#set document(title: "6.2 The Circular 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")) == 6.2#h(0.6em)The Circular Functions === Trigonometric Functions of Angles in Radians Measuring angles in radians has other applications besides calculating arclength, and we will need to evaluate trigonometric functions of angles in radians. The sine, cosine, or tangent of a particular angle is the same whether the angle is measured in radians or in degrees. In Section 4.1 we saw that the sine and cosine are related to points on a unit circle. #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")[$bold(italic(x)) = bold(cos) ( bold(italic(θ)) ) , " " " " " " " " " " " " bold(italic(y)) = bold(sin) ( bold(italic(θ)) )$] ] This fact follows easily from the definitions of sine and cosine, because on a unit circle #math.equation(block: false, alt: "r equals 1")[$r = 1$]. In the next Activity we'll use this relationship to get a feel for the trig values of angles in radians. #notebox("Activity", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Trigonometry in Radians] Here is a unit circle with arclengths labeled, measured counterclockwise from #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]. (Note that the distance around the whole circle is 6.28 units!) In Section 6.1 we learned that on a unit circle, the measure of a positive angle in radians is equal to the length of the arc that it spans. #figure(figph[circle on a grid, radians labeled], alt: "circle on a grid, radians labeled", caption: none) + Use the unit circle to estimate the sine, cosine, and tangent of each angle in radians.+ #math.equation(block: false, alt: "0.6")[$0.6$] + #math.equation(block: false, alt: "2.3")[$2.3$] + #math.equation(block: false, alt: "3.5")[$3.5$] + #math.equation(block: false, alt: "5.3")[$5.3$] + Use the unit circle to estimate two solutions to each equation. + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0.3")[$cos ( θ ) = 0.3$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0.7")[$sin ( θ ) = 0.7$] + Sketch the angle on the unit circle. Use symmetry to find three more angles with the same sine and cosine, up to sign. + #math.equation(block: false, alt: "1.8")[$1.8$] + #math.equation(block: false, alt: "5.2")[$5.2$] + #math.equation(block: false, alt: "3.7")[$3.7$] + #math.equation(block: false, alt: "0.3")[$0.3$] ] === Using a Calculator Your calculator can give you the trigonometric function values for angles expressed in radians. For example, #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] radians is the same as #math.equation(block: false, alt: "60 degrees")[$60^(∘)$], because #math.equation(block: false, alt: "the fraction π over 3 times the fraction 180 degrees over π equals 60 degrees")[$display(frac(π, 3)) ⋅ display(frac(180^(∘), π)) = 60^(∘)$], so #math.equation(block: true, alt: "sin ⁡ open parenthesis the fraction π over 3 close parenthesis equals sin ⁡ open parenthesis 60 degrees close parenthesis equals the fraction the square root of 3 over 2")[$sin ( display(frac(π, 3)) ) = sin ( 60^(∘) ) = display(frac(sqrt(3), 2))$] However, we don't have to convert radians to degrees in order to evaluate trig ratios. First, change the calculator setting from Degree mode to Radian mode. Then enter #math.equation(block: true, alt: "sin ⁡ open parenthesis π divided by 3 close parenthesis")[$sin " " ( π ÷ 3 )$] and the calculator will return #math.equation(block: false, alt: "0.8660254038")[$0.8660254038$]. You can check that this number is a decimal approximation for #math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]. #examplebox("Example 1")[][ Use your calculator to find the sine and cosine of the following angles in radians. Round your answers to four decimal places. + #math.equation(block: false, alt: "θ equals the fraction 7 π over 4")[$θ = display(frac(7 π, 4))$] + #math.equation(block: false, alt: "θ equals 3.5")[$θ = 3.5$] #solutionbox[ + With the calculator in radian mode, and rounding to four decimal places we find #math.equation(block: true, alt: "cos ⁡ open parenthesis 7 π / 4 close parenthesis, equals 0.7071; sin ⁡ open parenthesis 7 π / 4 close parenthesis, equals minus 0.7071")[$cos ( 7 π / 4 ) & = 0.7071 \ sin ( 7 π / 4 ) & = − 0.7071$] + Your calculator will also give you the values #math.equation(block: true, alt: "cos ⁡ open parenthesis 3.5 close parenthesis, equals minus 0.9365; sin ⁡ open parenthesis 3.5 close parenthesis, equals minus 0.3508")[$cos ( 3.5 ) & = − 0.9365 \ sin ( 3.5 ) & = − 0.3508$] rounded to four places. Note that 3.5 radians is a third-quadrant angle, so the signs of the trig values make sense. ] ] Use your calculator to find the tangents of the following angles in radians. Round your answers to four decimal places. + #math.equation(block: false, alt: "θ equals the fraction 5 π over 12")[$θ = display(frac(5 π, 12))$] + #math.equation(block: false, alt: "θ equals 5.2")[$θ = 5.2$] + #math.equation(block: false, alt: "3.7321")[$3.7321$] + #math.equation(block: false, alt: "minus 1.8856")[$− 1.8856$] If two angles are coterminal they have the same trig values, just as they do when measured in degrees. So adding or subtracting a multiple of #math.equation(block: false, alt: "2 π")[$2 π$] to any angle results in a new angle in the same standard position and with the same trig values. #examplebox("Example 2")[][ Find the sine and cosine of the following angles in radians. Round your answers to four decimal places. + #math.equation(block: false, alt: "θ equals the fraction 5 π over 2")[$θ = display(frac(5 π, 2))$] + #math.equation(block: false, alt: "θ equals minus 4")[$θ = − 4$] #solutionbox[ + Because #math.equation(block: false, alt: "the fraction 5 π over 2 equals 2 π plus the fraction π over 2 ,")[$" " display(frac(5 π, 2)) = 2 π + display(frac(π, 2)) , " "$] #math.equation(block: false, alt: "θ")[$θ$] is coterminal with #math.equation(block: false, alt: "the fraction π over 2 ,")[$display(frac(π, 2)) , " "$] and it has the same sine and cosine as #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]. And #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$] radians is equal to #math.equation(block: false, alt: "90 degrees")[$90^(∘)$], so #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals sin ⁡ open parenthesis the fraction π over 2 close parenthesis equals 1; cos ⁡ open parenthesis θ close parenthesis equals cos ⁡ open parenthesis the fraction π over 2 close parenthesis equals 0")[$sin ( θ ) = sin ( display(frac(π, 2)) ) = 1 \ cos ( θ ) = cos ( display(frac(π, 2)) ) = 0$] + Rounded to four places, your calculator will give you the values #math.equation(block: true, alt: "cos ⁡ open parenthesis minus 4 close parenthesis, equals minus 0.6536; sin ⁡ open parenthesis minus 4 close parenthesis, equals 0.7568")[$cos ( − 4 ) & = − 0.6536 \ sin ( − 4 ) & = 0.7568$] By adding #math.equation(block: false, alt: "6.2832")[$6.2832$] (that's approximately #math.equation(block: false, alt: "2 π")[$2 π$]) to #math.equation(block: false, alt: "minus 4")[$− 4$] we see that #math.equation(block: false, alt: "minus 4")[$− 4$] radians is coterminal with #math.equation(block: false, alt: "2.2832")[$2.2832$] radians, a second-quadrant angle. You can check that #math.equation(block: false, alt: "2.2832")[$2.2832$] radians has (approximately) the same trig values as #math.equation(block: false, alt: "minus 4")[$− 4$] radians. ] ] Use your calculator to find the tangent of the following angles in radians. Round your answers to four decimal places. + #math.equation(block: false, alt: "θ equals minus the fraction 3 π over 4")[$θ = − display(frac(3 π, 4))$] + #math.equation(block: false, alt: "θ equals 15")[$θ = 15$] + #math.equation(block: false, alt: "1")[$1$] + #math.equation(block: false, alt: "minus 0.8560")[$− 0.8560$] Recall the formula for arclength when the angle is measured in radians: #math.equation(block: false, alt: "s equals r θ")[$s = r θ$]. If we solve for #math.equation(block: false, alt: "θ")[$θ$], we see that #math.equation(block: false, alt: "θ equals the fraction s over r")[$θ = display(frac(s, r))$], a ratio of two lengths. The units of length cancel out, so that radian measure has no units; it is a "dimensionless" quantity. In other words, a radian measure is just a real number. We shall see that this property makes radians especially useful in applications. From now on we shall omit the tag "radians," and you may assume that any angle given without units is in radians. === Sine and Cosine of Real Numbers We have observed that the measure of an angle in radians is just a real number. This is an important observation, because it allows us to define the sine and cosine as functions of real numbers, instead of as functions of angles. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Sine and Cosine of Real Numbers] We define the trigonometric functions of the real number #math.equation(block: false, alt: "t")[$t$] by #math.equation(block: true, alt: "cos ⁡ open parenthesis t close parenthesis equals cos ⁡ open parenthesis θ close parenthesis and sin ⁡ open parenthesis t close parenthesis equals sin ⁡ open parenthesis θ close parenthesis")[$bold(cos) ( bold(italic(t)) ) = bold(cos) ( bold(italic(θ)) ) " " " " " " " " "and" " " " " " " " " bold(sin) ( bold(italic(t)) ) = bold(sin) ( bold(italic(θ)) )$] where #math.equation(block: false, alt: "θ equals t")[$θ = t " "$] is an angle measured in radians. ] We can think of the definition this way: to find the sine or cosine of a real number #math.equation(block: false, alt: "t")[$t$], we find the sine or cosine of the angle #math.equation(block: false, alt: "θ equals t")[$θ = t$] in radians. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ To understand the connection between real numbers and radians, you may find it helpful to remember that the measure of a positive angle in radians is equal to the length of the arc it spans on a unit circle. Consider the unit circle shown at right, and the angle #math.equation(block: false, alt: "θ")[$θ$] determined by the arc of length #math.equation(block: false, alt: "t")[$t$]. The radian measure of #math.equation(block: false, alt: "θ")[$θ$] is the same as the length #math.equation(block: false, alt: "t")[$t$] of the arc. (For example, in this figure #math.equation(block: false, alt: "θ equals t equals 2")[$θ = t = 2$].) #figure(figph[circle with arc], alt: "circle with arc", caption: none) In practice, there is no difference between finding the sine or cosine of the number 2 and the sine or cosine of an angle of 2 radians: in each case we set the calculator in Radian mode and evaluate #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 close parenthesis equals minus 0.4161 and sin ⁡ open parenthesis 2 close parenthesis equals 0.9093")[$cos " " ( 2 ) = − 0.4161 " " " " " " " " " " " " "and" " " " " " " " " " " sin " " ( 2 ) = 0.9093$] ] So, in some sense then, a trig function of a radian is the same as a trig function of a real number. This result is so important that it bears repeating: to find the sine or cosine of a real number #math.equation(block: false, alt: "t")[$t$], we find the sine or cosine of the angle #math.equation(block: false, alt: "t")[$t$] radians. We can now use the trigonometric functions to model periodic behavior as functions of time, or indeed, of any variable. #examplebox("Example 3")[][ The sunset time in Stockholm, Sweden, on the #math.equation(block: false, alt: "n")[$n$]th day of the year can be modeled by #math.equation(block: true, alt: "T equals 3.11 sin ⁡ open parenthesis 0.017 n minus 1.38 close parenthesis plus 6.03")[$T = 3.11 " " sin " " ( 0.017 n − 1.38 ) + 6.03$] where #math.equation(block: false, alt: "T")[$T$] is given in hours after noon. Find the sunset time on January 1 (day #math.equation(block: false, alt: "n equals 1")[$n = 1$]) and on July 1 (day #math.equation(block: false, alt: "n equals 182")[$n = 182$]). #figure(figph[Sinusoidal graph of Stockholm's sunset time vs days], alt: "Sinusoidal graph of Stockholm's sunset time vs days", caption: none) #solutionbox[ Evaluate the function for #math.equation(block: false, alt: "n equals 1")[$n = 1$] to find #math.equation(block: true, alt: "T equals 3.11 sin ⁡ open parenthesis 0.017 open parenthesis 1 close parenthesis minus 1.38 close parenthesis plus 6.03 equals 2.99")[$T = 3.11 " " sin " " ( 0.017 ( 1 ) − 1.38 ) + 6.03 = 2.99$] On January 1, sunset in Stockholm occurs about 2.99 hours after noon, or at 2:59 pm. Evaluate the function at #math.equation(block: false, alt: "n equals 182")[$n = 182$] to find #math.equation(block: true, alt: "T equals 3.11 sin ⁡ open parenthesis 0.017 open parenthesis 182 close parenthesis minus 1.38 close parenthesis plus 6.03 equals 9.11")[$T = 3.11 " " sin " " ( 0.017 ( 182 ) − 1.38 ) + 6.03 = 9.11$] On July 1, sunset occurs about 9.11 hours after noon, or at 9:07 pm. (Actually, 10:07 pm, because of daylight savings time.) ] ] Variable stars are important in astronomy because they are used to estimate distances. Their magnitude, or brightness, varies periodically and can be modeled by trigonometric functions. The star T Herculis reached its minimum magnitude on December 27, 2004, and #math.equation(block: false, alt: "t")[$t$] days later its magnitude is approximately #math.equation(block: true, alt: "M equals 10.2 minus 2.2 cos ⁡ open parenthesis 0.038 t close parenthesis")[$M = 10.2 − 2.2 " " cos " " ( 0.038 t )$] #figure(figph[Sinusoidal graph of star brightness], alt: "Sinusoidal graph of star brightness", caption: none) + What was the magnitude of T Herculis on December 27, 2004? + What was the magnitude of T Herculis be on December 27, 2006 (730 days later)? + #math.equation(block: false, alt: "8")[$8$] + #math.equation(block: false, alt: "12.09")[$12.09$] === The Tangent Function We can also define the tangent function for real numbers. Let #math.equation(block: false, alt: "P open parenthesis x , y close parenthesis")[$P ( x , y )$] be the terminal point of an arc of length #math.equation(block: false, alt: "t")[$t$] in standard position on a unit circle. Then #math.equation(block: true, alt: "tan ⁡ open parenthesis t close parenthesis equals the fraction y over x")[$bold(tan) ( bold(italic(t)) ) = display(frac(bold(italic(y)), bold(italic(x))))$] Of course, this definition agrees with our earlier definition of the tangent function for angles, because the point #math.equation(block: false, alt: "P")[$P$] lies on the terminal side of the angle #math.equation(block: false, alt: "θ equals t")[$θ = t$] radians. For example, we saw earlier that, rounded to four decimal places, #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 close parenthesis equals minus 0.4161 and sin ⁡ open parenthesis 2 close parenthesis equals 0.9093")[$cos ( 2 ) = − 0.4161 " " " " " " " " "and" " " " " " " " " sin ( 2 ) = 0.9093$] so the coordinates of point #math.equation(block: false, alt: "P")[$P$] on the unit circle in the figure at right are #math.equation(block: false, alt: "open parenthesis minus 0.4161 , 0.9093 close parenthesis")[$( − 0.4161 , 0.9093 )$]. Therefore, #math.equation(block: true, alt: "tan ⁡ open parenthesis 2 close parenthesis equals the fraction 0.9093 over minus 0.4161 equals minus 2.1853")[$tan ( 2 ) = display(frac(0.9093, − 0.4161)) = − 2.1853$] #figure(figph[circle with arc of length 2], alt: "circle with arc of length 2", caption: none) You can set your calculator in radian mode to verify that, to three decimal places, #math.equation(block: true, alt: "tan ⁡ open parenthesis 2 close parenthesis equals minus 2.185")[$tan ( 2 ) = − 2.185$] Because the trig values can be found as coordinates on a unit circle, our three trig functions are often called the #strong[circular functions]. We now have the following definitions for the circular functions of real numbers. Why they are called circular functions. The slider sets an arclength t, measured counterclockwise from (1, 0) around a unit circle; the marked terminal point has coordinates (cos t, sin t) — the cosine is how far right the point sits, the sine (the dashed drop) is exactly its height. Drag t through 1.57 (≈ π/2) to watch the sine reach 1 as the point passes the top of the circle, and on past 3.14 (≈ π) to see both coordinates go negative in the third quadrant. One full trip around the circle is t = 2π ≈ 6.28, the circumference — which is why every circular function repeats with period 2π.