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Pre-Class Problems 24 for Friday, April 21 These are the type of problems that you will be working on in class. 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: If given the value of a trigonometric function of an angle and information pertaining the location of the angle, then to use the half-angle formulas to find the values of the six trigonometric functions of the half angle. 3 2 , then find the exact value of the six 2 trigonometric functions of the angle . 2 5 8 1a. If cos 1b. If sin and 2 6 and trigonometric functions of the angle , then find the exact value of the six 2 . 2 , then find the exact value of the six 2 trigonometric functions of the angle . 2 1c. If tan 1d. If csc 17 8 and 3 3 and 2 , then find the exact value of the six 2 5 trigonometric functions of the angle . 2 Objective of the following problems: To use the half-angle formulas to find the value of a trigonometric function of a certain angle. 2. Find the exact value of the following. a. sin 12 b. cos 75 e. sec 7 8 f. 3 tan 8 c. cot 5 12 d. csc ( 165 ) Additional problems available in the textbook: Page 598 … 39 – 44, 47 – 50. Example 4 – 6 on page 595. Solutions: COMMENT: You will be given the following Half-Angle Formulas: cos 2 1 cos 2 sin 2 1 cos 2 tan 2 1 cos 1 cos These formulas are derived at the end of this set of pre-class problems. 1a. cos 5 3 2 and 8 2 Back to Problem 1. NOTE: The value of cos for the half-angle formulas has been given to us. In order to use the half-angle formulas, we also need to know what quadrant is in: 2 3 3 2 is in the second quadrant 2 4 2 2 NOTE: Cosine is negative, sine is positive, and tangent is negative in the second quadrant. cos 2 = cos 1 cos 2 5 8 1 = 2 = 5 8 8 = 2 8 1 8 5 16 13 4 2 13 sec 4 2 1 cos 2 1 sin 2 sin 3 4 csc 2 4 2 3 = 4 13 5 8 2 Using Basic Identities to find tan 5 8 8 = 2 8 1 = : 2 8 5 = 16 3 4 3 4 2 tan = 2 13 cos 2 4 sin Using the half-angle formula tan tan 2 = tan 1 cos 1 cos = 3 13 2 5 8 5 1 8 1 cos to find tan : 1 cos 2 1 = 3 13 3 13 cot 2 2 13 3 Answer: 13 sec , 2 4 cos 2 sin 2 tan 2 3 csc , 2 4 3 13 , cot 4 13 4 3 2 13 3 5 8 8 5 8 = 1 8 1 8 5 8 5 1b. sin 2 and 2 6 NOTE: Back to Problem 1. is in the third quadrant 2 In order to use the half-angle formulas, we will need to find cos . You can use the Pythagorean Identities or Right Triangle Trigonometry to find cos . 2 2 Using the Pythagorean Identity cos sin 1 to find cos : cos 2 sin 2 1 and sin cos 2 34 cos 36 2 2 1 cos 2 36 6 34 cos 6 34 since is the III 6 quadrant. NOTE: Cosine is negative in the third quadrant. Using Right Triangle Trigonometry to find cos : sin 2 sin ' 6 2 : 6 6 2 NOTE: sin ' opp hyp ' 34 cos cos ' 34 6 NOTE: cos ' adj and cosine is negative in the third quadrant. hyp In order to use the half-angle formulas, we also need to know what quadrant is in: 2 is in the fourth quadrant 2 2 2 4 2 NOTE: Cosine is positive, sine