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5.1 Angles8. 215π18=37.525 units 215π18=37.525 units 11. 1655 kilometers per hour 5.2 Unit Circle: Sine and Cosine Functions1. cos(t)=−2 2,sin(t)=22cos(t)=−22,sin(t)= 22 2. cos(π)=−1,cos(π)=−1, sin(π)=0sin(π)=0 3. sin(t)=−725 sin(t)=−725 4. approximately 0.866025403 6.
7. (12,−32)(12,−32) 5.3 The Other Trigonometric Functions1. sint=−22sint=−22 , cos t=22cost=22, tant=−1tant=−1, sect=2sect=2, csct=−2csct=−2, cott=−1cot t=−1 2. sinπ3=32sinπ3=32, cosπ3=12 cosπ3=12, tanπ3=3tanπ3=3, secπ3=2secπ3=2, cscπ3=233 cscπ3=233, cotπ3=33cotπ3=33 3. sin(
−7π4)=22,cos(−7π4)=22,tan(−7π4
)=1,sin(−7π4)=22,cos(−7π4)=2
2,tan(−7π4)=1, 7. cost=−817,sin
t=1517,tant=−158cost=−817,sint=
1517,tant=−158 8. sint= −1,cost=0,tant=Undefinedsect= Undefined,csct=−1,cott=0 sint=−1,cost=0,tant=Undefinedsect= Undefined,csc t=−1,cott=0 9. sect=2,csct =2,tant=1,cott=1sect=2,csct=2,tan t=1,cott=1 5.4 Right Triangle Trigonometry2. sint= 3365,cost=5665,tant=3356,sect=6556,csct=6533,cott=5633sint=3365,cost=5665,tant=3356, sect=6556,csct=6533,cott=5633 3. sin
(π4)=12,cos(π4)=12,tan(π4)=1,sin(π4)=12,cos(π4)=12,tan(π4)=1
, 5. adjacent =10;adjacent=10; opposite=103opposite=103; missing angle is π6π6 5.1 Section Exercises3. Whether the angle is positive or negative determines the direction. A positive angle is drawn in the counterclockwise direction, and a negative angle is drawn in the clockwise direction. 5. Linear speed is a measurement found by calculating distance of an arc compared to time. Angular speed is a measurement found by calculating the angle of an arc compared to time. 19. 4π3 4π3
21. 2π32π 3 23. 7π2≈11.00 in27π2≈11.00 in2 25. 81π20≈12.72 cm281π20≈12.72 cm2 41. 5.02π3≈5.265.02π3 ≈5.26 miles 43. 25π 9≈8.7325π9≈8.73 centimeters 45. 21π10≈6.60 21π10≈6.60 meters 61. 7 in./s, 4.77 RPM, 28.65 deg/s 63. 1,809,557.37 mm/min =30.16 m/s1,809,557.37 mm/min=30.16 m/s 5.2 Section Exercises1. The unit circle is a circle of radius 1 centered at the origin. 3. Coterminal angles are angles that share the same terminal side. A reference angle is the size of the smallest acute angle, t,t, formed by the terminal side of the angle tt and the horizontal axis. 5. The sine values are equal. 35. 60°,60°, Quadrant IV, sin(300°)=−32,cos(300°)=12 sin(300°)=−32,cos(300°)=12 37. 45°,45°, Quadrant II, sin(135°)=22,sin(135°)=22, cos(135°)=−22cos(135°)=−22 39. 60°, 60°, Quadrant II, sin(120°)=32,sin(120°)=32, cos(120°)=−12cos(120°)=− 12 41. 30°,30°, Quadrant II, sin(150°)=12,sin(150°)=12, cos(150°)=−32cos(150°)=−32 43. π6,π6 , Quadrant III, sin(7π6)=−12,sin(7π6)=−12, cos(7π6)=−32cos(7π6)=−32 45. π4,π4 , Quadrant II, sin(3π4)=22,sin(3π4 )=22, cos(4π3)=−22cos (4π3)=−22 47. π3,π3 , Quadrant II, sin(2π3)=32,sin(2π3 )=32, cos(2π3)=−12cos(2π3)=−12 49. π4,π4 , Quadrant IV, sin(7π4)=−22,sin(7π 4)=−22, cos(7π4)=22 cos(7π4)=22 55. (−10,103) (−10,103) 57. (–2.778,15.757)(–2.778,15.757) 61. sint=12,cost=− 32sint=12,cost=−32 63. sint=−22 ,cost=−22sint=−22,cost=−22 65. sin t=32,cost=−12sint=32,cost= −12 67. sint=−22,cost=22sint=−22 ,cost=22 69. sint=0,cost =−1sint=0,cost=−1 71. sint=−0.596,cos t=0.803sint=−0.596,cost=0.803 73. sint=12,cost=32sint=12,cost=32 75. sint=−1 2,cost=32sint=−12,cost=32 77. sin t=0.761,cost=−0.649sint=0.761,cost=−0.649 79. sint=1,cost =0sint=1,cost=0 103. 