Math 103 College Algebra and Trigonometry Exam 4 review answers Answer 1 Several of these answers may be written in different forms Two equivalent forms of the answer are given below where appropriate Other forms of these answers may be possible 3 c 1 d 1 b 3 a 2 p s 1 4 1 2 2 2 3 6 2 f g h i 2 2 2 4 2 6 2 1 23 j 45 4 n k 5 150 6 2 120 3 q o 2 2 1 2 2 60 m 7 3 2 1 p 2 2 p 2 2 r 2 l Answer 2 Quadrant IV Answer 3 sin 65 72 65 72 97 97 cos tan cot sec csc 97 97 72 65 72 65 Answer 4 Function Domain Range Even odd sin cos tan All real numbers except 2 k for integers k 1 1 1 1 Odd Even Odd cot sec csc All real numbers except 2 k for integers k All real numbers except 2 k for integers k All real numbers except k for integers k 1 1 1 1 Odd Even Odd Answer 5 While not strictly an error there was a typo in the problem on the question sheet Writing the inverse trigonometric functions as sin 1 cos 1 etc is misleading since usually represents an angle but the arguments to the inverse trigonometric functions are not angles The outputs of the inverse trigonometric functions are angles not the inputs It would have been better to have written sin 1 x cos 1 x etc Function Domain Range sin x cos 1 x tan 1 x 1 1 1 1 2 2 0 2 2 cot 1 x sec 1 x csc 1 x 1 1 1 1 2 2 0 2 2 2 0 0 2 1 Answer 6 The amplitude is 8 and the period is Page 1 6 5 Answer 7 An equation for the graph on the left is y 3 sin x graph on the right is y cos 3x 2 2 An equation for the Answer 8 Answer 9 Function y intercept sin x cos x tan x 0 1 0 cot x sec x csc x no y intercept cot x is undefined at x 0 1 no y intercept csc x is undefined at x 0 cos sec tan 1 sin cos2 1 1 sec cos2 cos Answer 11 Other methods are possible Answer 10 a cos tan cot cos sin cos cos sin cos2 sin2 sin cos sin cos 2 sin cos2 cos sin cos 1 cos sin cos cos sin cos 1 sin csc cos b c cos cos cos sin sin cos cos cos cos cos cos sin sin cos cos cos cos sin sin 1 cos cos 1 tan tan 9 sec2 5 tan2 4 sec2 5 sec2 5 tan2 4 sec2 5 sec2 tan2 4 sec2 5 1 5 4 sec2 Page 2 d sec2 u sin2 u sec2 u 2 sec2 u sin2 u sec2 u 2 sin2 u sec2 u 1 sin2 u 2 sin2 u 1 cos2 u 2 sin2 u cos2 u 2 cos u 2 sin2 u cos2 u 1 2 sin2 u e sin cos sin cos sin sin cos cos2 sin cos sin2 sin cos cos2 sin cos sin2 cos2 sin cos sin cos sin cos 2 1 cos sin cos2 sin2 sin2 sin2 2 cos sin 2 cos sin 2 sin cos 1 sin2 cos2 Answer 12 a Solution set b Solution set c Solution set d Solution set e Solution set 2 5 3 3 2 4 6 8 0 5 5 5 5 2 4 3 2 3 3 2 5 3 4 2 4 2 5 4 4 cos cos2 sin2 sin2 2 cos cos cos 3 g 2 cos 2 3 2 cos 3 2 2 cos 2 2 2 cos 4 2 cos 2 2 cos 2 2 cos 2 cos 2 cos 2 cos sin2 cos2 sin cos 1 sin cos 2 2 cos2 sin2 sin cos sin2 2 cos sin2 2 2 cos sin sin 2 sin2 cos 2 cos sin2 2 sin cos cos 2 sin 2 cot 2 sin2 cos2 sin cos cot 1 2 cot cos sin cos sin 2 f cos 2u tan cot tan cot 2 Page 3 70 9 Answer 13 9 sin 20 3 0782 20 9 cos 20 8 4572 Answer 14 Wichita is approximately 377 km from Grand Island A crow flying directly from Grand Island to Wichita must fly on a bearing of about S10 26 E that is 10 26 east of south Answer 15 To the nearest tenth of a square inch the area of the shaded region is 17 9 in 2 Answer 16 To the nearest foot the diameter of the wheel was 252 feet Answer 17 a B 110 b 3 6800 c 1 3394 b No solution c c 1 6905 A 65 B 65 d a 3 5128 A 43 78 C 36 22 e Two solutions C 74 62 A 65 38 a 2 8286 or C 105 38 A 34 62 a 1 7676 f A 36 34 B 26 38 C 117 28 g A 30 51 B 59 49 C 90 Answer 18 The area of the triangle is 3 255 11 9765 4 Answer 19 The area of the triangle is approximately 0 2939 m2 or approximately 2939 cm2 or approximately 455 52 in 2 or approximately 3 1633 ft2 3 91 square cubits or approximately 7 1545 square Answer 20 The area of the triangle is 4 cubits Answer 21 The area of the hendecagon is approximately 29 735 Note that the area of the circle is approximately 31 416 so our answer for the area of the hendecagon makes sense since it covers nearly all of the circle Bonus 1 Page 4