is negative, and tangent is negative in the fourth quadrant. cos 2 1 1 cos 2 = 34 6 6 = 2 6 18 3 = 34 1 6 = 2 6 34 12 = 34 6 1 2 6 34 3 = 12 3 = 18 3 34 36 34 6 cos 2 6 34 12 6 12 6 34 6 sec 34 34 = 2 12 6 34 12 ( 6 34 ) = 36 34 = 12 ( 6 34 ) = 2 6(6 sin 2 1 1 cos 2 18 3 34 36 sin 2 = = 6 34 12 6 12 6 34 6 6(6 34 34 1 6 = 2 = 34 6 6 = 2 6 36 6 34 ) = 6 34 12 18 3 = 34 6 1 2 = 6 34 3 = 12 3 34 6 csc 34 34 34 ) = 2 = 12 ( 6 34 ) = 36 34 = 36 6 Using Basic Identities to find tan 12 6 34 : 2 34 12 ( 6 34 ) 2 18 3 34 sin 6 2 tan = 2 18 3 34 cos 2 6 = 3( 6 34 ) 3( 6 34 ) (6 (6 6 34 6 34 34 ) 2 34 ) ( 6 70 12 34 2 = = 34 ) 1 cos = 2 1 cos 1 1 34 6 6 6 = 34 6 work above) = 70 12 34 2 Using the half-angle formula tan tan 18 3 34 18 3 34 = 6 34 6 34 = 6 34 6 34 36 12 34 34 = 36 34 = 35 6 34 1 cos to find tan : 2 1 cos 2 1 1 6 34 6 34 34 = 6 34 6 = 1 1 35 6 34 34 6 34 6 = (from our tan 2 6 34 6 34 cot 2 6 34 6 34 NOTE: I’ll let you verify that if you rationalize the denominator in the answer Answer: 1c. tan 6 34 6 34 cos 2 sin 2 tan 2 , you will obtain 18 3 34 , sec 6 18 3 2 36 6 34 , csc 2 36 6 34 34 , cot 2 35 6 34 34 6 35 6 35 6 34 . 17 and 8 2 Back to Problem 1. is in the second quadrant NOTE: 2 In order to use the half-angle formulas, we will need to find cos . You can use the Pythagorean Identities or Right Triangle Trigonometry to find cos . 2 2 Using the Pythagorean Identity sec tan 1 to find sec which will give us cos : sec 2 tan 2 1 and tan 17 17 1 sec 2 64 8 sec 2 81 9 9 sec sec since is the II 64 8 8 quadrant. NOTE: Secant is negative in the second quadrant. sec 9 8 cos 8 9 Using Right Triangle Trigonometry to find cos : tan 17 tan ' 8 17 : 8 17 9 NOTE: tan ' opp adj ' 8 cos cos ' NOTE: cos ' 8 9 adj and cosine is negative in the second quadrant. hyp In order to use the half-angle formulas, we also need to know what quadrant is in: 2 is in the first quadrant 2 4 2 2 2 NOTE: Cosine is positive, sine is positive, and tangent is positive in the first quadrant. cos 2 1 cos 2 9 8 = 18 cos sin 2 2 sin 2 8 9 1 2 = 18 2 17 = 18 = 1 2 8 9 17 2 = 18 2 34 csc 6 2 8 9 1 = 2 = 36 1 sec 2 18 1 cos 2 9 8 = 18 = 1 = 18 1 18 1 2 = 2 2 6 18 1 = 34 = 36 8 9 2 : 2 8 9 9 = 2 9 1 = 34 6 6 34 3 34 6 34 17 34 Using Basic Identities to find tan 8 9 9 = 2 9 1 2 tan = 2 cos 2 sin 34 6 2 6 34 2 Using the half-angle formula tan tan 2 1 cos 1 cos 9 8 = 9 8 tan 2 Answer: = 17 = 1 17 17 cot 2 2 1 1 17 1 cos to find tan : 1 cos 2 8 9 8 = 9 1 17 cos 2 2 , sec 2 6 sin 2 34 3 34 , csc 6 2 17 tan 2 17 , cot 2 18 1 17 8 9 8 1 9 1 8 9 9 8 9 = 1 9 1 = 1d. csc 3 3 and 2 2 5 NOTE: 2 Back to Problem 1. 3 is in the first quadrant 2 In order to use the half-angle formulas, we will need to find cos . You can use the Pythagorean Identities or Right Triangle Trigonometry to find cos . 