1. 8.1 RADIANS AND ARC LENGTH Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

2. Definition of a Radian An angle of 1 radian is defined to be the angle, in the counterclockwise direction, at the center of a unit circle which spans an arc of length 1. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally 1 radian Arc length = 1 radius =1

3. Relationship Between Radians and Degrees The circumference, C, of a circle of radius r is given by C = 2πr. In a unit circle, r = 1, so C = 2π. This means that the arc length spanned by a complete revolution of 360 ◦ is 2π, so 360 ◦ = 2 π radians. Dividing by 2π gives 1 radian = 360 ◦ /(2π) ≈ 57.296 ◦. Thus, one radian is approximately 57.296 ◦. One-quarter revolution, or 90 ◦, is equal to ¼(2 π) or π/2 radians. Since π ≈ 3.142, one complete revolution is about 6.283 radians and one-quarter revolution is about 1.571 radians. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Angle in degrees 0°30°45°60°90°120°135°150°180°270°360° 720° Angle in radians 0π/6π/4π/3π/22π/33π/45π/6π3π/22π2π 4π4π Equivalences for Common Angles Measured in Degrees and Radians

4. Converting Between Degrees and Radians To convert degrees to radians, or vice versa, we use the fact that 2π radians = 360 ◦. So 1 radian = 180 ◦ / π ≈ 57.296 ◦ and 1 ◦ = π/180 ≈ 0.01745 radians. Thus, to convert from radians to degrees, multiply the radian measure by 180 ◦ /π radians. To convert from degrees to radians, multiply the degree measure by π radians/180 ◦. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

5. Converting Between Degrees and Radians Example 3 (a)Convert 3 radians to degrees. (b)Convert 3 degrees to radians. Solution (a) 3 radians ・ 180 ◦ /(π radians) = 540 ◦ / (π radians) ≈ 171.887 ◦. (b) 3 ◦ ・ π radians/180 ◦ = π radians/60 ≈ 0.052 radians. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally The word “radians” is often dropped, so if an angle or rotation is referred to without units, it is understood to be in radians.

6. Arc Length in Circle of Radius r The arc length, s, spanned in a circle of radius r by an angle of θ in radians is s = r θ. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

7. Converting Between Degrees and Radians Example 6 You walk 4 miles around a circular lake. Give an angle in radians which represents your final position relative to your starting point if the radius of the lake is: (a) 1 mile (b) 3 miles Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Θ=4 rad Θ=4/3 rad Arc length 4 and radius 1, so angle θ = s/1= 4 radians Arc length 4 and radius 3, so angle θ = s/r= 4/3 radians ● ●● ● s = 4 miles 3 miles s = 4 miles 1 mile

8. Sine and Cosine of a Number Example 7 Evaluate: (a) cos 3.14 ◦ (b) cos 3.14 Solution (a) Using a calculator in degree mode, we have cos 3.14 ◦ = 0.9985. This is reasonable, because a 3.14 ◦ angle is quite close to a 0 ◦ angle, so cos 3.14 ◦ ≈ cos 0 ◦ = 1. (b) Here, 3.14 is not an angle measured in degrees; instead we interpret it as an angle of 3.14 radians. Using a calculator in radian mode, we have cos 3.14 = −0.99999873. This is reasonable, because 3.14 radians is extremely close to π radians or 180 ◦, so cos 3.14 ≈ cos π = −1. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

9. 8.2 SINUSOIDAL FUNCTIONS AND THEIR GRAPHS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

10. Amplitude and Midline The functions y = A sin t + k and y = A cos t + k have amplitude |A| and the midline is the horizontal line y = k. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

11. Amplitude and Midline Example 1 State the midline and amplitude of the following sinusoidal functions:(a) y = 3 sin t + 5 (b) y = (4 − 3 cos t)/20. SolutionRewrite (b) asy = 4/20 − 3/20 cos t = 0.2− 0.15 cos t (a) (b) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Midline y = 5 Midline y = 0.2 Graph of y = 3 sin t + 5Graph of y = 0.2 − 0.15 cos t Amplitude = 3 Amplitude = 0.15

12. Period The functions y = sin(Bt) and y = cos(Bt) have period P = 2π/|B|. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

