Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Pre-Class Problems 6 for Monday, February 13 These are the type of problems that you will be working on in class. These problems are from Lesson 5. 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 problem: To find the exact value of the six trigonometric functions for an angle if given the coordinates of a point on the terminal side of the angle. This will require the use of Radius r Trigonometry. 1. Find the exact value of the six trigonometric functions of the angle if the given point is on the terminal side of . a. ( 3, 7 ) b. (10 , 6 ) c. ( 5, 8) d. ( 9, 0) e. ( 5 , 2) f. ( 0, g. (12 , 17 ) 3) Objective of the following problems: To find the exact value of the six trigonometric functions for an angle whose terminal side lies on the graph of a line and the equation of the line is given. This will require the use of Radius r Trigonometry. 2. Find the exact value of the six trigonometric functions for the following angles. a. The terminal side of the angle is in the fourth quadrant and lies on 3 the line y x . 4 b. The terminal side of the angle is in the first quadrant and lies on the line 5 x 2 y 0 . c. The terminal side of the angle is in the third quadrant and lies on the 7 y x. line 3 d. The terminal side of the angle is in the second quadrant and lies on the line 8 x 14 y 0 . e. The terminal side of the angle is in the third quadrant and lies on the line 4 x y 0 . f. The terminal side of the angle is in the fourth quadrant and lies on x the line y . 6 Additional problems available in the textbook: Page 505 … 15 – 20, 33 - 36. Page 498 … Example 1. Solutions: 1a. ( 3, 7 ) r x2 y2 cos x r 3 58 sin y r tan y 7 x 3 7 58 9 49 58 58 3 sec r x csc r y cot x 3 y 7 58 7 NOTE: In addition to the point ( 3 , 7 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 58 . This is the circle whose center is at the origin and whose radius is 58 . Back to Problem 1. 1b. (10 , 6 ) r x2 y2 100 36 cos x 10 r 2 34 sin y 6 r 2 34 tan y 6 3 x 10 5 136 5 34 3 34 4 34 34 2 34 4 34 5 sec r x csc r y cot x 5 y 3 34 3 NOTE: In addition to the point (10 , 6 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 136 . This is the circle whose center is at the origin and whose radius is NOTE: Here is an easier way to simplify r x2 y2 4 34 2 34 Back to Problem 1. 100 36 136 2 34 . 136 : 4 ( 25 9 ) 4 ( 34 ) 1c. ( 5, 8) r x2 y2 25 64 89 cos x r 5 89 sec r x 89 5 sin y r 8 89 csc r y 89 8 tan y 8 x 5 cot x 5 y 8 NOTE: In addition to the point ( 5 , 8 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 89 . This is the circle whose center is at the origin and whose radius is 89 . Back to Problem 1. 1d. ( 9, 0) r x2 y2 81 0 81 9 cos x 9 1 r 9 sec r 1 x sin y 0 0 r 9 csc r undefined y tan y 0 0 x 9 cot x undefined y NOTE: In addition to the point ( 9 , 0 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 81 . This is the circle whose center is at the origin and whose radius is 9. Back to Problem 1. 1e. ( 5 , 2) r x2 y2 54 9 3 sec r x y 2 r 3 csc r 3 y 2 y x cot x y cos x r sin tan 5 3 2 5 3 5 5 2 NOTE: In addition to the point ( 5 , 2 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 9 . This is the circle whose center is at the origin and whose radius is 3. Back to Problem 1. 1f. ( 0, r 17 ) x2 y2 0 17 17 cos x r 0 0 17 sec r undefined x sin y r 17 csc r 1 y tan y x 17 undefined 0 cot x y 17 1 0 0 17 NOTE: In addition to the point ( 0 , 17 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 17 . This is the circle whose center is at the origin and whose radius is 17 . Back to Problem 1. 1g. (12 , r 3) x2 y2 144 3 147 3 49 49 3 7 3 cos 12 3 4 3 x 12 r 73 7 7 3 sec 7 3 r x 12 sin 3 y 1 r 7 7 3 csc r 7 y tan 3 y x 12 cot 12 3 x 12 4 3 y 3 3 NOTE: In addition to the point (12 , 3 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 147 . This is the circle whose center is at the origin and whose radius is 147 7 3 . Back to Problem 1. 2a. The terminal side of the angle is in the fourth quadrant and lies on the line 3 y x. 4 3 x in the 4 fourth quadrant. Since x-coordinates are positive in the fourth quadrant, then we will need to pick a positive number for x. We will find the y-coordinate 3 y x . Picking the number 4 for x, we of the point using the equation 4 have the following. NOTE: We will need to find a point that is on the line y x y 4 3 y 3 3 ( 4 ) (1) 3 4 1 NOTE: The point ( 4 , 3 ) is in the fourth quadrant and lies on the line 3 y x . Since the terminal side of the angle is in the fourth quadrant 4 and lies on this line, then the point ( 4 , 3 ) also lies on the terminal side of . r x2 y2 cos x 4 r 5 16 9 25 5 sec r 5 x 4 sin y 3 r 5 csc r 5 y 3 tan y 3 x 4 cot x 4 y 3 NOTE: In addition to the point ( 4 , 3 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 25 . This is the circle whose center is at the origin and whose radius is 5. Back to Problem 2. 