#set document(title: "8.3 The Reciprocal 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")) == 8.3#h(0.6em)The Reciprocal Functions === Three More Functions The three basic trigonometric functions occur so often as the denominator of a fraction that it is convenient to give names to their reciprocals. We define three new trigonometric functions as follows. #notebox("Definition", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Three More Functions] If #math.equation(block: false, alt: "θ")[$θ$] is an angle in standard position, and #math.equation(block: false, alt: "P open parenthesis x , y close parenthesis")[$P ( x , y )$] is a point on the terminal side, then we define the following functions. #math.equation(block: true, alt: "The secant :, sec ⁡ open parenthesis θ close parenthesis equals the fraction r over x; The cosecant :, csc ⁡ open parenthesis θ close parenthesis equals the fraction r over y; The cotangent :, cot ⁡ open parenthesis θ close parenthesis equals the fraction x over y")[$"The" " " " " "secant" : " " " " " " " " " " " " & bold(sec) ( bold(italic(θ)) ) = display(frac(bold(italic(r)), bold(italic(x)))) \ "The" " " " " "cosecant" : " " " " " " " " & bold(csc) ( bold(italic(θ)) ) = display(frac(bold(italic(r)), bold(italic(y)))) \ "The" " " " " "cotangent" : " " " " " " & bold(cot) ( bold(italic(θ)) ) = display(frac(bold(italic(x)), bold(italic(y))))$] #figure(figph[angle], alt: "angle", caption: none) ] We can find exact values for all six trig functions at a given angle if we know the value of any one of them. #examplebox("Example 1")[][ If #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis equals 3")[$" " sec ( θ ) = 3 " "$], and #math.equation(block: false, alt: "minus the fraction π over 2 less than or equal to θ less than or equal to 0")[$" " − display(frac(π, 2)) ≤ θ ≤ 0$], find exact values for the other five trig functions. #solutionbox[ Because #math.equation(block: false, alt: "minus the fraction π over 2 less than or equal to θ less than or equal to 0")[$" " − display(frac(π, 2)) ≤ θ ≤ 0$], we draw a reference triangle in the fourth quadrant, as shown at right. Because #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis equals 3 equals the fraction 3 over 1")[$" " sec ( θ ) = 3 = display(frac(3, 1))$], we label the horizontal leg with #math.equation(block: false, alt: "x equals 1")[$x = 1$] and the hypotenuse with #math.equation(block: false, alt: "r equals 3")[$r = 3$]. #figure(figph[fourth quadrant angle], alt: "fourth quadrant angle", caption: none) From the Pythagorean theorem, we find #math.equation(block: false, alt: "y equals minus the square root of 8 equals minus 2 the square root of 2")[$y = − sqrt(8) = − 2 sqrt(2)$]. We can now compute the values of the six trigonometric ratios. #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis, equals the fraction x over r equals the fraction 1 over 3, sec ⁡ open parenthesis θ close parenthesis equals the fraction r over x equals the fraction 3 over 1 equals 3; sin ⁡ open parenthesis θ close parenthesis, equals the fraction y over r equals the fraction minus 2 the square root of 2 over 3, csc ⁡ open parenthesis θ close parenthesis equals the fraction r over y equals the fraction 3 over minus 2 the square root of 2 equals the fraction minus 3 the square root of 2 over 4; tan ⁡ open parenthesis θ close parenthesis, equals the fraction y over x equals the fraction minus 2 the square root of 2 over 1 equals minus 2 the square root of 2, cot ⁡ open parenthesis θ close parenthesis equals the fraction x over y equals the fraction 1 over minus")[$cos ( θ ) & = display(frac(x, r)) = display(frac(1, 3)) & & sec ( θ ) = display(frac(r, x)) = display(frac(3, 1)) = 3 \ sin ( θ ) & = display(frac(y, r)) = display(frac(− 2 sqrt(2), 3)) & & csc ( θ ) = display(frac(r, y)) = display(frac(3, − 2 sqrt(2))) = display(frac(− 3 sqrt(2), 4)) \ tan ( θ ) & = display(frac(y, x)) = display(frac(− 2 sqrt(2), 1)) = − 2 sqrt(2) & & cot ( θ ) = display(frac(x, y)) = display(frac(1, − 2 sqrt(2))) = display(frac(− sqrt(2), 4))$] ] ] If #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis equals 4")[$" " csc ( θ ) = 4$], and #math.equation(block: false, alt: "90 degrees less than or equal to θ less than or equal to 180 degrees")[$" " 90^(∘) ≤ θ ≤ 180^(∘)$], find exact values for the other five trig functions. #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 15 over 4")[$cos ( θ ) = display(frac(− sqrt(15), 4))$], #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 4")[$" " sin ( θ ) = display(frac(1, 4))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 1 over the square root of 15")[$" " tan ( θ ) = display(frac(− 1, sqrt(15)))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction minus 4 over the square root of 15")[$" " sec ( θ ) = display(frac(− 4, sqrt(15)))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals minus the square root of 15")[$" " cot ( θ ) = − sqrt(15)$] By comparing the definitions of secant, cosecant, and cotangent to the three basic trigonometric functions, we find the following relationships. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Reciprocal Trigonometric Functions] #math.equation(block: true, alt: "The secant function : sec ⁡ open parenthesis θ close parenthesis equals the fraction 1 over cos ⁡ open parenthesis θ close parenthesis; The cosecant function : csc ⁡ open parenthesis θ close parenthesis equals the fraction 1 over sin ⁡ open parenthesis θ close parenthesis; The cotangent function : cot ⁡ open parenthesis θ close parenthesis equals the fraction 1 over tan ⁡ open parenthesis θ close parenthesis")[$"The" " " " " "secant" " " "function" : " " " " " " " " " " " " bold(sec) ( bold(italic(θ)) ) = display(frac(1, bold(cos) ( bold(italic(θ)) ))) \ "The" " " " " "cosecant" " " "function" : " " " " " " " " bold(csc) ( bold(italic(θ)) ) = display(frac(1, bold(sin) ( bold(italic(θ)) ))) \ "The" " " " " "cotangent" " " "function" : " " " " " " bold(cot) ( bold(italic(θ)) ) = display(frac(1, bold(tan) ( bold(italic(θ)) )))$] ] Calculators do not have keys for the secant, cosecant, and cotangent functions; instead, we calculate their values as