1. Angles and their measure  Terminal side Initial side Vertex x y Positive Angle in Standard Position Initial Side Terminal side Positive angles are generated by counterclockwise rotation. Negative angles are generated by clockwise rotation.

2. Radian measure r r s = r x y One radian is the measure of a central angle that intercepts an arc s equal in length to the radius r of the circle.

3. Radian measure, continued. It follows that a general angle has radian measure where s is the arc length intercepted by the angle. s = 2r r r x y Because the circumference of a circle is s = 2  r units, it follows that the angle corresponding to a complete counterclockwise revolution is 2  radians.

4. Quadrants and angles Quadrant I Quadrant III Quadrant II Quadrant IV

5. Degree measure Since a full counterclockwise revolution corresponds to 360º and also to 2  radians, we have To convert degrees to radians, multiply degrees by To convert radians to degrees, multiply radians by Using we can also convert. For example, to find the degree measure for simply divide by 3 to get

6. Arc length For a general angle, the radian measure is where s is the arc length intercepted by the angle and r is the radius of the circle. If we multiply through by r, we obtain This is the formula for arc length, but it only applies when is given in radians. Problem. What length of arc is cut off by an angle of 120 degrees on a circle of radius 12 cm? Solution. First convert to radians, then multiply by 12 to get an arc length of

7. Distance traveled by a rolling wheel of radius r = 1 Suppose the wheel rolls without slipping as shown.

8. Effect of tire wear on mileage The odometer in your car measures the mileage travelled. The odometer uses the angle that the axle turns to compute the mileage s in the formula Question. With worn tires, are the actual miles travelled more or less than the odometer miles?

9. Area of a sector of a circle For a circle of radius r, the area A of a sector of the circle with central angle is given by where is measured in radians. What area does A become when r Sector is shaded region

10. A pizza problem John has a slice of pizza from a 12 inch diameter pizza. The central angle formed by his slice measures 60 . Jane has a slice of pizza from a 14 inch diameter pizza. The central angle formed by her slice measures 45 . Who has more pizza? How much more?

11. The unit circle The equation of the unit circle is x x y y On the unit circle, arc length = radian measure since r =1. Also, we write t for arc length instead of s.

12. Definitions of Trigonometric Functions   

13. Coordinates on unit circle for special angles         Divide unit circle into eight equal arcs as shown. Since (a,b) is on the line y = x, b equals a. Since (a, a) is on the unit circle, That is, which implies that Since the point is in the first quadrant, The coordinates of the other points are determined by symmetry.

14. Coordinates on unit circle for special angles, continued         t0  /4  /23  /4  5  /43  /27  /4 sin t 010–1 cos t 10–10

15. Coordinates on unit circle for special angles, continued             Divide unit circle into twelve equal arcs as shown. dist((a,b),(1,0)) = dist((a,b),(b,a)) implies b = 1/2. Since (a,b) is on the unit circle, Therefore, and Since (a,b) is in the first quadrant,

16. Coordinates on unit circle for special angles, continued             t 0  /6  /3  /22  /35  /6  7  /64  /33  /25  /311  /6 sin t 010–1 cos t 10–10

17. The Unit Circle x y For any angle  : (x,y) = (cos , sin  )

18. Domain, range, and period of sine and cosine Domain of sine and cosine: All real numbers Range of sine and cosine: [–1, 1] A function f is periodic if there exists a positive real number c such that for all t in the domain of f. The smallest number c for which f is periodic is called the period of f. Since it follows that sine and cosine are periodic with period 2 .

19. More properties of trigonometric functions The cosine and secant functions are even: The sine, cosecant, tangent, and cotangent functions are odd: Problem. Evaluate Solution. Since Problem. Solution.