2b. The terminal side of the angle is in the first quadrant and lies on the line 5x 2 y 0. NOTE: We will need to find a point that is on the line 5 x 2 y 0 in the first quadrant. Since x-coordinates are positive in the first quadrant, then we will need to pick a positive number for x. We will find the y-coordinate of 5 the point using the equation y x . This is the equation that we obtain 2 when we solve the given equation 5 x 2 y 0 for y. 5x 2 y 0 5x 2 y y 5 x 2 Picking the number 2 for x, we have the following. x y 2 5 y 5 5 ( 2 ) (1) 5 2 1 NOTE: The point ( 2 , 5 ) is in the first quadrant and lies on the line 5 x 2 y 0 . Since the terminal side of the angle is in the first quadrant and lies on this line, then the point ( 2 , 5 ) also lies on the terminal side of . r x2 y2 4 25 29 cos x r 2 29 sec r x 29 2 sin y r 5 29 csc r y 29 5 tan y 5 x 2 cot x 2 y 5 NOTE: In addition to the point ( 2 , 5 ) lying on the terminal side of the angle , it also lies on the graph of the circle x 2 y 2 29 . This is the circle whose center is at the origin and whose radius is 29 . Back to Problem 2. 2c. The terminal side of the angle is in the third quadrant and lies on the line 7 y x. 3 7 x in the third 3 quadrant. Since x-coordinates are negative in the third quadrant, then we will need to pick a negative number for x. We will find the y-coordinate of the 7 y x . Picking the number 3 for x, we have point using the equation 3 the following. NOTE: We will need to find a point that is on the line y x 3 y 7 y 7 7 ( 3) (1) 3 1 7 NOTE: The point ( 3 , 7 ) is in the third quadrant and lies on the line 7 x . Since the terminal side of the angle is in the third quadrant and 3 lies on this line, then the point ( 3 , 7 ) also lies on the terminal side of . y r x2 y2 97 16 4 cos x 3 r 4 sec r 4 x 3 sin 7 y r 4 csc r 4 y 7 tan y x cot x y 7 3 3 7 NOTE: In addition to the point ( 3 , 7 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 16 . This is the circle whose center is at the origin and whose radius is 4. Back to Problem 2. 2d. The terminal side of the angle is in the second quadrant and lies on the line 8 x 14 y 0 . NOTE: We will need to find a point that is on the line 8 x 14 y 0 in the second quadrant. Since x-coordinates are negative in the second quadrant, then we will need to pick a negative number for x. We will find the y4 coordinate of the point using the equation y x . This is the equation 7 that we obtain when we solve the given equation 8 x 14 y 0 for y. 8 x 14 y 0 14 y 8 x y 8 4 x x 14 7 Picking the number 7 for x, we have the following. x y 7 4 y 4 4 ( 7 ) (1) 4 7 1 NOTE: The point ( 7 , 4 ) is in the second quadrant and lies on the line 8 x 14 y 0 . Since the terminal side of the angle is in the second quadrant and lies on this line, then the point ( 7 , 4 ) also lies on the terminal side of . r x2 y2 cos x r 7 65 sin y r tan y 4 x 7 4 65 49 16 65 65 7 sec r x csc r y cot x 7 y 4 65 4 NOTE: In addition to the point ( 7 , 4 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 65 . This is the circle whose center is at the origin and whose radius is Back to Problem 2. 65 . 2e. The terminal side of the angle is in the third quadrant and lies on the line 4 x y 0. NOTE: We will need to find a point that is on the line 4 x y 0 in the third quadrant. Since x-coordinates are negative in the third quadrant, then we will need to pick a negative number for x. We will find the y-coordinate of the point using the equation y 4 x . This is the equation that we obtain when we solve the given equation 4 x y 0 for y. 4x y 0 y 4x Picking the number 1 for x, we have the following. x y 1 4 y 4 ( 1) 4 NOTE: The point ( 1, 4 ) is in the third quadrant and lies on the line 4 x y 0 . Since the terminal side of the angle is in the third quadrant and lies on this line, then the point ( 1, 4 ) also lies on the terminal side of . r x2 y2 1 16 17 cos x r 1 17 sec r x 17 sin y r 4 17 csc r y 17 4 tan y 4 x cot x 1 y 4 NOTE: In addition to the point ( 1, 4 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 17 . This is the circle whose center is at the origin and whose radius is 17 . Back to Problem 2. 2f. The terminal side of the angle is in the fourth quadrant and lies on the line x y . 6 x in the fourth 6 quadrant. Since x-coordinates are positive in the fourth quadrant, then we will need to pick a positive number for x. We will find the y-coordinate of x the point using the equation y . Picking the number 6 for x, we have 6 the following. NOTE: We will need to find a point that is on the line y x y 6 1 y 6 1 6 NOTE: The point ( 6 , 1 ) is in the fourth quadrant and lies on the line x y . Since the terminal side of the angle is in the fourth quadrant and 6 lies on this line, then the point ( 6 , 1 ) also lies on the terminal side of . r x2 y2 cos x r 6 37 36 1 37 sec r x 37 6 sin y r 1 37 tan y 1 x 6 csc r y cot x 6 y 37 NOTE: In addition to the point ( 6 , 1 ) lying on the terminal side of the 2 2 angle , it also lies on the graph of the circle x y 37 . This is the circle whose center is at the origin and whose radius is 37 . Back to Problem 2. Solution to Problems on the Pre-Exam: 9. If the terminal side of passes through the point ( 4 , 7 ) , then find the exact value of sec and tan . tan r 10. Back to Page 1. y 7 x 4 x2 y2 sec r x sec 65 4 16 49 65 65 4 tan 7 4 If the terminal side of is in the III quadrant and lies on the line 15 x 6 y 0 , then find the exact value of sin and cot . 15 x 6 y 0 15 x 6 y y x y 2 5 cot r y 5 5 ( 2 ) (1) 5 2 1 x 2 y 5 x2 y2 sin 15 5 x x 6 2 y r sin 5 29 4 25 29 5 29 cot 2 5