1. Copyright © 2011 Pearson Education, Inc. Slide 9.4-1

2. Copyright © 2011 Pearson Education, Inc. Slide 9.4-2 Chapter 9: Trigonometric Identities and Equations 9.1Trigonometric Identities 9.2Sum and Difference Identities 9.3Further Identities 9.4The Inverse Circular Functions 9.5Trigonometric Equations and Inequalities (I) 9.6Trigonometric Equations and Inequalities (II)

3. Copyright © 2011 Pearson Education, Inc. Slide 9.4-3 9.4The Inverse Sine Function Summary of Inverse Functions 1.For 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 f -1. That is, if (a, b) is on the graph of f, then (b, a) is on the graph of f -1. 4.The graphs of f and f -1 are reflections of each other about the line y = x. (continued on next slide)

4. Copyright © 2011 Pearson Education, Inc. Slide 9.4-4 9.4The Inverse Sine Function Summary of Inverse Functions (continued) 5.To find f -1 (x) from f(x), follow these steps: Step1Replace f(x) with y and interchange x and y. Step 2Solve for y. Step 3Replace y with f -1 (x).

5. Copyright © 2011 Pearson Education, Inc. Slide 9.4-5 9.4The Inverse Circular Functions The Inverse Sine Function Apply the horizontal line test to show that y = sin x is not one-to-one. However, by restricting the domain over the interval a one-to-one function can be defined.

6. Copyright © 2011 Pearson Education, Inc. Slide 9.4-6 9.4The Inverse Sine Function The domain of the inverse sine function y = sin -1 x is [–1, 1], while the restricted domain of y = sin x, [–  /2,  /2], is the range of y = sin -1 x. We may think of y = sin -1 x as “y is the number in the interval whose sine is x. The Inverse Sine Function y = sin -1 x or y = arcsin x means that x = sin y, for

7. Copyright © 2011 Pearson Education, Inc. Slide 9.4-7 9.4Finding Inverse Sine Values ExampleFind y in each equation. Analytic Solution (a) y is the number in whose sine is Since sin  /6 = ½, and  /6 is in the range of the arcsine function, y =  /6. (b)Writing the alternative equation, sin y = –1, shows that y = –  /2 (c)Because –2 is not in the domain of the inverse sine function, y = sin -1 (–2) does not exist.

8. Copyright © 2011 Pearson Education, Inc. Slide 9.4-8 9.4 Finding Inverse Sine Values Graphical Solution To find the values with a graphing calculator, graph y = sin -1 x and locate the points with x-values ½ and –1. (a) The graph shows that when x = ½, y =  /6 .52359878. (b) The graph shows that when x = –1, y = –  /2  –1.570796. Caution It is tempting to give the value of sin -1 (–1) as 3  /2, however, 3  /2 is not in the range of the inverse sine function.

9. Copyright © 2011 Pearson Education, Inc. Slide 9.4-9 9.4Inverse Sine Function y = sin -1 x or y = arcsin x Domain: [–1, 1] Range: The inverse sine function is increasing and continuous on its domain [–1, 1]. Its x-intercept is 0, and its y-intercept is 0. Its graph is symmetric with respect to the origin.

10. Copyright © 2011 Pearson Education, Inc. Slide 9.4-10 9.4Inverse Cosine Function The function y = cos -1 x (or y = arccos x) is defined by restricting the domain of y = cos x to the interval [0,  ], and reversing the roles of x and y. y = cos -1 x or y = arccos x means that x = cos y, for 0  y  .

11. Copyright © 2011 Pearson Education, Inc. Slide 9.4-11 9.4 Finding Inverse Cosine Values ExampleFind y in each equation. Solution (a)Since the point (1, 0) lies on the graph of y = arccos x, the value of y is 0. Alternatively, y = arccos 1 means cos y = 1, or cos 0 = 1, so y = 0. (b)We must find the value of y that satisfies cos y = 0  y  . The only value for y that satisfies these conditions is 3  /4.

12. Copyright © 2011 Pearson Education, Inc. Slide 9.4-12 9.4 Inverse Cosine Function y = cos -1 x or y = arccos x Domain: [–1, 1] Range: [0,  ] The inverse cosine function is decreasing and continuous on its domain [–1, 1]. Its x-intercept is 1, and its y-intercept is  /2. Its graph is not symmetric with respect to the y-axis nor the origin.

