Precalculus Eleventh Edition Chapter 7 Analytic Trigonometry PDF

Precalculus Eleventh Edition Chapter 7 Analytic Trigonometry Copyright © 2020, 2016, 2012 Pearson Education, Inc....

Precalculus Eleventh Edition Chapter 7 Analytic Trigonometry Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 1 Section 7.1 The Inverse Sine, Cosine, and Tangent Functions Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 2 Objectives 1. Define the Inverse Sine Function 2. Find the Value of an Inverse Sine Function 3. Define the Inverse Cosine Function 4. Find the Value of an Inverse Cosine Function 5. Define the Inverse Tangent Function 6. Find the Value of an Inverse Tangent Function 7. Use Properties of Inverse Functions to Find Exact Values of Certain Composite Functions 8. Find the Inverse Function of a Trigonometric Function 9. Solve Equations Involving Inverse Trigonometric Functions Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 3 Copyright © 2017, 2013, 2009 Pearson Education, Inc. All Rights Reserved Slide - 4 Sine Function Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 5 Inverse Sine Function Definition Inverse Sine Function y sin  1 x if and only if x sin y   (1) where – 1  x 1 and  y  2 2   y sin  1 x,  1  x 1,  y  2 2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 6 Example 1: Finding the Exact Values of an Inverse Sine Function (1 of 3) 1 Find the exact value of: sin ( 1) 1   Let  sin ( 1). Then θ is the angle,    , 2 2 whose sine equals −1. 1    sin ( 1)    2 2   By definition of sin   1    2 2 y sin  1 x Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 7 Example 1: Finding the Exact Values of an Inverse Sine Function (2 of 3) Now look at the table and figure (next slide).    The only angle θ in the interval   ,  whose sine  2 2  3 3 is −1 is . (Notice that sin also equals −1, but 2 2 2    lies outside the interval   ,  , which is not allowed.)  2 2 1  So sin ( 1)  2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 8 Example 1: Finding the Exact Values of an Inverse Sine Function (3 of 3) Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 9 Example 2: Finding the Exact Value of an Inverse Sine Function (1 of 2)  3 Find the exact value of: sin 1   2   3   Let  sin  1 . Then θ is the angle    ,  2  2 2 3 whose sine equals. 2 1 3      sin     2  2 2 sin   3     2 2 2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 10 Example 2: Finding the Exact Value of an Inverse Sine Function (2 of 2)    The only angle in the interval   ,  whose sine  2 2 3  is is , so 2 3 1 3   sin  =  2  3 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 11 Example 3: Finding an Approximate Value of an Inverse Sine Function (1 of 2) Find an approximate value of each of the following functions: 1 1 1 1 a) sin b) sin    3  4 Express the answer in radians rounded to two decimal places. a) Because the angle is to be measured in radians, first set the mode of the calculator to radians. 1 1 Rounded to two decimal places, sin 0.34. 3 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 12 Example 3: Finding an Approximate Value of an Inverse Sine Function (2 of 2) b) The figure shows the solution using a TI-84 Plus C graphing calculator in radian mode.  1 Rounded to two decimal places, sin     0.25. 1  4 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 13 Cosine Function Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 14 Inverse Cosine Function Definition Inverse Cosine Function y cos  1 x if and only if x cos y (2) where –1  x 1 and 0  y  Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 15 Example 4: Find the Exact Value of an Inverse Cosine Function (1 of 3) 1 1 Find the exact value of: cos 2 1 1 Let  cos. Then θ is the angle 0   , 2 1 whose cosine equals. 2 1  cos 1 0   2 1 cos  0   2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 16 Example 4: Find the Exact Value of an Inverse Cosine Function (2 of 3) Look at the table and the figure. The only angle θ within the interval [0,  ] whose cosine is 1  is. 2 3 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 17 Example 4: Find the Exact Value of an Inverse Cosine Function (3 of 3)   5 1 [Note that cos    and cos also equal , but  3 3 2 they lie outside the interval [0,  ], so these values are not allowed.] Therefore, 1  1 cos  2 3 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 18 Example 5: Finding the Exact Value of an Inverse Cosine Function (1 of 2)  3 Find the exact value of: 1 cos     2   3 Let 1  cos   . Then θ is the angle, 0   ,  2  whose cosine equals  3. 2  31  cos    0    2  3 cos  0   2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 19 Example 5: Finding the Exact Value of an Inverse Cosine Function (2 of 2) Look at the table and figure. The only angle θ in the interval [0,  ], 3 5 whose cosine is  is , so 2 6  3  5 cos  1     2  6 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 20 Tangent Function Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 21 Inverse Tangent Function Definition Inverse Tangent Function y tan  1 x if and only if x tan y   (3) where –   x   and  y 2 2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 22 Example 6: Finding the Exact Value of an Inverse Tangent Function (1 of 3) Find the exact value of:  3  1 a) tan 1 b) tan   1   3       , a) Let  tan 1. Then θ is the angle, 1 2 2 whose tangent equals 1.  tan  1 1      2 2 tan  1      2 2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 23 Example 6: Finding the Exact Value of an Inverse Tangent Function (2 of 3)  3  b) Let  tan   1 . Then θ is the angle,  3        , whose tangent equals  3. 