Principles of Mathematics Answer Ke, Contents 85 Module : Section Trigonometr Trigonometric Functions Lesson The Trigonometric Values for θ, 0 θ 60 86 Lesson Solving Trigonometric Equations for 0 θ 60 8 Lesson Special Trigonometric Values 85 Lesson The Unit Circle 87 Lesson 5 The Circular Functions 89 Lesson 6 Radian Measures 9 Review 0 Section : Graphs of Trigonometric Functions Lesson Sine and Cosine Graphs 05 Lesson Transformations of the Sine and Cosine Functions 07 Lesson Graphs of the Remaining Circular Functions 5 Review 9 Section : Trigonometric Identities Lesson Elementar Identities 5 Lesson Using Elementar Identities Lesson Sum and Difference Identities 7 Lesson Double Angle Identities Review 55 Section : Problem Solving Lesson Modelling with Trigonometric Functions 6 Lesson Lesson Solving Trigonometric Equations Using Your Graphing Calculator 65 Solving Trigonometric Equations With General Solution 65 Review 66 Module
86 Answer Ke, Contents Principles of Mathematics Notes Module
Principles of Mathematics Section, Answer Ke, Lesson 87 Lesson Answer Ke.. 0 P(,) Q( 6,8) 8 r 6 r 0 θ θ 5 sinθ cosθ tanθ 8 sinθ 0 5 6 cosθ 0 5 8 tanθ 6 Module
88 Section, Answer Ke, Lesson Principles of Mathematics.. θ 5 r 5 or 5 θ S( 6, ) r 5 0 R(5, ) sinθ 5 cosθ tanθ 5 sinθ 5 5 6 cosθ 5 5 tanθ 6 Module
}r 5 Principles of Mathematics Section, Answer Ke, Lesson 89 5. 6. M(5,0) r { θ N(0,) 0 sinθ 0 5 5 cosθ 5 0 tanθ 0 5 sinθ 0 cosθ 0 tan θ which is undefined 0 7. 8. r L(,0) θ θ } r K(0, ) 0 sinθ 0 cosθ 0 tanθ 0 sinθ 0 cosθ 0 tan θ which is undefined 0 Module
90 Section, Answer Ke, Lesson Principles of Mathematics Notes Module
Principles of Mathematics Section, Answer Ke, Lesson 9 Lesson Answer Ke. a) Find the reference angles for 6, 65, 89,. 6, 85, 7, 7 b) Evaluate the three trigonometric ratios for the angles above. b) 6 65 89 sin 0.899 0.996 0.96 0.60 cos 0.8 0.087 0.6 0.799 tan.05..90 0.75 c) Evaluate the three trigonometric ratios for the angles above. i) 76 or ii) 65 or 95 iii) 69 or 7 iv) 68 or 7.. sinθ sinθ θr θ θ sin, º, 8º 5tanθ + tanθ 5tanθ tanθ 7 θ R tan 0º 7 θ 50º, θ 0º Module
9 Section, Answer Ke, Lesson Principles of Mathematics. sinθ + 0 sinθ θ R sin 9º θ 99º, θ º 5. 6. cos θ + cosθ 5 0 ( θ )( θ ) Case() cosθ + 5 0 cos + 5 cos 0 () cosθ 5 θ cos R () Case cosθ 0 cosθ θ R θ 7º, θ 89º cos 7º 7cosθ 8 cosθ 7 θ R 7 5 No Solution cos No Solution [ Remember that cos θ ] 7. sin θ sinθ 0 ; 80 or. (60 is not valid because of the allowed interval for θ.) 8. tan tan 76,, 56, Module
Principles of Mathematics Section, Answer Ke, Lesson 9 Lesson Answer Ke. 0 0 5 60 90 0 5 50 80 sin θ 0 0 cos θ 0 tan θ 0 undefined 0 0 5 0 70 00 5 0 60 sin θ 0 cos θ 0 tan θ undefined 0. a) b) + + c) + + d) e) f) + + g) + + 0 Module
9 Section, Answer Ke, Lesson Principles of Mathematics. a) 5, 5 b) 60, 0 o o o o c) 0, 0 d) 0 o o o o e) 0 f) cos θ o o, since 0 and 60 are not allowed. ( θ ) g) sin θ sin + 0 sin θ 0 or sin θ o o o θ 80, 0, 0 h) cos θ cos θ ± θ 60, 0, 0, 00 o o o o ( ) i) tan θ+ 0 j) sin θ sin θ 0 ( θ ) ( )( ) tan θ sin θ+ sin θ 0 θ θ+ θ o o 0, 00 sin 0 or sin 0 k) cos θ cos 0 cos θ 0 or cos θ θ o o 90, 70 sin θ or sin θ o o o θ 0, 0, 90 (Note: cos θ produces no solutions since 0 < θ < 60.) Module
Principles of Mathematics Section, Answer Ke, Lesson 95 Lesson Answer Ke. a) E b) I c) H d) J e) Q f) S g) L h) H i) A j) M k) D l) H m) K n) L o) G p) N q) M r) S. a) S b) H c) S k is a multiple of. H k is an even, non-multiple of. D or L k is an odd integer.. a) B, F, J, or N. b) S, D, H, or L.. a) b) c) 5 d) e) f) 6 g) h) Careful: 0 is not positive. It is non-negative. 7 i) j) 6 Module
96 Section, Answer Ke, Lesson Principles of Mathematics 5. a) 7 b) c) d) e) f) 6 g) h) 5 i) 5 j) 6 Note: This eercise was prett straight forward. However, just because it is simple does not mean it is not important. In fact, it is a ver important assignment as it forms the basis for the rest of the module. You must know where ou have stopped for ever special value of θ. Conversel, if ou know where ou are on the unit circle, ou must know how to get there! Module