13. Finding Formulas for Functions Example 3 Find possible formulas for the functions f and g in the graphs Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally This function has period 4πThis function has period 20 The graph of f resembles the graph of y = sin t except that its period is P = 4π. Using P = 2π/B Gives 4π = 2π/B so B = ½ and f(t) = sin(½ t) The graph of g resembles the graph of y = sin t except that its period is P = 20. Using P = 2π/B gives 20 = 2π/B so B = π/10 and g(t) = sin(π/10 t) y = f(t) y = g(t)

14. Horizontal Shift The graphs of y = sin(B(t − h)) and y = cos(B(t − h)) are the graphs of y = sin(Bt) and y = cos(Bt) shifted horizontally by h units. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

15. Horizontal Shift Example 7 Describe in words the graph of the function g(t) = cos (3t −π/4). Solution Write the formula for g in the form cos (B(t − h)) by factoring 3 out from the expression (3t −π/4) to get g(t) = cos (3(t − π/12)). The period of g is 2π/3 and its graph is the graph of f = cos 3t shifted π/12 units to the right, as shown Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally g(t) = cos (3(t − π/12)) f(t) = cos (3t ) Horizontal shift = π/12 Period = 2π/3

16. Summary of Transformations For the sinusoidal functions y = A sin(B(t − h)) + k and y = A cos(B(t − h)) + k, |A| is the amplitude 2π/|B| is the period h is the horizontal shift y = k is the midline |B|/(2π) is the frequency; that is, the number of cycles completed in unit time. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

17. Writing a Formula for the Ferris Wheel Height Example 7 Use the sinusoidal function f(t) = A sin(B(t − h)) + k to represent your height above ground at time t while riding the London Eye Ferris wheel. Solution The diameter of the this Ferris wheel is 450 feet, so the midline is k = 225 and the amplitude, A, is also 225. The period of the Ferris wheel is 30 minutes, so B=π/15. Because we reach y = 225 (the 3 o’clock position) when t = 7.5, the horizontal shift is h = 7.5, so the Ferris wheel height is: f(t) = 225 sin(π/15 (t − 7.5)) + 225. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally y (feet) t (minutes) Amplitude: Radius of wheel is 225 ft Period = 30 minutes Midline: Wheel’s hub is 225 ft above ground y = 225 Graph of the Ferris wheel height function

18. Phase Shift For sinusoidal functions written in the following form, φ is the phase shift: y = A sin(Bt + φ) and y = A cos(Bt + φ). Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally In the Ferris wheel height function, f(t) = 225 sin(π/15 (t − 7.5)) + 225, rewriting the function in the above form f(t) = 225 sin(π/15 t − π/2) + 225, the phase shift is π/2.

19. Phase Shift Example 8 (a) In the figure, by what fraction of a period is the graph of g(t) shifted from the graph of f(t)? (b) What is the phase shift? Solution (a) The period of f(t) is the length of the interval from A to B. The graph of g(t) appears to be shifted 1/4 period to the right. (b) The phase shift is 1/4 (2π) = π/2. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally f(t)f(t) g(t)g(t) AB

20. 8.3 TRIGONOMETRIC FUNCTIONS: RELATIONSHIPS AND GRAPHS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

21. Relationships Between the Graphs of Sine and Cosine Example 1 Use the fact that the graphs of sine and cosine are horizontal shifts of each other to find relationships between the sine and cosine functions. Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally The figure suggests that the graph of y = sin t is the graph of y = cos t shifted right by π/2 radians (or by 90 ◦ ). Likewise, the graph of y = cos t is the graph of y = sin t shifted left by π /2 radians. Thus, sin t = cos(t − π/2) cos t = sin(t + π/2) The graph of y = sin t can be obtained by shifting the graph of y = cos t to the right by π/2 y = sin t y = cos t

22. Relationships Between the Graphs of Sine and Cosine Example 2 Use the symmetry of the graph of cosine to obtain the following relationships:(a) sin t = cos(π/2 − t)(b) cos t = sin(π/2 − t) Solution (a)Since cosine has even symmetry, we can factor out −1 to write cos(π/2 − t) = cos(-(t − π/2)) = cos(t − π/2) And from Example 1, cos(t − π/2) = sin t. Putting these two facts together gives us what we wanted to show: cos(π/2 − t) = sin t. (b) Again from Example 1, we know cos t = sin (t + π /2). But since cosine has even symmetry, we can replace t with −t, giving us what we wanted to show: cos t = cos(−t) = sin(− t + π/2) = sin(π/2 − t). Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

