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LESSON 11 BASIC IDENTITIES Examples which are worked in this lesson (Click on the number): Examples Use Basic Identities to find the exact value of the other five trigonometric functions for the following. csc 5 and is in the III quadrant 1. 2 cos 2. and csc 0 3 17 tan 3. and sin 0 8 Examples Find all the exact solutions for the following equations. 2 cos 2 1 sin 7 2 cos 8 sin 2 1. 2. sec 2 9 7 tan 5 cot 2 x 6 csc x 5 3. 4. Recall the following Basic Identities from Lesson 2: 1. tan sin cos 3. cot 1 tan 5. csc 1 sin 2. cot cos sin 4. sec 1 cos Recall the following Pythagorean Identities from Lesson 2: 1. cos 2 sin 2 1 2. sec 2 tan 2 1 3. csc 2 cot 2 1 Examples Use Basic Identities to find the exact value of the other five trigonometric functions for the following. 1. csc 5 and is in the III quadrant Since csc 5 , then sin Back to Examples List 1 5 2 2 Since csc 5 and csc cot 1 by one of the Pythagorean Identities, then ( 5 ) 2 cot 2 1 25 cot 2 1 cot 2 24 cot 24 . Since is in the III quadrant, then cot 24 , then tan Since cot 24 . 1 24 . 1 and 5 cos 2 sin 2 1 by one of the Pythagorean Identities, then you could use this identity to find cos . However, you only need to use the Pythagorean Identities once to solve these problems. Now, we need to find cos and sec . Since cot Since sin cos , then cos cot sin . Since cot sin 1 sin , then cos 5 Since cos 24 5 1 24 5 , then sec 24 5 5 24 . . 24 and 24 Answers: cos 5 and cot 2. cos , sin 1 , tan 5 1 24 , sec 5 24 , 24 2 and csc 0 3 Back to Examples List First, determine what quadrant the angle is in. Using Method 1 from Lesson 6, we have: cos 0 the x-coordinate of the point of intersection of the terminal side of the angle with the Unit Circle is positive. That is, x > 0. csc 0 sin 0 the y-coordinate of the point of intersection of the terminal side of the angle with the Unit Circle is negative. That is, y < 0. Thus, we have that x > 0 and y < 0. Thus, the angle is in the IV quadrant. You may use Method 2 or Method 3 from Lesson 6 if you wish. Since cos Since cos 2 3 , then sec 3 2 2 2 2 and cos sin 1 by one of the Pythagorean 3 Identities, then 2 3 2 sin 2 1 sin 5 3 4 sin 2 1 9 sin 2 5 9 . Since is in the IV quadrant, then sin Since sin 5 3 , then csc 3 5 . 5 3 . Since tan Now, we need to find tan and cot . 5 2 sin and cos , then tan 3 3 Since tan 5 2 Answers: sin , then cot 5 3 and cot 3. tan 17 8 , tan sin cos and 5 3 2 3 5 2 . 2 5 5 2 . 3 3 , csc , 2 5 , sec 2 5 and sin 0 Back to Examples List First, determine what quadrant the angle is in. Using Method 1 from Lesson 6, we have: sin 0 the y-coordinate of the point of intersection of the terminal side of the angle with the Unit Circle is positive. That is, y > 0. The tangent of the angle is the y-coordinate of the point of intersection of the terminal side of the angle with the Unit Circle divided by the xy coordinate of the point of intersection. That is, tan . Since we have x that tan 0 and y > 0, then we have the following: ( ) tan () y x 0 x ? Thus, we have that x < 0 and y > 0. Thus, the angle is in the II quadrant. You may use Method 2 or Method 3 from Lesson 6 if you wish. Since tan 17 Since tan 17 8 8 8 , then cot 17 2 2 and sec tan 1 by one of the Pythagorean Identities, then sec 2 sec 17 8 2 1 17 81 1 sec 2 64 64 sec 2 9 9 . Since is in the II quadrant, then sec . 8 8 Since sec 9 8 , then cos . 