20. Evaluate without a calculator, give exact answers sin 4  cos 3  cos(7  /3) sin(9  /4) sin(19  /6) sin(  8  /3)

21. Definition of Sine and Cosine in Right Triangles Suppose P = (x, y) is the point on the unit circle specified by the angle We define the functions, cosine of, or, and the sine of, or by either pair of formulas: (-1,0) Hypotenuse Side Opposite Side Adjacent

22. Definition of the tangent function Suppose P = (x, y) is the point on the unit circle specified by the angle We define the tangent of, or tan, by tan = y/x, or by: Side Opposite Side Adjacent

23. Evaluating trigonometric functions Evaluate sine, cosine, and tangent for the triangle shown. First, 5 12 hyp

24. Another way to derive the trig functions of special angles Consider the following right triangle. Evaluate: 1 1 90º θ φ h

25. More on trig functions of special angles Consider the following equilateral triangle. Evaluate: 2 2 1 1 αβ h

26. Sines, cosines, and tangents of special angles

27. Fundamental Trigonometric Identities Reciprocal identities Quotient identities Pythagorean identities

28. Evaluating trigonometric functions using trig identities Evaluate cosine for the triangle shown. Using trig identities: 5 12 hyp

29. Solving Right Triangles One of the angles of a right triangle is 90°. Thus, one part is already known and the only additional information necessary is either two sides or an acute angle and a side. With this additional information, it is possible to solve for the remaining parts. Making a carefully labeled sketch of the triangle is important in solving these problems. The top of a 200-foot tower is to be anchored by cables that make an angle of 30° with the ground. How long must the cables be? How far from the base of the tower should the anchors be placed? sin 30° = 200/h => h = 200/sin 30° = 400 ft tan 30° = 200/x => x = 200/tan 30° = 346.4 ft 200 ft h x 30°

30. Height of church steeple At a point 65 ft. from the base of a church, the angles of elevation to the bottom of the steeple and the top of the steeple are 35º and 43º, resp. Find the height s of the steeple. s = h 2 – h 1 = 65tan 43º– 65tan 35º 35 º 43 º s h2h2 h1h1 65

31. Definitions of trigonometric functions of any angle Let be an angle in standard position with (x, y) a point on the terminal side of and x y

32. Interpretation of "any angle" definition The "any angle" definition is the same as the "unit circle" definition but on a circle of radius r   If we replace by similar right triangles may be used and Since both x and are negative Thus, cos has the same value by either definition.

33. Reference Angles Let θ be an angle in standard position. Its reference angle is the acute angle θ' formed by the terminal side of θ and the horizontal axis. Example. Find the reference angle of

34. Evaluating trigonometric functions of any angle θ 1.Determine the function value for the associated reference angle θ'. 2.Depending on the quadrant in which θ lies, affix the appropriate sign to the function value. Example. Find the value of cos 135º.

35. Evaluating trig functions Let θ be an angle in Quadrant III such that sin θ = Find the value of cos θ using trig identities. Using sin 2 θ + cos 2 θ = 1, we have It follows that Since θ is in Quadrant III, It is also possible to solve for cos θ using a reference angle. See the next slide.

36. Evaluating trig functions using a reference angle Find cos θ for the angle θ shown where sin θ = By Pythagoras on the triangle, So, |x| = 12. Therefore, cos θ' =12/13 and cos θ = –12/13 since θ is in QIII.

37. Trigonometric functions of quadrant angles Evaluate the cosine and tangent functions at the four quadrant angles 0,  /2, , 3  /2. Choose the points (1,0), (0,1), (  1,0), and (0,  1), resp., which correspond to these four quadrant angles. Thus, r = 1 for all four.

38. Evaluating trigonometric functions Find the value of the five remaining trigonometric functions with the given constraint.

39. Sines and cosines of special angles Using these values, we can plot the basic sine and cosine curves.

40. Basic sine curve: Any portion of the graph representing one period is called one cycle.      x y

41. Basic cosine curve: Any portion of the graph representing one period is called one cycle.      x y

42. Sketching basic sine and cosine graphs using five key points The five key points in one period of a graph: intercepts, maximum points, and minimum points.           x x y y

43. Transformations of sine and cosine curves Consider the functions defined by We will consider the effects of the constants a, b, c, and d on the shape of the basic sine and cosine curves. First, the constant a determines a vertical stretch (a > 1) or a vertical shrink (0 < a < 1) of the basic sine and cosine curves. |a| is the amplitude, which is half the distance between the maximum and minimum values of the functions *. *

44. Example for vertical stretch, vertical shrink Consider the functions defined by x y

45. Example for horizontal stretch Consider the function defined by The transformed graph will exhibit a horizontal stretch and the new period will be 4 . In general, will have a period of x y y

46. Horizontal translation of the sine and cosine curves The constant c in the general equations creates a horizontal translation of the sine and cosine curves. The graphs of have the following characteristics. (Assume b > 0.) The left and right endpoints of a one-cycle interval can be determined by solving the equations The graphs of and are shifted by an amount c/b in their respective general equation. The number c/b is called the phase shift.