23. Relationships Involving Triangles Converting from radians to degrees, we reinterpret the results of Example 2 in terms of the triangles in the figure. We see that: sin θ = a/c = cos φ and cos θ = b/c = sin φ Since the angles in any triangle add to 180 ◦, we have: θ + φ + 90 ◦ = 180 ◦ so θ = 90 ◦ − φ and φ = 90 ◦ − θ. We conclude, for θ in degrees, that sin θ = cos (90 ◦ − θ) and cos θ = sin (90 ◦ − θ), which corresponds to the results for radians we found in Example 2. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally φ θ c a b Relationship between sine and cosine of θ and φ

24. Relationships Involving The Tangent Function Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

25. Relationship Between The Graphs of the Sine, Cosine, and Tangent Functions The figure shows a graph of sin t and cos t along with a graph of tan t = sin t/cos t. Since a fraction equals zero where its numerator is zero, the tangent function has the same zeros as the sine function, at t = −2π,− π, 0, π, 2 π,.... The graphs of y = tan t and y = sin t cross the t-axis at the same points. Since a fraction equals one where its numerator equals its denominator, tan t = 1 where sin t = cos t, which happens where the graphs of sine and cosine intersect, at t = −3 π /4, π /4, 5 π /4,.... Since a fraction is undefined where its denominator is zero, tan t is undefined where cos t = 0, which happens at t = −3 π /2,− π /2, π /2, 3 π /2,..., where the graph of y = cos t crosses the t-axis. The graph of y = tan t has vertical asymptotes at these points. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally y = sin ty = cos t y = tan t

26. Relationships Involving Reciprocals of the Trigonometric Functions The reciprocals of the trigonometric functions are given special names. Where the denominators are not equal to zero, we define secant θ = sec θ = 1/cos θ. cosecant θ = csc θ = 1/sin θ. cotangent θ = cot θ = 1/tan θ = cos θ/sin θ. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

27. Relationships Between the Graphs of Secant and Cosine Example 3 Use a graph of g(θ) = cos θ to explain the shape of the graph of f(θ) = sec θ. Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally The figure shows the graphs of cos θ and sec θ. In the first quadrant cos θ decreases from 1 to 0, so sec θ increases from 1 toward +∞. The values of cos θ are negative in the second quadrant and decrease from 0 to −1, so the values of sec θ increase from −∞ to −1. The graph of y = cos θ is symmetric about the vertical line θ = π, so the graph of f(θ) = sec θ is symmetric about the same line. Since sec θ is undefined wherever cos θ = 0, the graph of f(θ) = sec θ has vertical asymptotes at θ = π/2 and θ = 3π/2. f(θ) = sec θ g(θ) = cos θ

28. Relationships Between the Graphs of Secant and Cosine The graphs of y = csc θ and y = cot θ are obtained in a similar fashion from the graphs of y = sin θ and y = tan θ, respectively. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Plots of y = csc θ and y = sin θPlots of y = cot θ and y = tan θ

29. The Pythagorean Identity We now see an extremely important relationship between sine and cosine. The figure suggests that no matter what the value of θ, the coordinates of the corresponding point P satisfy the following condition: x 2 + y 2 = 1. But since x = cos θ and y = sin θ, this means cos 2 θ + sin 2 θ = 1 Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally ● P = (x, y) = (cos θ, sin θ) x y θ 1

30. Summarizing the Trigonometric Relationships Sine and Cosine functions sin t = cos(t − π/2) = cos(π/2 − t) = −sin(−t) cos t = sin(t + π/2) = sin(π/2 − t) = cos(−t) Pythagorean Identity cos 2 θ + sin 2 θ = 1 Tangent and Cotangent tan θ = cos θ/sin θ and cot θ = 1/tan θ Secant and Cosecant sec θ = 1/cos θ and csc θ = 1/sin θ Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