reciprocals. #examplebox("Example 2")[][ Use a calculator to approximate #math.equation(block: false, alt: "sec ⁡ open parenthesis 47 degrees close parenthesis")[$sec ( 47^(∘) )$] to three decimal places. #solutionbox[ With the calculator in degree mode, enter $#h(2em) #h(2em) #h(2em)$ 1 COS 47 ENTER to obtain #math.equation(block: false, alt: "sec ⁡ open parenthesis 47 degrees close parenthesis approximately equals 1.466")[$sec ( 47^(∘) ) ≈ 1.466$]. Or we can calculate #math.equation(block: false, alt: "cos ⁡ open parenthesis 47 degrees close parenthesis")[$cos ( 47^(∘) )$] first, and then use the reciprocal key: $#h(2em) #h(2em) #h(2em)$COS 47 ENTER #math.equation(block: false, alt: "x to the power minus 1")[$#box(stroke: 0.6pt, inset: 3pt, outset: 0pt)[$display(x^(− 1))$]$] ENTER ] ] Use a calculator to approximate #math.equation(block: false, alt: "csc ⁡ open parenthesis 132 degrees close parenthesis")[$csc ( 132^(∘) )$] to three decimal places. #math.equation(block: true, alt: "1.346")[$1.346$] Of course, we can also evaluate the reciprocal trig functions for angles in radians, or for real numbers. Thus for example, #math.equation(block: true, alt: "csc ⁡ open parenthesis 3.5 close parenthesis equals the fraction 1 over sin ⁡ open parenthesis 3.5 close parenthesis equals minus 2.8508 and cot ⁡ open parenthesis minus 4 close parenthesis equals the fraction 1 over tan ⁡ open parenthesis minus 4 close parenthesis equals minus 0.8637")[$csc ( 3.5 ) = display(frac(1, sin ( 3.5 ))) = − 2.8508 " " " " " " " " " " " " "and" " " " " " " " " " " " " cot ( − 4 ) = display(frac(1, tan ( − 4 ))) = − 0.8637$] In particular, the exact values for the reciprocal trig functions of the special angles are easily obtained. #figure(table( columns: 4, align: left, inset: 6pt, [Exact Values for the Special Angles], [], [], [], [#math.equation(block: false, alt: "θ")[$" " θ " "$]], [#math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis")[$sec ( θ )$]], [#math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis")[$sec ( θ )$]], [#math.equation(block: false, alt: "cot ⁡ open parenthesis θ close parenthesis")[$cot ( θ )$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [undefined], [undefined], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction 2 the square root of 3 over 3")[$display(frac(2 sqrt(3), 3))$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$]], [#math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "the fraction 2 the square root of 3 over 3")[$display(frac(2 sqrt(3), 3))$]], [#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 π over 2")[$display(frac(π, 2))$]], [undefined], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], )) #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The reciprocal functions are not the same as the inverse trig functions! For example, #math.equation(block: false, alt: "sec ⁡ open parenthesis 0.8 close parenthesis")[$sec ( 0.8 )$] is not equal to #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis 0.8 close parenthesis")[$cos^(− 1) ( 0.8 )$]. Remember that #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis 0.8 close parenthesis")[$" " bold(cos)^(− 1) ( 0.8 ) " "$] is an angle, namely the angle whose cosine is 0.8, while #math.equation(block: false, alt: "sec ⁡ open parenthesis 0.8 close parenthesis")[$" " bold(sec) ( 0.8 ) " "$] is the reciprocal of the cosine of 0.8 radians, or #math.equation(block: false, alt: "the fraction 1 over cos ⁡ open parenthesis 0.8 close parenthesis")[$display(frac(1, cos ( 0.8 )))$]. You can check on your calculator that #math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis 0.8 close parenthesis equals 0.6435 radians , and sec ⁡ open parenthesis 0.8 close parenthesis equals 1.4353")[$cos^(− 1) ( 0.8 ) = 0.6435 " " "radians" , " " " " " " " " "and" " " " " " " " " sec ( 0.8 ) = 1.4353$] ] Each of the reciprocal functions is undefined when its denominator is equal to zero. For example, the secant is undefined when #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0")[$cos ( θ ) = 0$], or when #math.equation(block: false, alt: "θ")[$θ$] is an odd multiple of #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. #examplebox("Example 3")[][ For which angles is the cosecant undefined? #solutionbox[ The cosecant is undefined when its denominator, #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$], equals zero, and #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals 0")[$sin ( θ ) = 0$] when #math.equation(block: false, alt: "θ")[$θ$] is a multiple of #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. In radians, #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis")[$csc ( θ )$] is undefined if #math.equation(block: false, alt: "θ")[$θ$] is a multiple of #math.equation(block: false, alt: "π")[$π$]. ] ] For what angles is the cotangent undefined? Give your answers in degrees and in radians. Multiples of #math.equation(block: false, alt: "180 degrees")[$180^(∘)$], or multiples of #math.equation(block: false, alt: "π")[$π$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Although #math.equation(block: false, alt: "tan ⁡ open parenthesis the fraction π over 2 close parenthesis")[$tan ( display(frac(π, 2)) )$] is undefined, #math.equation(block: false, alt: "cot ⁡ open parenthesis the fraction π over 2 close parenthesis equals 0")[$cot ( display(frac(π, 2)) ) = 0$]. ] === Application to Right Triangles In Chapter 2 we defined three trigonometric ratios for an acute angle; namely, sine, cosine, and tangent. When we take the reciprocals of those ratios, we obtain expressions for the secant, cosecant, and cotangent. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Reciprocal Trigonometric Ratios] If #math.equation(block: false, alt: "θ")[$θ$] is one of the acute angles in a right triangle, #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis, equals the fraction hypotenuse over adjacent; csc ⁡ open parenthesis θ close parenthesis, equals the fraction hypotenuse over opposite; cot ⁡ open parenthesis θ close parenthesis, equals the fraction adjacent over opposite")[$bold(sec) ( bold(italic(θ)) ) & = display(frac("hypotenuse", "adjacent")) \ bold(csc) ( bold(italic(θ)) ) & = display(frac("hypotenuse", "opposite")) \ bold(cot) ( bold(italic(θ)) ) & = display(frac("adjacent", "opposite"))$] #figure(figph[triangle], alt: "triangle", caption: none) ] Although we can express any relationship between the sides of a right triangle using sine, cosine, and tangent, sometimes it is more convenient to use one of the reciprocal functions. #examplebox("Example 4")[][ The length, #math.equation(block: false, alt: "L")[$L$], of the shadow cast by a flagpole on a sunny day depends on the height, #math.equation(block: false, alt: "h")[$h$], of the flagpole and the angle, #math.equation(block: false, alt: "θ")[$θ$], that the sun's rays make with ground. #figure(figph[flagpole and shadow], alt: "flagpole and shadow", caption: none) + Write an expression for the length, #math.equation(block: false, alt: "L")[$L$], of the shadow cast by a flagpole of height #math.equation(block: false, alt: "h")[$h$] when the sun makes an angle of #math.equation(block: false, alt: "θ")[$θ$] from the ground. + Find the length (to the nearest 0.01 meter) of the shadow cast by a 3-meter flagpole when the sun makes an angle of #math.equation(block: false, alt: "20 degrees")[$20^(∘)$] from the ground. #solutionbox[ + From the figure, we see that #math.equation(block: false, alt: "the fraction L over h equals cot ⁡ open parenthesis θ close parenthesis")[$display(frac(L, h)) = cot ( θ )$], or #math.equation(block: false, alt: "L equals h cot ⁡ open parenthesis θ close parenthesis")[$L = h cot ( θ )$]. + Substiting #math.equation(block: false, alt: "3")[$3$] for #math.equation(block: false, alt: "h")[$h$] and #math.equation(block: false, alt: "20 degrees")[$20^(∘)$] for #math.equation(block: false, alt: "θ")[$θ$], we find #math.equation(block: true, alt: "L equals 3 cot ⁡ open parenthesis 20 degrees close parenthesis equals 3 open parenthesis 2.7475 close parenthesis equals 8.24")[$L = 3 cot ( 20^(∘) ) = 3 ( 2.7475 ) = 8.24$] The shadow is about 8.24 meters long. ] ] The area #math.equation(block: false, alt: "A")[$A$] of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides having perimeter #math.equation(block: false, alt: "L")[$L$] satisfies #math.equation(block: true, alt: "A equals the fraction L squared over 4 n cot ⁡ the fraction π over n")[$A = display(frac(L^(2), 4 n)) cot display(frac(π, n))$] Refer to the figure at right showing #math.equation(block: false, alt: "n equals 6")[$n = 6$] to prove this formula in the following steps. #figure(figph[hexagon], alt: "hexagon", caption: none) + Find an expression for the angle #math.equation(block: false, alt: "θ")[$θ$] in terms of #math.equation(block: false, alt: "n")[$n$]. + Find an expression for the base of the triangle shown. + Find an expression for the height of the triangle. + Write an expression for the area of the triangle, and then for the area of the entire polygon. + #math.equation(block: false, alt: "θ equals the fraction π over n")[$θ = display(frac(π, n))$] + #math.equation(block: false, alt: "b equals the fraction L over n")[$b = display(frac(L, n))$] + #math.equation(block: false, alt: "h equals the fraction L over 2 n cot ⁡ open parenthesis the fraction π over n close parenthesis")[$h = display(frac(L, 2 n)) cot ( display(frac(π, n)) )$] + #math.equation(block: false, alt: "A sub T equals the fraction L squared over 4 n squared cot ⁡ open parenthesis the fraction π over n close parenthesis , A sub P equals the fraction L squared over 4 n cot ⁡ open parenthesis the fraction π over n close parenthesis")[$A_(T) = display(frac(L^(2), 4 n^(2))) cot ( display(frac(π, n)) ) , " " " " A_(P) = display(frac(L^(2), 4 n)) cot ( display(frac(π, n)) )$] === Graphs of the Reciprocal Functions We can obtain graphs of the reciprocal trig functions by plotting points, as we did for the sine, cosine and tangent functions. However, it is more enlightening to construct these graphs as the reciprocals of the three basic functions. The dashed grey curve is y = cos x, the one you already know; the solid blue curve is its reciprocal, y = sec x = 1/cos x. The two dots sit on one vertical line at x = a, so the reciprocal relationship is readable straight off their heights — multiply them and you always get 1. Drag a out toward 1.4 and watch the pair separate: cos falls to about 0.17 while sec climbs to about 5.9, because the reciprocal of a small number is large. The slider deliberately stops short of 1.5708, and that is the lesson — there cos x reaches 0, so sec x has a vertical asymptote and simply does not exist, exactly as the section says of every reciprocal function where its denominator is zero. Drag back to a = 0 and both dots sit at height 1: wherever cos is 1 or −1 the two curves touch, because those are the only two numbers equal to their own reciprocal.{"functions":\[{"color":"\#6a6a6a","expression":"cos(x)","expression2":"","id":"cosine","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"cartesian","tMax":12,"tMin":-12,"variable":"x","visible":true},{"color":"\#1f4e79","expression":"1/cos(x)","expression2":"","id":"secant","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":12,"tMin":-12,"variable":"x","visible":true},{"color":"\#b8442a","expression":"a","expression2":"t","id":"readout-line","inequality":"lt","lineDash":\[4,4\],"lineWidth":1.5,"mode":"parametric","tMax":7,"tMin":-7,"variable":"t","visible":true},{"color":"\#b8442a","expression":"a + 0.13\*cos(t)","expression2":"cos(a) + 0.13\*sin(t)","id":"dot-cos","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"a + 0.13\*cos(t)","expression2":"1/cos(a) + 0.13\*sin(t)","id":"dot-sec","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-a","label":"Input a \\u2014 both dots read the two curves at this x","max":1.4,"min":-1.4,"name":"a","step":0.1,"value":1}\],"title":"y = sec x as the reciprocal of y = cos x","version":1,"viewport":{"centerX":0,"centerY":0,"scale":34}}#examplebox("Example 5")[][ Use the graph of #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$" " y = cos ( x ) " "$] to construct a graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sec ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = sec ( x )$]. #solutionbox[ Consider the graph of #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$" " y = cos ( x ) " "$] shown at left below. When #math.equation(block: false, alt: "x equals the fraction minus π over 2 , the fraction π over 2")[$x = display(frac(− π, 2)) , " " display(frac(π, 2))$] and #math.equation(block: false, alt: "the fraction 3 π over 2 , cos ⁡ open parenthesis x close parenthesis equals 0")[$display(frac(3 π, 2)) , " " cos ( x ) = 0$], so #math.equation(block: false, alt: "sec ⁡ open parenthesis x close parenthesis")[$sec ( x )$] is undefined at these #math.equation(block: false, alt: "x")[$x$]-values, and we insert vertical asymptotes at those #math.equation(block: false, alt: "x")[$x$]-values to start our graph of #math.equation(block: false, alt: "y equals sec ⁡ open parenthesis x close parenthesis")[$y = sec ( x )$], as shown at right below. #figure(figph[cosine], alt: "cosine", caption: none) #figure(figph[grid], alt: "grid", caption: none) To find some points on the graph, we look at points on the graph of #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$y = cos ( x )$]. At each #math.equation(block: false, alt: "x")[$x$]-value, the #math.equation(block: false, alt: "y")[$y$]-coordinate of the point on the graph of #math.equation(block: false, alt: "y equals sec ⁡ open parenthesis x close parenthesis")[$y = sec ( x )$] is the reciprocal of #math.equation(block: false, alt: "cos ⁡ open parenthesis x close parenthesis")[$cos ( x )$]. For example, at #math.equation(block: false, alt: "x equals 0")[$x = 0$] and #math.equation(block: false, alt: "x equals 2 π")[$x = 2 π$], we have #math.equation(block: false, alt: "cos ⁡ open parenthesis x close parenthesis equals 1")[$cos ( x ) = 1$], so #math.equation(block: true, alt: "sec ⁡ open parenthesis x close parenthesis equals the fraction 1 over 1 equals 1")[$sec ( x ) = frac(1, 1) = 1$] Thus, we plot the points #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$] and #math.equation(block: false, alt: "open parenthesis 2 π , 1 close parenthesis")[$( 2 π , 1 )$] on the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sec ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = sec ( x )$]. Similarly, at #math.equation(block: false, alt: "x equals minus π")[$x = − π$] and #math.equation(block: false, alt: "x equals π")[$x = π$], #math.equation(block: false, alt: "cos ⁡ open parenthesis x close parenthesis equals minus 1")[$cos ( x ) = − 1$], so the value of #math.equation(block: false, alt: "sec ⁡ open parenthesis x close parenthesis")[$sec ( x )$] is #math.equation(block: false, alt: "the fraction 1 over minus 1 equals minus 1")[$frac(1, − 1) = − 1$], and we plot the points #math.equation(block: false, alt: "open parenthesis minus π , minus 1 close parenthesis")[$( − π , − 1 )$] and #math.equation(block: false, alt: "open parenthesis π , minus 1 close parenthesis")[$( π , − 1 )$] on the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sec ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = sec ( x )$]. Finally, we notice that the values of #math.equation(block: false, alt: "cos ⁡ open parenthesis x close parenthesis")[$cos ( x )$] are decreasing toward #math.equation(block: false, alt: "0")[$0$] as #math.equation(block: false, alt: "x")[$x$] increases from #math.equation(block: false, alt: "0")[$0$] to #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$], so the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sec ⁡ open parenthesis x close parenthesis")[$f ( x ) = sec ( x )$] increases toward #math.equation(block: false, alt: "∞")[$upright(∞)$] on the same interval. By similar arguments, we fill in the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sec ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = sec ( x ) " "$] between each of the vertical asymptotes, to produce the graph below. #figure(figph[cosine and secant], alt: "cosine and secant", caption: none) ] ] Use the graph of #math.equation(block: false, alt: "y equals tan ⁡ open parenthesis x close parenthesis")[$y = tan ( x )$] to sketch a graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals cot ⁡ open parenthesis x close parenthesis")[$g ( x ) = cot ( x )$]. #figure(figph[tan and cotan], alt: "tan and cotan", caption: none) The graphs of the three new functions are shown below, with #math.equation(block: false, alt: "x")[$x$] in radians. Note that the secant function is undefined at odd multiples of #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$], the values at which #math.equation(block: false, alt: "cos ⁡ open parenthesis x close parenthesis equals 0")[$cos ( x ) = 0$]. The cosecant is undefined where #math.equation(block: false, alt: "sin ⁡ open parenthesis x close parenthesis equals 0")[$sin ( x ) = 0$], namely at multiples of #math.equation(block: false, alt: "π")[$π$]. The cotangent is also undefined at multiples of #math.equation(block: false, alt: "π")[$π$], because #math.equation(block: false, alt: "tan ⁡ open parenthesis x close parenthesis equals 0")[$tan ( x ) = 0$] at those values. #figure(figph[reciprocal trig functions], alt: "reciprocal trig functions", caption: none) #examplebox("Example 6")[][ State the domain and range of the secant function. #solutionbox[ Because the cosine is defined for all real numbers, the domain of the secant includes all real numbers except for values where the cosine is zero. These values are the odd multiples of #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$], that is, #math.equation(block: false, alt: "the fraction π over 2 , the fraction 3 π over 2 , the fraction 5 π over 2 , and so on")[$display(frac(π, 2)) , " " display(frac(3 π, 2)) , " " display(frac(5 π, 2)) , " " …$], and their opposites. Because the range of the cosine consists of all #math.equation(block: false, alt: "y")[$y$]-values with #math.equation(block: false, alt: "minus 1 less than or equal to y less than or equal to 1")[$− 1 ≤ y ≤ 1$], the range of the secant includes the reciprocals of those values, namely #math.equation(block: false, alt: "y greater than or equal to 1")[$y ≥ 1$] and #math.equation(block: false, alt: "y less than or equal to minus 1")[$y ≤ − 1$]. ] ] State the domain and range of the cosecant