8 9 Since tan Now, we need to find sin and csc . sin tan cos . sin Since 17 8 8 9 Since sin 17 9 tan 8 and cos 8 , then 9 17 9 . , then csc 9 8 , sin 9 17 Answers: cos 17 sin , then cos 17 9 . , sec 9 , csc 8 9 17 , and cot 8 17 Examples Find all the exact solutions for the following equations. 1. 2 cos 2 1 sin Back to Examples List This equation is neither quadratic in cos nor sin . We will use the 2 2 2 2 Pythagorean Identity cos sin 1 to replace cos by 1 sin . 2 2 The resulting equation will be quadratic in sin . Since cos sin 1, 2 2 then cos 1 sin . Thus, 2 cos 2 1 sin 2 (1 sin 2 ) 1 sin 2 2 sin 2 1 sin 0 2 sin 2 sin 1 ( sin 1) ( 2 sin 1) 0 Thus, either sin 1 0 or 2 sin 1 0 . Thus, there are two equations to be solved. NOTE: These types of equations were solved in Lesson 10. Please read that lesson in order to see how to solve these equations. Since sin 1 0 sin 1, then the solutions of the first equation 3 2 n , where n is an integer. are 2 1 , then the solutions of the second 2 5 2 n and 2 n , where n is an integer. equation are 6 6 Since 2 sin 1 0 sin Answers: 3 5 2 n , 2 n , and 2 n , where n 6 6 2 is an integer 2. 7 2 cos 8 sin 2 Back to Examples List This equation is neither quadratic in cos nor sin . We will use the 2 2 2 2 Pythagorean Identity cos sin 1 to replace sin by 1 cos . The resulting equation will be quadratic in cos . Since cos 2 sin 2 1 , 2 2 then sin 1 cos . Thus, 2 cos 7 8 sin 2 2 cos 7 8 (1 cos 2 ) 2 cos 7 8 8 cos 2 8 cos 2 2 cos 1 0 ( 2 cos 1) ( 4 cos 1) 0 Thus, either 2 cos 1 0 or 4 cos 1 0 . Thus, there are two equations to be solved. NOTE: These types of equations were solved in Lesson 10. Please read that lesson in order to see how to solve these equations. 1 , then the solutions of the second 2 4 2 2 n 2 n , where n is an integer. equation are and 3 3 Since 2 cos 1 0 cos To solve the second equation 4 cos 1 0 : 4 cos 1 0 4 cos 1 cos 1 4 1 is not the maximum 4 positive number for the cosine function, then the solutions do not occur at the coordinate axes. Thus, the solutions occur in two of the four quadrants. Since cosine is positive in the I and IV quadrants, the solutions for this equation occur in those quadrants. Determine where the solutions will occur. Since Find the reference angle ' for the solutions : cos 1 4 cos ' 1 4 ' cos 1 1 4 The solutions in the I quadrant: The one solution in the I quadrant, that is 1 1 between 0 and 2 , is cos . Now, all the other solutions in the I 4 quadrant are coterminal to this one solution. Thus, all the solutions in the I 1 1 cos 2 n , where n is an integer. quadrant are given by 4 The solutions in the IV quadrant: The one solution in the IV quadrant, that is 1 1 between 2 and 0, is cos . Now, all the other solutions in the 4 IV quadrant are coterminal to this one solution. Thus, all the solutions in the 1 1 2 n , where n is an integer. IV quadrant are given by cos 4 the solutions of the second equation 4 cos 1 0 are 1 1 cos 1 2 n and cos 1 2 n , where n is an integer. 4 4 Thus, Answers: c o s 1 cos 1 4 2 1 2 n , 2 n , 2 n and 3 4 3 1 2 n , where n is an integer 4 3. sec 2 9 7 tan Back to Examples List This equation is neither quadratic in sec nor tan . We will use the 2 2 2 2 Pythagorean Identity sec tan 1 to replace sec by 1 tan . The resulting equation will be quadratic in tan . Since sec 2 tan 2 1, 2 2 then sec 1 tan . Thus, sec 2 9 7 tan 1 tan 2 9 7 tan tan 2 7 tan 8 0 ( tan 1) ( tan 8 ) 0 Thus, either tan 1 0 or tan 8 0 . Thus, there are two equations to be solved. NOTE: These types of equations