47. Example for horizontal translation Consider the function defined by y = sin(  x–  /2). The amplitude is 1, the period is 2, and the phase shift is 1/2.  x–  /2 = 0 => x = 0.5,  x–  /2 = 2  => x = 2.5 so the interval [0.5, 2.5] corresponds to one cycle of the graph. x y = sin(  x–  /2) y = sin  x y

48. Sine with a phase shift of –π/2 plot([sin(t),sin(t+Pi/2)], t = –Pi..Pi, color = black); The shifted sine curve is the cosine. Likewise, we can shift the cosine curve right by π/2 to get the sine.

49. Vertical translation of the sine and cosine curves The constant d in the general equations creates a vertical translation of the sine and cosine curves. The line y = d is called the midline of the translated curve. Note that amplitude = distance from midline to max value = distance from midline to min value. Consider the function defined by y = 0.5+cos x. midline is y = 0.5 x amplitude = 1

50. London Eye Ferris wheel The London Eye Ferris wheel has a diameter of 450 feet. It completes one revolution every 30 minutes, and boarding is from ground level. As a function of angle θ, the height y in feet of a seat on this Ferris wheel is where θ is in radians. θ ground level y  

51. London Eye Ferris wheel, continued As a function of time t in minutes, the height y in feet of a seat on this Ferris wheel is ground level

52. Graphs of tangent and cotangent

53. Graphs of secant and cosecant

54. An application of cosecant An airplane, flying at an altitude of 6 miles, is on a flight path directly over an observer. Let in radians be the angle of elevation from the observer to the plane. Find the distance d from the observer to the plane for plane d observer 6 miles θ 

55. Another application of a trig function An airplane, flying at an altitude of 6 miles, is on a flight path directly over an observer. Let in radians be the angle of elevation from the observer to the plane. Find the signed distance x for plane observer 6 miles θ x Hint: Solve for x as a function of θ.

56. Sketch the graph of y = sec 2x First sketch the graph of y = cos 2x. Then use the fact that sec 2x is the reciprocal of cos 2x. Note that the zeros of y = cos 2x correspond to the vertical asymptotes of y = sec 2x at n an integer, and the period of y = sec 2x is . y = cos 2x y = sec 2x

57. Damped sine wave Consider the function f(x) = x∙sin x. It is not difficult to show that The graph of f touches the line y = –x or the line y = x at x =  /2 +n  and it has x-intercepts at x = n , n an integer. The factor x is called the damping factor.

58. Damped cosine wave The function could represent the damped oscillations of a weight hanging from a spring when friction is present. It is not difficult to prove that Here, e  t is the damping factor. y = e  t y = f(t) y =  e  t

59. Even, odd, or neither? sec x tan x cot x csc x Verify your answers both graphically and algebraically. Recall that sin x is odd and cos x is even.

60. Can you solve this? In traveling across flat land, you notice a mountain directly in front of you. Its angle of elevation (to the peak) is 3.5 . After you drive 19 miles closer to the mountain, the angle of elevation is 9 . Find y, which is the height of the mountain. Use the figure below (not drawn to scale). 3.5  y 99 19 x Answer:

61. Inverse Trigonometric Functions Solving equations involving trigonometric functions can often be reduced (after some algebraic manipulation) to an equation similar to the following example: Find a value of x radians satisfying cos x = 0.4. In order to find a value of x as in the above example, we can use a calculator which has support for inverse cosine. Such a calculator in radian mode allows us to find the required value of x by pressing the button labeled as shown below: Pressing this button yields cos -1 (0.4) 1.16, where cos(1.16) 0.4. cos -1

62. Inverse cosine Restricting the domain.   restricted cosine arccosine

63. The Inverse Cosine Function. The inverse cosine function, also called the arccosine function, is denoted by cos -1 x or arccos x. We define In other words, if y = arccos x, then y is the angle between 0 and whose cosine is x. The inverse cosine has domain In conjunction with the inverse cosine, we may have to use reference angles to find answers which are not in Example. cos -1 (0.4) 1.16 gives us an answer in the first quadrant. Suppose we want an answer in the fourth quadrant. We simply subtract the reference angle from to obtain t 5.12.