31. 8.4 TRIGONOMETRIC EQUATIONS AND INVERSE FUNCTIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

32. Solving Trigonometric Equations Graphically A trigonometric equation is one involving trigonometric functions. Consider, for example, the rabbit population of Example 6 on page 328: R = −5000 cos(π/6 t) + 10,000. Suppose we want to know when the population reaches 12,000.We need to solve the trigonometric equation −5000 cos(π/6 t) + 10,000 = 12,000. We use a graph to find approximate solutions to this trigonometric equation and see that two solutions are t ≈ 3.786 and t ≈ 8.214. This means the rabbit population reaches 12,000 towards the end of the month 3, April (since month 0 is January), and again near the start of month 8 (September). Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally R = −5000 cos(π/6 t) + 10,000 R = 12,000

33. Solving Trigonometric Equations Algebraically We can try to use algebra to find when the rabbit population reaches 12,000: −5000 cos(π/6 t) + 10,000 = 12,000 or −5000 cos(π/6 t) = 2000 cos(π/6 t) = − 0.4 Now we need to know the radian values having a cosine of −0.4. For angles in a right triangle, we would use the inverse cosine function, cos −1 : π/6 t = cos −1 0.4 ≈ 1.982 using a calculator in radian mode. So t = (6/ π) (1.982) ≈ 3.786 (We will call this t 1.) To get the other answer, we observe that the period is 12 and the graph is symmetric about the line t = 6. So t 2 = 12 – t 1 = 12 – 3.786 = 8.214. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally R = −5000 cos(π/6 t) + 10,000 R = 12,000 R (rabbits) t (months) t = 6 t1t1 t2t2

34. The Inverse Cosine Function In the figure, notice that on the part of the graph where 0 ≤ t ≤ π (solid blue), all possible cosine values from −1 to 1 occur once and once only. The calculator uses the following rule: cos −1 is the angle on the blue part of the graph in the figure whose cosine is y. On page 299, we defined cos −1 for right triangles. We now extend the definition as follows: cos −1 is the angle between 0 and π whose cosine is y. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally 0 ≤ t ≤ π y = cos t The solid portion of this graph, for 0 ≤ t ≤ π, represents a function that has only one input value for each output value 0

35. Terminology and Notation The inverse cosine function, also called the arccosine function, is written cos −1 y or arccos y. We define cos −1 y as the angle between 0 and π whose cosine is y. More formally, we say that t = cos −1 y provided that y = cos t and 0 ≤ t ≤ π. Note that for the inverse cosine function the domain is −1 ≤ y ≤ 1 and the range is 0 ≤ t ≤ π. The inverse cosine function, also called the arccosine function, is written cos −1 y or arccos y. We define cos −1 y as the angle between 0 and π whose cosine is y. More formally, we say that t = cos −1 y provided that y = cos t and 0 ≤ t ≤ π. Note that for the inverse cosine function the domain is −1 ≤ y ≤ 1 and the range is 0 ≤ t ≤ π. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

36. Evaluating the Inverse Cosine Function Example 1 Evaluate (a) cos −1 (0) (b) arccos(1) (c) cos −1 (−1) (d) (cos(−1)) −1 Solution: (a)cos −1 (0) means the angle between 0 and π whose cosine is 0. Since cos(π /2) = 0, we have cos −1 (0) = π /2. (b)arccos(1) means the angle between 0 and π whose cosine is 1. Since cos(0) = 1, we have arccos(1) = 0. (c)cos −1 (−1) means the angle between 0 and π whose cosine is −1. Since cos(π) = −1, we have cos −1 (−1) = π. (d)(cos(−1)) −1 means the reciprocal of the cosine of −1. Since (using a calculator) cos(−1) = 0.5403, we have (cos(−1)) −1 = (0.5403) −1 = 1.8508. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

37. The Inverse Sine and Inverse Tangent Functions The figure on the left shows that values of the sine function repeat on the interval 0 ≤ t ≤ π. However, the interval −π/2 ≤ t ≤ π/2 includes a unique angle for each value of sin t. This interval is chosen because it is the smallest interval around t = 0 that includes all values of sin t. The figure on the right shows why this same interval, except for the endpoints, is also used to define the inverse tangent function. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally y = sin t y = tan t - π/2 ≤ t ≤ π/2 - π/2 < t < π/2

