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Chapter 8

Chapter 8. Unit Circle and Radian Measures. Opening problem. 1. Consider an equilateral triangle with sides 2 cm long . Altitude [AN] bisects side [BC] and the vertical angle BAC.   Can you see from this figure that sin 30 o =1/2?

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Chapter 8

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  1. Chapter 8 Unit Circle and Radian Measures

  2. Opening problem 1. Consider an equilateral triangle with sides 2 cm long. Altitude [AN] bisects side [BC] and the vertical angle BAC.   Can you see from this figure that sin 30o =1/2? Use your calculator to find the values of sin 30o , sin 150o , sin 390o , sin 1110o and sin(-330o). What do you notice? Can you explain why this result occurs even though the angles are not between 0o and 90o ? A 30o 30o 2 60o 60o B C N 1

  3. 2. Degree, minute, second: One degree : 1o = 1/360 of a revolution One minute : 1’ = 1/60 of a degree One second: 1’’ = 1/60 of a minute

  4. 3. 1 radian (1c) ≈ 57.3o Radian is an abbreviation for “radial angle”

  5. Degree-radian conversion Degree Radian

  6. 5. Convert to radians, in terms of p a. 60o b. 80o c. 315o

  7. 6. Convert to degrees: a. p/5 b. 3 p/4 c. 5 p/6

  8. Parts of a circle minor arc radius sector chord center segment major arc *minor if it involves less than half the circle *major if it involves more than half the circle.

  9. Arc Length X l q Y O r For q in radians, arc length l = qr For q in degrees, arc length l= q/360 x 2pr

  10. q O r For q in radians, area of a sector A = ½ qr2 For q in degrees, area of a sector A= q/360 x pr2

  11. 9. Use radians to find the arc length and the area of a sector of a circle of Radius 9 cm and angle 7p/4 Arc Length: l = qr Area of a sector: A = ½ qr2

  12. 10. A sector has an angle of 1.19 radians and an area of 20.8 cm2. Find its radius and its perimeter. Area of a sector: A = ½ qr2 Perimeter: P =

  13. 11. A sector has an angle of 107.9oand an arc length of 5.92 m. Find its: For q in degrees, arc length l = q/360 x 2pr A= q/360 x pr2 a. radius b area.

  14. 12. Find, in radians, the angle of a sector of: Radius 4.3 m and arc length 2.95 m Arc Length: l = qr

  15. 13. (number 8 in Exercise 8B)

  16. 14. A nautical mile (nmi) is the distance on the Earth’s surface that subtends an angle of 1 minute (or1/60 of a degree) of the Great Circle arc measured from the center of the Earth. A knot is a speed of 1 nautical mile per hour. a. Given that the radius of the Earth is 6370 km, show that 1 nmi is approximately equal to 1.853 km. b. Calculate how long it would take a plane to fly from Perth to Adelaide (a distance of 2130 km) if the plane can fly at 480 knots.

  17. 14. A nautical mile (nmi) is the distance on the Earth’s surface that subtends an angle of 1 minute (or1/60 of a degree) of the Great Circle arc measured from the center of the Earth. A knot is a speed of 1 nautical mile per hour. a. Given that the radius of the Earth is 6370 km, show that 1 nmi is approximately equal to 1.853 km. ((1/60)/360) x 2 p (6370) = 1.853 km b. Calculate how long it would take a plane to fly from Perth to Adelaide (a distance of 2130 km) if the plane can fly at 480 knots. (2130/1.853) /480 = time 2.4 hours = time

  18. The unit circle is the circle with center (0, 0) and radius 1 unit. (0, 1) (1, 0) (0, 0) (-1, 0) (0, -1)

  19. x2 + y2 = r2 is the equation of a circle with center (0, 0) and radius r. The equation of the unit circle is x2 + y2 = 1.

  20. qis positive for anticlockwise rotations and negative for clockwise rotations. + - P

  21. Definition of sine and cosine SohCahToa P(cosq, sin q) q cosqis the x-coordinate of P sin qis the y-coordinate of P

  22. Pythagorean identity of sine and cosine cos2q+ sin2q= 1.

  23. Domain and range of sine and cosine of a unit circle. For all points on the unit circle, -1 <x <1 and -1 <y <1. So, -1 <cosq<1 and -1 <sin q<1 for all q. (0, 1) (1, 0) (0, 0) (-1, 0) (0, -1)

  24. Definition of tangent

  25. 22. PERIODICITY OF TRIGONOMETRIC RATIOS For qin radians and k єZ , cos (q+ 2kp) = cosqand sin (q+ 2kp) = sin q. For qin radians and k єZ , tan(q+ kp) = tan q.

  26. 24. sin (180 – q) = sin q and cos(180 – q) = -cosq

  27. 25. Find all possible values of cosq for sin q= 2/3. Illustrate your answer.

  28. 25. Find all possible values of cosqfor sin q= 2/3. Illustrate your answer. sin q = opp/hyp Use Pythagorean Theorem to solve for the 3rd side. cosq= √5/3 or -√5/3 3 2 3 2 q q √5 -√5 0

  29. 26. If sin q = -3/4 and p < q < 3p/2, find cosq and tan q without using a calculator.

  30. 26. If sin q = -3/4 and p < q < 3p/2, find cosq and tan q without using a calculator. the domain for q is in the between p and 3p/2 puts us in the third quadrant. cosq= -√7/4 tan q= -3/-√7 = (3√7) / 7 -√7 -3 4

  31. 27. If tan q = -2 and 3p/2 < q < 2p, find sin q and cosq.

  32. 27. If tan q = -2 and 3p/2 < q < 2p, find sin q and cosq. cosq = √5/5 sin q = -2√5/5 1 -2 √5

  33. 28. If qis a multiple of p/2, the coordinates of the points on the unit circle involve 0 and +1. 29. If qis a multiple ofp/4, but not a multiple ofp/2, the coordinates involve +√2 / 2. 30. If qis a multiple ofp/6, but not a multiple ofp/2, the coordinates involve +½ and + √3 / 2.

  34. 31. Use a unit circle diagram to find sin q, cosqand tan qfor qequal to: a. p/4 b. 5p/4 c. 7p/4 d. p e. -3p/4

  35. 31. Use a unit circle diagram to find sin q, cosqand tan qfor qequal to: a. p/4 b. 5p/4 c. 7p/4 d. p e. -3p/4

  36. 32. Without using a calculator, evaluate: a. sin260o b. sin 30ocos 60o c. 4sin 60ocos 30o

  37. 32. Without using a calculator, evaluate: a. sin260o b. sin 30ocos 60o c. 4sin 60ocos 30o Complete the rest and report out.

  38. 33. Equation of a straight line. If a straight line makes an angle of q with the positive x-axis then its gradient is m = tan q

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