1. Chapter 7 TRIGONOMETRY AND PERIODIC FUNCTIONS Section 7.1Section 7.1Introduction to Periodic Functions Section 7.2Section 7.2The Sine and Cosine Functions Section 7.3Section 7.3Radians and Arc Length Section 7.4Section 7.4Graphs of the Sine and Cosine Section 7.5Section 7.5Sinusoidal Functions Section 7.7Section 7.7Trigonometric Relationships and Identities 1 Section 7.6Section 7.6The Tangent Function Section 7.8Section 7.8Inverse Trigonometric Functions Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

2. 7.1 INTRODUCTION TO PERIODIC FUNCTIONS 2 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

3. Ferris Wheel Height As a Function of Time The London Eye Ferris Wheel measures 400 feet in diameter and turns continuously, completing a single rotation once every 30 minutes. Suppose you hop on the London Eye Ferris wheel at time t = 0 and ride it for two full turns. Let f(t) be your height above the ground, measured in feet as a function of t, the number of minutes you have been riding. We can figure out some values of f(t). Since the speed of the rotation is constant, you are at the top 15 minutes after boarding and one-quarter of the way up at 7.5 minutes and 22.5 minutes after boarding. Then you are back at the bottom after 30 minutes and the process continues. t (minutes)07.51522.53037.54552.5 f(t) (feet)02004002000 400200 t (minutes)6067.57582.59097.5105112.5120 f(t) (feet)02004002000 4002000 Values of f(t), your height above the ground t minutes after boarding the wheel 3 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

4. Graphing the Ferris Wheel Function Notice that the values of f(t) in the table begin repeating after 30 minutes. This is because the second turn is just like the first turn, except that it happens 30 minutes later. If you ride the wheel for more full turns, the values of f(t) continue to repeat at 30-minute intervals. We plot this data and fill in the blank spaces. t (minutes)07.51522.53037.54552.5 f(t) (feet)02004002000 400200 t (minutes)6067.57582.59097.5105112.5120 f(t) (feet)02004002000 4002000 Values of f(t), your height above the ground t minutes after boarding the wheel y (feet) t (minutes) 4 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally 400 200

5. Periodic Functions: Period, Midline, and Amplitude The period is the distance between consecutive peaks. The midline is the horizontal line midway between the maximum and minimum values. The amplitude is the vertical distance from midline to peak. The period is the distance between consecutive peaks. The midline is the horizontal line midway between the maximum and minimum values. The amplitude is the vertical distance from midline to peak. 5 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

6. The Ferris Wheel: Period, Midline, and Amplitude The Ferris Wheel function, f, is said to be periodic, because its values repeat on a regular interval or period. In the figure, the period is indicated by the horizontal gap between the first two peaks. The dashed horizontal line is the midline of the graph of f. The vertical distance shown between the first peak and the midline is called the amplitude. y (feet) t (minutes) Amplitude: Radius of wheel is 200 ft Period: One rotation takes 30 minutes Midline: Wheel’s hub is at 200 ft above ground y = 200 The graph of y = f(t) showing the amplitude, period, and midline 6 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally 400 200

7. 7.2 THE SINE AND COSINE FUNCTIONS 7 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

8. Using Angles to Measure Position On a Circle Conventions For Working With Angles We measure angles with respect to the horizontal, not the vertical, so that 0 ◦ describes the 3 o’clock position. Positive angles are measured in the counter-clockwise direction, negative angles in the clockwise direction. Large angles (greater than 360 ◦ or less than −360 ◦ ) wrap around a circle more than once. 8 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

9. Height on the Ferris Wheel as a Function of Angle Since we can measure position on the Ferris wheel using angles, we see that the height above the ground is a function of the angle position on the wheel. We can rewrite our table giving heights as a function of angle, instead of time. θ (degrees)-90°0°0°90°180°270°360°450°540° f(t) (feet)02004002000 400200 θ (degrees)630°720°810°900°990°1080°1170°1260°1350° f(t) (feet)02004002000 4002000 Your height above ground, y, as a function of the angle turned through by the wheel Recall how the y-values repeat every 30 minutes. Similarly, in the Table, the values of y repeat every 360 ◦. In both cases, the y-values repeat every time the wheel completes one full revolution. 9 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

10. The Unit Circle When we studied quadratic functions, we transformed a special “starting” function y = x 2 to get other quadratic functions. Similarly, we begin here with a special “starting” circle. This is the unit circle, the circle of radius one centered at the origin. The unit circle gets its name from the fact that its radius measures exactly one unit. (-1,0) (0,1) (1,0) (0,-1) P = (x, y) θ Origin (0,0) Radius is 1 10 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