and cotangent functions. Domain of cosecant: all real numbers except integer multiples of #math.equation(block: false, alt: "π")[$π$]; Range of cosecant: #math.equation(block: false, alt: "open parenthesis minus ∞ , minus 1 close bracket union open bracket 1 , ∞ close parenthesis")[$( − upright(∞) , − 1 ] ∪ [ 1 , upright(∞) )$] Domain of cotangent: all real numbers except integer multiples of #math.equation(block: false, alt: "π")[$π$]; Range of cotangent: all real numbers === Solving Equations From the graph of the secant function, we can see that the equation #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis equals k")[$" " sec ( θ ) = k " "$] has two solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$] if #math.equation(block: false, alt: "k greater than or equal to 1")[$k ≥ 1$] or #math.equation(block: false, alt: "k less than or equal to minus 1")[$k ≤ − 1$], but no solution for #math.equation(block: false, alt: "minus 1 less than k less than 1")[$− 1 < k < 1$]. The same is true of the cosecant function: the equation #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis equals k")[$" " csc ( θ ) = k " "$] has no solution for #math.equation(block: false, alt: "minus 1 less than k less than 1")[$− 1 < k < 1$]. #examplebox("Example 7")[][ Solve #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction 2 the square root of 3 over 3")[$" " " " csc ( θ ) = display(frac(2 sqrt(3), 3)) " " " "$] for #math.equation(block: false, alt: "θ")[$θ$] between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$]. #solutionbox[ We take the reciprocal of each side of the equation to obtain #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 2 the square root of 3 equals the fraction the square root of 3 over 2")[$sin ( θ ) = display(frac(3, 2 sqrt(3))) = display(frac(sqrt(3), 2))$] Because #math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$] is one of the special values, we recognize that one of the solutions is #math.equation(block: false, alt: "θ equals the fraction π over 3")[$θ = display(frac(π, 3))$]. The sine and the cosecant are also positive in the second quadrant, so the second solution is #math.equation(block: false, alt: "π minus the fraction π over 3 equals the fraction 2 π over 3")[$π − display(frac(π, 3)) = display(frac(2 π, 3))$]. ] ] Solve #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis equals minus 1.6")[$" " " " sec ( θ ) = − 1.6 " " " "$] for #math.equation(block: false, alt: "θ")[$θ$] between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$]. #math.equation(block: true, alt: "θ equals 2.25 , 4.04")[$θ = 2.25 , #h(0.167em) 4.04$] === Using Identities All six of the trigonometric ratios are related. If we know one of the ratios, we can use identities to find any of the others. #examplebox("Example 8")[][ If #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis equals 3")[$" " sec ( θ ) = 3$], and #math.equation(block: false, alt: "minus the fraction π over 2 less than or equal to θ less than or equal to 0")[$− display(frac(π, 2)) ≤ θ ≤ 0$], find an exact value for #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis")[$csc ( θ )$]. #solutionbox[ Because #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction 1 over cos ⁡ open parenthesis θ close parenthesis")[$" " sec ( θ ) = display(frac(1, cos ( θ )))$], we see that #math.equation(block: false, alt: "the fraction 1 over cos ⁡ open parenthesis θ close parenthesis equals 3")[$" " display(frac(1, cos ( θ ))) = 3$], or #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 3")[$" " cos ( θ ) = display(frac(1, 3))$]. We use the Pythagorean identity to find the sine. #math.equation(block: true, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis, equals 1, Substitute the fraction 1 over 3 for cos ⁡ open parenthesis θ close parenthesis .; open parenthesis the fraction 1 over 3 close parenthesis squared plus sin squared ⁡ open parenthesis θ close parenthesis, equals 1, Subtract open parenthesis the fraction 1 over 3 close parenthesis squared equals the fraction 1 over 9.; sin squared ⁡ open parenthesis θ close parenthesis, equals 1 minus the fraction 1 over 9 equals the fraction 8 over 9")[$cos^(2) ( θ ) + sin^(2) ( θ ) & = 1 & & "Substitute" " " frac(1, 3) " for " bold(cos) ( bold(italic(θ)) ) . \ attach(( display(frac(1, 3)) ), t: 2) + sin^(2) ( θ ) & = 1 & & "Subtract" " " attach(( frac(1, 3) ), t: 2) = frac(1, 9) . \ sin^(2) ( θ ) & = 1 − display(frac(1, 9)) = display(frac(8, 9))$] Because #math.equation(block: false, alt: "θ")[$θ$] lies in the fourth quadrant, where the sine function is negative, we choose the negative square root for #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]. Once we know #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$], we calculate its reciprocal to find #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis")[$csc ( θ )$]. #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals minus the square root of the fraction 8 over 9 equals the fraction minus 2 the square root of 2 over 3 , and csc ⁡ open parenthesis θ close parenthesis equals the fraction 1 over sin ⁡ open parenthesis θ close parenthesis equals the fraction minus 3 over 2 the square root of 2 equals the fraction minus 3 the square root of 2 over 4")[$sin ( θ ) = − sqrt(display(frac(8, 9))) = display(frac(− 2 sqrt(2), 3)) , " " " and " " " csc ( θ ) = display(frac(1, sin ( θ ))) = display(frac(− 3, 2 sqrt(2))) = display(frac(− 3 sqrt(2), 4))$] ] ] If #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 13 over 3")[$" " csc ( θ ) = display(frac(− sqrt(13), 3))$], and #math.equation(block: false, alt: "π less than or equal to θ less than or equal to the fraction 3 π over 2")[$π ≤ θ ≤ display(frac(3 π, 2))$], find an exact value for #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis")[$sec ( θ )$]. #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 13 over 2")[$sec ( θ ) = display(frac(− sqrt(13), 2))$] Identities are especially useful if the trig ratios are algebraic expressions, rather than numerical values. In the next example, we use the cotangent identity. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Cotangent Identity] #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction 1 over tan ⁡ open parenthesis θ close parenthesis equals the fraction cos ⁡ open parenthesis θ close parenthesis over sin ⁡ open parenthesis θ close parenthesis , sin ⁡ open parenthesis θ close parenthesis not equal to 0")[$bold(cot) ( bold(italic(θ)) ) = display(frac(1, bold(tan) ( bold(italic(θ)) ))) = display(frac(bold(cos) ( bold(italic(θ)) ), bold(sin) ( bold(italic(θ)) ))) , " " " " " " " " bold(sin) ( bold(italic(θ)) ) ≠ 0$] ] #examplebox("Example 9")[][ If #math.equation(block: false, alt: "csc ⁡ open parenthesis x close parenthesis equals w")[$" " csc ( x ) = w$] and #math.equation(block: false, alt: "0 less than x less than the fraction π over 2")[$0 < x < display(frac(π, 2))$], find an expression for #math.equation(block: false, alt: "cot ⁡ open parenthesis x close parenthesis")[$cot ( x )$]. #solutionbox[ Because the sine is the reciprocal of the cosecant, we have #math.equation(block: true, alt: "sin ⁡ open parenthesis x close parenthesis equals the fraction 1 over csc ⁡ open parenthesis x close parenthesis equals the fraction 1 over w")[$sin ( x ) = display(frac(1, csc ( x ))) = display(frac(1, w))$] . We substitute #math.equation(block: false, alt: "the fraction 1 over w")[$display(frac(1, bold(italic(w))))$] for #math.equation(block: false, alt: "sin ⁡ open parenthesis x close parenthesis")[$sin ( x )$] in the Pythagorean identity to find #math.equation(block: true, alt: "cos ⁡ open parenthesis x close parenthesis equals plus or minus the square root of 1 minus sin squared ⁡ open parenthesis x close parenthesis equals plus or minus the square root of 1 minus open parenthesis the fraction 1 over w close parenthesis squared")[$cos ( x ) = ± sqrt(1 − sin^(2) ( x )) = ± sqrt(1 − attach(( display(frac(1, bold(italic(w)))) ), t: 2))$] We choose the positive root because cosine is positive in the first quadrant, and simplify to get #math.equation(block: true, alt: "cos ⁡ open parenthesis x close parenthesis equals the square root of 1 minus the fraction 1 over w squared equals the square root of the fraction w squared minus 1 over w squared equals the fraction the square root of w squared minus 1 over vertical bar w vertical bar")[$cos ( x ) = sqrt(1 − display(frac(1, w^(2)))) = sqrt(display(frac(w^(2) − 1, w^(2)))) = display(frac(sqrt(w^(2) − 1), | w |))$] We can replace #math.equation(block: false, alt: "vertical bar w vertical bar")[$| w |$] by #math.equation(block: false, alt: "w")[$w$] in this last expression because #math.equation(block: false, alt: "w greater than 0")[$w > 0$]. (Do you see why #math.equation(block: false, alt: "w greater than 0")[$w > 0$]?) Finally, because the cotangent is the reciprocal of the tangent, we have #math.equation(block: true, alt: "cot ⁡ open parenthesis x close parenthesis equals the fraction cos ⁡ open parenthesis x close parenthesis over sin ⁡ open parenthesis x close parenthesis equals the fraction the fraction w squared minus 1 over w over the fraction 1 over w equals the square root of w squared minus 1")[$cot ( x ) = display(frac(cos ( x ), sin ( x ))) = display(frac(display(frac(w^(2) − 1, w)), display(frac(1, w)))) = sqrt(w^(2) − 1)$] ] ] If #math.equation(block: false, alt: "sec ⁡ open parenthesis t close parenthesis equals the fraction 2 over a")[$" " sec ( t ) = display(frac(2, a))$] and #math.equation(block: false, alt: "the fraction 3 π over 2 less than t less than 2 π")[$display(frac(3 π, 2)) < t < 2 π$], find expressions for #math.equation(block: false, alt: "csc ⁡ open parenthesis t close parenthesis")[$csc ( t )$] and #math.equation(block: false, alt: "cot ⁡ open parenthesis t close parenthesis")[$cot ( t )$]. #math.equation(block: true, alt: "csc ⁡ open parenthesis t close parenthesis equals the fraction minus 2 over the square root of 4 minus a squared")[$csc ( t ) = display(frac(− 2, sqrt(4 − a^(2))))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis t close parenthesis equals the fraction minus a over the square root of 4 minus a squared")[$" " cot ( t ) = display(frac(− a, sqrt(4 − a^(2))))$] We can often simplify trigonometric expressions by first converting all the trig ratios to sines and cosines. #examplebox("Example 10")[][ Simplify #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis minus tan ⁡ open parenthesis θ close parenthesis sin ⁡ open parenthesis θ close parenthesis")[$" " sec ( θ ) − tan ( θ ) sin ( θ )$]. #solutionbox[ We replace #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis")[$" " sec ( θ ) " "$] by #math.equation(block: false, alt: "the fraction 1 over cos ⁡ open parenthesis θ close parenthesis")[$" " display(frac(1, cos ( θ ))) " "$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$" " tan ( θ ) " "$] by #math.equation(block: false, alt: "the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis")[$" " display(frac(sin ( θ ), cos ( θ ))) " "$] to get #math.equation(block: true, alt: "the fraction 1 over cos ⁡ open parenthesis θ close parenthesis minus the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis times sin ⁡ open parenthesis θ close parenthesis, equals the fraction 1 over cos ⁡ open parenthesis θ close parenthesis minus the fraction sin squared ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis; equals the fraction 1 minus sin squared ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis equals the fraction cos squared ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis equals cos ⁡ open parenthesis θ close parenthesis")[$display(frac(1, cos ( θ ))) − display(frac(sin ( θ ), cos ( θ ))) ⋅ sin ( θ ) & = display(frac(1, cos ( θ ))) − display(frac(sin^(2) ( θ ), cos ( θ ))) \ & = display(frac(1 − sin^(2) ( θ ), cos ( θ ))) = display(frac(cos^(2) ( θ ), cos ( θ ))) = cos ( θ )$] ] ] In the previous example, you can verify that #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis minus tan ⁡ open parenthesis θ close parenthesis sin ⁡ open parenthesis θ close parenthesis equals cos ⁡ open parenthesis θ close parenthesis")[$sec ( θ ) − tan ( θ ) sin ( θ ) = cos ( θ )$] by graphing the functions #math.equation(block: false, alt: "Y sub 1 equals sec ⁡ open parenthesis θ close parenthesis minus tan ⁡ open