Principles of Mathematics Section, Answer Ke, Lesson 5 97 Lesson 5 Answer Ke. Question (a) (b) (c) θ P(θ) cos θ sin θ tan θ, 7, 6 7 (, 0) 0 0 (d) 5, (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) (r) 6, (, 0) 0 0 5 (0, ) 0 undefined 5 (, 0) 0 0, 6 7, (0, ) 0 undefined (, 0) 0 0, 5 6, 5, (0, ) 0 undefined 7, (, 0) 0 0 Module
98 Section, Answer Ke, Lesson 5 Principles of Mathematics. a) + + b) () c) () d) cos 0 + e) ( ) ( ) + + 6 f) cos g) () ()( ) h). a), b), c), 6 6 d) e) f) cos θ θ 0 ( θ ) g) sin θ sin + 0 sin θ 0 or sin θ 7 θ 0,,, 6 6 h) cos 0 θ ( cosθ )( cosθ+ ) 0 cos θ ± 5 θ,,, Module
Principles of Mathematics Section, Answer Ke, Lesson 5 99 ( θ ) i) tan + 0 tan θ 5 θ, j) sin sin 0 θ θ ( sin θ+ )( sin θ ) 0 sin θ, or sin θ 7 θ,, 6 6 ( θ ) k) cos θ cos 0 cos θ 0 or cos θ θ,,0. a) 0 b) + () + 5 c) ( ) d) ( ) e) ( ) f) () 8 5. a) cos θ 5 θ, b) sin θ θ, c) tan θ and sin θ < 0 7 θ ( θ) d) cot θ sin 0 cot θ 0 or sin θ cot θ 0 or θ θ, Module
00 Section, Answer Ke, Lesson 5 Principles of Mathematics ( θ )( θ ) e) tan tan + 0 tan θ ± 5 7 θ,,, f) sin θ 0 sin θ sin θ 0 ( sin θ)( + sin θ) 0 sin θ ± θ, Module
Principles of Mathematics Section, Answer Ke, Lesson 6 0.... 5. 5 5 7 ; o o Since 90 and 80 : F 5 80 0 50 50 H G I 80K J 8 The other base angle is also The third angle Lesson 6 Answer Ke F I HG K J F H G I K J F I HG K J F H G I K J 5 5 a) 5 b) 5 80 6 80 6 c) 60 d) 0 80 9 80 6 7 80 a) 0 b) 6 F H G I K J F80 H G I K J 65 F I HG K J F H G I K J c). 6 80 50 9 d) 0 985 80 56... 7 F H G I K J. 7. 7 6. 5. Therefore, or. 80 5 5 7. Use the CAST rule. S A T C Cos is positive in Quadrant IV. All are positive in Quadrant I. Sine is positive in Quadrant II. Tan is positive in Quadrant III. a) IV b) I or III c) II d) II or III Module
0 Section, Answer Ke, Lesson 6 Principles of Mathematics 8. Since s θr, it follows that. θ(8.). F Hence, θ or H G I 5. 5. 80 K J 85. 9. 9. Using the Pthagorean Theorem: AB AC + BC AB 6 + 8 AB 0 Circumference r (5) 0 cm. 0. Since C r 0 r 0 r 0 Since s θr, it follows that 5 θ ; therefore, θ or 60.. a) rev. ; therefore, setting up a ratio where measure in revolutions rev revolutions b) rev. 60 ; therefore, setting up a ratio where measure in revolutions rev o o 60 75 75 5 revolutions 60 Module
Principles of Mathematics Section, Answer Ke, Lesson 6 0. a). 76 b) 0. 9969 c) 0. 6756 d) b cos 5. g. 68 e) b tan g 7. 055 f) 0. 6 g) 0. 60 h) 0. 0589 i) k) sin 5 b b g tan 05.. 08 j) 0. 598 g. 809 l) 0. m) 0. 500 n) d sin o i. 559 o) 850.. a) Reference arc sin (0.) 0. Therefore, 0. to two decimal places. sin is positive in Quadrants I and II. 0. is the measure of the angle in Quadrant I. In Quadrant II, ( 0.).0 0. or.0 b) Reference arc cos (0.).75 Therefore,.5 to two decimal places. cos is negative in Quadrants II and III. In Quadrant II, (.5).69 In Quadrant III, ( +.5).59.69 or.59 Module
0 Section, Answer Ke, Lesson 6 Principles of Mathematics c) Reference arc tan.07 Therefore,. to two decimal places. tan is positive in Quadrants I and III.. in Quadrant I. In Quadrant III, ( +.).5. or.5 d) Reference arc sin 0.5 0.57 0.5 (rounded to two decimal places) sin is negative in Quadrants III and IV. In Quadrant III, ( + 0.5).9 In Quadrant IV, ( 0.5) 6.0.9 or 6.0 e) Reference arc cos (0.675) 0.898 0.8 cos is positive in Quadrants I and IV. In Quadrant I, 0.8 In Quadrant IV, ( 0.8) 5.5 0.8 or 5.5 Module
Principles of Mathematics Section, Answer Ke, Lesson 6 05 f) sec.5.5 cos cos 0..5 Reference arc cos (0.).59.6 cos is positive in Quadrants I and IV. In Quadrant I,.6 In Quadrant IV, (.6) 5..6 or 5. g) cot tan tan 0.5 Reference arc tan (0.5) 0.5 tan is negative in Quadrants II and IV. In Quadrant II, ( 0.5).90 In Quadrant IV, ( 0.5) 6.0.90 or 6.0 Module