{"functions":\[{"color":"\#6a6a6a","expression":"cos(t)","expression2":"sin(t)","id":"circle","inequality":"lt","lineDash":\[\],"lineWidth":2,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"x","visible":true},{"color":"\#1f4e79","expression":"t\*cos(s)","expression2":"t\*sin(s)","id":"spoke","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"x","visible":true},{"color":"\#b8442a","expression":"cos(s) + 0.06\*cos(t)","expression2":"sin(s) + 0.06\*sin(t)","id":"terminal-point","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"x","visible":true},{"color":"\#b8442a","expression":"cos(s)","expression2":"t\*sin(s)","id":"sine-height","inequality":"lt","lineDash":\[6,4\],"lineWidth":2,"mode":"parametric","tMax":1,"tMin":0,"variable":"x","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-s","label":"Arclength t","max":6.28,"min":0,"name":"s","step":0.01,"value":1}\],"title":"Unit circle: the terminal point (cos t, sin t)","version":1,"viewport":{"centerX":0,"centerY":0,"scale":150}}#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The Circular Functions] Let #math.equation(block: false, alt: "P")[$P$] be the terminal point of an arc of length #math.equation(block: false, alt: "t")[$t$] in standard position on a unit circle. The #strong[circular functions] of #math.equation(block: false, alt: "t")[$t$] are defined by #math.equation(block: true, alt: "cos ⁡ open parenthesis t close parenthesis, equals x; sin ⁡ open parenthesis t close parenthesis, equals y; tan ⁡ open parenthesis t close parenthesis, equals the fraction y over x , x not equal to 0")[$bold(cos) ( bold(italic(t)) ) " " & = bold(italic(x)) \ bold(sin) ( bold(italic(t)) ) " " & = bold(italic(y)) \ bold(tan) ( bold(italic(t)) ) " " & = display(frac(bold(italic(y)), bold(italic(x)))) , " " " " " " " " bold(italic(x)) ≠ 0$] #figure(figph[circle with arc], alt: "circle with arc", caption: none) ] #examplebox("Example 4")[][ Use the graph of the unit circle shown below to estimate #math.equation(block: false, alt: "cos ⁡ open parenthesis 2.5 close parenthesis , sin ⁡ open parenthesis 2.5 close parenthesis")[$cos ( 2.5 ) , " " sin ( 2.5 )$], and #math.equation(block: false, alt: "tan ⁡ open parenthesis 2.5 close parenthesis")[$tan ( 2.5 )$]. #figure(figph[Unit circle], alt: "Unit circle", caption: none) #solutionbox[ The circle is scaled in units of 0.1 radians, and an arc of 2.5 radians in standard position has its terminal point, #math.equation(block: false, alt: "P")[$P$], in the second quadrant. The coordinates of #math.equation(block: false, alt: "P")[$P$] are approximately #math.equation(block: false, alt: "open parenthesis minus 0.8 , 0.6 close parenthesis")[$( − 0.8 , 0.6 )$], so we have #math.equation(block: true, alt: "cos ⁡ open parenthesis 2.5 close parenthesis equals minus 0.8 and sin ⁡ open parenthesis 2.5 close parenthesis equals 0.6")[$cos ( 2.5 ) = − 0.8 " " " " " " " " "and" " " " " " " " " sin ( 2.5 ) = 0.6$] To find #math.equation(block: false, alt: "tan ⁡ open parenthesis 2.5 close parenthesis")[$tan ( 2.5 )$], we calculate #math.equation(block: false, alt: "the fraction y over x")[$display(frac(y, x))$]. #math.equation(block: true, alt: "tan ⁡ open parenthesis 2.5 close parenthesis equals the fraction y over x equals the fraction 0.6 over minus 0.8 equals minus 0.75")[$tan ( 2.5 ) = display(frac(y, x)) = display(frac(0.6, − 0.8)) = − 0.75$] ] ] Use the graph of the unit circle in the previous example to estimate #math.equation(block: false, alt: "cos ⁡ open parenthesis 4.2 close parenthesis , sin ⁡ open parenthesis 4.2 close parenthesis")[$cos ( 4.2 ) , " " sin ( 4.2 )$], and #math.equation(block: false, alt: "tan ⁡ open parenthesis 4.2 close parenthesis")[$tan ( 4.2 )$]. #math.equation(block: true, alt: "cos ⁡ open parenthesis 4.2 close parenthesis approximately equals minus 0.49")[$cos ( 4.2 ) ≈ − 0.49$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 4.2 close parenthesis approximately equals minus 0.87")[$sin ( 4.2 ) ≈ − 0.87$], #math.equation(block: true, alt: "tan ⁡ open parenthesis 4.2 close parenthesis approximately equals 1.78")[$tan ( 4.2 ) ≈ 1.78$] === The Special Values In Chapter 2 you learned the exact trig values for the "special angles" #math.equation(block: false, alt: "30 degrees , 45 degrees")[$30^(∘) , " " 45^(∘)$] and #math.equation(block: false, alt: "60 degrees")[$60^(∘)$]. Recall that the trig values for these angles can be used as benchmarks for estimating and mental calculation. Now we review those values when the angles are measured in radians. The easiest way to remember the special values is to refer to the sides of the two special triangles: #figure(figph[45-45-90 triangle], alt: "45-45-90 triangle", caption: none) #figure(figph[30-60-90 triangle], alt: "30-60-90 triangle", caption: none) #figure(table( columns: 5, align: left, inset: 6pt, table.header([Degrees], [Radians], [Sine], [Cosine], [Tangent]), [#math.equation(block: false, alt: "0 degrees")[$0^(∘)$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "30 degrees")[$30^(∘)$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 3")[$display(frac(1, sqrt(3)))$]], [#math.equation(block: false, alt: "45 degrees")[$45^(∘)$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "60 degrees")[$60^(∘)$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]], [#math.equation(block: false, alt: "90 degrees")[$90^(∘)$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [undefined], )) #figure(figph[circle with special angles], alt: "circle with special angles", caption: none) You can use these values to find trig values for the special angles in all four quadrants. Recall that we use reference angles to define the trigonometric ratios for angles greater than #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. (See Section 4.1 to review reference angles.) The figure below shows how to calculate reference angles in radians. Note that the formulas are the same as those for degrees if we replace #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] by #math.equation(block: false, alt: "π")[$π$]! #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Reference Angles in Radians] #figure(figph[angles with reference angles], alt: "angles with reference angles", caption: none) ] #examplebox("Example 5")[][ Give exact values for the following. + #math.equation(block: false, alt: "tan ⁡ open parenthesis the fraction 2 π over 3 close parenthesis")[$tan ( display(frac(2 π, 3)) )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction 5 π over 4 close parenthesis")[$cos ( display(frac(5 π, 4)) )$] #solutionbox[ + The reference angle for #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$] is #math.equation(block: false, alt: "π minus the fraction 2 π over 3 equals the fraction π over 3")[$π − display(frac(2 π, 3)) = display(frac(π, 3))$], and the tangent is negative in the second quadrant. (See the figure at right.) Thus, #math.equation(block: true, alt: "tan ⁡ open parenthesis the fraction 2 π over 3 close parenthesis equals minus tan ⁡ open parenthesis the fraction π over 3 close parenthesis equals minus the square root of 3")[$tan ( display(frac(2 π, 3)) ) = − tan ( display(frac(π, 3)) ) = − sqrt(3)$]#figure(figph[angles], alt: "angles", caption: none) + The reference angle for #math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$] is #math.equation(block: false, alt: "the fraction 5 π over 4 minus π equals the fraction π over 4")[$display(frac(5 π, 4)) − π = display(frac(π, 4))$], and the cosine is negative in the third quadrant, so #math.equation(block: true, alt: "cos ⁡ open parenthesis the fraction 5 π over 4 close parenthesis equals minus cos ⁡ open parenthesis the fraction π over 4 close parenthesis equals the fraction minus 1 over the square root of 2")[$cos ( display(frac(5 π, 4)) ) = − cos ( display(frac(π, 4)) ) = display(frac(− 1, sqrt(2)))$]#figure(figph[angles], alt: "angles", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ At this point, you may feel that expressions such as #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction 5 π over 4 close parenthesis equals the fraction minus 1 over the square root of 2")[$" " cos ( display(frac(5 π, 4)) ) = display(frac(− 1, sqrt(2))) " "$] look like hieroglyphics. Keep in mind that #math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$] and #math.equation(block: false, alt: "the fraction minus 1 over the square root of 2")[$display(frac(− 1, sqrt(2)))$] are just numbers, so the equation above says that "the cosine of an angle of 3.9 radians is about -0.7." ] Give exact values for the following. + #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction 5 π over 6 close parenthesis")[$sin ( display(frac(5 π, 6)) )$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis the fraction 7 π over 4 close parenthesis")[$tan ( display(frac(7 π, 4)) )$] + #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] + #math.equation(block: false, alt: "minus 1")[$− 1$] #notebox("Activity", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Special Angles in Radians] Here is a unit circle with arclengths labeled, measured counterclockwise from #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]. #figure(figph[circle on a grid], alt: "circle on a grid", caption: none) + Give a decimal approximation to two places for each angle, then the degree measure of each.#figure(table( columns: 10, align: left, inset: 6pt, table.header([Radians], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$]], [#math.equation(block: false, alt: "π")[$π$]]), [Decimal #linebreak() Approx. #linebreak()], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [Degrees], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) #figure(table( columns: 10, align: left, inset: 6pt, table.header([Radians], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "the fraction 11 π over 6")[$display(frac(11 π, 6))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [Decimal #linebreak() Approx. #linebreak()], [$#hide($000$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($000$)$], [Degrees], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], )) + On the unit circle above, plot the endpoint of each arc in standard position. + #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] + #math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$] + #math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$] + Use the unit circle to estimate the sine, cosine and tangent for each angle in part (2). Then compare your estimate with the exact trig value. ] #examplebox("Example 6")[][ Find the coordinates of the terminal point, #math.equation(block: false, alt: "P")[$P$], of an arc of length #math.equation(block: false, alt: "t equals the fraction 5 π over 6")[$t = display(frac(5 π, 6))$] starting at #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$] on a unit circle. #solutionbox[ The coordinates of #math.equation(block: false, alt: "P")[$P$] are #math.equation(block: false, alt: "open parenthesis cos ⁡ the fraction 5 π over 6 , sin ⁡ the fraction 5 π over 6 close parenthesis")[$( cos display(frac(5 π, 6)) , sin display(frac(5 π, 6)) )$], so #math.equation(block: true, alt: "x, equals cos ⁡ open parenthesis the fraction 5 π over 6 close parenthesis equals the fraction minus the square root of 3 over 2; y, equals sin ⁡ open parenthesis the fraction 5 π over 6 close parenthesis equals the fraction 1 over 2")[$x & = cos ( display(frac(5 π, 6)) ) = display(frac(− sqrt(3), 2)) " " " " " " \ y & = sin ( display(frac(5 π, 6)) ) = display(frac(1, 2))$] #figure(figph[circle with arc and labeled endpoint], alt: "circle with arc and labeled endpoint", caption: none) Thus, the coordinates of #math.equation(block: false, alt: "P")[$P$] are #math.equation(block: false, alt: "open parenthesis the fraction minus the square root of 3 over 2 , the fraction 1 over 2 close parenthesis")[$( display(frac(− sqrt(3), 2)) , display(frac(1, 2)) )$], as shown at right. ] ] Find the coordinates of the terminal point, #math.equation(block: false, alt: "P")[$P$], of an arc of length #math.equation(block: false, alt: "t equals the fraction π over 3")[$t = display(frac(π, 3))$] starting at #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$] on a unit circle. #math.equation(block: true, alt: "open parenthesis the fraction 1 over 2 , the fraction the square root of 3 over 2 close parenthesis")[$( display(frac(1, 2)) , display(frac(sqrt(3), 2)) )$] Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify. + + #math.equation(block: false, alt: "the fraction 2 over 3 plus the fraction 1 over 6")[$display(frac(2, 3)) + display(frac(1, 6))$] + #math.equation(block: false, alt: "the fraction 2 x over 3 plus the fraction x over 6")[$display(frac(2 x, 3)) + display(frac(x, 6))$] + + #math.equation(block: false, alt: "the fraction 3 over 4 minus the fraction 5 over 8")[$display(frac(3, 4)) − display(frac(5, 8))$] + #math.equation(block: false, alt: "the fraction 3 n over 4 minus the fraction 5 n over 8")[$display(frac(3 n, 4)) − display(frac(5 n, 8))$] + + #math.equation(block: false, alt: "2 minus the fraction 3 over 4")[$2 − display(frac(3, 4))$] + #math.equation(block: false, alt: "2 b minus the fraction 3 b over 4")[$2 b − display(frac(3 b, 4))$] + + #math.equation(block: false, alt: "1 plus the fraction 3 over 8")[$1 + display(frac(3, 8))$] + #math.equation(block: false, alt: "m plus the fraction 3 m over 8")[$m + display(frac(3 m, 8))$] + + #math.equation(block: false, alt: "the fraction 5 over 6 minus the fraction 7 over 6")[$display(frac(5, 6)) − display(frac(7, 6))$] + #math.equation(block: false, alt: "the fraction 5 q over 6 minus the fraction 7 q over 6")[$display(frac(5 q, 6)) − display(frac(7 q, 6))$] + + #math.equation(block: false, alt: "the fraction 2 over 3 minus the fraction 5 over 3")[$display(frac(2, 3)) − display(frac(5, 3))$] + #math.equation(block: false, alt: "the fraction 2 p over 3 minus the fraction 5 p over 3")[$display(frac(2 p, 3)) − display(frac(5 p, 3))$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + + #math.equation(block: false, alt: "the fraction 5 over 6")[$display(frac(5, 6))$] + #math.equation(block: false, alt: "the fraction 5 x over 6")[$display(frac(5 x, 6))$] + + #math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$] + #math.equation(block: false, alt: "the fraction n over 8")[$display(frac(n, 8))$] + + #math.equation(block: false, alt: "the fraction 5 over 4")[$display(frac(5, 4))$] + #math.equation(block: false, alt: "the fraction 5 b over 4")[$display(frac(5 b, 4))$] + + #math.equation(block: false, alt: "the fraction 11 over 8")[$display(frac(11, 8))$] + #math.equation(block: false, alt: "the fraction 11 m over 8")[$display(frac(11 m, 8))$] + + #math.equation(block: false, alt: "the fraction minus 1 over 6")[$display(frac(− 1, 6))$] + #math.equation(block: false, alt: "the fraction minus q over 6")[$display(frac(− q, 6))$] + + #math.equation(block: false, alt: "minus 1")[$− 1$] + #math.equation(block: false, alt: "minus p")[$− p$] ] === Section 6.2 Summary ==== Vocabulary - Circular functions ==== Concepts + The sine, cosine, or tangent of a particular angle is the same whether the angle is measured in radians or in degrees. + You should memorize the trig values of the special angles in radians.#figure(table( columns: 5, align: left, inset: 6pt, table.header([Degrees], [Radians], [Sine], [Cosine], [Tangent]), [#math.equation(block: false, alt: "0 degrees")[$0^(∘)$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "30 degrees")[$30^(∘)$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 3")[$display(frac(1, sqrt(3)))$]], [#math.equation(block: false, alt: "45 degrees")[$45^(∘)$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "60 degrees")[$60^(∘)$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]], [#math.equation(block: false, alt: "90 degrees")[$90^(∘)$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [undefined], )) + To find the sine or cosine of a real number #math.equation(block: false, alt: "t")[$t$], we find the sine or cosine of the angle #math.equation(block: false, alt: "θ equals t")[$θ = t$] in radians. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Coordinates on a Unit Circle] The #strong[coordinates] of the point #math.equation(block: false, alt: "P")[$P$] determined by an arc of length #math.equation(block: false, alt: "t")[$t$] in standard position on a unit circle are #math.equation(block: true, alt: "open parenthesis x , y close parenthesis equals open parenthesis cos ⁡ open parenthesis t close parenthesis , sin ⁡ open parenthesis t close parenthesis close parenthesis")[$( x , y ) = ( cos ( t ) , sin ( t ) )$]#figure(figph[circle], alt: "circle", caption: none) ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The Circular Functions] Let #math.equation(block: false, alt: "P")[$P$] be the terminal point of an arc of length #math.equation(block: false, alt: "t")[$t$] in standard position on a unit circle. The #strong[circular functions] of #math.equation(block: false, alt: "t")[$t$] are defined by #math.equation(block: true, alt: "cos ⁡ open parenthesis t close parenthesis, equals x; sin ⁡ open parenthesis t close parenthesis, equals y; tan ⁡ open parenthesis t close parenthesis, equals the fraction y over x , x not equal to 0")[$cos ( t ) & = x \ sin ( t ) & = y \ tan ( t ) & = display(frac(y, x)) , " " " " x ≠ 0$]#figure(figph[circle], alt: "circle", caption: none) ] ==== Study Questions + Write each statement using decimal approximations to four places.+ #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 6 close parenthesis equals the fraction the square root of 3 over 2")[$cos ( display(frac(π, 6)) ) = display(frac(sqrt(3), 2))$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction 5 π over 4 close parenthesis equals the fraction minus 1 over the square root of 2")[$sin ( display(frac(5 π, 4)) ) = display(frac(− 1, sqrt(2)))$] + Sketch a figure on a unit circle to illustrate each equation.+ #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction 3 π over 4 close parenthesis equals 0.7071")[$sin ( display(frac(3 π, 4)) ) = 0.7071$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2.5 close parenthesis equals minus 0.8011")[$cos ( 2.5 ) = − 0.8011$] + Write down the multiples of #math.equation(block: false, alt: "the fraction π over 12")[$display(frac(π, 12))$] from #math.equation(block: false, alt: "0")[$0$] to #math.equation(block: false, alt: "2 π")[$2 π$]. Reduce each fraction. + On a unit circle, sketch arcs in standard position with the following lengths. #math.equation(block: true, alt: "s equals the fraction π over 2 , s equals the fraction 3 π over 4 , s equals the fraction π over 3 , s equals the fraction 2 π over 3")[$s = display(frac(π, 2)) , " " s = display(frac(3 π, 4)) , " " s = display(frac(π, 3)) , " " s = display(frac(2 π, 3))$] ==== Skills + Know the trigonometric function values for the special angles in radians \#1–4, 46–48 + Use a unit circle to find trig values \#5–30, 45–58 + Find reference angles in radians \#33–45 + Evaluate trigonometric expressions \#31–32, 49–54 + Find coordinates on a unit circle \#55–60, 67–68 + Find an angle with a given terminal point on a unit circle \#61–66 + Use the tangent ratio to find slope \#69–74 + Find coordinates on a circle of radius \#77–80 === Homework 6.2 For Problems 1–4, each point on the unit circle is the terminal point of an angle in standard position. Give exact values for the radian measure, #math.equation(block: false, alt: "t")[$t$], of the angle, and the coordinates #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] of the point. #figure(figph[circle], alt: "circle", caption: none) #figure(table( columns: 5, align: left, inset: 6pt, table.header([$#hide($0000$)$], [a], [b], [c], [d]), [#math.equation(block: false, alt: "t")[$t$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "x")[$x$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "y")[$y$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(table( columns: 5, align: left, inset: 6pt, table.header([$#hide($0000$)$], [a], [b], [c], [d]), [#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction minus 1 over the square root of 2")[$display(frac(− 1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction minus 1 over the square root of 2")[$display(frac(− 1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]], [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 2")[$display(frac(1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction minus 1 over the square root of 2")[$display(frac(− 1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction minus 1 over the square root of 2")[$display(frac(− 1, sqrt(2)))$]], )) #figure(figph[circle], alt: "circle", caption: none) #figure(table( columns: 5, align: left, inset: 6pt, table.header([$#hide($0000$)$], [a], [b], [c], [d]), [#math.equation(block: false, alt: "t")[$t$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "x")[$x$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "y")[$y$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(figph[circle], alt: "circle", caption: none) #figure(table( columns: 5, align: left, inset: 6pt, table.header([$#hide($0000$)$], [a], [b], [c], [d]), [#math.equation(block: false, alt: "t")[$t$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "x")[$x$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "y")[$y$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(table( columns: 5, align: left, inset: 6pt, table.header([$#hide($0000$)$], [a], [b], [c], [d]), [#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$]], [#math.equation(block: false, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$]], [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], )) #figure(figph[circle], alt: "circle", caption: none) #figure(table( columns: 5, align: left, inset: 6pt, table.header([$#hide($0000$)$], [a], [b], [c], [d]), [#math.equation(block: false, alt: "t")[$t$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "x")[$x$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "y")[$y$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) For Problems 5–8, use the unit circle to estimate the sine, cosine, and tangent of each arc of given length. #figure(figph[Unit circle], alt: "Unit circle", caption: none) + #math.equation(block: false, alt: "0.4")[$0.4$] + #math.equation(block: false, alt: "1.2")[$1.2$] + #math.equation(block: false, alt: "2")[$2$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 0.4 close parenthesis approximately equals 0.39 , cos ⁡ open parenthesis 0.4 close parenthesis approximately equals 0.92 , tan ⁡ open parenthesis 0.4 close parenthesis approximately equals 0.42")[$sin ( 0.4 ) ≈ 0.39 , " " cos ( 0.4 ) ≈ 0.92 , " " tan ( 0.4 ) ≈ 0.42$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 1.2 close parenthesis approximately equals 0.93 , cos ⁡ open parenthesis 1.2 close parenthesis approximately equals 0.36 , tan ⁡ open parenthesis 1.2 close parenthesis approximately equals 2.6")[$sin ( 1.2 ) ≈ 0.93 , " " cos ( 1.2 ) ≈ 0.36 , " " tan ( 1.2 ) ≈ 2.6$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 close parenthesis approximately equals 0.91 , cos ⁡ open parenthesis 2 close parenthesis approximately equals minus 0.42 , tan ⁡ open parenthesis 2 close parenthesis approximately equals minus 2.2")[$sin ( 2 ) ≈ 0.91 , " " cos ( 2 ) ≈ − 0.42 , " " tan ( 2 ) ≈ − 2.2$] + #math.equation(block: false, alt: "0.8")[$0.8$] + #math.equation(block: false, alt: "2.6")[$2.6$] + #math.equation(block: false, alt: "4")[$4$] + #math.equation(block: false, alt: "2.8")[$2.8$] + #math.equation(block: false, alt: "3.5")[$3.5$] + #math.equation(block: false, alt: "5")[$5$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2.8 close parenthesis approximately equals 0.33 , cos ⁡ open parenthesis 2.8 close parenthesis approximately equals minus 0.94 , tan ⁡ open parenthesis 2.8 close parenthesis approximately equals minus 0.36")[$sin ( 2.8 ) ≈ 0.33 , " " cos ( 2.8 ) ≈ − 0.94 , " " tan ( 2.8 ) ≈ − 0.36$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 3.5 close parenthesis approximately equals minus 0.35 , cos ⁡ open parenthesis 3.5 close parenthesis approximately equals minus 0.94 , tan ⁡ open parenthesis 3.5 close parenthesis approximately equals 0.37")[$sin ( 3.5 ) ≈ − 0.35 , " " cos ( 3.5 ) ≈ − 0.94 , " " tan ( 3.5 ) ≈ 0.37$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis 5 close parenthesis approximately equals minus 0.96 , cos ⁡ open parenthesis 5 close parenthesis approximately equals 0.28 , tan ⁡ open parenthesis 5 close parenthesis approximately equals minus 3.3")[$sin ( 5 ) ≈ − 0.96 , " " cos ( 5 ) ≈ 0.28 , " " tan ( 5 ) ≈ − 3.3$] + #math.equation(block: false, alt: "3")[$3$] + #math.equation(block: false, alt: "4.3")[$4.3$] + #math.equation(block: false, alt: "5.5")[$5.5$] For Problems 9–14, use the unit circle to estimate two numbers with the given trig value. #figure(figph[Unit circle], alt: "Unit circle", caption: none) #math.equation(block: true, alt: "cos ⁡ open parenthesis t close parenthesis equals 0.3")[$cos ( t ) = 0.3$] #math.equation(block: false, alt: "t approximately