sin (0 +x) =sin (x) =0 where xis an integer. The measures of the angles for which the sine is equal to 1 are sin (____2+2x), where xis an integer.5. 45° or __47. ± 42.27°+(360°)k, where kis an integer9. =0.25 radians and 2.89 radians11. There is no angle with cosine -1.5 because -1.5 is not an element of the domain of cosine. 13. The student gave values for =cos -1(1__2), rather than sine. The correct answer is =__6+2nor 5__6+2n. 15. sec -1(1__2)does not exist, since sec x=1____cos xand cos xcannot equal 2. 17. sin = ±1 19. 0 x21. 60° or __323. 120or 2___325. 129.0°+(360°)k, 231.0°+(360°)k, where kis an integer 27. -32.2°+(180°)k, 147.8°+(360°)k, where kis an integer29. 4___3, 5___331. __6, 5___6, 3___233. a.t=1__cos Ũ±(h-6_____2)b.0.67 s 35. angle of elevation =42° 37. a.no b. yes c.yes d.yes e.yes f.no 39. Part A =65.4°Part B d=610.5 ftPart C 254.3 ftLesson 8-21. You can use the Law of Sines and the Law of Cosines in non-right triangles to find the measures of sides and angles when you have a side and its corresponding angle as well as another side or angle. 3. The negative square root is not valid because ais a side length and side lengths cannot be negative. 5. about 40.3° 7. about 32.3 9. about 10.9 11. about 97.6° 13. Angle-Side-Side 15. Neither student is incorrect; they reported their answers to different degrees of precision. Since the given values are in integers, 20 in. is probably precise enough. 17. Inverse sine gives an acute angle, and the supplement to that angle will always be obtuse. This causes a problem when the known angle is acute because both the acute angle and its obtuse supplement EQWNF DG XCNKF QRVKQPU HQT VJGmissing angle. However, if the known angle is obtuse, only the acute UQNWVKQP KU XCNKF DGECWUG C VTKCPINGECPPQV EQPVCKP VYQ QDVWUG CPINGU²19. Draw the altitude from E, label its length x. sin F=x__gand sin G=x__fgsin F=xand fsin G=xgsin F=fsin G. Therefore, sin F__f=sin G__genVision™Algebra 2|43| Selected Answers

Selected AnswersTopic 8PearsonRealize.com21. about 6.9 23. two: mE61.7°or 118.3°25. Sample: Student draws altitude outside triangle from vertex L. cos(180°-K) =-cos K=x__jx=-jcos K x 2+h 2=j 2and (l+x) 2+h 2=k 2l 2+2lx+x 2+h 2=k 2k 2=l 2+2l(-jcos K) +j 2k 2=l 2+j 2-2lj(cos K) 27. about 96.4°29. about 58.7°31. about 93.6 in. 33. about 55.6° 35. no; no; no; yes; yes; no 37. Part A240 ft Part Babout 55.2º Part Cabout 282.6 ftLesson 8-31. Relationships that exist between trigonometric functions can be verified and applied through the use of the unit circle and its definitions of sine, cosine, and tangent. 3. Because the equation cos(-x) =EQUxis true for all values of the variable for which both sides of the equation are defined. 5. The value of the tangent is undefined if the cosine is zero. The cosine is zero for =__2+ k, where kis an integer.7. sec cot csc 1_____cos cos _____sin csc 1____cos cos _____sin csc 1____sin =csc 9. tan 11. ū__2 -ū__6 _______413. Yes; cos 2=cos( +) =cos cos -sin sin =cos 2-sin 215. The student applied the cosine of sums formula rather than the cosine of differences.