38. The Inverse Sine and Inverse Tangent Functions The inverse sine function, also called the arcsine function, is denoted by sin −1 y or arcsin y. We define t = sin −1 y provided that y = sin t and −π/2 ≤ t ≤ π/2. The inverse sine has domain −1 ≤ y ≤ 1 and range −π/2 ≤ t ≤ π /2. The inverse tangent function, also called the arctangent function, is denoted by tan −1 y or arctan y. We define t = tan −1 y provided that y = tan t and −π/2 < t < π/2. The inverse tangent has domain −∞ < y < ∞and range −π/2 < t < π/2. The inverse sine function, also called the arcsine function, is denoted by sin −1 y or arcsin y. We define t = sin −1 y provided that y = sin t and −π/2 ≤ t ≤ π/2. The inverse sine has domain −1 ≤ y ≤ 1 and range −π/2 ≤ t ≤ π /2. The inverse tangent function, also called the arctangent function, is denoted by tan −1 y or arctan y. We define t = tan −1 y provided that y = tan t and −π/2 < t < π/2. The inverse tangent has domain −∞ < y < ∞and range −π/2 < t < π/2. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

39. Evaluating the Inverse Sine and Inverse Tangent Functions Example 1 Evaluate (a) sin −1 (1) (b) arcsin(−1) (c) tan −1 (0) (d) arctan(1) Solution: (a)sin −1 (1) means the angle between − π/2 and π/2 whose sine is 1. Since sin(π/2) = 1, we have sin −1 (1) = π/2. (b)arcsin(−1) = − π/2 since sin(− π/2) = −1. (c)tan −1 (0) = 0 since tan 0 = 0. (d)arctan(1) = π/4 since tan(π/4) = 1. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

40. Solving Trigonometric Equations Using the Unit Circle Example 4 Solve sin θ = 0.9063 for 0 ◦ ≤ θ ≤ 360 ◦. Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally x y = 0.9063 Using our calculator in degree mode, we know one solution is given by sin −1 (0.9063) = 65 ◦. Referring to the unit circle in the figure, we see that another angle on the interval 0 ◦ ≤ θ ≤360 ◦ also has a sine of 0.9063. By symmetry, we see that this second angle is 180 ◦ − 65 ◦ = 115 ◦. ●● 65° 115°

41. Finding Other Solutions Example 5 Solve sin θ = 0.9063 for −360 ◦ ≤ θ ≤ 1080 ◦. Solution We know that two solutions are given by θ = 65 ◦, 115 ◦. We also know that every time θ wraps completely around the circle (in either direction), we obtain another solution. This means that we obtain the other solutions: 65 ◦ + 1 · 360 ◦ = 425 ◦ wrap once around circle 65 ◦ + 2 · 360 ◦ = 785 ◦ wrap twice around circle 65 ◦ + (−1) · 360 ◦ = −295 ◦. wrap once around the other way For θ = 115 ◦, this means that we have the following solutions: 115 ◦ + 1 · 360 ◦ = 475 ◦ wrap once around circle 115 ◦ + 2 · 360 ◦ = 835 ◦ wrap twice around circle 115 ◦ + (−1) · 360 ◦ = −245 ◦. wrap once around the other way Thus, the solutions on the interval −360 ◦ ≤ θ ≤ 1080 ◦ are: θ = −295 ◦,−245 ◦, 65 ◦, 115 ◦, 425 ◦, 475 ◦, 785 ◦, 835 ◦. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

42. Reference Angles For an angle θ corresponding to the point P on the unit circle, the reference angle of θ is the angle between the line joining P to the origin and the nearest part of the x- axis. A reference angle is always between 0 ◦ and 90 ◦ ; that is, between 0 and π/2. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Angles in each quadrant whose reference angles are α xxxx yyyy P P PP αα αα

43. 8.5 POLAR COORDINATES Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

44. Relation Between Cartesian and Polar Coordinates From the right triangle in the figure, we see that The angle θ is determined by the equations Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally r x y ● θ P = (x, y)

45. Switching Between Cartesian and Polar Coordinates Example 1 (a) (a)Give Cartesian coordinates for the points with polar coordinates (r, θ) given by P = (7, π/3), Q = (5, 0), R = (5, π). Solution (a) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally x y P Q R U V The Cartesian coordinates of P are x = 7 cos π/3 = 7/2, y = 7 sin π/3 = 7 /2 So, P = (3.5, 6.062) For the point Q, x = 5 cos 0 = 5, y = 5 sin 0 = 0 So, Q= (5, 0) For the point R, x = 5 cos π =- 5, y = 5 sin π = 0 So, R= (-5, 0)