11. The Sine and Cosine Functions Suppose P = (x, y) in the figure is the point on the unit circle specified by the angle θ. We define the functions, cosine of θ, or cos θ, and sine of θ, or sin θ, by cos θ = x and sin θ = y. In other words, cos θ is the x-coordinate of the point P; and sin θ is the y-coordinate. Suppose P = (x, y) in the figure is the point on the unit circle specified by the angle θ. We define the functions, cosine of θ, or cos θ, and sine of θ, or sin θ, by cos θ = x and sin θ = y. In other words, cos θ is the x-coordinate of the point P; and sin θ is the y-coordinate. P = (x, y) θ x = cos(θ) y = sin(θ) Unit Circle 11 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

12. Values of the Sine and Cosine Functions Example 2 Find the values of sin θ and cos θ for θ = 0 ◦, 90 ◦, 180 ◦, 270 ◦. Solution We know that the angle θ = 0 ◦ specifies the point P = (1, 0). Since sin 0 ◦ is the y-coordinate of this point and cos 0 ◦ is the x- coordinate, this means sin 0 ◦ = 0 and cos 0 ◦ = 1. Likewise, we know that θ = 90 ◦ specifies the point P = (0, 1), so sin 90 ◦ = 1 and cos 90 ◦ = 0. Continuing, we can find the values of sin θ and cos θ for the other required angles as follows: At θ = 0 ◦, P = (1, 0) so cos 0 ◦ = 1 and sin 0 ◦ = 0 At θ = 90 ◦, P = (0, 1) so cos 90 ◦ = 0 and sin 90 ◦ = 1 At θ = 180 ◦, P = (−1, 0) so cos 180 ◦ = −1 and sin 180 ◦ = 0 At θ = 270 ◦, P = (0,−1) so cos 270 ◦ = 0 and sin 270 ◦ = −1. 12 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

13. 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 ◦. Angles in each quadrant whose reference angles are α xxxx yyyy P P PP αα αα 13 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

14. The coordinates of the point P on the unit circle in the figure are given by x = cos θ and y = sin θ. The coordinates (x, y) of the point Q in the figure are given by x = r cos θ and y = r sin θ. Coordinates of a Point on a Circle of Radius r Q =(x, y) P 1 ● θ x = r cosθ y = r sinθ r 14 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

15. Coordinates of Points on a Circle of Radius r = 5 Example 4 Find the coordinates of the points A, B, and C in the Figure to three decimal places. Solution With r = 5, the coordinates of point A are given by x = 5 cos 130 ◦ = 5(−0.6427) = −3.214, y = 5 sin 130 ◦ = 5(0.766) = 3.830. Point B corresponds to an angle of −70 ◦, (angle is measured clockwise), so B has coordinates x = 5 cos(−70 ◦ ) = 5(0.342) = 1.710, y = 5 sin(−70 ◦ ) = 5(−0.93969) = −4.698. For point C, we must first calculate the corresponding angle, since the 10 ◦ is not measured from the positive x-axis. The angle we want is 180 ◦ + 10 ◦ = 190 ◦, so x = 5 cos(190 ◦ ) = 5(−0.9848) = −4.924, y = 5 sin(190 ◦ ) = 5(−0.1736) = −0.868. A B C 130° 70° 10° ● r = 5 15 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

16. 7.3 RADIANS AND ARC LENGTH 16 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

17. 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. 1 radian Arc length = 1 radius =1 17 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

18. 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 ◦ = 2π 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(2π) or π/2 radians. Since π ≈ 3.1416, one complete revolution is about 6.283 radians and one-quarter revolution is about 1.571 radians. 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 18 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

19. 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 ◦. 19 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

20. 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. The word “radians” is often dropped, so if an angle or rotation is referred to without units, it is understood to be in radians. 20 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

21. 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. 21 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

22. Arc Length 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 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 22 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

23. 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. 23 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

24. Exact Values of Sine and Cosine 24 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally Angle in degrees 0°30°45°60°90°120°135°150°180°270°360° Angle in radians 0π/6π/4π/3π/22π/33π/45π/6π3π/22π2π Cosine 11/20−1/2−101 Sine 01/21 0−10 Triangles are used to compute the exact values of multiples of 30°, 45°, and 60°.