2 2 3  3   tan   1        3  2 2 tan   3      3 2 2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 24 Example 6: Finding the Exact Value of an Inverse Tangent Function (3 of 3) Look at the table. The only angle θ within the interval     ,  whose tangent is  3 is 2 2 3   , so 6 1 3   tan      3  6 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 25 Properties of Inverse Functions (1 of 3)   ff ( ( x)) sin (sin x)  x where  1 1 x  (4a) 2 2 f ( f  1 ( x)) sin (sin  1 x)  x where  1  x  1 (4b) Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 26 Example 7: Finding the Exact Value of Certain Composite Functions (1 of 3) Find the exact value, if any, of each composite function.    5  1 1 a) sin  sin  b) sin  sin   8  8  1  a) The composite function sin  sin  follows the form  8     ,  , of property (4a). Because is in the interval  8  2 2  use property (4a) to conclude 1   sin  sin    8 8 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 27 Example 7: Finding the Exact Value of Certain Composite Functions (2 of 3)  5  1 b) The composite function sin  sin  follows the  8  5 form of property (4a). But because is not in the 8     1 5  5 interval  , , sin  sin  .  2 2   8  8 1 5  To find sin  sin  , first find an angle θ in the  8      5 interval   ,  for which sin  sin.  2 2 8 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 28 Example 7: Finding the Exact Value of Certain Composite Functions (3 of 3) 5 See the figure. The angle 8 is in quadrant II. 5 3 The reference angle of is. 8 8 3 Since is in the interval 8    ,   , we can use 4(a).  2 2  1 5  1 3  3 sin  sin  sin  sin   8   8  8 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 29 Example 8: Finding the Exact Value of Certain Composite Functions (1 of 2) Find the exact value, if any, of each composite function. –1 a) sin (sin 0.5) b) sin[sin –1 (–2.5)] –1 a) The composite function sin (sin 0.5) follows the form of property (4b). Because 0.5 is in the interval [–1, 1], we can use 4(b). –1 sin (sin 0.5) 0.5 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 30 Example 8: Finding the Exact Value of Certain Composite Functions (2 of 2) –1 b) The composite function sin[sin (–2.5)] follows the form of (4b). But since −2.5 is not in the domain of –1 the inverse sine function, [–1, 1], [si n (–2.5)] is not –1 defined. Therefore, sin[sin (–2.5)] is also not defined. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 31 Properties of Inverse Functions (2 of 3) ff 1 ( ( x)) cos  1 (cos x)  x where 0  x  (5a) f ( f  1 ( x)) cos(cos  1 x)  x where  1  x 1 (5b) Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 32 Example 9: Using Properties of Inverse Functions to Find the Exact Value of Certain Composite Functions (1 of 4) Find the exact value, if any, of each composite function.    b) cos  cos  0.4  1 a) cos  cos  1   15    3   1 c) cos  cos     4  d) cos   cos 1  Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 33 Example 9: Using Properties of Inverse Functions to Find the Exact Value of Certain Composite Functions (2 of 4) 1    a) cos  cos    15  15  is in the interval [0,  ]; use property (5a). 15 b) cos   cos 1  0.4   0.4  0.4 is in the interval [ 1,1]; use property (5b). Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 34 Example 9: Using Properties of Inverse Functions to Find the Exact Value of Certain Composite Functions (3 of 4) 3 c) The angle  is not in the interval [0,  ], so 4 property (5a) cannot be used. However, because  3  3 the cosine function is even, cos    cos.  4  4 3 Because is in the interval [0,  ], property (5a) 4 can be used, and  1  3   1   3  3 cos  cos     cos  cos      4    4  4 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 35 Example 9: Using Properties of Inverse Functions to Find the Exact Value of Certain Composite Functions (4 of 4) d) Because π is not in the interval [ 1, 1], the domain of the inverse cosine function, cos –1 is not defined. This means the composite function cos    is cos 1 also not defined. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 36 Properties of Inverse Functions (3 of 3)   ff ( ( x)) tan ( tan x)  x where  1 1 x 2 2 f ( f  1 ( x)) tan ( tan  1 x)  x where –   x   Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 37 Example 10: Finding the Inverse Function of a Trigonometric Function (1 of 4) a) Find the inverse function ff of ( x) 2 sinx – 1, 1    x . 2 2 b) Find the range of f and the domain and range 1 of f. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 38 Example 10: Finding the Inverse Function of a Trigonometric Function (2 of 4) a) The function f is one-to-one and so has an inverse function. Follow the steps for finding the inverse function. y 2sin x – 1 x 2sin y – 1 Interchange x and y x  1 2sin y Solve for y x 1 sin y  2  1 x 1 Definition of inverse sine y sin function. 2  1 x 1 The inverse function is f ( x) sin 1. 2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 39 Example 10: Finding the Inverse Function of a Trigonometric Function (3 of 4) b) To find the range of f, use the fact that the 1 domain of f equals the range of f. Since the domain of the inverse sine function is the interval [ 1, 1], the argument x  1 must be in the interval 2 [ 1, 1]. x 1  1 1 2  2  x  1 2 Multiply by 2.  3  x 1 Subtract 1 from each part. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 40 Example 10: Finding the Inverse Function of a Trigonometric Function (4 of 4) The domain of f  1 is {x | –3 x 1}, or the interval [ 3, 1]. So the range of f is the interval [ 3, 1]. 1 The range of f equals the domain of f. So the    range of f is   , . 1  2 2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 41 Example 11: Solving an Inverse Trigonometric Equation (1 of 2) Solve the equation: 12 sin –1 x 3 To solve an equation involving a single inverse trigonometric function, first isolate the inverse trigonometric function. 12 sin –1 x 3 1 3 Divide both sides by 12. sin x  12 1  sin x  Simplify. 4  x sin y sin –1 x if and only if x sin y 4 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 42 Example 11: Solving an Inverse Trigonometric Equation (2 of 2)  x sin 4 2 x 2  2 The solution set is  .  2  Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Slide - 43