Chapter 11. Graphs of Trigonometric Functions

Transcription

1 Chater. Grahs of Trigonometric Functions - Grah of the Sine Function (ages 0 ). Yes, since for each (, ) on the grah there is also a oint (, ) on the grah.. Yes. The eriod of 5 sin is. Develoing Skills. Grah a. 7 0 #,,,, 5,, # 5,,, 7,, c. ccles No. It fails the horizontal line test. 8. a. cos 5 n or n c. h 5 d. 5 sin 5.5 ft. a. < # # # # c. 5 sin Y () Y () u Aling Skills 9. a! < P Y alwas gives the better aroimation. u u u is the reflection of u about the -ais. sin u 5 sin(u) c. Yes. For all angles in the four quadrants, sin u 5 sin(u). d. Yes. sin 0 5 sin (0 ) 5 sin 80 5 sin (80 ) 5 0 sin 90 5 sin (90 ) 5 sin 70 5 sin (70 ) 5 e. Yes, since for all in the domain, f() 5 f(). 0. a. # u # 5 h(u) 5 5 sin u P9 R - Grah of the Cosine Function (ages 5 7). Yes. For ever (, ) on the grah there is also a oint (, ) on the grah.. Yes. The eriod of 5 sin is. Develoing Skills. Grah a.,,,,, 0,,,,, c. ccles No. It fails the horizontal line test.

2 Aling Skills 8. a u is the reflection of u about the -ais. cos u 5 cos (u) c. Yes. For all angles in the four quadrants, cos u 5 cos (u). d. Yes. cos 0 5 cos (0 ) 5 cos 80 5 cos (80 ) 5 cos 90 5 cos (90 ) 5 cos 70 5 cos (70 ) 5 0 e. Yes. For all in the domain, f() 5 f(). 9. a. d(u) 5 cos u # u # 8 c. d 0. a. d. cos 8 < 5.90 feet c # # # # u u 8 P P9 cos Y () Y ()! < ! < R u Amlitude, Period, and Phase Shift (ages 5 55). Yes. The first grah is shifted to the left and the second is shifted to the right, resulting in the grahs starting units aart. Since this is equal to the eriod of each curve, their grahs will comletel overla.. No. If we factor out a from the second equation, we see that its grah is shifted 8 units, in contrast to a shift of units for the first grah. Develoing Skills Y alwas gives the better aroimation. 5

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4 8. Aling Skills 9. Since sine and cosine are cofunctions, it follows that sin a. 0.0 volt. a h c. The eriod of middle C aears to be onehalf the eriod of C.. a. f e cos A B a No. The eriod of this function is not months, but 0.5 <.5 months. Etending this model would shift all the temeratures b more than half a month each ear. 9 t t t Hands-n Activit sin. Maimum 5 5, minimum 5 5. The amlitude is equal to one-half the difference between the maimum and the minimum values: A () Grah () 5 sin () Maimum 5, minimum 5 (5) The amlitude is equal to one-half the difference between the maimum and () () minimum values: A Writing the Equation of a Sine or Cosine Grah (ages 57 59). No. The equation that Tler wrote has eriod and hase shift. Thus, it is equivalent to 5 5 cos. The maimum of this curve is at n and the minimum is at n.. Yes. The hase shift of the first grah is equal to the eriod of both equations. Develoing Skills. a.. a. 5. a.. a. 7. a. 8. a. 5 sin 5 cos A B 5 sin A B 5 cos 5 sin A B 5 cos 5 sin 5 cos A B 5 sin 5 cos ( ) 5 sin A B 5 cos 7

5 9. a. 0. a.. a.. a.. a.. a. 5 cos Aling Skills 5. a m 0 s c ccle er second d. e. No; if the amlitude is 0.75, then the maimum height is 0.75 meter. -5 The Grah of the Tangent Function (ages ). The tangent grah has no maimum or minimum values, the eriod is rather than, it has vertical asmtotes, and the range is all real numbers rather than [, ].. No; the range is (`, `). Develoing Skills. Grah a. c. (`, `) U: nv. a c. 5 sin A B 5 cos A B 5 sin A B 5 cos A B 5 sin A B 5 cos 5 sin A B 5 cos A B 5 sin 5 cos A B 5 sin A B h(t) cos A 5 B 5. a. b c. d. The are the same. Aling Skills. a. h 5 r tan u 7. a. () () () AP 5 r tan () AB 5 s 5 r tan (5) Perimeter 5 ()r tan () 5 () 5 V 5 r tan u () AP 5 r tan 5 () AB 5 s 5 r tan 5 (5) Perimeter 5 (5)r tan 5 c. For an regular olgon with n sides circumscribing a circle of radius r, the erimeter is nr tan n. - Grahs of the Recirocal Functions (ages 7). Cotangent is the recirocal of tangent. As the value of tan increases, its recirocal decreases, so cot decreases for all values of for which it is defined. 8

