Math Chapter 4 Test Review

Chapter 4 Test Review Name:_______________________________________ Hour:_______ HPC For exercises #1-21, find the exact value without the use of a calculator. 1. sin 330° 2. cot 120° 4. sin(−780°) 5. tan (− 13𝜋 ) 3 7𝜋 ) 4 5𝜋 6. csc ( 6 ) 8. csc[tan−1 2.5] 7. cos ( 9. tan(arctan 35) exact answer! 10. sin[arcsin(−1.1)] 13. cos −1 (tan 16. tan−1 (− 3. sec 135° 3𝜋 ) 4 √3 ) 3 3 19. sec (arcsin 5) 11. arccos (cos 7𝜋 ) 2 14. cos[arctan(−1)] 17. sin−1 (− √2 ) 2 20. csc [arctan (− 12. sin−1 (sin 7𝜋 ) 4 𝜋 15. arccos [sin (− 6 )] 18. arccos 0 12 )] 5 21. sec[tan−1 (−1.5)] exact answer! Write an algebraic expression that is equivalent to the expression. 𝑥 22. tan (arccos 2) 23. sec[arcsin(𝑥 − 1)] 24. Find the exact values of the six trigonometric functions of 𝜃 = − Simplify the expression. 25. (sec 2 𝜃 − 1)(csc 2 𝜃 − 1) 26. Graph at least one full cycle of the function. 𝜋 2 28. 𝑦 = 2 cos (𝑥 − 1) Period: Amplitude: Phase Shift: Vertical Shift: 29. 𝑦 = −2 csc 2𝑥 − 3 Period: Amplitude: Phase Shift: Vertical Shift: Asymptotes: tan2 𝜃 sec 𝜃+1 +1 5𝜋 . 6 27. sin 𝜃 cos 𝜃 1−cos2 𝜃 30. 𝑦 = −3 sin 4𝜋 (𝑥 + 1) + 1 Period: Amplitude: Phase Shift: Vertical Shift: 1 𝜋 31. 𝑦 = 2 sec 2 (𝑥 − 2 ) Period: Amplitude: Phase Shift: Vertical Shift: Asymptotes: 32. 𝑦 = −2 cot 2𝑥 + 1 Period: Vert. Stretch/Shrink: Phase Shift: Vertical Shift: Asymptotes: 33. 𝑦 = tan(2𝑥 + 𝜋) − 2 Period: Vert. Stretch/Shrink: Phase Shift: Vertical Shift: Asymptotes: 34. Find a possible trig equation in terms of sine and cosine. Sine: _________________________ Cosine:_________________________ 35. Find a possible trig equation in terms of sine and cosine. Sine: _________________________ Cosine:_________________________ 36. Find the altitude of an isosceles triangle if the base is 10 feet long and the base angle is 42°. 37. From a point 45 feet in front of a church, the angles of elevation to the base of the steeple and the top of the steeple are 35° and 46°20′, respectively. Find the height of the steeple. 38. When an airplane leaves the runway, its angle of climb is 16° and its speed is 300 feet per second. Find the plane’s altitude relative to the runway (in feet) after one minute. Round your answer to the nearest foot. 39. A ship leaves port at noon and has a bearing of 𝑆 27° 𝑊. The ship sails at 25 knots. How many nautical miles south will the ship have traveled by 4:00 p.m.? Round your answer to the nearest hundredth. 40. A plane is 57 miles west and 42 miles north of an airport. The pilot wants to fly directly to the airport. What bearing should the pilot take? 41. A sailboat leaves a pier and heads due west at 8 knots. After 15 minutes, the sailboat tacks, changing course to 𝑁 16° 𝑊 at 10 knots. Find the sailboat’s bearing and distance from the pier after 12 minutes on this course. 42. A regular hexagon is inscribed in a circle of radius 24 inches. Find the length of the sides of the hexagon. 43. An airplane flying at 600 miles per hour has a bearing of 52. After flying for 1.5 hours, how far north and how far east has the plane traveled from its point of departure? 44. A plane is 160 miles north and 85 miles east of an airport. The pilot wants to fly directly to the airport. What bearing should be taken? 45. A passenger in an airplane flying at an altitude of 10 km sees two towns directly to the east of the plane. The angles of depression to the towns are 28° and 55°. How far apart are the towns? Answer Key 1 2 3 3 3.  2 4.  8. 29 5 9. 35 2.  2 7. 1 2 11.  2 12.   4 13.  14. 16.  17.   4 18.  2 19. 1.  6.  6 13 2 21. 25. 1 34. Sine: Cosine: 22. 26. 4 x x 2 23. sec 27. y  2 sin 2 x 4.5 feet 37. 40. S 53.7 E or BEARING 126.3 42. 24 in. 44. 207.98 or S 27.98 W 43. 15.64 feet  x 2  2x cot  5 4 1 2 24. 3 2 3 tan  3 cos    5. 1 10. No solution 15. 2 3 20.  13 12 csc   2 2 3 3 cot   3 sec    28 through 33. see online key for graph 1 5  y  2 sin  x   3 2  1 Cosine: y  2 cos x    3 35. Sine: 38. 4961 feet 41. N 53 W , 3.2 nautical miles (nm) 554.10 miles north, 709.21 miles east 45. 2 2 sin    1   y  2 cos 2 x   4  36. 3 2 11.81 km 39. 89.10 nm