46. Switching Between Cartesian and Polar Coordinates Example 1 (b) (b) Give polar coordinates for the points with Cartesian coordinates (x, y) given by U = (3, 4) and V = (0,−5). Solution (b) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally x y P Q R U V For U = (3, 4), and tan θ = 4/3. A possible value for θ is θ = arctan 4/3 = 0.927 ≈ 53 ◦. The polar coordinates of U are (5, 0.927). The point V falls on the y-axis, so we can choose r = 5, θ = 3π/2 for its polar coordinates. In this case, we cannot use tan θ = y/x to find θ, because tan θ = y/x = −5/0 is undefined.

47. Graphs in Polar and Cartesian Coordinates Example 3 (a) Describe in words the graphs of the equation y = 1 (in Cartesian coordinates) and the equation r = 1 (in polar coordinates). (b) Write the equation r = 1 using Cartesian coordinates. Write the equation y = 1 using polar coordinates. Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally y = 1 ( a line) r = 1 ( a circle) In polar coordinates, the equation of the line y = 1 is r sinθ = 1 or r = 1/sinθ. In Cartesian coordinates, the equation of the circle r = 1 is

48. Graphs in Polar Coordinates Example 5 For an integer n, curves of the form r = a sin nθ or r = a cos nθ are called roses. Graph the roses(a) r = 3 sin 2θ (b) r = 4 cos 3θ Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally 3 The graph of r = 3 sin 2θ is a rose with four petals, each of length 3. The graph of r = 4 cos 3θ is a rose with three petals, each of length 4. 4

49. Inequalities in Polar Coordinates Example 8 An 18 inch pizza is cut into 12 slices. Use inequalities to describe one of the slices. Solution The pizza has radius 9 inches; the angle at the center is 2π/12 = π/6. Thus, if the origin is at the center of the original pizza, the slice is represented by0 ≤ r ≤ 9 and 0 ≤ θ ≤ π/6. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally 9’’ π/6

50. 8.6 COMPLEX NUMBERS AND POLAR COORDINATES Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

51. Definition of a Complex Number A complex number is defined as any number that can be written in the form z = a + bi, where a and b are real numbers and. The real part of z is the number a; the imaginary part is the number b. A complex number is defined as any number that can be written in the form z = a + bi, where a and b are real numbers and. The real part of z is the number a; the imaginary part is the number b. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

52. Algebra of Complex Numbers Two complex numbers are called conjugates if their real parts are equal and if their imaginary parts differ only in sign. The complex conjugate of the complex number z = a + bi is denoted, so = a − bi. (Note that z is real if and only if z =.) Two complex numbers, z = a + bi and w = c + di, are equal only if a = c and b = d. Adding two complex numbers is done by adding real and imaginary parts separately:(a + bi) + (c + di) = (a + c) + (b + d)i. Subtracting is similar: (a + bi) − (c + di) = (a − c) + (b − d)i. Multiplication works just like for polynomials, using i 2 = −1 to simplify: (a + bi)(c + di) = a(c + di) + bi(c + di) = ac + adi + bci + bdi 2 = ac + adi + bci − bd = (ac − bd) + (ad + bc)i. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

53. Algebra of Complex Numbers Powers of i: We know that i 2 = −1; then, i 3 = i · i 2 = −i, and i 4 = (i 2 ) 2 = (−1) 2 = 1. Then i 5 = i · i 4 = i, and so on. Thus we have in The product of a number and its conjugate is always real and nonnegative: z · = (a + bi)(a − bi) = a 2 −abi + abi − b 2 i 2 = a 2 + b 2. Dividing is done by multiplying the numerator and denominator by the conjugate of the denominator, thereby making the denominator real: Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

54. Computations with Complex Numbers Example 1 Compute the sum (3 + 6i) + (5 − 2i). Solution Adding like terms gives (3 + 6i) + (5 − 2i) = (3 + 5) + (6i − 2i) = 8 + 4i. Example 2 Simplify (2 + 7i)(4 − 6i) − i. Solution Multiplying gives (2+7i)(4−6i)−i = 8+28i−12i−42i 2 −i = 8+16i+42 = 50+16i. Example 3 Compute (2 + 7i)/(4 − 6i). Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