06 Section, Answer Ke, Lesson 6 Principles of Mathematics h) csc 6 6 sin sin 0.667 6 Reference arc sin (0.6667) 0.675 0.7 sin is positive in Quadrants I and II. In Quadrant I, 0.7 In Quadrant II, ( 0.7).97 0.7 or.97. a) sin sin 0 ( ) sin sin 0 sin 0 or sin 0, 0, 50, 80 sin b), provided cos 0 cos tan o o 5, or 5 o o c) sin θ when θ 90 or (60 + 90) 50 90 or 50 5 or 5 Note: Since < 60 in solving the equation, ou use < 70 (twice around the unit circle). Module
Principles of Mathematics Section, Answer Ke, Lesson 6 07 d) cos cos when θ 60, 00, 0, 660 (twice around again) 60, 00, 0, 660 o o o o 0, 50, 0, 0 o o o o o o o o e) csc sin 0, 50 f) tan tan 0 ( )( ) tan + tan 0 tan or tan Related tan 6.57 o tan is negative in Quadrants II and IV. (80 6.57), (60 6.57), or 5, 5 5.,., 5, 5 o o o o o o o o 5. ( ) a) cos cos 0 cos 0 or cos, cos b) (twice around) 5 9,,, 5 9,,, 8 8 8 8 Module
08 Section, Answer Ke, Lesson 6 Principles of Mathematics c) sin cos sin cos tan 5, d) sin ( ) 5, 6 6 7, (both within range) 6 6 e),, (not allowed) Use. 0, ( ) f) sin tan + 0 sin 0 or tan 7 0,,, Module
Principles of Mathematics Section, Answer Ke, Review 09. Review Answer Ke a) b) c) d) 0 e F I HG K J F H G I K J bg j e j b g b g e) f) + 0. a) 5 7 7, b), c) 5 7, d) 6 e) f) cos θ, θ, 6 6 6 ( θ ) g) cos θ cos + 0 cos θ 0, cos θ θ,,, h) cos θ cos θ ± 5 θ,,, i) cot θ 5 θ, 6 6 ( θ ) k) tan θ tan 0 tan θ 0, 5 θ 0,,, j) cos cos 0 θ+ θ ( cos θ )( cos θ+ ) 0 cos θ, 5 θ,, Module
0 Section, Answer Ke, Review Principles of Mathematics.. F I HG K J F H G I K J F I HG K J F H G I K J 5 a) 5 b) 50 80 80 6 5 0 c) 500 d) 600 80 9 80 a) 5 80 80 50 b) 65 6 c) 80 80.8 7.7 d) 0.5. 5. a) sin sin 0 θ θ sin θ( sin θ ) 0 sin θ 0, 5 θ 0,,, 6 6 b) sin cos tan 5, 5 c) θ, θ, d) sin 0, 50, 90, 50 o o o o 5, 75, 95, 55 o o o o e) csc θ f) sin θ 7 θ, 6 6 tan tan 0 + ( )( ) tan tan + 0 tan, 6.565, 06.565, 5, 5 o o o o Module
Principles of Mathematics Section, Answer Ke, Review g) cos θ h) 5 θ...,,, 5 θ...,,, 6 6 6 5 θ, 6 6 5 θ +,,,... θ,,,... θ, i) ( ) cos tan + 0 cos 0, tan 90, 70, 5, 5 o o o o Module
Section, Answer Ke, Review Principles of Mathematics Notes Module
Principles of Mathematics Section, Answer Ke, Lesson Lesson Answer Ke θ. Domain is R.. -intercept. Zeros are the odd integral multiples of This is the same as the sequence... 5. Period [ ]{ R } Range is, 6. Cos () cos ( ); the transformation of cos () to cos ( ) is a reflection across the -ais, and ields the same graph. This kind of smmetr is called "even." Cos () cos ( ); when lines are drawn through the origin to the cosine graph, the origin is rarel the midpoint of those lines. Note that this is the opposite case from the sine curve. Thus, the cosine curve is not smmetric about the origin, but the sine curve is. We ll ask ou to distinguish "even" from "odd" smmetr several times in the course. How can ou remember that cos() has "even" and sin() has "odd" or ( k+ ) k Ι. 5,,,,... The graph of f() The graph of f() Module
Section, Answer Ke, Lesson Principles of Mathematics smmetr? One trick is to know where the even-odd terms come from. The parabola f() is familiar to ou, and ou can see right awa that f() f( ). Like the cosine graph, the parabola graph has even smmetr. And it just so happens that eponent in f() is an even number. Now consider the graph of f(). It has smmetr like the sine curve-through the origin, which we call odd smmetrjust as odd as the in its equation! In fact, the graphs of,, and 5 all displa "odd" smmetr like the sine curve, while the graphs of 0,, and all have "even" smmetr like the cosine curve. You ma use our graphing calculator to see this. B recalling the shapes of the and curves, ou can use the smmetr terms "even" and "odd" with confidence. 7. Amplitude. 8. In one revolution, cos θ > 0 in Quadrants I and IV; cos θ < 0 in Quadrants II and III. 9. In one revolution, the cosine curve is increasing in Quadrants III and IV and decreasing in Quadrants I and II. Module
Principles of Mathematics Section, Answer Ke, Lesson 5 Lesson Answer Ke. Question Domain Range Amplitude -intercept Period (a) R [, ] 0 (b) R [, ] 0 (c) R [, ] 0 (d) R [, ] (e) R [, ] (f) R [, 0] (g) R [0, ] not a wave 0 (h) R [, ] 0 (i) R [, ] (j) R [0, ] not a wave (k) R [, ] (l) R [, ] 0 / a) b) Module