were solved in Lesson 10. Please read that lesson in order to see how to solve these equations. Since tan 1 0 ta n 1 , then the solutions of the first equation 5 2 n and 2 n , where n is an integer. are 4 4 To solve the second equation tan 8 0 : tan 8 0 tan 8 Determine where the solutions will occur. Since tangent is negative in the II and IV quadrants, the solutions for this equation occur in those quadrants. Find the reference angle ' for the solutions : tan 8 tan ' 8 ' tan 1 8 The solutions in the II quadrant: The one solution in the II quadrant, that is 1 between 0 and 2 , is ' tan 8 . Now, all the other solutions in the II quadrant are coterminal to this one solution. Thus, all the 1 solutions in the II quadrant are given by tan 8 2 n = tan 1 8 ( 2 n 1) , where n is an integer. The solutions in the IV quadrant: The one solution in the IV quadrant, that is 1 between 2 and 0, is tan 8 . Now, all the other solutions in the IV quadrant are coterminal to this one solution. Thus, all the solutions in the 1 IV quadrant are given by tan 8 2 n , where n is an integer. Thus, the solutions of the second equation tan 8 0 are tan 1 8 ( 2 n 1) and tan 1 8 2 n , where n is an integer. NOTE: Since the period of the tangent function is , then this 1 answer may also be written as tan 8 n , where n is an integer. Answers: 5 2 n , 2 n , and tan 1 8 n , 4 4 where n is an integer 4. 5 cot 2 x 6 csc x 5 Back to Examples List This equation is neither quadratic in cot x nor csc x . We will use the 2 2 2 2 Pythagorean Identity csc x cot x 1 to replace cot x by csc x 1 . 2 2 The resulting equation will be quadratic in csc x . Since csc x cot x 1 , 2 2 then cot x csc x 1 . Thus, 5 cot 2 x 6 csc x 5 5 ( csc 2 x 1) 6 csc x 5 5 csc 2 x 5 6 csc x 5 5 csc 2 x 6 csc x 0 5 csc 2 x 6 csc x csc x ( 5 csc x 6 ) 0 Thus, either csc x 0 or 5 csc x 6 0 . Thus, there are two equations to be solved. To solve the first equation csc x 0 : Since csc x 0 and csc x 1 , sin x 1 0 . A fraction is zero if and only if its numerator is zero. The sin x 1 numerator of the fraction will always be 1. Thus, the equation sin x csc x 0 has no solutions. then To solve the second equation 5 csc x 6 0 : 5 csc x 6 0 5 csc x 6 csc x 6 5 sin x 5 6 5 is not the minimum 6 negative number for the sine function, then the solutions do not occur at the coordinate axes. Thus, the solutions occur in two of the four quadrants. Since sine is negative in the III and IV quadrants, the solutions for this equation occur in those quadrants. Determine where the solutions x will occur. Since Find the reference angle x ' for the solutions x : sin x 5 6 sin x ' 5 6 x ' sin 1 5 6 The solutions in the III quadrant: The one solution in the III quadrant, that is 1 5 between 0 and 2 , is x x ' sin . Now, all the other 6 solutions in the III quadrant are coterminal to this one solution. Thus, all the 1 5 2 n = solutions in the III quadrant are given by x sin 6 5 x sin 1 ( 2 n 1 ) , where n is an integer. 6 The solutions in the IV quadrant: The one solution in the IV quadrant, that is 5 . Now, all the other solutions in the 6 IV quadrant are coterminal to this one solution. Thus, all the solutions in the 1 5 2 n , where n is an integer. IV quadrant are given by x sin 6 between 2 and 0, is x sin 1 the solutions of the second equation 5 csc x 6 0 are 5 5 x sin 1 ( 2 n 1 ) and x sin 1 2 n , where n is an 6 6 integer. Thus, 1 Answers: x sin is an integer 5 ( 2 n 1 ) and x sin 6 1 5 2 n , where n 6