25. 7.4 GRAPHS OF SINE AND COSINE 25 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

26. Tabulating and Graphing Values of Sine and Cosine Values of sin θ and cos θ for 0 ◦ ≤ θ < 360 ◦ θsin θcos θθsin θcos θθsin θcos θθsin θcos θ 0◦0◦ 1090 ◦ 01180 ◦ 0270 ◦ 0 30 ◦ 0.870.5120 ◦ -0.50.87210 ◦ -0.87-0.5300 ◦ 0.5-0.87 45 ◦ 0.71 135 ◦ -0.710.71225 ◦ -0.71 315 ◦ 0.71-0.71 60 ◦ 0.50.87150 ◦ -0.870.5240 ◦ -0.5-0.87330 ◦ 0.87-0.5 θ in degrees y = cos θ y = sin θ y 26 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

27. Properties of Sine and Cosine Properties of the sine and cosine functions that are apparent from the graph include: Domain: All values of θ, since any angle, positive or negative, specifies a point on the unit circle. Range: Since values of the sine and cosine are coordinates of points on the unit circle, they lie between −1 and 1. So the range of the sine and cosine are −1 ≤ sin θ ≤ 1 and − 1 ≤ cos θ ≤ 1. Odd/Even Symmetry: The sine function is odd and the cosine function is even:sin(− θ) = −sin θ and cos(− θ) = cos θ. Period: Both sine and cosine are periodic functions, because the values repeat regularly. The smallest interval over which the function values repeat—here 360 ◦ —is called the period. We have sin(θ + 360 ◦ ) = sin θ and cos(θ + 360 ◦ ) = cos θ. 27 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

28. 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, 5th Edition, 2014, Connally 28

29. Amplitude and Vertical Stretches Example 3 Compare the graph of y = sin t to the graphs of y = 2 sin t and y = −0.5 sin t, for 0 ◦ ≤ t ≤ 360 ◦. How are these graphs similar? How are they different? What are their amplitudes? Solution The graphs are shown. The amplitude of y = sin t is 1, the amplitude of y = 2 sin t is 2 and the amplitude of y = −0.5 sin t is 0.5. The graph of y = −0.5 sin t is “upside-down” relative to y = sin t. These observations are consistent with the fact that the constant A in the equation y = A sin t stretches or shrinks the graph vertically, and reflects it about the t-axis if A is negative. The amplitude of the function is |A|. The t axis is the midline for all three functions. 29 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

30. Midlines and Vertical Shifts Example Compare the graph of y = cos t to the graphs of y = cos t + 3 and y = cos t - 2, for 0 ◦ ≤ t ≤ 360 ◦. How are these graphs similar? How are they different? What are their midlines? Solution The graphs are shown. The midline of y = cos t is the t axis (y = 0), the midline of y = cos t + 3 is y = 3 and the midline of y = cos t - 2 is y = -2. Recall that when the function f(t) is shifted vertically by a distance k, the new function is f(t) + k. Similarly, the midline is shifted vertically by that same distance k. Generalizing, we conclude that the graphs of y = sin t + k and y = cos t + k have midlines y = k. The t axis is the midline for all three functions. 30 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally y = cos t + 3, midline: y=3 y = cos t, midline: y=1 y = cos t – 2, midline: y=-2

31. Amplitude and Midline Example 5 State the midline and amplitude of the following sinusoidal functions:(a) y = 3 sin t + 5 (b) y = (4 − 3 cos t)/20. Solution Rewrite (b) as y = 4/20 − 3/20 cos t = 0.2− 0.15 cos t (a) (b) Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, 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 31

32. Height on the Ferris Wheel as a Function of Angle Example 6 The Ferris wheel has a radius of 200 feet. Graph the height function and find the amplitude and midline. Solution Position the origin at the center of the Ferris wheel, so that the positive x-axis extends horizontally to the right, through the 3 o’clock position P = (x, y) θ Height = 200 + y = 200 + 200 sin θ (0,0) 200 y = 200 sinθ ● 32 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

33. Height on the Ferris Wheel as a Function of Angle Example 6 continued Graph the Ferris wheel function giving your height, h = f(θ), in feet, above ground as a function of the angle θ : f(θ) = 200 + 200 sin θ. Solution A calculator gives the values to graph f(θ). The period of this function is 2  or 360 ◦, because 360 ◦ is one full rotation, so the function repeats every 360 ◦. The midline is h = 200 feet, since the values of h oscillate about this value. The amplitude is also 200 feet, since the maximum value of h is 400 feet. 33 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

34. 7.5 SINUSOIDAL FUNCTIONS 34 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

35. Periodic Functions A function f is periodic if its values repeat at regular intervals. Then if the graph of f is shifted horizontally by c units, for some constant c, the new graph is identical to the original graph. In function notation, periodic means that, for all t in the domain of f, f(t + c) = f(t). The smallest positive constant c for which this relationship holds for all values of t is called the period of f. A function f is periodic if its values repeat at regular intervals. Then if the graph of f is shifted horizontally by c units, for some constant c, the new graph is identical to the original graph. In function notation, periodic means that, for all t in the domain of f, f(t + c) = f(t). The smallest positive constant c for which this relationship holds for all values of t is called the period of f. 35 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