6 Section, Answer Ke, Lesson Principles of Mathematics c) d) e) f) Module
Principles of Mathematics Section, Answer Ke, Lesson 7 g) makes all the values of < 0 behave as values of > 0. h) i) j) Module
8 Section, Answer Ke, Lesson Principles of Mathematics k) Cosine sine with period. l) Stretch b a factor of in -direction and -direction; followed b a reflection in the -ais.. Some possible answers. There are others. a) a b c 0 b) a b c. Some possible answers. There are others. a) a b c b) a b c 0 a b c Module
Principles of Mathematics Section, Answer Ke, Lesson 9. a) b) c) d) e) f) 5. Note: Be sure ou know how to denote various sets of integers. All integers are denoted b k I. Even integers are denoted b k. Odd integers are denoted b k + or k, since odd integers are one greater or one less than an even integer denoted b k. This question is equivalent to solving equations that ou are familiar with ecept that now the answers are not restricted to an interval. a) 0 sin...,,0,,,,......,, 0,,,... k k I b) 0 cos 5...,,,,,... 5...,,,,,... ( k ) + k I Module
0 Section, Answer Ke, Lesson Principles of Mathematics c) 0 sin...,,0,,,,......, +,,,,......,,,,,... ( k ) + k I d) 0 sin...,,0,,,,......,,0,,,,... I e) 0 cos 5...,,,,,......,,,,... ( k ) + k I ( ) f) 0 sin...,,0,,,,......,,0,,,,... k k I Module
Principles of Mathematics Section, Answer Ke, Lesson Etra for Eperts 6. cos cos cos cos 7. sin sin Module
Section, Answer Ke, Lesson Principles of Mathematics Notes Module
Principles of Mathematics Section, Answer Ke, Lesson Lesson Answer Ke. csc sec Module
Section, Answer Ke, Lesson Principles of Mathematics cot Note: cot 0 where tan is undefined and tan 0 where cot is undefined.. Function Domain Range -intercept Zeros cot { k k I} R none k+ k I ( ) csc { k k I} (, ] [, ) none none sec ( k+ ), k I (, ] [, ) none Function Asmptotes Period Smmetr Quadrant Sign cot { k k I} odd csc { k k I} odd sec ( k+ ), k I even + in I & III in II & IV + in I & II in III & IV + in I & IV in II & III Increasing or Decreasing alwas decreasing inreasing: II & III decreasing: I & IV inreasing: I & II decreasing: III & IV Module
Principles of Mathematics Section, Answer Ke, Lesson 5. a) tan b) sec + c) csc d) cot Module
6 Section, Answer Ke, Lesson Principles of Mathematics e) tan f) sec Module
Principles of Mathematics Section, Answer Ke, Review 7 Review Answer Ke. a) b) c). a) b) c) Module
8 Section, Answer Ke, Review Principles of Mathematics d) e) f) g) Module
Principles of Mathematics Section, Answer Ke, Review 9 h) i) j) k) Module
0 Section, Answer Ke, Review Principles of Mathematics l) The sketches can be used to fill in man parts of the following table. Question Domain Range Amplitude -intercept Period (a) R [, ] 0 (b) R [, ] (c) R [, ] 0 (d) R [, ] 0 (e) R [, ] 0 (f) R [, ] (g) R [0, ] none 0 (h) R [, ] 0 none or (i) R [, ] (j) ( k + ) k I R none 0 (k) ( k + ) k I (, ] [0, ) none 0 (l) k k [, ) none none I In question (h): the period is mostl but has a break at the -intercept. So strictl speaking, there is not a pattern that repeats across the whole function. Consider either none or as correct. Module
Principles of Mathematics Section, Answer Ke, Review. ( ) a) sin + 5 9 ( ) b) 5cos + + 8 sin ( ) + ( ) Range R R Domain [, ] [, 8] period amplitude 5 phase shift / -intercept 5cos + + Module
Section, Answer Ke, Review Principles of Mathematics. a) sin sin 5 b) cos cos 5 Compare our answers to question 5 with the graphs of sin and cos, ou should notice that: sin cos cos sin Module
Principles of Mathematics Section, Answer Ke, Lesson Lesson Answer Ke. Using the CAST rule: a) II b) IV c) III d) IV. a) The third side is a + 5 or a. Furthermore, since sin θ > 0 and tan θ < 0, it follows that θ II. Therefore, cos θ tan θ csc θ sec θ cot θ 5 5 5 5 a θ b) The third side is a + 5 or a. Furthermore, since sin < 0 and cos < 0, it follows that III. Therefore, sin tan csc sec cot 5 5 5 5 a, 5 5 Module
