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Circular Trigonometric Functions

Circular Trigonometric Functions. Special Angles. *Special Angles. 30 ° , 45°, and 60° → common reference angles Memorize their trigonometric functions. Use the Pythagorean Theorem; triangles below. 60 °. 45 °. 2. 1. 1. 45 °. 30 °. 1. * Special Angles. * Special Angles.

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Circular Trigonometric Functions

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  1. Circular Trigonometric Functions

  2. Special Angles

  3. *Special Angles 30°, 45°, and 60° → common reference angles Memorize their trigonometric functions. Use the Pythagorean Theorem; triangles below. 60° 45° 2 1 1 45° 30° 1

  4. *Special Angles

  5. *Special Angles

  6. Find trig functions of 300° without calculator. Reference angle is 60°[360° - 300°]; IV quadrant sin 300° = - sin 60° cos 300° = cos 60° tan 300° = - tan 60° csc 300° = - csc 60° sec 300° = sec 60° cot 300° = - cot 60° 60° 300° Use special angle chart.

  7. sin 300° =- sin 60° = - 0.8660 = cos 300° = cos 60° = 0.5000 = 1/ 2 tan 300° = - tan 60° = -1.732 = csc 300° = - csc 60° = -1.155 = sec 300° = sec 60° = 2.000 = 2/1 cot 300° = - cot 60° = - 0.5774 =

  8. Quadrant Angles

  9. *Quadrant Angles • Reference angles cannot be drawn for • quadrant angles 0°, 90°, 180°, and 270° • Determined from the unit circle; r = 1 • Coordinates of points (x, y) • correspond to (cos θ, sin θ)

  10. *Quadrant Angles 90° (0,1) → (cos θ, sin θ) r = 1 180° (-1,0) 0° (1,0) 270° (0,-1)

  11. *Quadrant Angles

  12. Find trig functions for - 90°. Reference angle is (360° - 90°) → 270° sin 270° = -1 cos 270° = 0 tan 270° undefined csc 270° = -1 sec 270° undefined cot 270° = 0 -90° Use quadrant angle chart. 270°

  13. Coterminal Angles

  14. *Coterminal Angle The angle between 0º and 360º having the same terminal side as a given angle. Ex. 405º- 360º = coterminal angle 45º θ1 = 405º θ2 = 45º

  15. *Coterminal Angles Used with anglesgreater than 360°, or angles less than 0°. • Example • cos 900° = • cos (900° - 720°) = • cos 180° = -1 • (See quadrant angles chart)

  16. Example • tan (-135° ) = • tan (360° -135°) = • tan 225° = • LOOK→ tan (225° - 180°) • tan 45° = 1 • (See special angles chart)

  17. Find the value of sec 7π / 4 • Convert from radian to degrees: • sec [(7π/ 4)(180/ π)] = sec 315°

  18. Angle in IV quadrant: sec →positive • sec (360° - 315°) = sec 45° • = 1 /(cos 45°) = √2 = 1.414 • Look how this problem was worked • in previous lesson.

  19. Practice

  20. Express as a function of the reference angle and find the value. tan 210° sec 120 °

  21. Express as a function of the reference angle and find the value. sin (- 330°) csc 225°

  22. Express as a function of the reference angle and find the value. cot (9π/2) cos (-5π)

  23. Inverse Trigonometric Functions

  24. Inverse Trig Functions • Used to find the angle • when two sides of right triangle are known... • or if its trigonometric functions are known Notation used: Read: “angle whose sine is …” Also,

  25. Inverse trig functions have only one principal value of the angle defined for each value of x: 0° <arccos< 180° -90° <arcsin< 90° -90° <arctan< 90°

  26. Example: Given tan θ = -1.600, find θ to the nearest 0.1° for 0° <θ< 360° • Tan is negative in II & IV quadrants

  27. θ = 180° - 58.0° = 122° II θ = 360° - 58.0° = 302° IV Note: On the calculator entering results in -58.0°

  28. More Practice

  29. Given sin θ = 0.3843, find θ to the nearest 0.1° for 0° <θ< 360°

  30. Given cos θ = - 0.0157, find θ to the nearest 0.1° for 0° <θ< 360°

  31. Given sec θ = 1.553 where sin θ < 0, find θ to the nearest 0.1° for 0° <θ< 360°

  32. Given the terminal side of θ passes through point (2, -1), find θthe nearest 0.1° for 0° <θ< 360°

  33. Application

  34. The voltage of ordinary house current is • expressed asV = 170 sin2πft , where • f = frequency = 60 Hzand t = time in seconds. • Find the angle2πft in radians when • V = 120 volts and 0 <2πft <2π

  35. Find t when V = 120 volts

  36. The angleβof a laser beam is expressed as: where w = width of the beam (the diameter) and d = distance from the source. Findβif w = 1.00m and d = 1000m.

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