parenthesis θ close parenthesis sin ⁡ open parenthesis θ close parenthesis")[$Y_(1) = sec ( θ ) − tan ( θ ) sin ( θ )$] and #math.equation(block: false, alt: "Y sub 2 equals cos ⁡ open parenthesis θ close parenthesis")[$Y_(2) = cos ( θ )$] to see that they are the same. Show that #math.equation(block: false, alt: "sin squared ⁡ open parenthesis x close parenthesis open parenthesis 1 plus cot squared ⁡ open parenthesis x close parenthesis close parenthesis equals 1")[$" " sin^(2) ( x ) ( 1 + cot^(2) ( x ) ) = 1$]. #math.equation(block: true, alt: "sin squared ⁡ open parenthesis x close parenthesis open parenthesis 1 plus cot squared ⁡ open parenthesis x close parenthesis close parenthesis, equals sin squared ⁡ open parenthesis x close parenthesis open parenthesis 1 plus the fraction cos squared ⁡ open parenthesis x close parenthesis over sin squared ⁡ open parenthesis x close parenthesis close parenthesis; equals sin squared ⁡ open parenthesis x close parenthesis plus cos squared ⁡ open parenthesis x close parenthesis; equals 1")[$sin^(2) ( x ) ( 1 + cot^(2) ( x ) ) & = sin^(2) ( x ) ( 1 + display(frac(cos^(2) ( x ), sin^(2) ( x ))) ) \ & = sin^(2) ( x ) + cos^(2) ( x ) \ & = 1$] There are two alternate versions of the Pythagorean identity which involve the reciprocal trig functions. These identities are useful when we know the value of #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] or #math.equation(block: false, alt: "cot ⁡ open parenthesis θ close parenthesis")[$cot ( θ )$] and want to find the other trig values. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Two More Pythagorean Identities] #math.equation(block: true, alt: "1 plus tan squared ⁡ open parenthesis θ close parenthesis equals sec squared ⁡ open parenthesis θ close parenthesis 1 plus cot squared ⁡ open parenthesis θ close parenthesis equals csc squared ⁡ open parenthesis θ close parenthesis")[$1 + bold(tan)^(2) ( bold(italic(θ)) ) = bold(sec)^(2) ( bold(italic(θ)) ) " " " " " " " " " " " " " " " " " " " " 1 + bold(cot)^(2) ( bold(italic(θ)) ) = bold(csc)^(2) ( bold(italic(θ)) )$] ] You can memorize these identities, but they are easy to derive from the original Pythagorean identity, #math.equation(block: false, alt: "sin squared ⁡ open parenthesis θ close parenthesis plus cos squared ⁡ open parenthesis θ close parenthesis equals 1")[$sin^(2) ( θ ) + cos^(2) ( θ ) = 1$]. We will prove them in the Homework problems. #examplebox("Example 11")[][ If #math.equation(block: false, alt: "tan ⁡ open parenthesis α close parenthesis equals the fraction 3 over 5")[$" " tan ( α ) = display(frac(3, 5))$] and #math.equation(block: false, alt: "α")[$α$] lies in the third quadrant, find exact values for #math.equation(block: false, alt: "sec ⁡ open parenthesis α close parenthesis")[$sec ( α )$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis")[$cos ( α )$]. #solutionbox[ We cannot find the sine and cosine of an angle directly from the value of the tangent; in particular, it is not true that #math.equation(block: false, alt: "sin ⁡ open parenthesis α close parenthesis equals 3")[$sin ( α ) = 3$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis equals 5")[$cos ( α ) = 5$]! (Do you see why?) Instead, we begin with the Pythagorean identity for the tangent. #math.equation(block: true, alt: "sec squared ⁡ open parenthesis α close parenthesis, equals 1 plus tan squared ⁡ open parenthesis α close parenthesis equals 1 plus open parenthesis the fraction 3 over 5 close parenthesis squared; equals the fraction 25 over 25 plus the fraction 9 over 25 equals the fraction 34 over 25; sec ⁡ open parenthesis α close parenthesis, equals plus or minus the square root of the fraction 34 over 25 equals the fraction plus or minus the square root of 34 over 5")[$sec^(2) ( α ) & = 1 + tan^(2) ( α ) = 1 + ( display(frac(3, 5)) )^(2) \ & = display(frac(25, 25)) + display(frac(9, 25)) = display(frac(34, 25)) \ sec ( α ) & = ± sqrt(display(frac(34, 25))) = display(frac(± sqrt(34), 5))$] Because #math.equation(block: false, alt: "α")[$α$] is in the third quadrant, both its sine and cosine are negative. Therefore the reciprocals of cosine and sine, namely secant and cosecant, must also be negative, and hence #math.equation(block: false, alt: "sec ⁡ open parenthesis α close parenthesis equals the fraction minus the square root of 34 over 5")[$sec ( α ) = display(frac(− sqrt(34), 5))$]. The cosine of #math.equation(block: false, alt: "α")[$α$] is the reciprocal of the secant, so #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis equals the fraction minus 5 over the square root of 34")[$cos ( α ) = display(frac(− 5, sqrt(34)))$]. ] ] If #math.equation(block: false, alt: "cot ⁡ open parenthesis ϕ close parenthesis equals the fraction minus 3 over the square root of 2")[$" " cot ( ϕ ) = display(frac(− 3, sqrt(2)))$] and #math.equation(block: false, alt: "ϕ")[$ϕ$] lies in the second quadrant, find exact values for #math.equation(block: false, alt: "csc ⁡ open parenthesis ϕ close parenthesis")[$csc ( ϕ )$] and #math.equation(block: false, alt: "sin ⁡ open parenthesis ϕ close parenthesis")[$sin ( ϕ )$]. #math.equation(block: true, alt: "csc ⁡ open parenthesis ϕ close parenthesis equals the square root of the fraction 11 over 2")[$csc ( ϕ ) = sqrt(display(frac(11, 2)))$], #math.equation(block: true, alt: "sin ⁡ open parenthesis ϕ close parenthesis equals the square root of the fraction 2 over 11")[$" " sin ( ϕ ) = sqrt(display(frac(2, 11)))$] #notebox("Activity", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Reciprocal Ratios] Part 1 The figure shows a unit circle centered at #math.equation(block: false, alt: "O")[$O$]. The line through #math.equation(block: false, alt: "A C")[$A C$] is tangent to the circle at #math.equation(block: false, alt: "A")[$A$]. (Recall that a tangent to a circle is perpendicular to the radius that meets it.) #figure(figph[unit circle], alt: "unit circle", caption: none) + What is the radius of the circle? + Find line segments whose lengths are #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")[$sin ( θ )$]. + Explain why #math.equation(block: false, alt: "△ O A C")[$upright(△) O A C$] is similar