55. The Complex Plane and Polar Coordinates It is often useful to picture a complex number z = x + i y in the plane, with x along the horizontal axis and y along the vertical. The xy-plane is then called the complex plane. From the right triangle in the figure, we see that Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally r x = r cosθ y = r sinθ ● θ P: z = x + iy z = x + i y = r cosθ + i r sinθ

56. Euler’s Formula Based on what we have seen so far, there is no reason to expect a connection between the exponential function e x and the trigonometric functions sin x and cos x. However, in the eighteenth century, the Swiss mathematician Leonhard Euler discovered a surprising connection between these functions that involves complex numbers. This result, called Euler’s formula, states that, for real in radians, Example 6 Evaluate e i π. Solution Using Euler’s formula, e i π = cos π + i sin π = −1. This statement, known as Euler’s Identity, is sometimes written e i π + 1 = 0. It is famous because it relates five of the most fundamental constants in mathematics: 0, 1, e, π, and i. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally e iθ = cos θ + i sin θ

57. Polar Form of a Complex Number If the point representing z has polar coordinates (r, θ), then z = r(cos θ + i sin θ) = re iθ. The expression re iθ is called the polar form of the complex number z. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

58. Polar Form of a Complex Number Example 8 Express the complex number represented by the point with polar coordinates r = 8 and θ = 3π/4, in Cartesian form, a + bi, and in polar form, z = re iθ. Solution Using Cartesian coordinates, the complex number is The polar form isz = 8e i 3π/4. Example 9 Show that the polar coordinates (8, 3π/4), (8, 11π/4), (8, 19π/4 ) all represent the same complex number. Solution Since 11π/4 = 3π/4+2π and 19π/4 = 3π/4+4π, these polar coordinates all represent the point P. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

59. Products of Complex Numbers in Polars If z 1 = r 1 e i θ1 and z 2 = r 2 e i θ2 then z 1 ˑz 2 = (r 1 r 2 )e i(θ1 + θ2) and z 1 / z 2 = (r 1 / r 2 )e i(θ1 - θ2) In words, To multiply two complex numbers, multiply the r coordinates and add the θ coordinates. To divide two complex numbers, divide the r coordinates and subtract the θ coordinates Example 10 Let z 1 = 10e i π/3 and z 2 = 2e i π/4. Find the Cartesian form of z 1 ˑz 2 and z 1 / z 2. Solution z 1 ˑ z 2 = (10ˑ2)e i(π/3 + π/4) = 20 e i(7π/12) = 20 (cos(7π/12) + iˑsin(7π/12) ) = 20 cos(7π/12) + iˑ20sin(7π/12) z1 / z2 = (10/2)e i(π/3 - π/4) = 5 e i(π/12) = 5 (cos(π/12) + iˑsin(π/12) ) = 5 cos(π/12) + iˑ5sin(π/12) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

60. Powers of Complex Numbers in Polar Form de Moivre’s formula For p an integer (cos θ + i sin θ) p = cos (pθ) + i sin(pθ). Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

61. Polar Form of a Complex Number Example 13 Find the Cartesian form of all cube roots of the complex number z = 8e i 3π/4 Solution The point with polar coordinates (8, 3π/4) also has polar coords. (8, 3π/4 + 2 π) = (8, 11π/4) and (8, 3π/4 + 4π) = (8, 19 π/4). The three cube roots of z, in polar form, are (8e i3π/4 ) 1/3 = 8 1/3 e i(3π/4)·(1/3) = 2e iπ/4 = 1.414 + 1.414i (8e i11π/4 ) 1/3 = 8 1/3 e i(11π/4)·(1/3) = 2e i11π/12 = −1.932 + 0.518i (8e i19π/4 ) 1/3 = 8 1/3 e i(19π/4)·(1/3) = 2e i19π/12 = 0.518 − 1.932i. Using more polar representations of z does not produce more cube roots. For example, computing with (r, θ) = (8, 3π/4 + 6π) = (8, 27π/4) leads to a cube root that we have already found: (8e i27π/4 ) 1/3 = 8 1/3 e i(27π/4)·(1/3) = 2e i9π/4 = 2e iπ/4. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally