyright © 2005 Pearson Education, Inc. Slide 6-5 Inverse Functions Review 1. In a one-to-one function, each x-value corresponds to only one y-value and each y-value corresponds to only one x-value. 2. If a function f is one-to-one, then f has an inverse function f -1. 3. The domain of f is the range of f -1 and the range of f is the domain of f -1. 4. The graphs of f and f -1 are reflections of each other about the line y = x. 5. To find f -1 (x) from f(x), follow these steps.  Replace f(x) with y and interchange x and y.  Solve for y.  Replace y with f -1 (x).

6. Copyright © 2005 Pearson Education, Inc. Slide 6-6 Inverse Sine Function means that x = sin y, for Example: Find

7. Copyright © 2005 Pearson Education, Inc. Slide 6-7 Examples sin  1 2 Not possible to evaluate because there is no angle whose sine is 2.

8. Copyright © 2005 Pearson Education, Inc. Slide 6-8 Inverse Sine Function

9. Copyright © 2005 Pearson Education, Inc. Slide 6-9 Inverse Cosine Function means that x = cos y, for Example: Find

10. Copyright © 2005 Pearson Education, Inc. Slide 6-10 Inverse Cosine Function

11. Copyright © 2005 Pearson Education, Inc. Slide 6-11 Inverse Tangent Function means that x = tan y, for

12. Copyright © 2005 Pearson Education, Inc. Slide 6-12 Inverse Tangent Function

13. Copyright © 2005 Pearson Education, Inc. Slide 6-13 Other Inverse Functions

14. Copyright © 2005 Pearson Education, Inc. Slide 6-14 Examples Find the degree measure of  in the following. a)  = arctan 1b)  = sec  1 2 a)  must be in (  90 , 90  ), since 1 > 0,  must be in quadrant I. The alternative statement, tan  = 1, leads to  = 45 . b) sec  = 2, for sec  1 x,  is in quadrant I or II. Because 2 is positive,  is in quadrant I and sec 60  = 2.

15. Copyright © 2005 Pearson Education, Inc. Slide 6-15 Example Find y in radians if y = arctan(  6.24).  Calculator in radian mode  Enter tan  1 (  6.24) y   1.411891065 Find y in radians if y = arccos 2.  Calculator in radian mode  Enter cos  1 (2) (error message since the domain of the inverse cosine function is [  1, 1].

16. Copyright © 2005 Pearson Education, Inc. Slide 6-16 Example: Evaluate the expression without using a calculator. Let The inverse tangent function yields values only in quadrants I and III, since 3/2 is positive,  is in quadrant I.

17. Copyright © 2005 Pearson Education, Inc. Slide 6-17 Example: Evaluate the expression without using a calculator continued Sketch and label the triangle. The hypotenuse is

18. Copyright © 2005 Pearson Education, Inc. Slide 6-18 Example Evaluate the expression without using a calculator. Let arcsin 2/5 = B Since arcsin 2/5 = B, sin B = 2/5. Sketch a triangle in quadrant I, find the length of the third side, then find tan B.

19. Copyright © 2005 Pearson Education, Inc. Slide 6-19 Example continued

20. Copyright © 2005 Pearson Education, Inc. Slide 6-20 Homework Do pp. 246-249, 1-6 all, 8, 14-52 even, 58-62 all, 64-78 even, 79-82 all, & 84-96 even