equals 1.27")[$t ≈ 1.27$] or #math.equation(block: false, alt: "t approximately equals 5")[$t ≈ 5$] #math.equation(block: true, alt: "sin ⁡ open parenthesis t close parenthesis equals 0.1")[$sin ( t ) = 0.1$] #math.equation(block: true, alt: "sin ⁡ open parenthesis t close parenthesis equals minus 0.7")[$sin ( t ) = − 0.7$] #math.equation(block: false, alt: "t approximately equals 3.92")[$t ≈ 3.92$] or #math.equation(block: false, alt: "t approximately equals 5.5")[$t ≈ 5.5$] #math.equation(block: true, alt: "cos ⁡ open parenthesis t close parenthesis equals minus 0.6")[$cos ( t ) = − 0.6$] #math.equation(block: true, alt: "tan ⁡ open parenthesis t close parenthesis equals the fraction minus 4 over 9")[$tan ( t ) = display(frac(− 4, 9))$] #math.equation(block: false, alt: "t approximately equals 2.72")[$t ≈ 2.72$] or #math.equation(block: false, alt: "t approximately equals 5.87")[$t ≈ 5.87$] #math.equation(block: true, alt: "tan ⁡ open parenthesis t close parenthesis equals the fraction 8 over 6")[$tan ( t ) = display(frac(8, 6))$] Each of Problems 13–20 describes an arc in standard position on the unit circle. In which quadrant does the terminal point of the arc lie? #math.equation(block: true, alt: "sin ⁡ open parenthesis s close parenthesis greater than 0 , cos ⁡ open parenthesis s close parenthesis less than 0")[$sin ( s ) > 0 , " " cos ( s ) < 0$] II #math.equation(block: true, alt: "sin ⁡ open parenthesis s close parenthesis less than 0 , cos ⁡ open parenthesis s close parenthesis greater than 0")[$sin ( s ) < 0 , " " cos ( s ) > 0$] #math.equation(block: true, alt: "cos ⁡ open parenthesis t close parenthesis less than 0 , tan ⁡ open parenthesis t close parenthesis less than 0")[$cos ( t ) < 0 , " " tan ( t ) < 0$] II #math.equation(block: true, alt: "cos ⁡ open parenthesis t close parenthesis greater than 0 , tan ⁡ open parenthesis t close parenthesis less than 0")[$cos ( t ) > 0 , " " tan ( t ) < 0$] #math.equation(block: true, alt: "sin ⁡ open parenthesis x close parenthesis less than 0 , tan ⁡ open parenthesis x close parenthesis greater than 0")[$sin ( x ) < 0 , " " tan ( x ) > 0$] III #math.equation(block: true, alt: "sin ⁡ open parenthesis x close parenthesis greater than 0 , tan ⁡ open parenthesis x close parenthesis less than 0")[$sin ( x ) > 0 , " " tan ( x ) < 0$] For Problems 21–26, without using a calculator, decide whether the quantity is positive or negative. #math.equation(block: true, alt: "cos ⁡ open parenthesis 2.7 close parenthesis")[$cos ( 2.7 )$] Negative #math.equation(block: true, alt: "sin ⁡ open parenthesis 4.1 close parenthesis")[$sin ( 4.1 )$] #math.equation(block: true, alt: "tan ⁡ open parenthesis 3.8 close parenthesis")[$tan ( 3.8 )$] Positive #math.equation(block: true, alt: "tan ⁡ open parenthesis 5.4 close parenthesis")[$tan ( 5.4 )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis 2.2 close parenthesis")[$sin ( 2.2 )$] Positive #math.equation(block: true, alt: "cos ⁡ open parenthesis 4.9 close parenthesis")[$cos ( 4.9 )$] For Problems 27–30, place the trig values in order from smallest to largest. Use the figure to help you, but try not to use a calculator! #figure(figph[Unit circle], alt: "Unit circle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis 0.5 close parenthesis")[$sin ( 0.5 )$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 1.5 close parenthesis")[$sin ( 1.5 )$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 2.5 close parenthesis")[$sin ( 2.5 )$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 3.5 close parenthesis")[$sin ( 3.5 )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis 3.5 close parenthesis")[$sin ( 3.5 )$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 0.5 close parenthesis")[$sin ( 0.5 )$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 2.5 close parenthesis")[$sin ( 2.5 )$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 1.5 close parenthesis")[$sin ( 1.5 )$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 1.6 close parenthesis")[$cos ( 1.6 )$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 2.6 close parenthesis")[$cos ( 2.6 )$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 3.6 close parenthesis")[$cos ( 3.6 )$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 5.6 close parenthesis")[$cos ( 5.6 )$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 close parenthesis")[$cos ( 2 )$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 3 close parenthesis")[$cos ( 3 )$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 4 close parenthesis")[$cos ( 4 )$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 5 close parenthesis")[$cos ( 5 )$] #math.equation(block: true, alt: "cos ⁡ open parenthesis 3 close parenthesis")[$cos ( 3 )$],#math.equation(block: true, alt: "cos ⁡ open parenthesis 4 close parenthesis")[$cos ( 4 )$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 close parenthesis")[$cos ( 2 )$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 5 close parenthesis")[$cos ( 5 )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis 2.8 close parenthesis")[$sin ( 2.8 )$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 3.8 close parenthesis")[$sin ( 3.8 )$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 4.8 close parenthesis")[$sin ( 4.8 )$], #math.equation(block: true, alt: "sin ⁡ open parenthesis 5.8 close parenthesis")[$sin ( 5.8 )$] The sunrise time in Wellington, New Zealand, on the #math.equation(block: false, alt: "n")[$n$]th day of the year can be modeled by #math.equation(block: true, alt: "S equals 1.93 sin ⁡ open parenthesis 0.016 n minus 1.13 close parenthesis plus 6.14")[$S = 1.93 " " sin " " ( 0.016 n − 1.13 ) + 6.14$] where #math.equation(block: false, alt: "S")[$S$] is given in hours after midnight. Find the sunrise time on January 1 (day #math.equation(block: false, alt: "n equals 1")[$n = 1$]), April 1 (day #math.equation(block: false, alt: "n equals 91")[$n = 91$]), July 1 (day #math.equation(block: false, alt: "n equals 182")[$n = 182$]), and on October 1 (day #math.equation(block: false, alt: "n equals 274")[$n = 274$]). January 1: 4:24, April 1: 6:45, July 1: 8:02, October 1: 5:55 The variable star RT Cygni reached its maximum magnitude on May 22, 2004, and #math.equation(block: false, alt: "t")[$t$] days later its magnitude is given by #math.equation(block: true, alt: "M equals 9.55 minus 2.25 cos ⁡ open parenthesis 0.033 t close parenthesis")[$M = 9.55 − 2.25 " " cos " " ( 0.033 t )$] Find the magnitude of RT Cygni on days #math.equation(block: false, alt: "t equals 0 , t equals 48 , t equals 95 , t equals 142 ,")[$t = 0 , " " t = 48 , " " t = 95 , " " t = 142 ,$] and #math.equation(block: false, alt: "t equals 190")[$t = 190$]. (Note that smaller values of #math.equation(block: false, alt: "M")[$M$] denote brighter magnitudes.) For Problems 33–38, find the reference angle in radians, rounded to two decimal places. #math.equation(block: true, alt: "1.8")[$1.8$] #math.equation(block: true, alt: "1.34")[$1.34$] #math.equation(block: true, alt: "4.9")[$4.9$] #math.equation(block: true, alt: "minus 2.3")[$− 2.3$] #math.equation(block: true, alt: "0.84")[$0.84$] #math.equation(block: true, alt: "minus 6.0")[$− 6.0$] #math.equation(block: true, alt: "9.4")[$9.4$] #math.equation(block: true, alt: "0.02")[$0.02$] #math.equation(block: true, alt: "7.1")[$7.1$] For Problems 39–44, find the reference angle in radians, expressed as a multiple of #math.equation(block: false, alt: "π")[$π$]. #math.equation(block: true, alt: "the fraction 11 π over 12")[$display(frac(11 π, 12))$] #math.equation(block: true, alt: "the fraction 1 over 12 π")[$display(frac(1, 12)) π$] #math.equation(block: true, alt: "the fraction 11 π over 8")[$display(frac(11 π, 8))$] #math.equation(block: true, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$] #math.equation(block: true, alt: "the fraction 1 over 3 π")[$display(frac(1, 3)) π$] #math.equation(block: true, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$] #math.equation(block: true, alt: "the fraction 13 π over 4")[$display(frac(13 π, 4))$] #math.equation(block: true, alt: "the fraction 1 over 4 π")[$display(frac(1, 4)) π$] #math.equation(block: true, alt: "the fraction 8 π over 3")[$display(frac(8 π, 3))$] Find three angles in radians between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$] with