2 2 Using the Pythagorean Identity cos sin 1 to find cos : csc 3 sin 5 5 3 cos 2 sin 2 1 and sin cos 2 5 5 1 cos 2 9 3 4 2 2 cos cos since is the I 9 3 3 quadrant. NOTE: Cosine is positive in the first quadrant. Using Right Triangle Trigonometry to find cos : 5 5 sin sin ' : 3 3 3 5 NOTE: sin ' opp hyp ' 2 cos cos ' 2 3 NOTE: cos ' adj and cosine is positive in the first quadrant. hyp In order to use the half-angle formulas, we also need to know what quadrant is in: 2 2 3 3 is in the third quadrant 2 2 4 2 NOTE: Cosine is negative, sine is negative, and tangent is positive in the third quadrant. cos 1 cos 2 2 = cos sin 5 = 6 2 2 5 6 5 6 = = sec 1 cos 2 1 2 3 2 = 2 3 3 = 2 3 3 2 6 2 3 3 = 2 3 3 2 6 1 30 6 2 = 6 5 1 2 2 3 30 5 = 1 6 1 1 = = 6 6 6 sin 1 csc 2 2 6 6 Using Basic Identities to find tan 6 6 2 tan = 2 30 cos 2 6 sin 6 30 Using the half-angle formula tan = 1 5 tan 1 cot 2 2 5 2 2 2 3 2 1 3 2 cos 1 tan Answer: 1 cos 1 cos : 2 = 1 5 1 cos tan to find : 1 cos 2 2 3 3 2 3 = 1 3 1 = 5 30 , sec 6 2 30 5 3 2 = 3 2 1 5 2a. sin sin 6 , csc 2 2 6 tan 1 , cot 2 2 5 6 5 12 Back to Problem 2. is one-half of or is twice . 12 12 6 6 NOTE: We will use the half-angle formula sin with , which means that . 2 12 6 2 1 cos 2 to find sin 12 NOTE: The angle is in the first quadrant. Sine is positive in the first 12 quadrant. sin 12 2 = 1 cos 6 2 3 2 NOTE: cos 6 3 2 3 2 1 = 2 1 = 3 2 2 = 2 2 2 3 4 2 Answer: 2b. 3 2 cos 75 Back to Problem 2. NOTE: 75 is one-half of 150 or 150 is twice 75 . We will use the half-angle formula cos 2 75 , which means that 150 . with 2 1 cos to find cos 75 2 NOTE: The angle 75 is in the first quadrant. Cosine is positive in the first quadrant. cos 75 1 3 1 2 = 2 1 cos 150 = 2 3 2 2 = 2 2 2 2 3 4 = NOTE: cos 150 cos 30 2 Answer: 2 3 2 3 2 3 3 2 1 2 = 2c. cot 5 12 NOTE: Back to Problem 2. 5 5 5 5 is one-half of or is twice . 6 6 12 12 Since cotangent is the reciprocal of tangent, then we will use the half-angle 1 cos 5 5 tan formula to find tan with , which 2 1 cos 2 12 12 5 means that . 6 NOTE: The angle 5 is in the first quadrant. Tangent is positive in the first 12 quadrant. tan 5 12 1 1 5 6 5 1 cos 6 1 cos 3 2 2 2 = 3 2 NOTE: cos 3 1 2 3 = 1 2 = 2 3 2 3 3 5 cos 6 6 2 1 1 3 2 3 2 = tan 5 12 2 3 2 3 (2 = (2 7 4 5 12 3) 2 3 2 3 4 4 3 3 = 4 3 3 )2 3 )(2 = 2 3 2 3 7 4 1 2 3 2 3 3 = 3 Answer: 2d. cot 7 4 3 csc ( 165 ) Back to Problem 2. NOTE: 165 is one-half of 330 or 330 is twice 165 . Since cosecant is the reciprocal of sine, then we will use the half-angle 1 cos 165 , formula sin to find sin ( 165 ) with 2 2 2 which means that 330 . NOTE: The angle 165 is in the third quadrant. Sine is negative in the third quadrant. sin ( 165 ) 1 1 cos ( 330 ) = 2 3 2 2 = 2 2 2 3 4 = 3 2 1 2 2 2 3 = 3 2 1 2 = NOTE: cos ( 330 ) cos 30 2 sin ( 165 ) 2 2 2 3 2 3 4 3 Answer: 2 2e. sec 3 3 2 3 = 2 2 = 2 1 3 2 (2 = 2 2 2 3 = 3 3 )(2 2 3) = 3 3 7 8 NOTE: 2 csc ( 165 ) 2 2 3 2 Back to Problem 2. 7 7 7 7 is one-half of or is twice . 8 4 4 8 Since secant is the reciprocal of cosine, then we will use the half-angle 1 cos 7 7 formula cos to find sec with , which means 2 8 8 2 2 7 that . 