#set document(title: "4.3 Using Trigonometric Functions", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 4.3#h(0.6em)Using Trigonometric Functions === Solving Trigonometric Equations One of the main reasons we learned about reference angles is to help us solve trigonometric equations. Remember that there are always two angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] with a given sine ratio between 0 and 1. #examplebox("Example 1")[][ You are riding on the Ferris wheel we modeled in Section 4.2, where your height above ground is given as a function of #math.equation(block: false, alt: "θ")[$θ$] by #math.equation(block: true, alt: "h equals 100 plus 100 sin ⁡ open parenthesis θ close parenthesis")[$h = 100 + 100 sin ( θ )$] When your height is at least 136 feet, you can see the snow on a distant mountain range. Between what angles of rotation can you see snow? #solutionbox[ Look at the diagram of the Ferris wheel to see that when your height is 136 feet, the sine of the angle #math.equation(block: false, alt: "θ")[$θ$] is #math.equation(block: false, alt: "the fraction 36 over 100")[$display(frac(36, 100))$], or 0.36. The angle with sine 0.36 is #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis 0.36 close parenthesis equals 21.1 degrees")[$sin^(− 1) ( 0.36 ) = 21.1^(∘)$] #figure(figph[ferris wheel where height is 136], alt: "ferris wheel where height is 136", caption: none) And you will be able to see snow until you descend again below 136 feet. That will happen at the angle in the second quadrant with sine 0.36, namely, the second quadrant angle with reference angle 21.1°. And that angle is #math.equation(block: true, alt: "180 degrees minus 21.1 degrees equals 158.9 degrees")[$180^(∘) − 21.1^(∘) = 158.9^(∘)$] . ] ] As another example, the two solutions to the equation #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.4226")[$" " sin ( θ ) = 0.4226 " "$] are #math.equation(block: false, alt: "25 degrees")[$25^(∘)$] and #math.equation(block: false, alt: "155 degrees")[$155^(∘)$]. That is, #math.equation(block: true, alt: "sin ⁡ open parenthesis 25 degrees close parenthesis equals sin ⁡ open parenthesis 155 degrees close parenthesis equals 0.4226")[$sin ( 25^(∘) ) = sin ( 155^(∘) ) = 0.4226$] These two angles are supplementary, so they have the same reference angle, as shown below on the left. #figure(figph[supplementary angles], alt: "supplementary angles", caption: none) #figure(figph[third and fourth quadrant angles], alt: "third and fourth quadrant angles", caption: none) But now consider the equation #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals minus 0.4226")[$sin ( θ ) = − 0.4226$]. There are two solutions between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$] to this equation also. They are the angles in the third and fourth quadrants whose reference angle is #math.equation(block: false, alt: "25 degrees")[$25^(∘)$], namely #math.equation(block: false, alt: "205 degrees")[$205^(∘)$] and #math.equation(block: false, alt: "335 degrees")[$335^(∘)$], as shown above on the right. But your calculator will not give you one of these angles! When you ask your calculator for an angle whose sine is negative, it will give you a negative angle. For example, you will find #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis minus 0.4226 close parenthesis equals minus 25 degrees")[$sin^(− 1) ( − 0.4226 ) = − 25^(∘)$] Of course, the reference angle for #math.equation(block: false, alt: "minus 25 degrees")[$− 25^(∘)$] is #math.equation(block: false, alt: "25 degrees")[$25^(∘)$], so you can find the solutions between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$] with that reference angle. Solve the equation #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals minus 0.6428")[$" " sin ( θ ) = − 0.6428 " "$] for angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$]. Round your answers to the nearest degree. Use your calculator to evaluate #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis minus 0.6428 close parenthesis equals minus 40 degrees")[$sin^(− 1) ( − 0.6428 ) = − 40^(∘)$] The angle #math.equation(block: false, alt: "minus 40 degrees")[$− 40^(∘)$] is in the fourth quadrant, but it is not between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$]. We need an angle that is coterminal with #math.equation(block: false, alt: "minus 40 degrees")[$− 40^(∘)$], so we add #math.equation(block: false, alt: "360 degrees")[$360^(∘)$]. #math.equation(block: true, alt: "minus 40 degrees plus 360 degrees equals 320 degrees")[$− 40^(∘) + 360^(∘) = 320^(∘)$] Thus, one of the solutions is #math.equation(block: false, alt: "320 degrees")[$320^(∘)$]. The sine is also negative in the third quadrant, so there should also be a solution in the third quadrant. The reference angle for #math.equation(block: false, alt: "320 degrees")[$320^(∘)$] is #math.equation(block: false, alt: "40 degrees")[$40^(∘)$], and the third-quadrant angle with reference angle #math.equation(block: false, alt: "40 degrees")[$40^(∘)$] is #math.equation(block: false, alt: "220 degrees")[$220^(∘)$], as shown at right. You can check that, rounded to four decimal places, both angles satisfy the equation, that is, #figure(figph[angles in 3rd and 4th quadrants], alt: "angles in 3rd and 4th quadrants", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis 320 degrees close parenthesis equals minus 0.6428 and sin ⁡ open parenthesis 220 degrees close parenthesis equals minus 0.6428")[$sin ( 320^(∘) ) = − 0.6428 " " " " " " " " "and" " " " " " " " " sin ( 220^(∘) ) = − 0.6428$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the previous Exercise we found two solutions of the equation #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals minus 0.6428")[$sin ( θ ) = − 0.6428$]. Actually, there are infinitely many solutions — namely, all the angles coterminal with #math.equation(block: false, alt: "320 degrees")[$320^(∘)$] or #math.equation(block: false, alt: "220 degrees")[$220^(∘)$]. Because we can easily find coterminal solutions by adding or subtracting multiples of #math.equation(block: false, alt: "360 degrees")[$360^(∘)$], when solving an equation we usually list only the solutions between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$]. ] What about equations involving the cosine? If the cosine is positive, the solutions will lie in the first and fourth quadrants. #examplebox("Example 2")[][ Refer to the Example in Section 4.2. about the Knee Over Pedal Stem Rule. The distance between the cyclist's foot and the KOPS line is #math.equation(block: true, alt: "d equals 18 minus 18 cos ⁡ open parenthesis θ close parenthesis")[$d = 18 − 18 cos ( θ )$] At what angles is the cyclist's knee within 9 centimeters of the KOPS line? #solutionbox[ We would like to find all angles for which #math.equation(block: false, alt: "d")[$d$] is less than 9. Consider the diagram of the bicycle wheel at right to see that the cosine of the angle at #math.equation(block: false, alt: "d equals 9")[$d = 9$] cm is #math.equation(block: false, alt: "the fraction 9 over 18 equals the fraction 1 over 2")[$display(frac(9, 18)) = display(frac(1, 2))$]. So we need two angles whose cosine is #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$], one in the first quadrant and one in the fourth quadrant. #figure(figph[knee within 9cm of KOPS line], alt: "knee within 9cm of KOPS line", caption: none) You may recognize #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] as the cosine of the special angle 60°, or you can evaluate #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis the fraction 1 over 2 close parenthesis")[$cos^(− 1) ( display(frac(1, 2)) )$]. The fourth quadrant angle with reference angle 60° is 300°. So #math.equation(block: false, alt: "d")[$d$] is less than 9 when the angle of the crank is between 0° and 60°, and again between 300° and 360°. ] ] If the cosine of an angle is negative, your calculator will give you the second quadrant angle with that cosine. For equations involving a negative tangent, your calculator will give you a negative angle. Solve the equation #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus 0.4")[$" " tan ( θ ) = − 0.4 " "$] for angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$]. You can check that #math.equation(block: false, alt: "tan to the power minus 1 minus open parenthesis 0.4 close parenthesis equals minus 21.8 degrees")[$" " tan^(− 1) − ( 0.4 ) = − 21.8^(∘)$]. The tangent is negative in the second and fourth quadrants, and the reference angle is #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis 0.4 close parenthesis equals 21.8 degrees")[$tan^(− 1) ( 0.4 ) = 21.8^(∘)$], so the solutions are #math.equation(block: false, alt: "158.2 degrees")[$158.2^(∘)$] and #math.equation(block: false, alt: "338.2 