the given reference angle. Sketch all the angles on a unit circle. + #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$] + #math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$] + #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] + #math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$], #math.equation(block: false, alt: "the fraction 7 π over 6")[$" " display(frac(7 π, 6))$], #math.equation(block: false, alt: "the fraction 11 π over 6")[$" " display(frac(11 π, 6))$]#figure(figph[circle], alt: "circle", caption: none) + #math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$], #math.equation(block: false, alt: "the fraction 5 π over 4")[$" " display(frac(5 π, 4))$], #math.equation(block: false, alt: "the fraction 7 π over 4")[$" " display(frac(7 π, 4))$]#figure(figph[circle], alt: "circle", caption: none) + #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$], #math.equation(block: false, alt: "the fraction 4 π over 3")[$" " display(frac(4 π, 3))$], #math.equation(block: false, alt: "the fraction 5 π over 3")[$" " display(frac(5 π, 3))$]#figure(figph[circle], alt: "circle", caption: none) Complete the table. #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$" " θ " "$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$" " " " " " sin ( θ ) " " " " " "$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$" " " " " " cos ( θ ) " " " " " "$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$" " " " " " tan ( θ ) " " " " " "$]]), [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) Complete the table. #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$" " θ " "$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$" " " " " " sin ( θ ) " " " " " "$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$" " " " " " cos ( θ ) " " " " " "$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$" " " " " " tan ( θ ) " " " " " "$]]), [#math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$" " θ " "$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$" " " " " " sin ( θ ) " " " " " "$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$" " " " " " cos ( θ ) " " " " " "$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$" " " " " " tan ( θ ) " " " " " "$]]), [#math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 3")[$display(frac(1, sqrt(3)))$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction minus 1 over the square root of 2")[$display(frac(− 1, sqrt(2)))$]], [#math.equation(block: false, alt: "the fraction minus 1 over the square root of 2")[$display(frac(− 1, sqrt(2)))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]], )) Complete the table. #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$" " θ " "$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$" " " " " " sin ( θ ) " " " " " "$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$" " " " " " cos ( θ ) " " " " " "$]], [#math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$" " " " " " tan ( θ ) " " " " " "$]]), [#math.equation(block: false, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 11 π over 6")[$display(frac(11 π, 6))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) For Problems 49–54, evaluate the expression exactly. #math.equation(block: true, alt: "cos ⁡ open parenthesis the fraction π over 3 close parenthesis sin ⁡ open parenthesis the fraction π over 6 close parenthesis")[$cos ( display(frac(π, 3)) ) sin ( display(frac(π, 6)) )$] #math.equation(block: true, alt: "the fraction 1 over 4")[$display(frac(1, 4))$] #math.equation(block: true, alt: "sin ⁡ open parenthesis the fraction π over 4 close parenthesis tan ⁡ open parenthesis the fraction π over 3 close parenthesis")[$sin ( display(frac(π, 4)) ) tan ( display(frac(π, 3)) )$] #math.equation(block: true, alt: "tan ⁡ open parenthesis the fraction 5 π over 6 close parenthesis plus tan ⁡ open parenthesis the fraction 7 π over 4 close parenthesis")[$tan ( display(frac(5 π, 6)) ) + tan ( display(frac(7 π, 4)) )$] #math.equation(block: true, alt: "minus the fraction 3 plus the square root of 3 over 3")[$− display(frac(3 + sqrt(3), 3))$] #math.equation(block: true, alt: "cos ⁡ open parenthesis the fraction 3 π over 4 close parenthesis minus cos ⁡ open parenthesis the fraction 5 π over 3 close parenthesis")[$cos ( display(frac(3 π, 4)) ) − cos ( display(frac(5 π, 3)) )$] #math.equation(block: true, alt: "cos squared ⁡ open parenthesis the fraction 11 π over 6 close parenthesis minus 3 cos ⁡ open parenthesis the fraction 11 π over 6 close parenthesis")[$cos^(2) ( display(frac(11 π, 6)) ) − 3 cos ( display(frac(11 π, 6)) )$] #math.equation(block: true, alt: "the fraction 3 minus 6 the square root of 3 over 4")[$display(frac(3 − 6 sqrt(3), 4))$] #math.equation(block: true, alt: "2 sin ⁡ open parenthesis the fraction 4 π over 3 close parenthesis minus sin squared ⁡ open parenthesis the fraction 4 π over 3 close parenthesis")[$2 sin ( display(frac(4 π, 3)) ) − sin^(2) ( display(frac(4 π, 3)) )$] Starting at #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$], you walk #math.equation(block: false, alt: "s")[$s$] units around a unit circle. For Problems 55–58, sketch a unit circle showing your position. What are your coordinates? #math.equation(block: true, alt: "s equals 2.5")[$s = 2.5$] #math.equation(block: true, alt: "open parenthesis cos ⁡ open parenthesis 2.5 close parenthesis , sin ⁡ open parenthesis 2.5 close parenthesis close parenthesis approximately equals open parenthesis minus 0.8 , 0.6 close parenthesis")[$( cos ( 2.5 ) , sin ( 2.5 ) ) ≈ ( − 0.8 , 0.6 )$] #math.equation(block: true, alt: "s equals 4.3")[$s = 4.3$] #math.equation(block: true, alt: "s equals 8.5")[$s = 8.5$] #math.equation(block: true, alt: "open parenthesis cos ⁡ open parenthesis 8.5 close parenthesis , sin ⁡ open parenthesis 8.5 close parenthesis close parenthesis approximately equals open parenthesis minus 0.6 , 0.8 close parenthesis")[$( cos ( 8.5 ) , sin ( 8.5 ) ) ≈ ( − 0.6 , 0.8 )$] #math.equation(block: true, alt: "s equals 11")[$s = 11$] City Park features a circular jogging track of radius 1 mile, centered on the open-air bandstand. You start jogging on the track 1 mile due east of the bandstand and proceed counterclockwise. What are your coordinates, relative to the bandstand, when you have jogged five miles? #math.equation(block: false, alt: "cos ⁡ open parenthesis 5 close parenthesis approximately equals 0.28")[$cos ( 5 ) ≈ 0.28$] mi east, #math.equation(block: false, alt: "sin ⁡ open parenthesis 5 close parenthesis approximately equals minus 0.96")[$sin ( 5 ) ≈ − 0.96$] mi north, or about 0.96 mi south Silver Reservoir is a circular man-made lake of radius 1 kilometer. If you start at the easternmost point on the reservoir and walk counterclockwise for 4 kilometers, how far south of your intial position are you? For Problems 61–66, find the angle in radians between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$] determined by the terminal point on the unit circle. Round your answer to hundredths. #math.equation(block: true, alt: "open parenthesis minus 0.1782 , 0.9840 close parenthesis")[$( − 0.1782 , 0.9840 )$] #math.equation(block: true, alt: "1.75")[$1.75$] #math.equation(block: true, alt: "open parenthesis minus 0.8968 , minus 0.4425 close parenthesis")[$( − 0.8968 , − 0.4425 )$] #math.equation(block: true, alt: "open parenthesis 0.8855 , minus 0.4646 close parenthesis")[$( 0.8855 , − 0.4646 )$] #math.equation(block: true, alt: "5.8")[$5.8$] #math.equation(block: true, alt: "open parenthesis 0.9801 , 0.1987 close parenthesis")[$( 0.9801 , 0.1987 )$] #math.equation(block: true, alt: "open parenthesis minus 0.7659 , minus 0.6430 close parenthesis")[$( − 0.7659 , − 0.6430 )$] #math.equation(block: true, alt: "3.84")[$3.84$] #math.equation(block: true, alt: "open parenthesis 0.9602 , minus 0.2794 close parenthesis")[$( 0.9602 , − 0.2794 )$] + Sketch a unit circle and the line #math.equation(block: false, alt: "y equals x")[$y = x$]. Find the coordinates of the two points where the line and the circle intersect. + State your answers to part (a) using trigonometric functions. + #figure(figph[circle], alt: "circle", caption: none) #linebreak() Intersections: #math.equation(block: false, alt: "open parenthesis the fraction 1 over the square root of 2 , the