13. Copyright © 2011 Pearson Education, Inc. Slide 9.4-13 9.4Inverse Tangent Function The function y = tan -1 x (or y = arctan x) is defined by restricting the domain of y = tan x to the interval and reversing the roles of x and y. y = tan -1 x or y = arctan x means that x = tan y, for

14. Copyright © 2011 Pearson Education, Inc. Slide 9.4-14 9.4 Inverse Tangent Function y = tan -1 x or y = arctan x Domain: (– ,  ) Range: The inverse tangent function is increasing and continuous on its domain (– ,  ). Its x-intercept is 0, and its y-intercept is 0. Its graph is symmetric with respect to the origin and has horizontal asymptotes y =

15. Copyright © 2011 Pearson Education, Inc. Slide 9.4-15 9.4The Inverse Sine Function Inverse Cotangent, Secant, and Cosecant Functions y = cot -1 x or y = arccot x means that x = cot y, for 0 < y < . y = sec -1 x or y = arcsec x means that x = sec y, for 0 < y < , y   /2. y = csc -1 x or y = arccsc x means that x = csc y, for –  /2 < y <  /2, y  0.

16. Copyright © 2011 Pearson Education, Inc. Slide 9.4-16 9.4Remaining Inverse Trigonometric Functions Inverse trigonometric functions are formally defined with real number values. Sometimes we want the degree-measured angles equivalent to these real number values. FunctionDomainIntervalQuadrants y = sin -1 x y = cos -1 x y = tan -1 x y = cot -1 x y = sec -1 x y = csc -1 x [–1, 1] (– ,  ) (– , –1]  [1,  ) [0,  ] (0,  ) [0,  ], y  I and IV I and II I and IV I and II I and IV

17. Copyright © 2011 Pearson Education, Inc. Slide 9.4-17 9.4Finding Inverse Function Values ExampleFind the degree measure of  in each of the following. Solution (a)Since 1 > 0 and –90° <  < 90°,  must be in quadrant I. So tan  = 1 leads to  = 45°. (b)Write the equation as sec  = 2. Because 2 s positive,  must be in quadrant I and  = 60° since sec 60° = 2.

18. Copyright © 2011 Pearson Education, Inc. Slide 9.4-18 9.4Finding Inverse Functions with a Calculator Inverse trigonometric function keys on the calculator give results for sin -1, cos -1, and tan -1. Finding cot -1 x, sec -1 x, and csc -1 x with a calculator is not as straightforward. –e.g. If y = sec -1 x, then sec y = x, must be written as follows: From this statement, Note: Since we take the inverse tangent of the reciprocal of x to find cot -1 x, the calculator gives values of cot -1 with the same range as tan -1, (–  /2,  /2), which is incorrect. The proper range must be considered and the results adjusted accordingly.

19. Copyright © 2011 Pearson Education, Inc. Slide 9.4-19 9.4Finding Inverse Functions with a Calculator Example (a)Find y in radians if y = csc -1 (–3). (b)Find  in degrees if  = arccot(–0.3541). Solution (a)In radian mode, enter y = csc -1 (–3) as sin -1 ( ) to get y  –0.3398369095. (b)In degree mode, the calculator gives inverse tangent values of a negative number as a quadrant IV angle. But  must be in quadrant II for a negative number, so we enter arccot(–0.3541) as tan -1 (1/ –0.3541) +180°,   109.4990544°.

20. Copyright © 2011 Pearson Education, Inc. Slide 9.4-20 9.4Finding Function Values ExampleEvaluate each expression without a calculator. Solution (a)Let  = tan -1 so that tan  =. Since is positive,  is in quadrant I. We sketch the figure to the right, so (b)Let A = cos -1 ( ). Then cos A =. Since cos -1 x for a negative x is in quadrant II, sketch A in quadrant II.

21. Copyright © 2011 Pearson Education, Inc. Slide 9.4-21 9.4Writing Function Values in Terms of u ExampleWrite each expression as an algebraic expression in u. Solution (a)Let  = tan -1 u, so tan  = u. Sketch  in quadrants I and IV since (b)Let  = sin -1 u, so sin  = u.

22. Copyright © 2011 Pearson Education, Inc. Slide 9.4-22 9.4Finding the Optimal Angle of Elevation of a Shot Put ExampleThe optimal angle of elevation  a shot putter should aim for to throw the greatest distance depends on the velocity of the throw and the initial height of the shot. One model for  that achieves this goal is Figure 32 pg 9-73

23. Copyright © 2011 Pearson Education, Inc. Slide 9.4-23 9.4Finding the Optimal Angle of Elevation of a Shot Put Suppose a shot putter can consistently throw a steel ball with h = 7.6 feet and v = 42 ft/sec. At what angle should he throw the ball to maximize distance? SolutionSubstitute into the model and use a calculator in degree mode.