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Pre-Class Problems 13 for Thursday, March 20 These are the type of problems that you will be working on in class. These problems are from Lesson 9. Solution to Problems on the Pre-Exam. You can go to the solution for each problem by clicking on the problem letter. Objective of the following problems: To find the value of the inverse sine of a given number. 1. Find the exact value of the following. a. sin 1 2 d. 3 sin 1 2 g. Arc sin 1 2 2 2 2 b. 3 Arc sin 2 c. sin e. Arc sin 1 f. sin 1 0 h. sin i. Arc sin 1 ( 1) 1 1 2 Objective of the following problems: To find the value of the inverse tangent of a given number. 2. Find the exact value of the following. 1 1 3 a. tan d. tan 1 ( 3) b. Arc tan e. Arc tan 0 3 c. Arc tan ( 1 ) f. tan 1 1 g. 1 Arc tan 3 Additional problems available in the textbook: Page 186 … 5, 6, 9, 10, 12, 13, 14, 16, 17, 18. Examples 1ab and 3bc on page 181. Solutions: Definition The inverse sine function, denoted by sin if and only if sin y x , where 1 x 1 and Notation: Sometimes, sin sin 1 x . 1 1 , is defined by y sin 1 x y . 2 2 x is denoted by Arc sin x . That is, Arc sin x = The restriction that is put on the sine function in order to make it be one-to-one means that your angle answer must be in the closed interval , . NOTE: 2 2 The angles in the open interval 0 , are in the first quadrant. The angles in the 2 open interval , 0 are in the fourth quadrant. The angle 0 is on the positive x 2 axis, the angle is on the positive y-axis, and the angle is on the negative y2 2 axis. 1a. sin 1 1 2 Answer: 6 1 is positive and not maximum positive of 1, then the angle 2 answer is in the first quadrant. That is, the angle answer is in the open NOTE: Since interval 0 , . 2 1 either by Unit Circle 6 2 1 Trigonometry or Right Triangle Trigonometry. Note that the point , is 6 2 We know that sin on the graph of the sine function. Back to Problem 1. 1b. Arc sin 3 2 Answer: 3 3 is positive and not maximum positive of 1, then the angle 2 answer is in the first quadrant. That is, the angle answer is in the open 3 interval 0 , . We know that sin either by Unit Circle 3 2 2 Trigonometry or Right Triangle Trigonometry. Note that the point 3 , 3 2 is on the graph of the sine function. NOTE: Since Back to Problem 1. 1c. 2 sin 1 2 Answer: 4 2 is negative and not minimum negative of 1 , then the 2 angle answer is in the fourth quadrant. That is, the angle answer is in the open interval , 0 . 2 NOTE: Since 2 1 , then ' sin If we let 2 1 ' . That is, sin 4 2 1 NOTE: We will show that sin 2 sin 1 2 . Thus, 2 4 . 4 in Problem 1g. 4 2 2 2 We know that sin sin using reference angles. Note 4 2 4 2 is on the graph of the sine function. , that the point 4 2 Back to Problem 1. 1d. 3 sin 1 2 Answer: 3 3 is negative and not minimum negative of 1 , then the 2 angle answer is in the fourth quadrant. That is, the angle answer is in the open interval , 0 . 2 NOTE: Since 3 3 1 1 sin ' sin If we let . Thus, 2 , then 2 3 3 1 . sin ' . That is, 2 3 3 3 1 NOTE: We showed that sin in Problem 1b. 2 3 3 sin sin We know that using reference angles. Note 3 2 3 3 is on the graph of the sine function. , that the point 3 2 Back to Problem 1. 1e. Arc sin 1 Answer: 2 NOTE: Since 1 is the maximum positive of 1, then the angle answer is not in the first quadrant. So, the angle answer is on one of the coordinate axes, namely the positive y-axis. The angle which is in the closed interval 2 , 2 and lies on the positive y-axis is the angle 2 . We know that sin 1 by Unit Circle Trigonometry. Note that the point 2 , 1 is on the 2 graph of the sine function. Back to Problem 1. 1f. sin 1 0 Answer: 0 NOTE: Since 0 is not positive, then the angle answer is not in the first quadrant. Since 0 is not negative, then the angle answer is not in the fourth quadrant. So, the angle answer is on one of the coordinate axes, namely the positive x-axis. The angle which is in the closed interval , and lies 2 2 on the positive x-axis is the angle 0. We know that sin 0 0 by Unit Circle Trigonometry. Note that the point 0 , 0 is on the graph of the sine function. Back to Problem 1. 1g. Arc sin 2 2 Answer: 4 2 is positive and not maximum positive of 1, then the angle 2 answer is in the first quadrant. That is, the angle answer is in the open 2 interval 0 , . We know that sin either by Unit Circle 4 2 2 Trigonometry or Right Triangle Trigonometry. Note that the point 2 , 4 2 is on the graph of the sine function. NOTE: Since Back to Problem 1. 1h. sin 1 ( 1) Answer: 2 NOTE: Since 1 is the minimum negative of 1 , then the angle answer is not in the fourth quadrant. So, the angle answer is on one of the coordinate axes, namely the negative y-axis. The angle which is in the closed interval , and lies on the negative y-axis is the angle . We know that 2 2 2 sin 1 by Unit Circle Trigonometry. Note that the point 2 , 1 is on the graph of the sine function. 2 Back to Problem 1. 1i. 1 Arc sin 2 Answer: 6 1 is negative and not minimum negative of 1 , then the 2 angle answer is in the fourth quadrant. That is, the angle answer is in the open interval , 0 . 