6 . sec increases from to ` in the interval S0, B and increases from ` to in the interval A, T. Develoing Skills. (). (8) 5. (). (7) 7. () 8. (5) 9. () 0. (). a.. a.. a.,,,,, 0,, cot and 5 csc 9. 5 tan and 5 sec 0. a. dd dd c. Even d. dd Aling Skills. a. a 5 0 sec u c. No d. 5. ft -7 Grahs of Inverse Trigonometric Functions (ages 7 7). Since sin (0 ) 5, arcsin A B The reference angle of 0 is 0.. No, tan (0 ) Develoing Skills !. 0.5 In 5, art a, answers will be grahs.. 5 arcsin Domain 5 5: # #.,,,,, 0,,,,,,, 0,, u 8 Range 5 U: # # V 5 sin Domain 5 U: # # V Range 5 5: # # 5 arccos Domain 5 5: # # Range 5 5: 0 # # 5 cos Domain 5 5: 0 # # Range 5 5: # # 9 9 a ,,, 9

7 5. 5 arctan Domain 5 5: is a real number Range 5 U:,, V 5 tan Domain 5 U:,, V Range 5 5: is a real number 5. Aling Skills. a. u 5 tan d A 00B u 5. or u 5 0. radians or u 5 9 c Sketching Trigonometric Grahs (ages 7 75) 5 tan A B. Yes. is the tangent grah with a. hase shift of. Therefore, the asmtotes also have a hase shift of.. No.. The 5 sin A B 5 sin A 8B 7. hase shift is 8, not. Develoing Skills

8 a.. a. 5 sin A B 5 cos A B 5 sin A B 5 cos A B 7. a. 5 sin 5 cos A B. 8. a. 5 sin A B 5 cos 9. a. 0. a. 5 sin A B 5 cos A B 5 sin A B 5 cos A B. a..

9 . a. f.. a.. a. No amlitude c. d. U: nv e. {all real numbers} f. Review Eercises (ages 7 79). a. c. d. {all real numbers} e. [, ] f.. a. c. d. {all real numbers} e. [, ] f.. a. c. d. {all real numbers} e. [, ] 5. a. c. d. {all real numbers} e. [, ] f.

10 . a. c. d. {all real numbers} e. [, ] f. 7. a. 5 sin 8. a. 9. a. 0. a. 5 sin. (). (). (). () cos A B 5 sin 5 cos A B 5 sin A 5 B 5 cos A B 5 cos A B. a. S, T [, ] c. S, T d. Grah. a c. 0. 5, 5, 5, 5. a. 5. No. The function 5 csc is undefined at 5 n for integer values of n.. a. c. 5 d. 0 mmhg e. 0 mmhg 7. a..5 ft hr c sin 7 Eloration (ages 78 79) Answers will var. Cumulative Review (ages 79 8) Part I Part II P 5, 7 7, 5, 7,,,, 0 5 t

11 . Answer:, 0 tan u 5.5 u 5 tan (.5) u<578 Tangent is negative in the second and fourth quadrants. Therefore, u and u Part III. Answer: 5 0, i ( 8 5) 5 0 Therefore, 5 0 is one solution. Use the quadratic formula to find the roots of the quadratic factor: 5 8! 5 i 5. a () n 5 () () 0 () n50 () () () Part IV 5. a. Since BG is the diagonal of a square, m GBC 5 5. tan u 5 AC GC 5! 5! u 5 tan A! B < 58. Answer: A, B and (, ) Substitute the linear equation into the quadratic and solve for : ( )( ) 5, 5 A B () Basic Identities (age 85). No. We also need to know the quadrant in which the terminal side of the angle lies to determine the sign of the other trigonometric functions.. a. To derive tan u 5 sec u, divide each term of sin u cos u 5 b cos u. To derive cot u 5 csc u, divide each term of sin u cos u 5 b sin u. No; tan u and sec u are not defined for u 5 n and cot u and csc u are undefined for u 5 n, so the identities are not defined for those values of u. Develoing Skills sin u cos u. cos u. sin u 5. cos u. sin u 7. sin u 8. cos u 9. sin 0. cos u. sin u cos u u cos u sin u cos u. cos. sin u. u cos u Chater.Trigonometric Identities - Proving an Identit (ages 87 88). No. The equation is conditional. If u is an angle whose terminal side lies in quadrant III or IV, then the equation is false.. Yes. The fraction is equal to. When the left side cos is multilied and simlified it becomes u sin u. Develoing Skills.. 5. sin u csc u cos u 5? cos u sin u? sin u? cos u 5? cos u cos u 5 cos u tan u sin u cos u 5? sin u sin u cos u? sin u? cos u 5? sin u sin u 5 sin u cot u sin u cos u 5? cos u cos u sin u? sin u? cos u 5? cos u cos u 5 cos u