36. Period The functions y = sin(Bt) and y = cos(Bt) have period P = 2π/|B|. 36 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

37. Finding Formulas for Functions Example 2 Find possible formulas for the functions f and g in the graphs Solution This function has period 4π.This 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). 37 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally y = f(t) y = g(t)

38. 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. 38 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

39. Example 4 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 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 Horizontal Shift 39 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

40. 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. 40 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

41. Writing a Formula for the Ferris Wheel Height Example 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 400 feet, so the midline is k = 200 and the amplitude, A, is also 200. The period of the Ferris wheel is 30 minutes, so B=π/15. Because we reach y = 200 (the 3 o’clock position) when t = 7.5, the horizontal shift is h = 7.5, so the Ferris wheel height is f(t) = 200 sin(π/15 (t − 7.5)) + 200. 41 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

42. 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). In the Ferris wheel height function, f(t) = 200 sin(π/15 (t − 7.5)) + 200, rewriting the function in the above form f(t) = 200 sin(π/15 t − π/2) + 200, the phase shift is π/2. 42 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

43. Phase Shift Example 6 (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. f(t)f(t) g(t)g(t) AB 43 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

44. 7.6 THE TANGENT FUNCTION 44 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

45. The Tangent Function Suppose P = (x, y) in the figure is the point on the unit circle specified by the angle θ. We define the function, tangent of θ, or tan θ by tan θ = y / x for x ≠ 0. Since x = cos θ and y = sin θ, we see that tan θ = sin θ/cos θ for cos θ ≠ 0. Suppose P = (x, y) in the figure is the point on the unit circle specified by the angle θ. We define the function, tangent of θ, or tan θ by tan θ = y / x for x ≠ 0. Since x = cos θ and y = sin θ, we see that tan θ = sin θ/cos θ for cos θ ≠ 0. P = (x, y) θ x y 1 45 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

46. Interpreting the Tangent Function as Slope We can think about the tangent function in terms of slope. In the Figure, the line passing from the origin through P has In words, tan θ is the slope of the line passing through the origin and point P. P = (x, y) Line has slope y/x = tan θ (0, 0) x y θ 46 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

47. Graphing the Tangent Function For values of θ between  and 2  observe that tan(θ +  ) = tan θ, because the angles θ and θ +  determine the same line through the origin, and hence the same slope. Thus, y = tan θ has period  or 180 ◦. Since the tangent is not defined when the x-coordinate of P is zero, the graph of the tangent function has a vertical asymptote at θ = −3π/2,− π/2, π/2, 3π/2, etc. Graph of the tangent function 47 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

48. 7.7 TRIGONOMETRIC RELATIONSHIPS AND IDENTITIES 48 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

49. 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 ● P = (x, y) = (cos θ, sin θ) x y θ 1 49 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

50. Relationships Involving The Tangent Func tion 50 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

51. 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 = −3π/4, π/4, 5π/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. y = sin ty = cos t y = tan t 51 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

52. 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. 52 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

53. Relationships Between the Graphs of Secant and Cosine 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. 53 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally g(θ) = cos θ Example 3 Use a graph of g(θ) = cos θ to explain the shape of the graph of f(θ) = sec θ. Solution f(θ) = sec θ

54. Relationships Between the Graphs of Cosecant/Sine and Cotangent/Tangent 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. Plots of y = csc θ and y = sin θPlots of y = cot θ and y = tan θ 54 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

55. Relationships Between the Graphs of Sine and Cosine Example 4 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 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. 55 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally y = cos t y = sin t

56. Relationships Between the Graphs of Sine and Cosine Example 5 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). 56 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

57. 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 57 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

58. 7.8 INVERSE TRIGONOMETRIC FUNCTIONS 58 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

59. 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. 59 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

60. 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. 60 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

61. 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. y = sin t y = tan t - π/2 ≤ t ≤ π/2 - π/2 < t < π/2 61 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

62. 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. 62 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

63. Evaluating the Inverse Sine and Inverse Tangent Functions Example 2 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. 63 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

64. 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. Angles in each quadrant whose reference angles are α xxxx yyyy P P PP αα αα 64 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally

65. Solving Trigonometric Equations Using the Unit Circle Example 3 Solve sin θ = 0.9063 for 0 ◦ ≤ θ ≤ 360 ◦. Solution 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° 65 Functions Modeling Change: A Preparation for Calculus, 5th Edition, 2014, Connally