Section, Answer Ke, Lesson Principles of Mathematics c) The third side is 5 + a or a 6. Furthermore, since tan θ > 0 and cos θ < 0, it follows that θ III. Therefore, sin θ cos θ csc θ sec θ cot θ 5 6 6 6 5 6 5 a 5, 5 θ 6 6 d) Since csc θ, it follows that sin θ The third side is. a + or a. Furthermore, since sin θ > 0, it follows that θ I or II. Therefore, sin θ cos θ tan θ sec θ cot θ ± ± ± ±,, θ a Module
Principles of Mathematics Section, Answer Ke, Lesson 5 e) 5 Since sec, it follows that cos. The thir d side 5 + > is a 5 or a. Furthermore, since cos 0 and tan > 0 then I. Therefore, sin cos tan csc cot 5 5 5 5, a 5 f) Since cot 8 it follows that tan 5 The third 5, 8. side is 5 + 8 a or a 7. Furtherfore, since tan < 0, it follows that θ II or IV. Therefore, sin cos tan csc sec ± 5 7 ± 8 7 5 8 ± 7 5 ± 7 8 8, 5 7 7 a 5 8 8, 7 5 7 Module
6 Section, Answer Ke, Lesson Principles of Mathematics. One solution is provided but there are other solutions. F a) LHS H G I cos K J RHS F cos b) LHS RHS H G I K J sin sin F c) LHS H G I tan K J RHS tan cos d) LHS sin cos RHS sin sin e) LHS cos sin RHS cos f) LHS cos sin sin cos sin cos tan RHS g) LHS ( sin ) sin h) LHS sin RHS (cos + sin )(cos sin ) (( sin ) sin ) sin RHS i) LHS ( + sin θ) + ( sin θ) ( + sin θ)( sin θ) sin θ cos θ sec θ RHS j) LHS (tan ) (cot ) θ + θ + tan cot θ + θ tan cot RHS θ θ Module
Principles of Mathematics Section, Answer Ke, Lesson 7 k) LHS sin θ sin θ sin θ sin θ sin θ sin θ sin θ sin θ sin θ cos θ RHS l) LHS sec θ( + tan θ) sec θ(sec θ) sec θ RHS m) RHS sin cos + cos sin sin + cos cos sin cos sin ( ) sin + cos cos sin cos sin LHS n) RHS sin sin sin sin ( cos ) sin + cos LHS sin Module
8 Section, Answer Ke, Lesson Principles of Mathematics. sin cos csc sin a) + + cos csc csc sin sin + + sin cos sin cos cos sin cos sin b) sin cos cos sin c) sin cos cos sin cos sin + d) csc sin sin e) cos (sin cos ) cos () cos + + cot f) + cot Module
Principles of Mathematics Section, Answer Ke, Lesson 9 Lesson Answer Ke. a) b) c). a) tan + ( + tan ) tan tan ± 5, 5, 5, 5 sin provided cos 0 cos tan o o o o 5, 5, 05, 585.5,.5, 0.5, 9.5 tan sin tan 0 tan ( sin ) 0 tan 0 or sin o o o 0, 80, 90 ( sin ) sin o o o o o o o o 0 sin sin 0 ( sin + )(sin ) sin or sin 7,, 6 6 b) cos ± 5 7,,, 6 6 6 6 Module
0 Section, Answer Ke, Lesson Principles of Mathematics c) sin sin ± 5 7...,,,,,,... 6 6 6 6 6 5,,, (Note: 7 is out of range.). a) sin cos sin cos + cos ( cos ) cos + cos 0 cos cos 0 cos (cos ) cos 0 or cos (square both sides) Possible values of 0,,. Check: 0:sin 0 0 and cos0 0 : sin and cos 0 :sin and cos 0 Therefore, is not a root. Solution is: 0,. b) Be careful! Since the LHS is 0, this equation can be done quite easil. Module
Principles of Mathematics Section, Answer Ke, Lesson sin cos sin cos tan 7 Therefore,,. c) sin sin + cos sin sin ( sin ) + sin sin 0 + (square both sides) The discriminant (b ac) of the quadratic is ( ) () < 0. Hence, no solution! Observe the original equation as sin cos. It should be apparent that no solution eists since sin and cos, but the are never equal to at the same time.. a) cos(90 0 ) cos 90 cos 0 + sin 90 sin 0 o o o o o o 0 +. o o o But, cos(90 0 ) cos 60. The are equal! b) cos(50 0 ) cos 50 cos 0 + sin 50 sin 0 o o o o o o + o o o But, cos(50 0 ) cos 0. The are equal! c) cos cos cos sin sin + + 0 But, cos cos. The are equal! The identit seems to be true since it was valid in all the above cases. Module
Section, Answer Ke, Lesson Principles of Mathematics 5. a) cos cos cos sin sin + 0(cos ) + sin sin b) Use part(a), replacing with. sin cos cos sin cos c) tan cot sin cos Note: 90. Therefore, the identities are valid with either measure 90. or cos(90 ) sin o d) sec(90 ) csc o sin(90 ) cos o e) csc(90 ) sec o tan(90 ) cot o f) cot(90 ) tan o 6. The form of the identities is A B( ) where A and B have names which differ b the prefi co. For eample, sine and cosine are two such functions. Module
Principles of Mathematics Section, Answer Ke, Lesson 7. a) sin ( ) cos( ) cos b) tan ( ) cot( ) cot c) cos + cos ( ) sin( ) sin cos d) sin( )( cot ) sin cot sin cos sin sin e) (sin ) ( csc ) (sin ) sin f) csc sin 8. a) cos ( ) sin co-function related arc: + ( k) k I b) sin cos (co-function) related arc: 0 { ( k) k I} Module
Section, Answer Ke, Lesson Principles of Mathematics c) sin 0 sin 0 cos 0 related arc:, + k k I There are other forms of these answers. This answer ma also be stated in degrees since there were no instructions on the values of. In degrees, the answers matching the above form are: { o + k o k I} { k o k I} { o + k o k I} a) 90 (60 ) b) (60 ) c) 90 (80 ) Module
Principles of Mathematics Section, Answer Ke, Lesson 5 Lesson Answer Ke... 5 a) sin + sin 9 6 b) sin + sin 0 5 5 7 c) cos cos 5 d) cos + cos 9 9 7 sin sin + sin cos + cos sin + + 6 + or 7 cos cos + cos cos sin sin 6 or 7 6 + 6 Therefore, the coordinates are: P,. Module
6 Section, Answer Ke, Lesson Principles of Mathematics. a) cos 05 cos(60 + 5 ) cos 60 cos 5 sin 60 sin 5 o o o o o o o 6 Note: The answers appear to is the rationalized form of. be different. However, the are not; hence, ou can use either form. b) sin 05 sin(60 + 5 ) sin 60 cos 5 + cos 60 sin 5 o o o o o o o 6 + + 6 + o sin 05 6 + c) tan 05 cos 05 6 6 (This is the eas wa after (a) and (b). ) o o o tan 05 tan(60 5 ) The answers are equal! o o tan 60 + tan 5 + o o tan 60 tan 5 + 5. 7 Since sin α and P( α) QI, it follows that 5 P( α) QII, so cos α < 0. 7 Since sin α + cos α, + cos α. 5 Therefore, cos α. 5 9 Similarl, since cos β and P( β) QI, it foll ows that P( β) QIV, so sin β < 0. 9 Since sin β + cos β, sin β +. 0 Therefore, sin β. Module
Principles of Mathematics Section, Answer Ke, Lesson 7 a) sin( α + β) sinαcos β + cos αsin β 7 9 0 6 + 960 0 5 + 5 05 05 b) cos( α + β) cosαcos β sin αsin β 9 7 0 6 + 80 6 5 5 05 05 0 sin( α + β) c) tan( α + β) 05 0 cos( α + β) 6 6 05 05 d) sec( α + β) cos( α + β) 6 6 05 e) In Quadrant I, since both sin( α + β) > 0 and cos( α + β) > 0. 6. 7. 8. cos + θ cos cos θ sin sin θ cos θ sin θ tan θ tan tan θ 6 tanθ tan θ 6 + tanθ tan + tanθ + tan θ 6 a) sin( α + β) + sin( α β) sinαcos β + cosαsin β + sinαcos β cosαsin β sinαcosβ b) Etra for Eperts cos( + )cos( ) (cos cos sin sin )(cos cos + sin sin ) cos cos sin sin cos cos + 0 sin sin cos cos + (cos sin cos sin ) sin sin cos (cos sin ) sin (cos sin ) + + cos () sin cos sin () Module
8 Section, Answer Ke, Lesson Principles of Mathematics 9. a) sin(+ ) sin 5 b) tan( + ) tan c) sin( α + 6 α) sin 7α d) cos cos sin sin cos( + ) cos 6 e) sin 5α cos 5α + sin 5α cos 5α sin(5α + 5 α) sin0α f) cos(a a) cos( a) cos a, since cos is an even f unction Module
Principles of Mathematics Section, Answer Ke, Lesson 9 Lesson Answer Ke. sin A sin A cos A ( sin A cos A)(cos A sin A) The above is one possibilit.. cos θ cos (θ) sin (θ) ( 0.09) (0.95) 0.809 sin θ sin θ cos θ (0.95)( 0.09) 0.588. Therefore, P(θ) ( 0.809, 0.588).. Use the following two forms of cos θ, and replace θ b θ: θ cos θ becomes cos sin θ θ and now solve for cos and sin. θ θ cos θ cos cos θ sin θ θ 0.588 cos 0.588 sin.588 θ θ.588 cos sin θ θ cos 0.89 sin 0.5 cos cos θ θ sin θ θ (Note: Both positive since θ θ I, then I.). a) cos 8 (Using the identit cosα cos α sin α.) b) sin (Using the identit sinα sin α cos α.) c) sin d) cos 0α e) cos 0α sin α cosα sin α f) sin α cos α Module
50 Section, Answer Ke, Lesson Principles of Mathematics 5. 6. 7. 5 0 cos cos (0.95) 0.805 Since 0 < θ < and sin θ, it follows that 5 cos +, so cos with θ θ θ I. 5 5 sin θ sin θ cos θ and 5 5 5 8 7 cos θ sin θ. cos θ cos θ (cos θ ) ( p ) + ( p p ) + 8p 8p 5 5 5 8. Since sin α > 0 and not in Quadrant I, it follows P(α) is in Quadrant II. Similarl, since cos β > 0 and not in Quadrant I, it follows that P(β) is in Quadrant IV. cos α. Hence, in II cos α and + 5 5 5 sin β. Hence, in IV sin β. + a) sin α sin α cos α 5 5 5 5 b) cos( α β) cos α cos β + sin α sin β + 5 5 5 8 6 65 65 5 5 9 c) cos β cos β sin β 69 69 Module
Principles of Mathematics Section, Answer Ke, Lesson 5 d) sin( α + β) sin α cos β + cos α sin β 5 0+ 6 56 5 + 5 65 65 9. a) (sin + cos ) sin + sin cos + cos + sincos + sin b) sin cos sin cos cos sin 0. sin cos sin cos + a) LHS + cos sin cos sin cos sin csc cos sin sin b) LHS (cos sin )(cos sin ) + ()(cos sin ) cos sin c) LHS sec sin cos cos d) LHS cos cos cos cos cos cos sec cos e) sin( + ) sin cos + cos sin + ( sin cos ) cos ( sin )sin sin cos + sin sin sin ( sin ) + sin sin sin sin + sin sin sin sin Module
5 Section, Answer Ke, Lesson Principles of Mathematics f) cos( + ) cos cos sin sin ( cos ) cos sin cos sin cos cos cos sin cos cos cos ( cos ) cos cos cos + cos cos cos Module
Principles of Mathematics Section, Answer Ke, Review 5 Review Answer Ke. sin a) sin cos sin sin cos b) sin (sin + cos ) sin () sin c) sin cos sin cos d) cos cos ( sin ) sin cos + sin e) (sin cos ) sin8 f) sin(5 ) sin. a) LHS cos + sin + sin + sin RHS b) LHS (cos sin )(cos sin ) + (cos ) cos RHS c) LHS + sin θ + sin θ ( + sin θ)( sin θ) sin θ cos θ sec θ (tan θ + ) tan θ + RHS d) LHS θ + + θ (tan ) (csc ) θ + θ tan csc RHS Module
5 Section, Answer Ke, Lesson Principles of Mathematics e) LHS ( sin cos )sin cos sin sin cos sin sin ( cos ) sin cos RHS f) RHS sin ( sin ) + sin cos LHS. a) The third side is a + 5 or a. Furthermore, since sin θ < 0 and tan θ < 0, it follows that θ QIV. Therefore, cos θ tan θ csc θ sec θ cot θ 5 5 5 b) The third side is a + 5 or a. Furthermore, since cos > 0 and csc < 0, it follows that QIV. Therefore, sin tan csc sec cot 5 5 5 c) The third side is + a or a 5. Furthermore, since tan θ < 0 and cos θ < 0, it follows that θ QII. Therefore, cos θ sin θ csc θ sec θ cot θ 5 5 5 5 d) The third side is 8 + 5 a or a 7. Furthermore, since cot > 0, it follows that QI or QIII. Therefore, sin cos csc sec tan ± 5 7 ± 8 7 ± 7 5 ± 7 8 5 8 Module
Principles of Mathematics Section, Answer Ke, Lesson 55. Since sin α and P( α) QI, it follows that P( α) QII, 5 so cos α < 0. Since sin α + cos α, + cos α. 5 Therefore, cos α. 5 5 Similarl, since cos β and P( β) QI, it foll ows that P( β) QIV, so sin β < 0. Since sin β + cos β, sin β +. Therefore, sin β. 5 5 a) sin( α + β) sin α cos β + cos α sin β + 5 5 5 + 8 6 65 65 5 b) cos( α + β) cos α cos β sin α sin β 5 5 0 + 6 6 65 65 c) sin α sin α cos α 5 5 5 d) Since both sin( α + β) and cos( α + β) are positive, it follows that ( α + β) terminates in Quadrant I. 5. a) sin cos sin 0 sin ( cos ) 0 sin 0, cos o o o o 0, 80, 60, 00 Module
56 Section, Answer Ke, Lesson Principles of Mathematics b) sin cos + cos 0 cos ( sin + ) 0 cos 0, sin 90, 70, 0, 0 o o o o 6. a) sin sin sin cos cos sin 6 6 6 6 b) Use the identit cos cos where. 8 8 8 8 + cos cos cos cos cos 8 cos + cos 8 + cos 8 (positive, since is in Quadrant I) 8 8 Module
Principles of Mathematics Section, Answer Ke, Review 57 7. Etra for Eperts 5 9 The sketch can be used to fill in the following: Domain Range Amplitude -intercept Period R, cos + sin r + cos + sin cos Module
58 Section, Answer Ke, Review Principles of Mathematics Notes Module
Principles of Mathematics Section, Answer Ke, Lesson 59 Lesson Answer Ke. Question Period Frequenc Phase Shift (a) (b) (c) none none (d). a) 6 6 b) 5 5
60 Section, Answer Ke, Lesson Principles of Mathematics c) 6 5 d). a) d 8. 9..5 6.5 6.5.5 t b) Period is:.5 hours. 0.6 Module
Principles of Mathematics Section, Answer Ke, Lesson 6 c).5 cos(0.6 t) +.7.5.5 cos(0.6 t) 0.8 0.8 cos(0.6 t) 0.7778.5 Related arc is: cos (0.7778).9. Therefore, (.9) 0.6 t ( +.9).8 t 9.0. Hence, the carrier can dock safel for (9. 0.8) 5.5 hours. d 8..5 9..8 9.0 t t 0.7778 Module
6 Section, Answer Ke, Lesson Principles of Mathematics Lesson Answer Ke. 5 cos 6 cos 8, 0 < Y 5[cos()] 6 cos() 8 Y Y Y [ 0., 7] [, ] Solutions:.50,.79 X min X ma Y min Y ma. tan tan +, 0 < Y [tan()] tan() Y Y Y [ 0., 7] [ 0, 0] Solutions: 0.9,.6,.07, 5.50 X min X ma Y min Y ma. The sstem: sin tan and, 0 < Y [sin()] tan() Y Y Y Solutions: (0.7, ) (.5, ) [ 0., 7] [ 0, 0] The net solution (6.65, ) X min X ma Y min Y ma is out of the specified domain. Module