2 NOTE: Since 1 1 . Thus, If we let Arc sin , then ' Arc sin 2 6 2 1 ' . That is, Arc sin . 6 6 2 1 NOTE: We showed that Arc sin in Problem 1a. 2 6 1 sin sin We know that using reference angles. Note that 6 2 6 1 the point , is on the graph of the sine function. 2 6 Back to Problem 1. 1 Definition The inverse tangent function, denoted by tan , is defined by y tan 1 x if and only if tan y x , where x is any real number and y . 2 2 1 Notation: Sometimes, tan x is denoted by Arc tan x . tan 1 x . That is, Arc tan x = The restriction that is put on the tangent function in order to make it be one-to-one means that your angle answer must be in the open interval , . NOTE: 2 2 The angles in the open interval 0 , are in the first quadrant. The angles in the 2 open interval , 0 are in the fourth quadrant. The angle 0 is on the positive x 2 axis. From Lesson 2, you had the following diagram to help you find the value of the tangent of the three Special Angles in the I quadrant. 6 30 45 4 3 60 | | | 1 3 1 3 Tangent Now, reverse the arrows to have a diagram to help you find the value of the inverse 1 tangent of these three numbers of 1, 3 , and : 3 6 30 45 4 3 60 | | | 1 3 1 Inverse Tangent 3 2a. tan 1 1 3 Answer: 6 1 NOTE: Since is positive, then the angle answer is in the first quadrant. 3 That is, the angle answer is in the open interval 0 , . We know that 2 1 tan either by memorization or Right Triangle Trigonometry. Note 6 3 1 is on the graph of the tangent function. that the point 6 , 3 Back to Problem 2. 2b. Arc tan 3 Answer: 3 3 is positive, then the angle answer is in the first quadrant. That is, the angle answer is in the open interval 0 , . We know that 2 tan 3 either by memorization or Right Triangle Trigonometry. Note 3 that the point , 3 is on the graph of the tangent function. 3 NOTE: Since Back to Problem 2. 2c. Arc tan ( 1 ) Answer: 4 NOTE: Since 1 is negative, then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval , 0 . 2 If we let Arc tan ( 1 ) , then ' Arc tan 1 . Thus, 4 . That is, Arc tan ( 1 ) . 4 4 NOTE: We will show that Arc tan 1 in Problem 2f. 4 ' 1 using reference angles. Note that We know that tan tan 4 4 the point , 1 is on the graph of the tangent function. 4 Back to Problem 2. 2d. tan 1 ( Answer: 3) NOTE: Since 3 3 is negative, then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval , 0 . 2 1 If we let tan ( ' 1 3 3 ) , then ' tan 1 . That is, tan ( 3 1 3 NOTE: We showed that tan 3) . 3 . Thus, 3 in Problem 2b. 3 We know that tan 3 that the point , 3 3 tan 3 3 using reference angles. Note is on the graph of the tangent function. Back to Problem 2. 2e. Arc tan 0 Answer: 0 NOTE: Since 0 is not positive, then the angle answer is not in the first quadrant. Since 0 is not negative, then the angle answer is not in the fourth quadrant. So, the angle answer is on one of the coordinate axes, namely the positive x-axis. The angle which is in the open interval , and lies on 2 2 the positive x-axis is the angle 0. We know that tan 0 0 by Unit Circle Trigonometry. Note that the point 0 , 0 is on the graph of the tangent function. Back to Problem 2. 2f. tan 1 1 Answer: 4 NOTE: Since 1 is positive, then the angle answer is in the first quadrant. That is, the angle answer is in the open interval 0 , . We know that 2 1 either by memorization or Right Triangle Trigonometry. Note 4 that the point , 1 is on the graph of the tangent function. 4 tan Back to Problem 2. 2g. 1 Arc tan 3 Answer: 6 1 is negative, then the angle answer is in the fourth 3 quadrant. That is, the angle answer is in the open interval , 0 . 2 NOTE: Since 1 , then ' Arc tan 1 . Thus, Arc tan If we let 6 3 3 1 . ' . That is, Arc tan 6 3 6 1 NOTE: We showed that Arc tan in Problem 2a. 6 3 1 tan tan We know that using reference angles. Note 6 3 6 1 , that the point 6 is on the graph of the tangent function. 3 Back to Problem 2. Solution to Problems on the Pre-Exam: 16. Back to Page 1. Find the exact value of each of the following. (3 pts. each) a. sin 1 1 Answer: 2 NOTE: Since 1 is the maximum positive of 1, then the angle answer is not in the first quadrant. So, the angle answer is on one of the coordinate axes, namely the positive y-axis. The angle which is in the closed interval , and lies on the positive y-axis is the angle . 2 2 2 We know that sin 1 by Unit Circle Trigonometry. Note that the 2 point , 1 is on the graph of the sine function. 2 c. 3 Arc sin 2 Answer: 3 3 is negative and not minimum negative of 1 , 2 then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval , 0 . 2 NOTE: Since 3 3 Arc sin ' Arc sin If we let , then . Thus, 2 2 3 3 . Arc sin ' . That is, 2 3 3 3 sin sin We know that using reference angles. 3 2 3 3 is on the graph of the sine function. , Note that the point 3 2 d. Answer: tan 1 ( 1) 4 NOTE: Since 1 is negative, then the angle answer is in the fourth quadrant. That is, the angle answer is in the open interval , 0 . 2 1 1 If we let tan ( 1) , then ' tan 1 ' 1 . That is, tan ( 1) . 4 4 . Thus, 4 1 using reference angles. We know that tan tan 4 4 Note that the point , 1 is on the graph of the tangent function. 4