64. Inverse sine Restricting the domain   restricted sine arcsine

65. The Inverse Sine Function. The inverse sine function, also called the arcsine function, is denoted by sin -1 x or arcsin x. We define The inverse sine has domain In conjunction with the inverse sine, we may have to use reference angles to find answers which are not in its range. Example. sin -1 (0.707) = gives us an answer in the first quadrant. Suppose we want an answer in the second quadrant. We simply subtract the reference angle from to obtain t =.

66. Inverse tangent Restricting the domain. restricted tangent arctangent

67. The Inverse Tangent Function. The inverse tangent function, also called the arctangent function, is denoted by tan -1 x or arctan x. We define The inverse tangent has domain In conjunction with the inverse tangent, we may have to use reference angles to find answers which are not in its range. Example. tan -1 (1.732) = gives us an answer in the first quadrant. Suppose we want an answer in the third quadrant. We simply add the reference angle to to obtain t =

68. Inverse properties of trigonometric functions for angles in radians

69. Inverse properties of trigonometric functions for angles in degrees  Example. What can happen if the stated conditions don’t hold.

70. Right Triangles: Inverse Trigonometric Functions If we know the sides of a right triangle, what are the angles? Example. Consider the 3-4-5 right triangle. Value of Ө? 3 4 5 Ө Ө = arctan 0.75 = 0.64 Ө = arcsin 0.6 = 0.64 Ө = arccos 0.8 = 0.64

71. Writing an expression. Write an algebraic expression that is equivalent to the expression Hint: Draw a right triangle. y 4 r

72. Problem: Writing an expression Write an algebraic expression that is equivalent to the expression Hint: Draw a right triangle. Get the following as functions of r: Now, write an algebraic expression for: 1 r y

73. Example problem for inverse trig functions The grade of a road is 5.8%. What angle does the road make with the horizontal? We have tan Ө = 5.8/100 = 0.058. Using a calculator in degree mode, we have Ө = tan –1 (0.058) = 3.319º. 5.8 ft 100 ft Ө

74. London Eye, revisited The London Eye Ferris wheel makes one revolution every 30 minutes. The height y of a seat on this wheel in feet is given by y =, where t is in minutes. The ride starts when t = 0 and f(0) = 0. During the first 30 minutes of the ride, when will a person be 112.5 feet above ground? During the first 30 minutes of the ride, when will a person be 337.5 feet above ground? In the first 30 minutes of the ride, how much time will a person spend at 400 feet or more?

75. Simple harmonic motion Oscillations of a ball on a spring.    A point that moves on a coordinate line is in simple harmonic motion when its displacement d from the origin at time t is given by either d = a∙sin  t or d = a∙cos  t where a and  are real numbers such that  > 0. The motion has amplitude |a|, period 2  / , and frequency  /2 . 2 sec 4 sec 10 cm –10 cm

76. A problem involving simple harmonic motion Given the equation for simple harmonic motion (a) find the maximum displacement from equilibrium (d = 0) (b) the frequency (c) the least positive value of t for which d = 0. The maximum displacement is given by the amplitude = 6. The frequency = For (c), we must solve for t. This equation is satisfied when or when t = so the least positive value of t is t = See the next slide for a graphical approach.

77. A problem involving simple harmonic motion: graphical approach The graph of is shown below. maximum displacement = 6 period = 1/frequency =  

78. The displacement of an object in simple harmonic motion The time t is in seconds and the distance d is in centimeters. Which equation applies: or What is the value of a? What is the value of: the period? the frequency? What is the value of

79. Solving a right triangle for all unknown sides and angles Given that a = 25 and c = 35. Solve for the remaining parts. Since it follows that We have θ = 90º– 45.58º We could use Pythagoras' theorem to find b, or so that a = 25 c = 35 b = ? θ φ 90º

80. Angle of depression A satellite orbits 12,500 miles above earth's surface (see figure). Find the angle of depression from the satellite to the horizon. The answer is 75.97º. Can you solve for it? satellite angle of depression 4000

81. Can you solve this? While watching a softball game, Sean notices a blimp straight ahead and above him at an angle of elevation of 48 . Three minutes later, he notices the blimp is still straight ahead, but now at an angle of elevation of 35 . If the blimp maintained an altitude of 2000 feet, how far did the blimp travel in those three minutes? Draw a picture and label it. The answer is 1055.5 feet.