Principles of Mathematics Section, Answer Ke, Lesson 6 Lesson Answer Ke cos cos 0. Solving over the domain cos (cos ) 0 0 θ < cos 0, and cos 5, Then add multiples of the period until ou reach 6 5 7 + + 7 and 5 + + 9 + + + + 5 7 + + + +. Solve tan 0 over the domain 0 θ < tan tan, then add multiples of the period. + n + n where n integers. These two answers can be combined to one answer + n, where n integers. Module
6 Section, Answer Ke, Lesson Principles of Mathematics. Solve tan + tan 0 over the domain 0 θ < tan (tan + ) 0 tan 0 0 and tan 7, Add multiples of the period 0 + n n + n + n 7 + n. Solve cos sin 0 over the domain ( ) sin sin sin sin 0 sin 0 sin 0 ( sin + )( sin ) 0 0 θ < sin + 0 sin sin 7, 6 6 and sin 0 sin which has no solution Add multiples of the period where n integers. 7 + n 6 + n 6 Module
5. Solve a simpler version of the problem b solving over the domain Now replace so the solutions become and Lastl, add multiples of the period. Since the period of is, the final solutions are where n integers. 6. b n b n. 0.76 + + n n 5 + + cos 5 5 5 θ 5, cos cos 0 cos θ θ θ θ + 0 θ < 0 cos + θ Principles of Mathematics Section, Answer Ke, Lesson 65 Module
66 Section, Answer Ke, Review Principles of Mathematics. a) b) c) cos A csc A tan A 7 7 Review Answer Ke 7. a).6 80 b) 85 c).. a) ( ) 5 5 5 5 b) or 90. θ 60 58.9 0. 5. a).87 b) 5 cos cot sin cos sin 5 5 ( + 6 ) ( + ) 6 6 6 6 + 6 cos cos sin sin sin cos cos sin + 6 + 6 Module
Principles of Mathematics Section, Answer Ke, Review 67 c) 6. a) b) csc sin sin cos cos 0 ( ) cos cos 0 cos 0 or cos, cos is undefined. sin θ sinθ 0 ( θ )( θ ) sin + sin 0 sinθ or θ.8 or 8. or 90 θ.8 + 60 nor 8. + 60 n or 90 + 60 n where n integer c) tan sin 0 sin sin 0 cos sin 0 cos sin 0 or 0 cos sin 0 or cos 0,,, or 6 6 Module
68 Section, Answer Ke, Review Principles of Mathematics 7. a) tan θ cos θ + sin θ sin θ cos θ + sin θ cos θ sin θ + sin sin θ b) sin θcos θ is a version of the double angle identit sin A cos A sin A so sinθcosθ ( ( θ) cos ( θ) ) ( θ ) sin sin θ sinθ csc θ sinθ c) cotθ sinθ cosθ sinθ sinθ Multipl numerator and denominator b si nθ sin θ cos θ cos θ cosθ cosθ cosθ sinθ d) + + tan θ cot θ sin θ cos θ cos θ + sin θ sinθcosθ sinθcosθ Note: It is possible to simplif further using a double angle identit. cscθ sinθ cosθ sin θ ( ) Module
Principles of Mathematics Section, Answer Ke, Review 69 8. amplitude period phase shift vertical displacement 5 domain R [ ] range,8 8 5 8 9. period phase shift 5 cos has zeros at,,, etc. 8 8 8 5 sec has asmtotes at,,, etc. 8 8 8 ( k + ) domain, k 8 range (, ] [, ) 0. a) sin75 cos5 cos75 sin5 ( ) sin 75 5 sin50 Module
70 Section, Answer Ke, Review Principles of Mathematics b) sin cos cos 6. a) Prove RHS Factoring cos as a difference of squares. sec sec + cos sin + cos cos + cos cos cos + cos cos cos cos cos cos sec cos b) Prove + cot θ cosθ LHS + sin ( + )( ) ( ) ( Double angle identit ) θ + sin θ sin θ csc θ ( ) + cot θ Pthagorean Identit RHS Module
Principles of Mathematics Section, Answer Ke, Review 7.a) ma 7. m min.8 m ais occurs at 7. +.8 5.6 m 7. +.8 amplitude.8 m period (:5 5:0) (6:5) :0.5 hours b 0.6 period.5 5 Since the ma occurs at 5:0 a.m., we can use a cosine function to model the tide flow. phase shift 5.5 hours b) 0:0 p.m. is.5 hours from midnight h(.5).8 cos 0.6 (.5 5.5) + 5.6.5 m ( t ) ht ( ).8cos 0.6 5.5 + 5.6 c) ( t ).8 cos 0.6 5.5 + 5.6 6.0 0. cos 0.6 ( t 5.5).8 0. t 5.5 + cos 0.6.8 8.8 hours or 8: a.m. h 7. 5.6.8 5:0 :5 7:00 :00 a.m. t Module
7 Section, Answer Ke, Review Principles of Mathematics Using the graphing calculator Set Y.8 cos (0.6( 5.5)) + 5.6 Set Y 6.0 Solving we get 8.8 or 5. or 0.68 The net time after.5 a.m. is.9 p.m. (5 hours. 0. hours 9 min).set Y tan θ sin θ Set Y + tan θ.85, 6.078 [ 0., 7] [, 0] Solutions: 0.89,.85, X min X ma Y min Y ma Another solution,using one graph: Set Y tan θ sin θ tan θ.85, 6.078 [ 0., 7] [, 0] Solutions: 0.89,.85, X min X ma Y min Y ma Module