degrees")[$338.2^(∘)$]. === The Unit Circle Now that we are thinking of angles as rotations, we'll introduce a tool called the #strong[unit circle] that will be useful as we proceed. Figure (a) below shows an angle of #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] in standard position in a circle of radius 2. The hypotenuse of its reference triangle is the radius of the circle, so the legs of the triangle have lengths 1 and #math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]. The coordinates of the point #math.equation(block: false, alt: "P")[$P$] where the terminal side meets the circle are thus #math.equation(block: false, alt: "open parenthesis the square root of 3 , 1 close parenthesis")[$( sqrt(3) , 1 )$]. (You can check that these coordinates satisfy the equation of the circle, #math.equation(block: false, alt: "x squared plus y squared equals 4")[$x^(2) + y^(2) = 4$].) #figure(figph[angle on circle of radius 2], alt: "angle on circle of radius 2", caption: none) #figure(figph[angle in unit circle], alt: "angle in unit circle", caption: none) Now consider the circle of radius 1 in Figure (b). A circle of radius 1 is called a #strong[unit circle]. In this figure, the hypotenuse of the reference triangle for #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] has length 1. What are the coordinates of the point #math.equation(block: false, alt: "Q")[$Q$] where the terminal side meets the circle? Each side of this triangle is #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$] the length of the sides of the similar triangle in Figure (a), so the coordinates of #math.equation(block: false, alt: "Q")[$Q$] are #math.equation(block: false, alt: "open parenthesis the fraction the square root of 3 over 2 , the fraction 1 over 2 close parenthesis")[$( frac(sqrt(3), 2) , frac(1, 2) )$]. (Once again, you should check that these coordinates satisfy the equation of the circle, #math.equation(block: false, alt: "x squared plus y squared equals 1")[$x^(2) + y^(2) = 1$].) Perhaps you recognize the coordinates of the point #math.equation(block: false, alt: "Q")[$Q$]. Because #math.equation(block: false, alt: "r equals 1")[$r = 1$] in this circle, the definitions of the sine and cosine are #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over r equals the fraction x over 1 equals x and sin ⁡ open parenthesis θ close parenthesis equals the fraction y over r equals the fraction y over 1 equals y")[$cos ( θ ) = display(frac(x, r)) = display(frac(x, 1)) = x " " " " " " " " "and" " " " " " " " " sin ( θ ) = display(frac(y, r)) = display(frac(y, 1)) = y$] We see that the coordinates #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] of #math.equation(block: false, alt: "Q")[$Q$] are given by #math.equation(block: false, alt: "open parenthesis cos ⁡ open parenthesis θ close parenthesis , sin ⁡ open parenthesis θ close parenthesis close parenthesis")[$( cos ( θ ) , sin ( θ ) )$]. We have discovered an important property of unit circles. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Angles in a Unit Circle] Let #math.equation(block: false, alt: "P")[$P$] be a point on a unit circle determined by the terminal side of an angle #math.equation(block: false, alt: "θ")[$θ$] in standard position. Then the coordinates #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] of #math.equation(block: false, alt: "P")[$P$] are given by #math.equation(block: true, alt: "x equals cos ⁡ open parenthesis θ close parenthesis , y equals sin ⁡ open parenthesis θ close parenthesis")[$bold(italic(x)) = bold(cos) ( bold(italic(θ)) ) , " " " " " " " " " " " " bold(italic(y)) = bold(sin) ( bold(italic(θ)) )$] ] #examplebox("Example 3")[][ Find the coordinates of point #math.equation(block: false, alt: "P")[$P$] on the unit circle shown at right. #figure(figph[unit circle with angle], alt: "unit circle with angle", caption: none) #solutionbox[ The coordinates of #math.equation(block: false, alt: "P")[$P$] are given by #math.equation(block: true, alt: "x equals cos ⁡ open parenthesis 315 degrees close parenthesis , y equals sin ⁡ open parenthesis 315 degrees close parenthesis")[$x = cos ( 315^(∘) ) , " " " " " " y = sin ( 315^(∘) )$] . The reference angle for #math.equation(block: false, alt: "315 degrees")[$315^(∘)$] is #math.equation(block: false, alt: "360 degrees minus 315 degrees equals 45 degrees")[$" " 360^(∘) − 315^(∘) = 45^(∘)$], so the trig ratios of #math.equation(block: false, alt: "315 degrees")[$315^(∘)$] are the same as the ratios for #math.equation(block: false, alt: "45 degrees")[$45^(∘)$], up to sign. Because #math.equation(block: false, alt: "315 degrees")[$315^(∘)$] is in the fourth quadrant, we have #math.equation(block: false, alt: "cos ⁡ open parenthesis 315 degrees close parenthesis equals the fraction 1 over the square root of 2")[$cos ( 315^(∘) ) = frac(1, sqrt(2))$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis 315 degrees close parenthesis equals the fraction minus 1 over the square root of 2")[$sin ( 315^(∘) ) = frac(− 1, sqrt(2))$]. Thus, the coordinates of #math.equation(block: false, alt: "P")[$P$] are #math.equation(block: false, alt: "open parenthesis the fraction 1 over the square root of 2 , the fraction minus 1 over the square root of 2 close parenthesis")[$( frac(1, sqrt(2)) , frac(− 1, sqrt(2)) )$]. ] ] Find the sine, cosine, and tangent of the angle #math.equation(block: false, alt: "φ")[$φ$] shown at right. #figure(figph[angles], alt: "angles", caption: none) #math.equation(block: false, alt: "y equals minus 0.7")[$y = − 0.7$], so #math.equation(block: false, alt: "sin ⁡ open parenthesis φ close parenthesis equals minus 0.7000")[$sin ( φ ) = − 0.7000$]. Then #math.equation(block: true, alt: "x equals minus the square root of 1 minus y squared equals minus 0.7141 equals cos ⁡ open parenthesis φ close parenthesis")[$x = − sqrt(1 − y^(2)) = − 0.7141 = cos ( φ )$] , and #math.equation(block: false, alt: "tan ⁡ open parenthesis φ close parenthesis equals the fraction y over x equals 0.9802")[$" " tan ( φ ) = display(frac(y, x)) = 0.9802$]. #notebox("Activity", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Unit Circles] #figure(figph[grid with circles], alt: "grid with circles", caption: none) + Use a protractor to draw an angle #math.equation(block: false, alt: "36 degrees")[$36^(∘)$] in standard position. + Estimate the coordinates of the point #math.equation(block: false, alt: "P")[$P$] where the terminal side of the angle intersects the circle of radius #math.equation(block: false, alt: "r equals 2")[$r = 2$]. + Calculate approximate values for #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] using the coordinates of #math.equation(block: false, alt: "P")[$P$]. + Estimate the coordinates of the point #math.equation(block: false, alt: "Q")[$Q$] where the terminal side of the angle intersects the circle of radius #math.equation(block: false, alt: "r equals 1")[$r = 1$]. + Calculate approximate values for #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] using the coordinates of #math.equation(block: false, alt: "Q")[$Q$]. + Use a protractor to draw an angle #math.equation(block: false, alt: "107 degrees")[$107^(∘)$] in standard position. Repeat parts (a)-(d) for this new angle. + Use a protractor to draw an angle #math.equation(block: false, alt: "212 degrees")[$212^(∘)$] in standard position. Repeat parts (a)-(d) for this new angle. + Use a protractor to draw an angle #math.equation(block: false, alt: "325 degrees")[$325^(∘)$] in standard position. Repeat parts (a)-(d) for this new angle. + What do you notice about the coordinates of the point located on the unit circle by an angle and the values of the trig ratios of that angle? ] === Location by Coordinates One of the most useful applications of the trigonometric ratios allows us to find distances or locations specified by angles. Starting with the definitions of sine and cosine, #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over r and sin ⁡ open parenthesis θ close parenthesis equals the fraction y over r")[$cos ( θ ) = display(frac(x, r)) " " " " " " " " "and" " " " " " " " " sin ( θ ) = display(frac(y, r))$] we can solve for #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$], the coordinates of points on the terminal side of the angle, and obtain the following results. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Coordinates] If point #math.equation(block: false, alt: "P")[$P$] is located at a distance #math.equation(block: false, alt: "r")[$r$] from the origin in the direction specified by angle #math.equation(block: false, alt: "θ")[$θ$] in standard position, then the coordinates of #math.equation(block: false, alt: "P")[$P$] are #math.equation(block: true, alt: "x equals r cos ⁡ open parenthesis θ close parenthesis and y equals r sin ⁡ open parenthesis θ close parenthesis")[$bold(italic(x)) = bold(italic(r)) bold(cos) ( bold(italic(θ)) ) " " " " " " " " "and" " " " " " " " " bold(italic(y)) = bold(italic(r)) bold(sin) ( bold(italic(θ)) )$] ] These formulas make sense when we think of the unit circle. On a unit circle, the coordinates of a point designated by angle #math.equation(block: false, alt: "θ")[$θ$] are #math.equation(block: false, alt: "open parenthesis cos ⁡ open parenthesis θ close parenthesis , sin ⁡ open parenthesis θ close parenthesis close parenthesis")[$( cos ( θ ) , sin ( θ ) )$], as shown at right. On a circle of radius #math.equation(block: false, alt: "r")[$r$], the angle #math.equation(block: false, alt: "θ")[$θ$] forms a similar triangle whose dimensions are scaled up by a factor of #math.equation(block: false, alt: "r")[$r$]. In particular, the legs of the new triangle are #math.equation(block: false, alt: "r")[$r$] times larger than the original triangle. #figure(figph[points on circles], alt: "points on circles", caption: none) #examplebox("Example 4")[][ Point #math.equation(block: false, alt: "P")[$P$] is located 6 centimeters from the origin in the direction of #math.equation(block: false, alt: "292 degrees")[$292^(∘)$]. Find the coordinates of #math.equation(block: false, alt: "P")[$P$], rounded to hundredths. #solutionbox[ The location of point #math.equation(block: false, alt: "P")[$P$] is shown at right. We see that #math.equation(block: false, alt: "r equals 6")[$r = 6$], and we can use a calculator to evaluate #math.equation(block: false, alt: "cos ⁡ open parenthesis 292 degrees close parenthesis")[$cos ( 292^(∘) )$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis 292 degrees close parenthesis")[$sin ( 292^(∘) )$]. #math.equation(block: true, alt: "x, equals r cos ⁡ open parenthesis 292 degrees close parenthesis, and, y, equals r sin ⁡ open parenthesis 292 degrees close parenthesis; equals 6 open parenthesis 0.3746 close parenthesis, equals 6 open parenthesis minus 0.9272 close parenthesis; equals 2.2476, equals minus 5.5632")[$x & = r cos ( 292^(∘) ) & & "and" & y & = r sin ( 292^(∘) ) \ & = 6 ( 0.3746 ) & & & & = 6 ( − 0.9272 ) \ & = 2.2476 & & & & = − 5.5632$] #figure(figph[point on angle], alt: "point on angle", caption: none) The coordinates of #math.equation(block: false, alt: "P")[$P$] are approximately #math.equation(block: false, alt: "open parenthesis 2.25 , minus 5.56 close parenthesis")[$( 2.25 , − 5.56 )$]. ] ] You find an old map that shows a buried treasure located 500 yards from the big oak tree in the direction #math.equation(block: false, alt: "215 degrees")[$215^(∘)$], as shown below. You don't have anything with you to measure angles, but you have your calculator. + Find the cosine of #math.equation(block: false, alt: "215 degrees")[$215^(∘)$]. How far west should you walk from the big oak in order to be directly north of the treasure? + Find the sine of #math.equation(block: false, alt: "215 degrees")[$215^(∘)$]. How far south should you walk from your present location before you begin digging? #figure(figph[angle], alt: "angle", caption: none) + #math.equation(block: false, alt: "500 cos ⁡ open parenthesis 35 degrees close parenthesis equals 409.58")[$500 cos ( 35^(∘) ) = 409.58$] yds + #math.equation(block: false, alt: "500 sin ⁡ open parenthesis 35 degrees close parenthesis equals 286.79")[$500 sin ( 35^(∘) ) = 286.79$] yds === Bearings Navigational directions for ships and planes are sometimes given as #strong[bearings], which are angles measured clockwise from north. For example, a bearing of #math.equation(block: false, alt: "110 degrees")[$110^(∘)$] is equivalent to an angle of #math.equation(block: false, alt: "minus 20 degrees")[$− 20^(∘)$] in standard position, or to its coterminal angle #math.equation(block: false, alt: "340 degrees")[$340^(∘)$], as shown at right. #figure(figph[Bearing of 110 degrees], alt: "Bearing of 110 degrees", caption: none) From this example, we see that to convert a bearing to an angle #math.equation(block: false, alt: "θ")[$θ$] in standard position, we can subtract the bearing from #math.equation(block: false, alt: "90 degrees")[$90^(∘)$], or #math.equation(block: true, alt: "θ equals minus bearing plus 90 degrees")[$bold(italic(θ)) = − "bearing" + 90^(∘)$] #examplebox("Example 5")[][ Francine leaves the airport at a bearing of #math.equation(block: false, alt: "245 degrees")[$245^(∘)$] and flies 60 miles. How far south of the airport is she at that time? #solutionbox[ A bearing of #math.equation(block: false, alt: "245 degrees")[$245^(∘)$] is in the same direction as an angle of #math.equation(block: true, alt: "minus 245 degrees plus 90 degrees equals minus 155 degrees")[$− 245^(∘) + 90^(∘) = − 155^(∘)$] in standard position, as shown at right, or as the coterminal angle #math.equation(block: true, alt: "minus 155 degrees plus 360 degrees equals 205 degrees")[$− 155^(∘) + 360^(∘) = 205^(∘)$] . #figure(figph[Bearing of 245 degrees, 60 miles distance], alt: "Bearing of 245 degrees, 60 miles distance", caption: none) We would like the #math.equation(block: false, alt: "y")[$y$]-coordinate of Francine's position, so we calculate #math.equation(block: false, alt: "y equals r sin ⁡ open parenthesis 205 degrees close parenthesis")[$y = r sin ( 205^(∘) )$]. #math.equation(block: true, alt: "y, equals r sin ⁡ open parenthesis 205 degrees close parenthesis; equals 60 open parenthesis minus 0.4226 close parenthesis equals minus 25.36")[$y & = r sin ( 205^(∘) ) \ & = 60 ( − 0.4226 ) = − 25.36$] Francine is about 25.4 miles south of the airport. ] ] Delbert leaves the airport and flies 150 miles at a bearing of #math.equation(block: false, alt: "132 degrees")[$132^(∘)$]. How far east of the airport is he at that time? #math.equation(block: false, alt: "150 cos ⁡ open parenthesis 42 degrees close parenthesis equals 111.5")[$150 cos ( 42^(∘) ) = 111.5$] mile === Angle of Inclination The tangent function also has applications in measurement. The figure at right shows a line in the #math.equation(block: false, alt: "x y")[$x y$]-plane. The angle #math.equation(block: false, alt: "α")[$α$] measured in the positive direction from the positive #math.equation(block: false, alt: "x")[$x$]-axis to the line is called the #strong[angle of inclination] of the line. #figure(figph[angle of inclination], alt: "angle of inclination", caption: none) Recall that the #strong[slope] of a line is given by the ratio #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x equals the fraction change in y over change in x")[$m = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac("change in" " " y, "change in" " " x))$] as we move from one point to another on the line. So, if we create a right triangle by dropping a perpendicular segment from the line to the #math.equation(block: false, alt: "x")[$x$]-axis, the ratio of sides #math.equation(block: false, alt: "the fraction opposite over adjacent")[$display(frac("opposite", "adjacent"))$] gives the slope of the line. #figure(figph[slope of angle of inclination], alt: "slope of angle of inclination", caption: none) But the ratio #math.equation(block: false, alt: "the fraction opposite over adjacent")[$display(frac("opposite", "adjacent"))$] is also the tangent of the angle #math.equation(block: false, alt: "α")[$α$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Angle of Inclination] The #strong[angle of inclination] of a line is the angle #math.equation(block: false, alt: "α")[$α$] measured in the positive direction from the positive #math.equation(block: false, alt: "x")[$x$]-axis to the line. If the slope of the line is #math.equation(block: false, alt: "m")[$m$], then #math.equation(block: true, alt: "tan ⁡ open parenthesis α close parenthesis equals m")[$bold(tan) ( bold(italic(α)) ) = bold(italic(m))$] where #math.equation(block: false, alt: "0 degrees less than or equal to α less than or equal to 180 degrees")[$0^(∘) ≤ α ≤ 180^(∘)$]. ] #examplebox("Example 6")[][ Find the angle of inclination of the line #math.equation(block: false, alt: "y equals the fraction 3 over 4 x minus 3")[$" " y = frac(3, 4) x − 3$]. #solutionbox[ The slope of the line is #math.equation(block: false, alt: "the fraction 3 over 4")[$frac(3, 4)$]. Therefore, #math.equation(block: true, alt: "tan ⁡ open parenthesis α close parenthesis, equals the fraction 3 over 4; α, equals tan to the power minus 1 ⁡ open parenthesis the fraction 3 over 4 close parenthesis equals 36.9 degrees")[$tan ( α ) & = display(frac(3, 4)) \ α & = tan^(− 1) ( display(frac(3, 4)) ) = 36.9^(∘)$] The angle of inclination is #math.equation(block: false, alt: "36.9 degrees")[$36.9^(∘)$]. #figure(figph[graph y = 3/4 x - 3], alt: "graph y = 3/4 x - 3", caption: none) ] ] Find the angle of inclination of the line shown at right, #math.equation(block: true, alt: "y equals the fraction minus 6 over 