4 NOTE: The angle 7 is in the second quadrant. Cosine is negative in the 8 second quadrant. cos 7 8 2 4 cos 2 2 2 2 2 2 4 2 2(2 2 2 2 = 2 ) 4 2 2 2 sec 2 2 Answer: 2 2 = 2 2 2 2 = 2 2 2 2 7 8 1 2 7 cos 4 4 2 2 = 2 2 1 = 2 2 NOTE: cos 7 4 1 cos 2 7 8 = 2 2 2 = 2 4 2 2 2 2 2 (2 = 2 = 2 2 2 )(2 2 2 2 2) 2 = 2 2 = 2f. 3 tan 8 NOTE: Back to Problem 2. 3 3 3 3 is one-half of or is twice . 8 4 4 8 1 cos 3 to find tan 2 1 cos 8 3 3 with , which means that . 4 2 8 We will use the half-angle formula tan NOTE: The angle 3 is in the fourth quadrant. Tangent is negative in the 8 fourth quadrant. 3 tan 8 = 1 1 1 cos 1 cos 2 2 2 2 = 2 2 (2 2 )2 (2 2 )(2 3 2 2 2) 3 4 = 3 4 2 2 2 2 = 2 1 2 2 = 1 2 = 2 2 2 2 4 4 2 2 = 4 2 2 2 2 2 1 1 2 2 2 2 6 4 2 = 2 = 2 3 cos cos NOTE: 4 2 4 Answer: 3 2 2 Solution to Problems on the Pre-Exam: 23. Find the exact value of cos Back to Page 1. 7 3 2 . Put if cot and 2 3 2 a box around your answer. (4 pts.) From the Formula Sheet, we have that cos 2 1 cos . 2 3 3 2 . Thus, the angle is in the II quadrant. 2 4 2 2 Thus, cos 2 1 cos 2 since cosine is negative in the second quadrant. We will need to find cos . You can use the Pythagorean Identities or Right Triangle Trigonometry to find cos . cot 7 3 tan 3 7 3 2 is the IV quadrant 2 2 2 Using the Pythagorean Identity sec tan 1 to find sec which will give us cos : sec tan 1 and tan 2 sec 2 2 3 9 sec 2 1 7 7 4 16 sec sec 7 7 4 since is the IV 7 quadrant. NOTE: Secant is positive in the fourth quadrant. sec 4 cos 7 7 4 Using Right Triangle Trigonometry to find cos : tan 3 tan ' 7 3 : 7 4 NOTE: tan ' opp adj 3 ' 7 cos cos ' NOTE: cos ' 7 4 adj and cosine is positive in the fourth quadrant. hyp cos 2 24. 4 1 cos 2 7 = 8 8 2 = 4 7 8 7 4 1 2 = 2 8 2 7 16 7 4 4 = 2 4 = 7 4 4 7 8 or Find the exact value of tan NOTE: 8 2 7 4 5 . (5 pts.) Put a box around your answer. 8 5 5 5 5 is one-half of or is twice . 8 4 4 8 We will use the half-angle formula tan with = 2 1 1 cos 5 to find tan 2 1 cos 8 5 5 , which means that . 2 8 4 NOTE: The angle second quadrant. 5 is in the second quadrant. Tangent is negative in the 8 tan = 5 1 cos 4 5 1 cos 4 5 8 2 2 2 2 = 2 2 1 1 (2 (2 2 )(2 3 2 2 NOTE: cos 2 )2 2 2 2 2 2) = 2 1 2 2 = 1 2 2 2 2 2 = = 1 1 2 2 2 2 4 4 2 2 = 4 2 2 2 2 2 2 2 2 2 6 4 2 = 2 = 2 5 cos 4 4 2 or 3 2 2 Deriving the Half-Angle Formulas 2 Using the double angle formula cos 2 2 cos 1 and solving for cos , we obtain the following: 2 cos 2 2 cos 2 1 1 cos 2 2 cos 2 cos 1 cos 2 2 cos 1 cos 2 . 2 Setting , we obtain that cos 2 2 1 cos 2 2 Using the double angle formula cos 2 1 2 sin and solving for sin , we obtain the following: 2 cos 2 1 2 sin 2 2 sin 2 1 cos 2 sin sin 1 cos 2 . 2 Setting , we obtain that sin 2 2 1 cos 2 1 cos 2 sin 1 cos 2 2 tan 2 tan 1 cos 2 cos 2 1 cos 2 2 2 Setting , we obtain that tan 2 2 Back to the half-angles formulas above. 1 cos 2 2 1 cos 1 cos 1 cos 2 1 cos 2