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Alpha Trigonometry Solutions 2013 FAMAT State Convention 5 2 2 1. sin . arcsin since the range of the arcsine function is , . A 4 2 4 2 2 2 2. If is an acute angle in a right triangle then the length of the adjacent side is 5 while the length of the hypotenuse is 8. The length of the opposite side is therefore 39 . The cotangent of is the length of the adjacent side divided by the length of the opposite side, and hence 5 5 39 . B cot 39 39 3. The area of the triangle will be maximized when the side of length 2 is the smallest of the three sides. This will occur when this side is the side opposite the angle of 30 . The length of 1 the other leg is therefore 2 3 and hence the area is A 2 2 3 2 3 . E 2 1 1 4. y csc 2 x will have a vertical asymptote whenever 2 3 2sin 2 x 3 k . sin 2 x 0 . This will occur when 2 x k , where k is an integer. Then x 3 3 5 Of the choices, this will only occur when x . D 6 3 5. 1 sec 2 tan 2 sec tan sec tan sec tan and hence 2 2 sec tan . B 3 3 6. If 2 sin 2 x 3 , then sin 2 x . Either 2 x 2 k and hence 6 3 6 6 2 2 x k or 2 x 2 k and hence x k . The solutions in 0, 2 are 12 6 3 4 13 5 13 5 1 3 13 15 32 8 , therefore x , , . . D 12 4 12 4 12 4 12 4 12 12 3 3 7. If the terminal side of 6 lies in the third quadrant, then 6 and hence . 2 6 4 1 2 0.707 . The only choice in this interval is 0.59. C Then sin 2 2 2013 2013 i i 2013 2013 8. cos i sin e 6 e 6 cos i sin 6 6 6 6 2013 2013 3 3 cos i sin cos i sin i . A 6 6 2 2 13 20 13 1 400 20 2 9. 213'20'' is equal to 2 degrees. Converting this to 60 3600 60 180 180 9 20 . B radians, 9 180 81 10. The graphs of y cos x and y x intersect at only one point. B 2013 Page 1 of 3 Alpha Trigonometry Solutions 11. The largest angle is the angle opposite the side of length 2013 FAMAT State Convention 5 . Using the Law of Cosines, 2 and hence C 135 . C 2 12. The point P x, y on the unit circle is represented in terms of as cos , sin , so 5 2 12 2 2 2 1 2 cos C . cos C y sin . A 13. is perpendicular to the radius of the circle, which has length 1. The radius and the line segment PQ therefore form the legs of a right triangle. tan PQ 1 and hence the length of PQ is given by tan . D 14. x r cos and y r sin , where r 2 x 2 y 2 . We than have that r 2 4 12 16 and y hence r 4 . Notice that tan 3 . Noticing that 2, 2 3 is in the third quadrant, we x have 240 . The polar coordinates are therefore 4, 240 . E 15. The horizontal asymptotes of y arctan x correspond to the vertical asymptotes of y tan x whose domain is restricted to , . These vertical asymptotes are x and 2 2 2 x . y is therefore a horizontal asymptote. D 2 2 16. The range of sin 3 x is [1,1] , so the largest 4sin 3 x can be is 4. The largest 12 12 possible value of f x 4sin 3 x 2 is therefore 2. C. 12 sin cos sin 2 2sin cos cos2 sin 2 sin 2 cos2 . Since 2 2 5 sin 2 and sin 2 cos 2 1 , sin cos . Therefore 3 3 5 15 . However, since is a first quadrant angle, sin and cos must be sin cos 3 3 positive and hence we disregard the negative solution. B 2 18. If tan , the angle is in an acute triangle with an opposite side of length 2 and an 7 adjacent side of length 7, and hence a hypotenuse of length 53 . The secant of this angle would 17. 2 53 , however since is a second quadrant angle, the secant of will be negative. A 7 19. If two angles of the triangle 60 and 45 , then the third is 180 60 45 75 . The longest side will be opposite the largest angle and the shortest side will be opposite the smallest 2 c angle. By the Law of Sines , and hence c 2 2 sin 75 . Notice that sin 45 sin 75 2 6 . 75 30 45 , so sin 75 sin 30 45 sin 30 cos 45 sin 45 cos 30 4 Therefore c 1 3 . C be Page 2 of 3 Alpha Trigonometry Solutions 20. 89 89 k 1 k 1 sin k 2013 FAMAT State Convention 89 tan k cos k k 1 cos 90 k cos89 cos88 cos k cos1 cos 2 21. sec2 3sec 2 sec 2sec 1 0 . sec 1 0 0 . cos 2 cos1 1. D cos88 cos89 sec 2 0 3 and B 1 1 . B 3 3 2 2 1 cos 2sin 2 . D 23. versin 1 cos 2 2 2 22. versin 24. 1 cos 1 csc2 tan 2 1 csc2 sec2 cot csc tan sin cos cos cos 2 2 csc2 csc csc cos 1 tan 2 cos sec 2 sec . D 16 . As varies the curve is therefore a circle of 2 2 radius 4 starting at 4, 0 moving counterclockwise to the point 4,0 . The curve is therefore 25. Notice that x 2 y 2 16 cos 2 16 sin 2 half of a circle of radius 4 and hence the arc length is 4 . C u v 26. Let be the acute angle u and v. Then cos . Since u v 2 , u 0 , and u v 3 . u lies in the fourth quadrant. v 0 we must have that cos 0 . As a result, 2 2 Therefore v can lie in either the first, second, or third quadrants but not the fourth. D 2 2 3 4x 27. The period of 7sin is while the period of 2cos 9x is . The period of 4 9 2 3 3 the sum of these two functions will be the least common multiple of these individual periods. 4x Since the denominators of the periods are relatively prime, the period of 7sin 2cos 9 x 3 will occur at an integer multiple of . This will occur at 6 . C 28. The range of the arcsine function is the domain of the (restricted) sine function. The restricted sine function has domain , , and hence the domain of f ( x) arcsin x is also 2 2 2 , 2 . E 3 1 cos x sin x 2 3 cos x , the maximum 29. Notice that 3cos x 3 sin x 2 3 2 6 2 value of which is 2 3 . C 30. Since the cosecant has a period of 360 , csc 2130 csc 2130 6 360 csc 30 2 . A Page 3 of 3