LHS CAT4e ISM-07-final

# LHS CAT4e ISM-07-final - Chapter 7 TRIGONOMETRIC IDENTITIES...

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624 Chapter 7 TRIGONOMETRIC IDENTITIES AND EQUATIONS Section 7.1: Fundamental Identities 1. By a negative-angle identity, ( ) tan tan . θθ −= Thus, if tan 2.6, θ = then ( ) tan 2.6. 2. By a negative angle identity, ( ) cos cos . Thus, if cos .65, =− then ( ) cos .65. 3. By a reciprocal identity, 1 cot . tan = Thus, if tan 1.6, = then cot .625. = 4. By a quotient identity, sin tan . cos = By a negative angle identity ( ) sin sin and ( ) cos cos . Thus, if cos .8 and sin .6, xx == then () ( ) sin sin .6 tan .75. cos cos .8 x x x −− = = = 5. By a negative angle identity ( ) sin sin . Thus, ( ) ( ) sin sin sin = = . So, if 2 sin 3 = , then 2 sin . 3 = 6. By a negative angle identity ( ) cos cos . Thus ( ) cos cos = . So, if 1 cos 5 , then 1 cos 5 = . 7. 3 cos 4 = , is in quadrant I. An identity that relates sine and cosine is 22 sin cos 1. ss += 2 2 2 3 sin cos 1 sin 1 4 97 7 sin 1 sin 16 16 4 ⎛⎞ ⇒+ = ⎜⎟ ⎝⎠ =− = ⇒ Since is in quadrant I, 7 sin . 4 = 8. 1 cot 3 , in quadrant IV Use the identity 1c o t c s c since 1 sin . csc = 2 2 1 o t c s c 1 c s c 3 11 0 s c c s c 99 10 csc 3 +=⇒ + =⇒ ⇒ = Since is in quadrant IV, csc < 0, so 10 csc . 3 Thus, 133 1 0 3 1 0 sin csc 10 10 10 10 = = 9. 5 cos , 5 tan 0 < Since 5 cos , 5 we have 5 cos 5 = by a negative angle identity. An identity that relates sine and cosine is sin cos 1. 2 2 2 5 sin cos 1 sin 1 5 51 sin 1 sin 1 25 5 14 sin 1 55 5 2 5 sin 5 5 = + = Since tan < 0 and cos > 0, is in quadrant IV, so sin < 0. Thus, 25 sin . 5 10. 7 tan , sec 0 2 > 2 2 2 7 tan 1 sec 1 sec 2 11 11 sec sec 42 ⇒− = ±

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Section 7.1: Fundamental Identities 625 Since sec 0, θ > 11 sec . 2 = Also, 11 2 since cos , cos . sec 11 11 2 θθ == = Now, use the identity 22 sin cos 1: += 2 2 24 sin 1 sin 1 11 11 47 sin 1 11 11 77 sin 11 11 71 1 7 7 11 11 11 ⎛⎞ + = ⎜⎟ ⎝⎠ =− = ⇒ Since tan < 0 and sec > 0, is in quadrant IV, so sin < 0. Thus, 7 sin 11 11 =− . 11. 11 sec , tan 0 4 =< Since 4 cos ,cos . 11 sec 11 4 = Use the identity sin cos 1, to obtain 2 2 2 4 sin cos 1 sin 1 11 16 16 sin 1 sin 1 121 121 105 105 sin sin 121 11 ⇒+= = −⇒ =⇒ = ± Since tan < 0 and sec > 0, is in quadrant IV, so sin < 0. Thus, 105 sin . 11 12. 8 csc 5 5 sin , so sin . 8 csc 8 5 = 13. The quadrants are given so that one can determine which sign ( ) or +− sin will take. Since 1 sin , csc = the sign of sin will be the same as csc . 14. The range of cosine is [-1, 1], thus there is no number or angle whose cosine is 3. 15. ( ) sin sin xx −= 16. Since ( ) ( ) ( ) sin sin , fx x x f x −= −= = ( ) sin x = is odd. 17. ( ) cos cos 18. Since ( ) ( ) ( ) cos cos , x xf x = ( ) cos x = is even 19. ( ) tan tan 20. Since ( ) ( ) ( ) tan tan , x x = ( ) tan x = is odd. 21. This is the graph of ( ) sec . x = It is symmetric about the y ±axis. Moreover, since () () () sec sec , cos cos x x = = = ( ) ( ) . fx f x 22. This is the graph of ( ) csc . x = It is symmetric about the origin. Moreover, since csc sin sin csc , x x = −− ( ) ( ) . f x 23. This is the graph of ( ) cot . x = It is symmetric about the origin. Moreover, since ( ) cos cos cot sin sin cos cot sin x x x x x x = ( ) ( ) . f x 24. This is the graph of ( ) tan . x = It is symmetric about the origin. Moreover, since ( ) sin sin tan cos cos sin tan cos x x x x x x = ( ) ( ) . f x
626 Chapter 7: Trigonometric Identities and Equations 25.

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LHS CAT4e ISM-07-final - Chapter 7 TRIGONOMETRIC IDENTITIES...

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