fraction 1 over the square root of 2 close parenthesis")[$( display(frac(1, sqrt(2))) , display(frac(1, sqrt(2))) )$] and #math.equation(block: false, alt: "open parenthesis the fraction minus 1 over the square root of 2 , the fraction minus 1 over the square root of 2 close parenthesis")[$( display(frac(− 1, sqrt(2))) , display(frac(− 1, sqrt(2))) )$] + #math.equation(block: false, alt: "open parenthesis cos ⁡ open parenthesis the fraction π over 4 close parenthesis , sin ⁡ open parenthesis the fraction π over 4 close parenthesis close parenthesis")[$( cos ( display(frac(π, 4)) ) , sin ( display(frac(π, 4)) ) )$] and #math.equation(block: false, alt: "open parenthesis cos ⁡ open parenthesis the fraction 5 π over 4 close parenthesis , sin ⁡ open parenthesis the fraction 5 π over 4 close parenthesis close parenthesis")[$( cos ( display(frac(5 π, 4)) ) , sin ( display(frac(5 π, 4)) ) )$] + Sketch a unit circle and the line #math.equation(block: false, alt: "y equals minus x")[$y = − x$]. Find the coordinates of the two points where the line and the circle intersect. + State your answers to part (a) using trigonometric functions. + Sketch a line that passes through the origin and the point #math.equation(block: false, alt: "open parenthesis 8 , 3 close parenthesis")[$( 8 , 3 )$]. What is the slope of the line? + What is the angle of inclination of the line in radians, measured from the positive #math.equation(block: false, alt: "x")[$x$]-axis? #figure(figph[grid], alt: "grid", caption: none) + #figure(figph[graph], alt: "graph", caption: none) #linebreak() #math.equation(block: false, alt: "m equals the fraction 3 over 8")[$m = display(frac(3, 8))$] + #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis the fraction 3 over 8 close parenthesis approximately equals 0.3588")[$tan^(− 1) ( frac(3, 8) ) ≈ 0.3588$] + Sketch a line that passes through the origin and the point #math.equation(block: false, alt: "open parenthesis 3 , 8 close parenthesis")[$( 3 , 8 )$]. What is the slope of the line? + What is the angle of inclination of the line in radians, measured from the positive #math.equation(block: false, alt: "x")[$x$]-axis? #figure(figph[grid], alt: "grid", caption: none) For Problems 71–74, find an equation for the line with the given angle of inclination, passing through the given point. (See Section 4.3 to review angle of inclination.) #math.equation(block: true, alt: "α equals the fraction π over 3 , open parenthesis 4 , 2 close parenthesis")[$α = display(frac(π, 3)) , " " ( 4 , 2 )$] #math.equation(block: true, alt: "y minus 2 equals the square root of 3 open parenthesis x minus 4 close parenthesis")[$y − 2 = sqrt(3) ( x − 4 )$] #math.equation(block: true, alt: "α equals the fraction 5 π over 6 , open parenthesis minus 6 , 3 close parenthesis")[$α = display(frac(5 π, 6)) , " " ( − 6 , 3 )$] #math.equation(block: true, alt: "α equals 2.4 , open parenthesis 5 , minus 8 close parenthesis")[$α = 2.4 , " " ( 5 , − 8 )$] #math.equation(block: false, alt: "y plus 8 equals open parenthesis tan ⁡ open parenthesis 2.4 close parenthesis close parenthesis open parenthesis open parenthesis x minus 5 close parenthesis")[$y + 8 = \( tan \( 2.4 \) \) \( \( x − 5 \)$] or #math.equation(block: false, alt: "y plus 8 equals minus 0.916 open parenthesis x minus 5 close parenthesis")[$y + 8 = − 0.916 ( x − 5 )$] #math.equation(block: true, alt: "α equals 0.6 , open parenthesis minus 2 , minus 3 close parenthesis")[$α = 0.6 , " " ( − 2 , − 3 )$] Use similar triangles to show that the coordinates of point #math.equation(block: false, alt: "P")[$P$] on the unit circle shown at right are #math.equation(block: false, alt: "open parenthesis cos ⁡ open parenthesis t close parenthesis , sin ⁡ open parenthesis t close parenthesis close parenthesis")[$( cos ( t ) , sin ( t ) )$]. #figure(figph[circle], alt: "circle", caption: none) Any point #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] on the terminal side of #math.equation(block: false, alt: "θ")[$θ$] satisfies #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over r")[$cos ( θ ) = display(frac(x, r))$], #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction y over r")[$" " sin ( θ ) = display(frac(y, r))$]. For the point #math.equation(block: false, alt: "P")[$P$] where #math.equation(block: false, alt: "r equals 1")[$r = 1$], #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals x")[$" " cos ( θ ) = x$], #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals y")[$" " sin ( θ ) = y$]. The arc of length #math.equation(block: false, alt: "t")[$t$] is spanned by an angle #math.equation(block: false, alt: "θ")[$θ$] in standard position. Because arclength is #math.equation(block: false, alt: "r θ")[$r θ$] and #math.equation(block: false, alt: "r equals 1")[$r = 1$], #math.equation(block: false, alt: "t equals θ ,")[$" " t = θ ,$] so #math.equation(block: false, alt: "x equals cos ⁡ open parenthesis t close parenthesis")[$x = cos ( t )$], #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis t close parenthesis")[$" " y = sin ( t )$]. Use similar triangles to show that #math.equation(block: false, alt: "S T equals tan ⁡ open parenthesis t close parenthesis")[$S T = tan ( t )$]. #figure(figph[circle], alt: "circle", caption: none) Use the results of Problem 77 for for Problems 78–80. Use similar triangles to show that the coordinates of a point #math.equation(block: false, alt: "P")[$P$] determined by angle #math.equation(block: false, alt: "θ")[$θ$] on a circle of radius #math.equation(block: false, alt: "r")[$r$] are #math.equation(block: false, alt: "x equals r cos ⁡ open parenthesis θ close parenthesis , y equals r sin ⁡ open parenthesis θ close parenthesis")[$x = r cos ( θ ) , " " y = r sin ( θ )$]. (See the figure at right.) #figure(figph[circles], alt: "circles", caption: none) The two right triangles shown are similar, so their sides are proportional. The hypotenuse of the large triangle is #math.equation(block: false, alt: "r")[$r$] times the hypotenuse of the small triangle, so the two legs of the large triangle must be #math.equation(block: false, alt: "r")[$r$] times the legs of the small triangle. Thus, because the coordinates of the vertex on the unit circle are #math.equation(block: false, alt: "open parenthesis cos ⁡ open parenthesis θ close parenthesis , sin ⁡ open parenthesis θ close parenthesis close parenthesis")[$( cos ( θ ) , sin ( θ ) )$], the coordinates of #math.equation(block: false, alt: "P")[$P$] must be #math.equation(block: false, alt: "open parenthesis r cos ⁡ open parenthesis θ close parenthesis , r sin ⁡ open parenthesis θ close parenthesis close parenthesis")[$( r cos ( θ ) , r sin ( θ ) )$]. The Astrodome in Houston has a diameter of 710 feet. If you start at the easternmost point and walk counterclockwise around its perimeter for a distance of 250 feet, how far north of your starting point are you? The Barringer meteor crater near Winslow, Arizona is 1182 meters in diameter. You start at the easternmost point on the rim of the crater and walk counterclockwise around the edge. After walking for 1 kilometer, what is your position relative to your starting point 71 m west, 587 m north One of the most intriguing features of Stonehenge is the position of the four Station Stones. #figure(figph[Stonehenge], alt: "Stonehenge", caption: none) They form the corners of a rectangle inscribed in the Aubrey Circle on the perimeter of the henge, which has diameter 288 feet. A line from the center of the circle and perpendicular to the long edge of the rectangle points through the Slaughter Stones at the entrance of the henge and out to the Heel Stone. If you stood in the center of the circle on the summer solstice, you would see the sun rise directly over the Heel Stone. #figure(figph[Stonehenge], alt: "Stonehenge", caption: none) + The sun rises #math.equation(block: false, alt: "48.6 degrees")[$48.6^(∘)$] east of north on the summer solstice at Stonehenge. If the positive #math.equation(block: false, alt: "y")[$y$]-axis points north, find the coordinates of the henge entrance relative to its center. + The northernmost station stone is located #math.equation(block: false, alt: "66.6 degrees")[$66.6^(∘)$] of arc counterclockwise from the entrance. Find its coordinates relative to the center of the henge.