5 x plus 2")[$y = display(frac(− 6, 5)) x + 2$] #figure(figph[graph y= -6/5 x +2], alt: "graph y= -6/5 x +2", caption: none) #math.equation(block: true, alt: "180 degrees plus tan to the power minus 1 ⁡ open parenthesis the fraction minus 6 over 5 close parenthesis equals 129.8 degrees")[$180^(∘) + tan^(− 1) ( display(frac(− 6, 5)) ) = 129.8^(∘)$] Thinking about slope and the angle of inclination can help us understand the graph of the tangent function. #examplebox("Example 7")[][ + What happens to the slope of a line as its angle of inclination #math.equation(block: false, alt: "α")[$α$] increases from #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] toward #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]? + What happens to the slope of a line as its angle of inclination #math.equation(block: false, alt: "α")[$α$] decreases from #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] toward #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]? #solutionbox[ + Consider the slopes of the lines shown at right, as #math.equation(block: false, alt: "α")[$α$] increases. #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "α")[$α$]], [#math.equation(block: false, alt: "0 degrees")[$0^(∘)$]], [#math.equation(block: false, alt: "30 degrees")[$30^(∘)$]], [#math.equation(block: false, alt: "45 degrees")[$45^(∘)$]], [#math.equation(block: false, alt: "60 degrees")[$60^(∘)$]], [#math.equation(block: false, alt: "75 degrees")[$75^(∘)$]], [#math.equation(block: false, alt: "85 degrees")[$85^(∘)$]]), [#math.equation(block: false, alt: "m")[$m$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.6")[$0.6$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1.7")[$1.7$]], [#math.equation(block: false, alt: "3.7")[$3.7$]], [#math.equation(block: false, alt: "11.4")[$11.4$]], )) #figure(figph[lines with different angles of inclination], alt: "lines with different angles of inclination", caption: none) The slopes increase towards infinity as #math.equation(block: false, alt: "α")[$α$] increases. But the values of the slope #math.equation(block: false, alt: "m")[$m$] are the values of #math.equation(block: false, alt: "tan ⁡ open parenthesis α close parenthesis")[$tan ( α )$], and if we plot those values against #math.equation(block: false, alt: "α")[$α$], we see the graph of the tangent function emerging. #figure(figph[tangent graph], alt: "tangent graph", caption: none) + Now consider the slopes of the lines as #math.equation(block: false, alt: "α")[$α$] decreases from #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] toward #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "180 degrees")[$180^(∘)$]], [#math.equation(block: false, alt: "150 degrees")[$150^(∘)$]], [#math.equation(block: false, alt: "135 degrees")[$135^(∘)$]], [#math.equation(block: false, alt: "120 degrees")[$120^(∘)$]], [#math.equation(block: false, alt: "105 degrees")[$105^(∘)$]], [#math.equation(block: false, alt: "95 degrees")[$95^(∘)$]]), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 0.6")[$− 0.6$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus 1.7")[$− 1.7$]], [#math.equation(block: false, alt: "minus 3.7")[$− 3.7$]], [#math.equation(block: false, alt: "minus 11.5")[$− 11.5$]], )) #figure(figph[lines with different angles of inclination], alt: "lines with different angles of inclination", caption: none) The slopes decrease towards negative infinity as #math.equation(block: false, alt: "α")[$α$] decreases. But, as before, the values of the slope #math.equation(block: false, alt: "m")[$m$] are the values of #math.equation(block: false, alt: "tan ⁡ open parenthesis α close parenthesis")[$tan ( α )$], and if we plot those values against #math.equation(block: false, alt: "α")[$α$], we see the graph of the tangent function. #figure(figph[tangent graph], alt: "tangent graph", caption: none) ] ] + What happens to the angle of inclination #math.equation(block: false, alt: "α")[$α$] of a line as its slope increases from #math.equation(block: false, alt: "0")[$0$] toward #math.equation(block: false, alt: "∞")[$upright(∞)$]? + What happens to the angle of inclination #math.equation(block: false, alt: "α")[$α$] of a line as its slope decreases through negative values from #math.equation(block: false, alt: "0")[$0$] toward #math.equation(block: false, alt: "minus ∞")[$− upright(∞)$]? + The angle increases from #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] toward #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. + The angle decreases from #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] toward #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] === Sinusoidal Functions Many interesting functions have graphs shaped like sines or cosines, even though they may not be functions of angles. #examplebox("Example 8")[][ Imagine a grandfather clock. As the minute hand sweeps around, the height of its tip changes with time. Which of the graphs shown below best represents the height of the tip of the minute hand as a function of time? #figure(figph[3 graphs], alt: "3 graphs", caption: none) #solutionbox[ Figure (a) is not the graph of a function at all: It does not pass the vertical line test. That is, some values of #math.equation(block: false, alt: "t")[$t$], such as #math.equation(block: false, alt: "t equals 0")[$t = 0$], correspond to more than one value of #math.equation(block: false, alt: "h")[$h$], which is not possible in the graph of a function. Figure (b) shows the height of the minute hand varying between a maximum and minimum value. The height decreases at a constant rate (the graph is straight and the slope is constant) until the minimum is reached, and then increases at a constant rate. But notice that during the 10 minutes from 12:10 to 12:20 the height of the minute hand decreases about half the diameter of the clock, while from 12:20 to 12:30 the height decreases only about a quarter of the diameter of the clock, as shown at right. #figure(figph[clock], alt: "clock", caption: none) Thus the height of the minute hand does not decrease at a constant rate. Figure (c) is the best choice. The graph is curved because the slopes are not constant. The graph is steep when the height is changing rapidly, and the graph is nearly horizontal when the height is changing slowly. The height changes slowly near the hour and the half-hour, and more rapidly near the quarter-hours. ] ] As the moon revolves around the earth, the percent of the disk that we see varies sinusoidally with a period of approximately 30 days. There are eight phases, starting with the new moon, when the moon's disk is dark, followed by waxing crescent, first quarter, waxing gibbous, full moon, waning gibbous, last quarter, and waning crescent. Which graph best represents the phases of the moon? + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) (b), because it has the correct period. #examplebox("Example 9")[][ The table shows the number of hours of daylight in Glasgow, Scotland on the first of each month. #figure(table( columns: 13, align: left, inset: 6pt, table.header([Month], [Jan], [Feb], [Mar], [Apr], [May], [Jun], [Jul], [Aug], [Sep], [Oct], [Nov], [Dec]), [Daylight Hours], [#math.equation(block: false, alt: "7.1")[$7.1$]], [#math.equation(block: false, alt: "8.7")[$8.7$]], [#math.equation(block: false, alt: "10.7")[$10.7$]], [#math.equation(block: false, alt: "13.1")[$13.1$]], [#math.equation(block: false, alt: "15.3")[$15.3$]], [#math.equation(block: false, alt: "17.2")[$17.2$]], [#math.equation(block: false, alt: "17.5")[$17.5$]], [#math.equation(block: false, alt: "16.7")[$16.7$]], [#math.equation(block: false, alt: "13.8")[$13.8$]], [#math.equation(block: false, alt: "11.5")[$11.5$]], [#math.equation(block: false, alt: "9.2")[$9.2$]], [#math.equation(block: false, alt: "7.5")[$7.5$]], )) + Sketch a sinusoidal graph of daylight hours as a function of time, with #math.equation(block: false, alt: "t equals 1")[$t = 1$] in January. + Estimate the period, amplitude, and midline of the graph. #solutionbox[ + Plot the data points and fit a sinusoidal curve by eye, as shown below.#figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + The period of the graph is 12 months. The midline is approximately #math.equation(block: false, alt: "y equals 12.25 ,")[$y = 12.25 ,$] and the amplitude is approximately 5.25. ] ] The figure shows the number of daylight hours in Jacksonville, Florida, in Anchorage, Alaska, at the Arctic Circle, and at the Equator. #figure(figph[graph of daylight hours], alt: "graph of daylight hours", caption: none) + Which graph corresponds to each location? + What are the maxium and minimum number of daylight hours in Jacksonville? + For how long are there 24 hours of daylight per day at the Arctic Circle? + At the Equator there is no variation in the hours of daylight, so it is the constant graph at height 12. The greatest variation in the hours of daylight occurs farthest north, at the Arctic Circle, so it is the green graph, then red for Anchorage and blue for Jacksonville. + 14 hours and 10 hours + Four months === Other Periodic Functions There are other periodic functions besides sinusoidal functions. Any function that repeats a pattern at intervals of fixed length is periodic. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Periodic Function] The function #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] is #strong[periodic] if there is a smallest value of #math.equation(block: false, alt: "p")[$p$] such that #math.equation(block: true, alt: "f open parenthesis x plus p close parenthesis equals f open parenthesis x close parenthesis")[$bold(italic(f)) ( bold(italic(x)) + bold(italic(p)) ) = bold(italic(f)) ( bold(italic(x)) )$] for all #math.equation(block: false, alt: "x")[$x$]. The constant #math.equation(block: false, alt: "p")[$p$] is called the #strong[period] of the function. ] #examplebox("Example 10")[][ Which of the functions shown below are periodic? If the function is periodic, give its period. #figure(figph[Arccos graph], alt: "Arccos graph", caption: none) #figure(figph[polynomial graph], alt: "polynomial graph", caption: none) #figure(figph[periodic graph], alt: "periodic graph", caption: none) #solutionbox[ + This graph is periodic with period 360. + This graph is not periodic. + This graph is periodic with period 8. ] ] Which of the functions shown below are periodic? If the function is periodic, give its period. #figure(figph[periodic graph], alt: "periodic graph", caption: none) #figure(figph[periodic graph], alt: "periodic graph", caption: none) #figure(figph[not periodic graph], alt: "not periodic graph", caption: none) + Period: 2 + Period: 3 + Not periodic #examplebox("Example 11")[][ A patient receives regular doses of medication to maintain a certain level of the drug in his body. After each dose, the patient's body eliminates a certain percent of the medication before the next dose is administered. The graph shows the amount of the drug, in milliliters, in the patient's body as a function of time in hours. #figure(figph[Piecewise decreasing exponential graph], alt: "Piecewise decreasing exponential graph", caption: none) + How much of the medication is administered with each dose? + How often is the medication administered? + What percent of the drug is eliminated from the body between doses? #solutionbox[ + The medication level increase from 30 ml to 50 ml at each cycle of the graph, so 20 ml of medication are administerd at each dose. + The medication level peaks sharply evey four hours, when each new dose is administered. + The medication level declines by 20 ml between doses, or #math.equation(block: false, alt: "the fraction 20 over 50 equals 0.4")[$display(frac(20, 50)) = 0.4$], or 40% ] ] You are sitting on your front porch late one evening, and you see a light coming down the road tracing out the path shown below, with distances in inches. You realize that you are seeing a bicycle light, fixed to the front wheel of the bike. #figure(figph[Curtate troichoid traced by fixed point inside circle as circle rolls without slipping], alt: "Curtate troichoid traced by fixed point inside circle as circle rolls without slipping", caption: none) + Approximately what is the period of the graph? + How far above the ground is the light? + What is the diameter of the bicycle wheel? + 75 in + 4 in + 24 in Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Sketch the graph. Give the equations of any vertical or horizontal asymptotes. + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x minus 2")[$f ( x ) = display(frac(1, x − 2))$] + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction 1 over x plus 3")[$g ( x ) = display(frac(1, x + 3))$] + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals the fraction 1 over x squared minus 1")[$h ( x ) = display(frac(1, x^(2) − 1))$] + #math.equation(block: false, alt: "H open parenthesis x close parenthesis equals the fraction minus 1 over open parenthesis x plus 1 close parenthesis squared")[$H ( x ) = display(frac(− 1, ( x + 1 )^(2)))$] + #math.equation(block: false, alt: "G open parenthesis x close parenthesis equals the fraction 2 x plus 1 over x")[$G ( x ) = display(frac(2 x + 1, x))$] + #math.equation(block: false, alt: "F open parenthesis x close parenthesis equals the fraction x over x minus 1")[$F ( x ) = display(frac(x, x − 1))$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #figure(figph[graph], alt: "graph", caption: none) #linebreak() #math.equation(block: false, alt: "x equals 2 , y equals 0")[$x = 2 , " " " " y = 0$] + #figure(figph[graph], alt: "graph", caption: none) #linebreak() #math.equation(block: false, alt: "x equals minus 3 , y equals 0")[$x = − 3 , " " " " y = 0$] + #figure(figph[graph], alt: "graph", caption: none) #linebreak() #math.equation(block: false, alt: "x equals minus 1 , x equals 1 y equals 0")[$x = − 1 , " " " " x = 1 " " " " y = 0$] + #figure(figph[graph], alt: "graph", caption: none) #linebreak() #math.equation(block: false, alt: "x equals minus 1 , y equals 0")[$x = − 1 , " " " " y = 0$] + #figure(figph[graph], alt: "graph", caption: none) #linebreak() #math.equation(block: false, alt: "x equals 0 , y equals 2")[$x = 0 , " " " " y = 2$] + #figure(figph[graph], alt: "graph", caption: none) #linebreak() #math.equation(block: false, alt: "x equals 1 , y equals 1")[$x = 1 , " " " " y = 1$] ] === Section 4.3 Summary ==== Vocabulary - Unit Circle - Angle of Inclination - Bearings - Period - Periodic ==== Concepts + To solve an equation of the form #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals k")[$sin ( θ ) = k$], or #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals k")[$cos ( θ ) = k$], or #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals k")[$tan ( θ ) = k$], we can use the appropriate inverse trig key on a calculator to find one solution (or a coterminal angle.) We use reference angles to find a second solution between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$]. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Angles in a Unit Circle] Let #math.equation(block: false, alt: "P")[$P$] be a point on a unit circle determined by the terminal side of an angle #math.equation(block: false, alt: "θ")[$θ$] in standard position. Then the coordinates #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] of #math.equation(block: false, alt: "P")[$P$] are given by #math.equation(block: true, alt: "x equals cos ⁡ open parenthesis θ close parenthesis , y equals sin ⁡ open parenthesis θ close parenthesis")[$x = cos ( θ ) , " " " " " " " " " " " " y = sin ( θ )$] ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Coordinates] If point #math.equation(block: false, alt: "P")[$P$] is located at a distance #math.equation(block: false, alt: "r")[$r$] from the origin in the direction specified by angle #math.equation(block: false, alt: "θ")[$θ$] in standard position, then the coordinates of #math.equation(block: false, alt: "P")[$P$] are #math.equation(block: true, alt: "x equals r cos ⁡ open parenthesis θ close parenthesis and y equals r sin ⁡ open parenthesis θ close parenthesis")[$x = r cos ( θ ) " " " " " " " " "and" " " " " " " " " y = r sin ( θ )$] ] + Navigational directions for ships and planes are sometimes given as #strong[bearings], which are angles measured clockwise from north. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Angle of Inclination] The #strong[angle of inclination] of a line is the angle #math.equation(block: false, alt: "α")[$α$] measured in the positive direction from the positive #math.equation(block: false, alt: "x")[$x$]-axis to the line. If the slope of the line is #math.equation(block: false, alt: "m")[$m$], then #math.equation(block: true, alt: "tan ⁡ open parenthesis α close parenthesis equals m")[$tan ( α ) = m$]where #math.equation(block: false, alt: "0 degrees less than or equal to α less than or equal to 180 degrees")[$0^(∘) ≤ α ≤ 180^(∘)$]. ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Periodic Function] The function #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] is #strong[periodic] if there is a smallest value of #math.equation(block: false, alt: "p")[$p$] such that #math.equation(block: true, alt: "f open parenthesis x plus p close parenthesis equals f open parenthesis x close parenthesis")[$f ( x + p ) = f ( x )$]for all #math.equation(block: false, alt: "x")[$x$]. The constant #math.equation(block: false, alt: "p")[$p$] is called the #strong[period] of the function. ] ==== Study Questions + If the angle of inclination of a line is greater than #math.equation(block: false, alt: "45 degrees")[$45^(∘)$], what can you say about its slope? + Sketch two examples of a function with period 8: one that is sinusoidal, and one that is not. + Explain why the coordinates of points on a unit circle are given by the cosine and sine of an angle in standard position. + How are bearings measured? ==== Skills + Solve trigonometric equations, graphically and algebraically \#1-20 + Find coordinates of points on circles \#21-36 + Use bearings to determine position \#37-42 + Find and use the angle of inclination of a line \#43-50 + Identify periodic functions and give their periods \#51–54 + Sketch periodic functions \#55-58 + Sketch graphs to model sinusoidal functions \#59-68 + Analyze periodic graphs \#69-76 === Homework 4.3 For Problems 1–8, use the graphs to estimate the solutions to the equations. Show your work on the graph. #figure(figph[sine graph], alt: "sine graph", caption: none) #figure(figph[cosine graph], alt: "cosine graph", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.6")[$sin ( θ ) = 0.6$] #math.equation(block: true, alt: "36.9 degrees , 143.1 degrees")[$36.9^(∘) , " " 143.1^(∘)$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals minus 0.8")[$sin ( θ ) = − 0.8$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0.3")[$cos ( θ ) = 0.3$] #math.equation(block: true, alt: "72.5 degrees , 287.5 degrees")[$72.5^(∘) , " " 287.5^(∘)$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals minus 0.4")[$cos ( θ ) = − 0.4$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals minus 0.2")[$sin ( θ ) = − 0.2$] #math.equation(block: true, alt: "191.5 degrees , 348.5 degrees")[$191.5^(∘) , " " 348.5^(∘)$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals 1.2")[$sin ( θ ) = 1.2$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals minus 0.9")[$cos ( θ ) = − 0.9$] #math.equation(block: true, alt: "154.2 degrees , 205.8 degrees")[$154.2^(∘) , " " 205.8^(∘)$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals minus 1.1")[$cos ( θ ) = − 1.1$] For Problems 9–14, find all solutions between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$]. Round to the nearest degree. #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals 8.1443")[$tan ( θ ) = 8.1443$] #math.equation(block: true, alt: "83 degrees , 263 degrees")[$83^(∘) , " " 263^(∘)$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.7880")[$sin ( θ ) = 0.7880$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0.9205")[$cos ( θ ) = 0.9205$] #math.equation(block: true, alt: "23 degrees , 337 degrees")[$23^(∘) , " " 337^(∘)$] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus 3.4874")[$tan ( θ ) = − 3.4874$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals minus 0.9962")[$sin ( θ ) = − 0.9962$] #math.equation(block: true, alt: "265 degrees , 275 degrees")[$265^(∘) , " " 275^(∘)$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals minus 0.0349")[$cos ( θ ) = − 0.0349$] For Problems 15–20, find exact values for all solutions between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "360 degrees")[$360^(∘)$]. #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals minus cos ⁡ open parenthesis 24 degrees close parenthesis")[$cos ( θ ) = − cos ( 24^(∘) )$] #math.equation(block: true, alt: "156 degrees , 204 degrees")[$156^(∘) , " " 204^(∘)$] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus tan ⁡ open parenthesis 9 degrees close parenthesis")[$tan ( θ ) = − tan ( 9^(∘) )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals minus sin ⁡ open parenthesis 66 degrees close parenthesis")[$sin ( θ ) = − sin ( 66^(∘) )$] #math.equation(block: true, alt: "246 degrees , 294 degrees")[$246^(∘) , " " 294^(∘)$] #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals minus cos ⁡ open parenthesis 78 degrees close parenthesis")[$cos ( θ ) = − cos ( 78^(∘) )$] #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus tan ⁡ open parenthesis 31 degrees close parenthesis")[$tan ( θ ) = − tan ( 31^(∘) )$] #math.equation(block: true, alt: "149 degrees , 329 degrees")[$149^(∘) , " " 329^(∘)$] #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals minus sin ⁡ open parenthesis 42 degrees close parenthesis")[$sin ( θ ) = − sin ( 42^(∘) )$] For Problems 21–24, + Use a calculator to find the coordinates of the point #math.equation(block: false, alt: "P")[$P$]. Round to hundredths. + Find the coordinates of the point #math.equation(block: false, alt: "Q")[$Q$] on the circle of radius 2. #figure(figph[circle], alt: "circle", caption: none) + #math.equation(block: false, alt: "open parenthesis minus 0.94 , minus 0.34 close parenthesis")[$( − 0.94 , − 0.34 )$] + #math.equation(block: false, alt: "open parenthesis minus 1.88 , minus 0.68 close parenthesis")[$( − 1.88 , − 0.68 )$] #figure(figph[circle], alt: "circle", caption: none) #figure(figph[circle], alt: "circle", caption: none) + #math.equation(block: false, alt: "open parenthesis minus 0.94 , 0.34 close parenthesis")[$( − 0.94 , 0.34 )$] + #math.equation(block: false, alt: "open parenthesis minus 1.88 , 0.68 close parenthesis")[$( − 1.88 , 0.68 )$] #figure(figph[circle], alt: "circle", caption: none) For Problems 25–30, find exact values for the coordinates of the point. #figure(figph[circle], alt: "circle", caption: none) #math.equation(block: true, alt: "open parenthesis 4 the square root of 2 , minus 4 the square root of 2 close parenthesis")[$( 4 sqrt(2) , − 4 sqrt(2) )$] #figure(figph[circle], alt: "circle", caption: none) #figure(figph[circle], alt: "circle", caption: none) #math.equation(block: true, alt: "open parenthesis minus 10 , minus 10 the square root of 3 close parenthesis")[$( − 10 , − 10 sqrt(3) )$] #figure(figph[circle], alt: "circle", caption: none) #figure(figph[circle], alt: "circle", caption: none) #math.equation(block: true, alt: "open parenthesis the fraction minus 15 the square root of 3 over 2 , the fraction 15 over 2 close parenthesis")[$( display(frac(− 15 sqrt(3), 2)) , display(frac(15, 2)) )$] #figure(figph[circle], alt: "circle", caption: none) For Problems 31–36, find the coordinates of the point, rounded to hundredths. #figure(figph[circle], alt: "circle", caption: none) #math.equation(block: true, alt: "open parenthesis minus 1.25 , minus 5.87 close parenthesis")[$( − 1.25 , − 5.87 )$] #figure(figph[circle], alt: "circle", caption: none) #figure(figph[circle], alt: "circle", caption: none) #math.equation(block: true, alt: "open parenthesis 5.70 , minus 11.86 close parenthesis")[$( 5.70 , − 11.86 )$] #figure(figph[circle], alt: "circle", caption: none) #figure(figph[circle], alt: "circle", caption: none) #math.equation(block: true, alt: "open parenthesis 9.46 , minus 3.26 close parenthesis")[$( 9.46 , − 3.26 )$] #figure(figph[circle], alt: "circle", caption: none) For Problems 37–41, a ship sails from the seaport on the given bearing for the given distance. + Make a sketch showing the ship's current location relative to the seaport. + How far east or west of the seaport is the ship's present location? How far north or south? #math.equation(block: false, alt: "36 degrees")[$36^(∘)$], 26 miles + #figure(figph[angle], alt: "angle", caption: none) + 15.3 mi east, 21 mi north #math.equation(block: false, alt: "124 degrees")[$124^(∘)$], 80 km #math.equation(block: false, alt: "230 degrees")[$230^(∘)$], 120 km + #figure(figph[angle], alt: "angle", caption: none) + 91.9 km west, 77.1 km south #math.equation(block: false, alt: "318 degrees")[$318^(∘)$], 75 miles #math.equation(block: false, alt: "285 degrees")[$285^(∘)$], 32 km + #figure(figph[angle], alt: "angle", caption: none) + 30.9 km west, 8.3 km north #math.equation(block: false, alt: "192 degrees")[$192^(∘)$], 260 miles For Problems 43–46, find the angle of inclination of the line. #math.equation(block: true, alt: "y equals the fraction 5 over 4 x minus 3")[$y = display(frac(5, 4)) x − 3$] #math.equation(block: true, alt: "51.34 degrees")[$51.34^(∘)$] #math.equation(block: true, alt: "y equals 6 plus the fraction 2 over 9 x")[$y = 6 + display(frac(2, 9)) x$] #math.equation(block: true, alt: "y equals minus 2 minus the fraction 3 over 8 x")[$y = − 2 − display(frac(3, 8)) x$] #math.equation(block: true, alt: "159.44 degrees")[$159.44^(∘)$] #math.equation(block: true, alt: "y equals the fraction minus 7 over 2 x plus 1")[$y = display(frac(− 7, 2)) x + 1$] For Problems 47–50, find an equation for the line passing through the given point with angle of inclination #math.equation(block: false, alt: "α")[$α$]. #math.equation(block: true, alt: "open parenthesis 3 , minus 5 close parenthesis , α equals 28 degrees")[$( 3 , − 5 ) , " " α = 28^(∘)$] #math.equation(block: false, alt: "y plus 5 equals open parenthesis tan ⁡ 28 degrees close parenthesis open parenthesis x minus 3 close parenthesis")[$y + 5 = ( tan 28^(∘) ) ( x − 3 ) " "$] or #math.equation(block: false, alt: "y plus 5 equals 0.532 open parenthesis x minus 3 close parenthesis")[$" " y + 5 = 0.532 ( x − 3 )$] #math.equation(block: true, alt: "open parenthesis minus 2 , 6 close parenthesis , α equals 67 degrees")[$( − 2 , 6 ) , " " α = 67^(∘)$] #math.equation(block: true, alt: "open parenthesis minus 8 , 12 close parenthesis , α equals 112 degrees")[$( − 8 , 12 ) , " " α = 112^(∘)$] #math.equation(block: false, alt: "y minus 12 equals open parenthesis tan ⁡ 112 degrees close parenthesis open parenthesis x plus 8 close parenthesis")[$y − 12 = ( tan 112^(∘) ) ( x + 8 ) " "$] or #math.equation(block: false, alt: "y minus 12 equals minus 2.475 open parenthesis x plus 8 close parenthesis")[$" " y − 12 = − 2.475 ( x + 8 )$] #math.equation(block: true, alt: "open parenthesis minus 4 , minus 1 close parenthesis , α equals 154 degrees")[$( − 4 , − 1 ) , " " α = 154^(∘)$] Which of the graphs in Problems Problems 51–54 are periodic? If the graph is periodic, give its period. #figure(figph[graph], alt: "graph", caption: none) not periodic #figure(figph[graph], alt: "graph", caption: none) #figure(figph[graph], alt: "graph", caption: none) Periodic with period 4 #figure(figph[graph], alt: "graph", caption: none) For Problems 55–58, sketch a periodic function that models the situation. At a ski slope, the lift chairs take 5 minutes to travel from the bottom, at an elevation of 3000 feet, to the top, at elevation 4000 feet. The cable supporting the ski lift chairs is a loop turning on pulleys at a constant speed. At the top and bottom, the chairs are at a constant elevation for a few seconds to allow skiers to get on and off. + Sketch a graph of #math.equation(block: false, alt: "h open parenthesis t close parenthesis")[$h ( t )$], the height of one chair at time #math.equation(block: false, alt: "t")[$t$]. Show at least two complete trips. + What is the period of #math.equation(block: false, alt: "h open parenthesis t close parenthesis")[$h ( t )$]? + #figure(figph[piecewise linear graph], alt: "piecewise linear graph", caption: none) + 10 minutes The heater in Paul's house doesn't have a thermostat; it runs on a timer. It uses 300 watts when it is running. Paul sets the heater to run from 6 am to noon, and again from 4 pm to 10 pm in the evening. + Sketch a graph of #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$], the power drawn by the heater as a function of time. Show at least two days of heater use. + What is the period of #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$]? Francine adds water to her fish pond once a week to keep the depth at 30 centimeters. During the week the water evaporates at a constant rate of 0.5 centimeters per day. + Sketch a graph of #math.equation(block: false, alt: "D open parenthesis t close parenthesis")[$D ( t )$], the depth of the water as a function of time. Show at least two weeks. + What is the period of #math.equation(block: false, alt: "D open parenthesis t close parenthesis")[$D ( t )$]? + #figure(figph[piercewise linear graph], alt: "piercewise linear graph", caption: none) + 1 week Erin's fox terrier, Casey, is very energetic and bounces excitedly at dinner time. Casey can jump 30 inches high, and each jump takes him 0.8 seconds. + Sketch a graph of Casey's height, #math.equation(block: false, alt: "h open parenthesis t close parenthesis")[$h ( t )$], as a function of time. Show at least two jumps. + What is the period of #math.equation(block: false, alt: "h open parenthesis t close parenthesis")[$h ( t )$]? For Problems 59–64, sketch a sinusoidal function that models the situation. Delbert's bicycle wheel is 24 inches in diameter, and he has a light attached to the spokes 10 inches from the center of the wheel. It is dark, and he is cycling home slowly from work. The bicycle wheel makes one revolution every second. + At #math.equation(block: false, alt: "t equals 0")[$t = 0$], the light is at its highest point the bicycle wheel. Sketch a graph of the light's height as a function of #math.equation(block: false, alt: "t")[$t$]. + Give the period, midline, and amplitude of your graph. + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + period 1 sec, midline #math.equation(block: false, alt: "y equals 12")[$y = 12$], amp 10 inches The paddlewheel on the Delta Queen steamboat is 28 feet in diameter, and is rotating once every ten seconds. The bottom of the paddlewheel is 4 feet below the surface of the water. + The ship's logo is painted on the center of one of the paddlewheel blades. At #math.equation(block: false, alt: "t equals 0")[$t = 0$], the logo is at the top of the wheel. Sketch a graph of the logo's height above the water as a function of #math.equation(block: false, alt: "t")[$t$]. + Give the period, midline, and amplitude of your graph. The population of mosquitoes at Marsh Lake is a sinusoidal function of time. The population peaks around June 1 at about 6000 mosquitoes per square kilometer, and is smallest on December 1, at 1000 mosquitoes per square kilometer. + Sketch a graph of #math.equation(block: false, alt: "M open parenthesis t close parenthesis")[$M ( t )$], the number of mosquitoes as a function of the month, where #math.equation(block: false, alt: "t equals 0")[$t = 0$] on January 1. + Give the period, midline, and amplitude of your graph. + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + period 1 year, midline #math.equation(block: false, alt: "y equals 3500")[$y = 3500$], amp 2500 The height of the tide in Cabot Cove can be approximated by a sinusoidal function. At 5 am on July 23, the water level reached its high mark at the 20-foot line on the pier, and at 11 am, the water level was at its lowest at the 4-foot line. + Sketch a graph of #math.equation(block: false, alt: "W open parenthesis t close parenthesis")[$W ( t )$], the water level as a function of time, from midnight on July 23 to midnight on July 24. + Give the period, midline, and amplitude of your graph. The average daily maximum temperature in Stockholm, Sweden is #math.equation(block: false, alt: "30 degrees")[$30^(∘)$]F in January and #math.equation(block: false, alt: "72 degrees")[$72^(∘)$] F in July. + Sketch a sinusoidal graph of #math.equation(block: false, alt: "S open parenthesis t close parenthesis")[$S ( t )$], the average maximum temperature in Stockholm as a function of time, for one year. + Give the period, midline, and amplitude of your graph. + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + period 1 year, midline #math.equation(block: false, alt: "y equals 51")[$y = 51$], amp 21 The average daily maximum temperature in Riyadh, Saudi Arabia is #math.equation(block: false, alt: "86 degrees")[$86^(∘)$]F in January and #math.equation(block: false, alt: "113 degrees")[$113^(∘)$]F in July. + Sketch a sinusoidal graph of #math.equation(block: false, alt: "R open parenthesis t close parenthesis")[$R ( t )$], the average maximum temperature in Riyadh as a function of time, for one year. + Give the period, midline, and amplitude of your graph. Each situation describes a periodic function. Match each situation with the appropriate graph. + When the heart contracts, blood pressure in the arteries rises rapidly to a peak (systolic blood pressure) and then falls off quickly to a minimum (diastolic blood pressure). Blood pressure is a function of time. + After an injection is given to a patient, the amount of the drug present in his bloodstream decreases over time. The patient receives injections at regular intervals to restore the drug level to the prescribed level. The amount of the drug present is a function of time. + The monorail shuttle train between the north and south terminals at Gatwick Airport departs from the south terminal every 12 minutes. The distance from the train to the south terminal is a function of time. + Delbert gets a haircut every two weeks. The length of his hair is a function of time. #figure(figph[graph], alt: "graph", caption: none) a. IV b. III c. II d. I Match each of the following situations with an appropriate graph below. + The number of hours of daylight in Salt Lake City varies from a minimum of 9.6 hours on the winter solstice to a maximum of 14.4 hours on the summer solstice. + A weight is 6.5 feet above the floor, suspended from the ceiling by a spring. The weight is pulled down to 5 feet above the floor and released, rising past 6.5 feet in 0.5 seconds before attaining its maximum height of feet. Neglecting the effects of friction, the height of the weight will continue to oscillate between its minimum and maximum height. + The voltage used in U.S. electrical current changes from #math.equation(block: false, alt: "155")[$155$]V to #math.equation(block: false, alt: "minus 155")[$− 155$]V and back 60 times each second. + Although the moon is spherical, what we see from earth looks like a disk, sometimes only partly visible. The percentage of the moon's disk that is visible varies between 0 (at new moon) to 100 (at full moon). #figure(figph[graph], alt: "graph", caption: none) The table shows sunrise and sunset times in Los Angeles on the fifteenth of each month. #figure(table( columns: 7, align: left, inset: 6pt, table.header([Month], [Oct], [Nov], [Dec], [Jan], [Feb], [Mar]), [Sunrise], [#math.equation(block: false, alt: "5 : 58")[$5 : 58$]], [#math.equation(block: false, alt: "6 : 26")[$6 : 26$]], [#math.equation(block: false, alt: "6 : 51")[$6 : 51$]], [#math.equation(block: false, alt: "6 : 59")[$6 : 59$]], [#math.equation(block: false, alt: "6 : 39")[$6 : 39$]], [#math.equation(block: false, alt: "6 : 04")[$6 : 04$]], [Sunset], [#math.equation(block: false, alt: "17 : 20")[$17 : 20$]], [#math.equation(block: false, alt: "16 : 50")[$16 : 50$]], [#math.equation(block: false, alt: "16 : 45")[$16 : 45$]], [#math.equation(block: false, alt: "17 : 07")[$17 : 07$]], [#math.equation(block: false, alt: "17 : 37")[$17 : 37$]], [#math.equation(block: false, alt: "18 : 01")[$18 : 01$]], )) #figure(table( columns: 7, align: left, inset: 6pt, table.header([Month], [Apr], [May], [Jun], [Jul], [Aug], [Sep]), [Sunrise], [#math.equation(block: false, alt: "5 : 22")[$5 : 22$]], [#math.equation(block: false, alt: "4 : 52")[$4 : 52$]], [#math.equation(block: false, alt: "4 : 42")[$4 : 42$]], [#math.equation(block: false, alt: "4 : 43")[$4 : 43$]], [#math.equation(block: false, alt: "5 : 15")[$5 : 15$]], [#math.equation(block: false, alt: "5 : 37")[$5 : 37$]], [Sunset], [#math.equation(block: false, alt: "18 : 25")[$18 : 25$]], [#math.equation(block: false, alt: "18 : 48")[$18 : 48$]], [#math.equation(block: false, alt: "19 : 07")[$19 : 07$]], [#math.equation(block: false, alt: "19 : 05")[$19 : 05$]], [#math.equation(block: false, alt: "18 : 40")[$18 : 40$]], [#math.equation(block: false, alt: "18 : 00")[$18 : 00$]], )) + Use the left-hand grid to plot the sunrise times and sketch a sinusoidal graph through the points. + Use the right-hand grid to plot the sunset times and sketch a sinusoidal graph through the points. #figure(figph[grid], alt: "grid", caption: none) #figure(figph[two graphs], alt: "two graphs", caption: none) + Use the data from Problem 67 to complete the table with the hours of sunlight in Los Angeles on the fifteenth of each month.#figure(table( columns: 7, align: left, inset: 6pt, table.header([Month], [Oct], [Nov], [Dec], [Jan], [Feb], [Mar]), [Hours of Daylight], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(table( columns: 7, align: left, inset: 6pt, table.header([Month], [Apr], [May], [Jun], [Jul], [Aug], [Sep]), [Hours of Daylight], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Plot the daylight hours and sketch a sinusoidal graph through the points.#figure(figph[grid], alt: "grid", caption: none) Many people who believe in astrology also believe in biorhythms. The graph shows an individual's three biorhythms, physical, emotional, and intellectual, for 36 days, from #math.equation(block: false, alt: "t equals 0")[$t = 0$] on September 30 to November 5. #figure(figph[grid], alt: "grid", caption: none) + Find the dates of highest and lowest activity for each biorhythm during the month of October. + Find the period of each biorhythm in days. + On the day of your birth, all three biorhythms are at their maximum. How old will you be before all three are again at the maximum level? + Emotional high: Oct 5 and Nov 3, low: Oct 19; Physical high: Sep 30 and Oct 23, low: Oct 12 and Nov 4; Intellectual high: Oct 10, low: Oct 26 + Emotional: 28 days, physical: 23 days, intellectual: 32 days + 5152 days The path of a satellite orbiting above the earth makes a sinusoidal graph on a map of the earth, with its midline at the equator. On the map below, sketch a graph for a satellite that orbits the earth every 90 minutes, and strays no farther than 4000 km from the equator. (One degree of latitude is equal to 111 kilometers.) The satellite passes over the spot #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] latitude and #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] longitude at time #math.equation(block: false, alt: "t equals 0")[$t = 0$]. Label a scale on the equator to serve as a time axis for your graph. #figure(figph[map], alt: "map", caption: none) + Is the function shown periodic? If so, what is its period? If not, explain why not.#figure(figph[graph], alt: "graph", caption: none) + Compute the difference between the maximum and minimum function values. Sketch in the midline of the graph. + Find the smallest positive value of #math.equation(block: false, alt: "k")[$k$] for which #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals f open parenthesis x plus k close parenthesis")[$f ( x ) = f ( x + k )$] for all #math.equation(block: false, alt: "x")[$x$]. + Find the smallest positive values of #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] for which #math.equation(block: false, alt: "f open parenthesis b close parenthesis minus f open parenthesis a close parenthesis")[$f ( b ) − f ( a )$] is a maximum. + periodic, period 8 + 4, midline: #math.equation(block: false, alt: "y equals 3")[$y = 3$] + #math.equation(block: false, alt: "k equals 8")[$k = 8$] + #math.equation(block: false, alt: "a equals 3 , b equals 7")[$a = 3 , " " b = 7$] + Find the period, the maximum and minimum values, and the midline of the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] shown. #figure(figph[graph], alt: "graph", caption: none) + Sketch a graph of #math.equation(block: false, alt: "y equals 2 f open parenthesis x close parenthesis")[$y = 2 f ( x )$]. + Sketch a graph of #math.equation(block: false, alt: "y equals 2 plus f open parenthesis x close parenthesis")[$y = 2 + f ( x )$]. + Modify the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] so that the period is twice its current value. The graph shows arterial blood pressure, measured in millimeters of mercury (mmHg), as a function of time. #figure(figph[graph], alt: "graph", caption: none) + What are the maximum (systolic) and minimum (diastolic) pressures? The pulse pressure is the difference of systolic and diastolic pressures. What is the pulse pressure? + The mean arterial pressure is the diastolic pressure plus one-third of the pulse pressure. Calculate the mean arterial pressure, and draw a horizontal line on the graph at that pressure. + The blood pressure graph repeats its cycle with each heartbeat. What is the heart rate, in beats per minute, of the person whose blood pressure is shown in the graph? + systolic 120 mm Hg, diastolic 80 mm Hg, pulse pressure 40 mm Hg. + #math.equation(block: false, alt: "93 the fraction 1 over 3")[$93 frac(1, 3)$] + 72 beats per minute Here is a tide chart for Los Angeles for the week of December 17–23, 2000. The horizontal axis shows time in hours, with #math.equation(block: false, alt: "t equals 12")[$t = 12$] corresponding to noon on December 17. The vertical axis shows the height of the tide in feet above mean sea level. #figure(figph[graph], alt: "graph", caption: none) + High tides occurred at 3:07 am and 2:08 pm on December 17, and low tides at 8:41 am and 9:02 pm. Estimate the heights of the high and low tides on that day. + Is tide height a periodic function of time? Use the information from part (a) to justify your answer. + Make a table showing approximate times and heights for the high tides throughout the week. Make a similar table for the low tides. + Describe the trend in the heights of the high tides over the week. Describe the trend in the heights of the low tides. + What is the largest height difference between consecutive high and low tides during the week shown? When does it occur? The apparent magnitude of a star is a measure of its brightness as seen from earth. Smaller values of apparent magnitude correspond to brighter stars. The graph below, called a light curve, shows the apparent magnitude of the star Algol as a function of time. Algol is actually a system of two stars, a bright principal star and its dimmer companion, in orbit around each other. As each star passes in front of the other it eclipses some of the light that reaches earth. #figure(figph[graph], alt: "graph", caption: none) + The light curve is periodic. What is its period? + What is the range of apparent magnitudes of the Algol system? + Explain the large and small dips in the light curve. What is happening to cause the dips? + 69 hours. + 2.2 to 3.5 + The larger dip corresponds to when the brighter star is eclipsed, the smaller dip corresponds to when the dimmer star is eclipsed. Some stars, called Cepheid variable stars, appear to pulse, getting brighter and dimmer periodically. The graph shows the light curve for the star Delta Cephei. #figure(figph[graph], alt: "graph", caption: none) + What is the period of the graph? + What is the range of apparent magnitudes for Delta Cephei?