37.5 seconds, 97.5 seconds, 157.5 seconds, 217.5 seconds, 277.5 seconds, 337.5 seconds 5.3 Section Exercises1. Yes, when the reference angle is π4 π4 and the terminal side of the angle is in quadrants I and III. Thus, at x=π4,5π4,x=π4 ,5π4, the sine and cosine values are equal. 3. Substitute the sine of the angle in for yy in the Pythagorean Theorem x2+y2=1.x2+y2=1. Solve for x x and take the negative solution. 5. The outputs of tangent and cotangent will repeat every ππ units. 39. If sint=−223 sint=−223, sect=−3sect =−3, csct=−324csct=−324, tant=22tant=22, cott=24 cott=24 41. sect=2sec t=2, csct=233csct=233 , tant=3tant=3, cott=3 3cott=33 49. sint=22,cos t=22,tant=1,cott=1,sect=2,csct=2 sint=22,cost=22,tant=1,cott=1,sec t=2,csct=2 51. sint=−32 sint=−32, cost=−12cost =−12, tant=3,cott=33tant=3 ,cott=33, sect=−2sect=−2, csct=−233csct=−233
63. sin(t)≈0.79 sin(t)≈0.79 71. sintcost=tantsintcost= tant 73. 13.77 hours, period: 1000π1000π 5.4 Section Exercises3. The tangent of an angle is the ratio of the opposite side to the adjacent side. 5. For example, the sine of an angle is equal to the cosine of its complement; the cosine of an angle is equal to the sine of its complement. 11. b=2033,c =4033b=2033,c=4033 13. a=10,000,c=10,000.5a=10,000,c=10,000.5 15. b=533,c=10 33b=533,c=1033 29. c=14,b=73 c=14,b=73 33. b=9.9970,c=12.2041b=9.9970,c=12.2041 35. a=2.0838,b=11.8177a=2.0838,b=11.8177 37. a=55.9808,c=57.9555a =55.9808,c=57.9555 39. a=46.6790,b=17.9184a=46.6790,b=17.9184 41. a=16.4662,c=16.8341a =16.4662,c=16.8341 Review Exercises15. 1036.73 miles per hour 35. 2 222 or −22−22 37. sine, cosecant, tangent, cotangent 45. 11157157 11157157 Practice Test9. 3.351 feet per second, 2π752π 75 radians per second 23. a=92,b=932a=92,b=932 What is the formula for 30 60 90 Triangle?The sides of a 30-60-90 triangle are always in the ratio of 1:√3: 2. This is also known as the 30-60-90 triangle formula for sides y: y√3: 2y. Let us learn the derivation of this ratio in the 30-60-90 triangle proof section. This formula can be verified using the Pythagoras theorem.
What is SOH CAH TOA?"SOHCAHTOA" is a helpful mnemonic for remembering the definitions of the trigonometric functions sine, cosine, and tangent i.e., sine equals opposite over hypotenuse, cosine equals adjacent over hypotenuse, and tangent equals opposite over adjacent, (1) (2)
How do you find the missing side of a triangle?Given two sides. if leg a is the missing side, then transform the equation to the form when a is on one side, and take a square root: a = √(c² - b²). if leg b is unknown, then. b = √(c² - a²). for hypotenuse c missing, the formula is. c = √(a² + b²). |