to #math.equation(block: false, alt: "△ O B D")[$upright(△) O B D$]. + Find a line segment in #math.equation(block: false, alt: "△ O A C")[$upright(△) O A C$] whose length is #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]. Explain why your choice is correct. + Find a line segment in #math.equation(block: false, alt: "△ O A C")[$upright(△) O A C$] whose length is #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis")[$sec ( θ )$]. Explain why your choice is correct. + Find a line segment in #math.equation(block: false, alt: "△ O A C")[$upright(△) O A C$] whose length is #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis")[$csc ( θ )$]. Explain why your choice is correct. Part 2 Here is another unit circle, with tangent #math.equation(block: false, alt: "B D")[$B D$]. #figure(figph[unit circle], alt: "unit circle", caption: none) + What is the measure of the angles #math.equation(block: false, alt: "∠ O C D")[$upright(∠) O C D$] and #math.equation(block: false, alt: "∠ O C B")[$upright(∠) O C B$]? + Explain why #math.equation(block: false, alt: "△ O C B")[$upright(△) O C B$] is similar to #math.equation(block: false, alt: "△ O A C")[$upright(△) O A C$]. + Find a line segment in #math.equation(block: false, alt: "△ O C B")[$upright(△) O C B$] whose length is #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]. Explain why your choice is correct. + Find a line segment in #math.equation(block: false, alt: "△ O C B")[$upright(△) O C B$] whose length is #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis")[$sec ( θ )$]. Explain why your choice is correct. + Explain why #math.equation(block: false, alt: "△ O C B")[$upright(△) O C B$] is similar to #math.equation(block: false, alt: "△ D C O")[$upright(△) D C O$]. + Find a line segment in #math.equation(block: false, alt: "△ D C O")[$upright(△) D C O$] whose length is #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis")[$csc ( θ )$]. Explain why your choice is correct. ] Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Solve #math.equation(block: false, alt: "vertical bar 2 x minus 6 vertical bar equals 4")[$| 2 x − 6 | = 4$] + Solve #math.equation(block: false, alt: "vertical bar 1 minus 3 x vertical bar equals 10")[$| 1 − 3 x | = 10$] + Simplify #math.equation(block: false, alt: "the square root of open parenthesis x minus 4 close parenthesis squared")[$sqrt(( x − 4 )^(2))$] + Simplify #math.equation(block: false, alt: "the square root of open parenthesis 1 minus x close parenthesis squared")[$sqrt(( 1 − x )^(2))$] + For what values of #math.equation(block: false, alt: "x")[$x$] is #math.equation(block: false, alt: "vertical bar x plus 2 vertical bar equals x plus 2")[$| x + 2 | = x + 2$] ? + For what values of #math.equation(block: false, alt: "x")[$x$] is #math.equation(block: false, alt: "vertical bar x minus 3 vertical bar equals 3 minus x")[$| x − 3 | = 3 − x$] ? + Graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction x over vertical bar x vertical bar")[$f ( x ) = display(frac(x, | x |))$]. + Explain the difference between the graphs of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis the square root of x close parenthesis squared")[$f ( x ) = ( sqrt(x) )^(2)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the square root of x squared")[$g ( x ) = sqrt(x^(2))$]. #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #math.equation(block: false, alt: "x equals 1 , 5")[$x = 1 , " " 5$] + #math.equation(block: false, alt: "x equals minus 3 , the fraction 11 over 3")[$x = − 3 , " " display(frac(11, 3))$] + #math.equation(block: false, alt: "vertical bar x minus 4 vertical bar")[$| x − 4 |$] + #math.equation(block: false, alt: "vertical bar 1 minus x vertical bar")[$| 1 − x |$] + #math.equation(block: false, alt: "x greater than or equal to minus 2")[$x ≥ − 2$] + #math.equation(block: false, alt: "x less than or equal to 3")[$x ≤ 3$] + #figure(figph[graph], alt: "graph", caption: none) + The domain of #math.equation(block: false, alt: "f")[$f$] is #math.equation(block: false, alt: "open bracket 0 , ∞ close parenthesis")[$[ 0 , upright(∞) )$], and the domain of #math.equation(block: false, alt: "g")[$g$] is #math.equation(block: false, alt: "open parenthesis minus ∞ , ∞ close parenthesis")[$( − upright(∞) , upright(∞) )$]. ] === Section 8.3 Summary ==== Vocabulary - Reciprocal - Secant - Cosecant - Cotangent ==== Concepts + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Three More Functions] If #math.equation(block: false, alt: "θ")[$θ$] is an angle in standard position, and #math.equation(block: false, alt: "P open parenthesis x , y close parenthesis")[$P ( x , y )$] is a point on the terminal side, then we define the following functions. #math.equation(block: true, alt: "The secant : sec ⁡ open parenthesis θ close parenthesis equals the fraction r over x")[$"The" " " " " "secant" : " " " " " " " " " " " " sec ( θ ) = display(frac(r, x))$]#math.equation(block: true, alt: "The cosecant : csc ⁡ open parenthesis θ close parenthesis equals the fraction r over y")[$"The" " " " " "cosecant" : " " " " " " " " csc ( θ ) = display(frac(r, y))$]#math.equation(block: true, alt: "The cotangent : cot ⁡ open parenthesis θ close parenthesis equals the fraction x over y")[$"The" " " " " "cotangent" : " " " " " " cot ( θ ) = display(frac(x, y))$]#figure(figph[angles], alt: "angles", caption: none) ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Reciprocal Trigonometric Ratios] If #math.equation(block: false, alt: "θ")[$θ$] is one of the acute angles in a right triangle, #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis, equals the fraction hypotenuse over adjacent; csc ⁡ open parenthesis θ close parenthesis, equals the fraction hypotenuse over opposite; cot ⁡ open parenthesis θ close parenthesis, equals the fraction adjacent over opposite")[$sec ( θ ) & = display(frac("hypotenuse", "adjacent")) \ csc ( θ ) & = display(frac("hypotenuse", "opposite")) \ cot ( θ ) & = display(frac("adjacent", "opposite"))$]#figure(figph[triangle], alt: "triangle", caption: none) ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Reciprocal Trigonometric Functions] #math.equation(block: true, alt: "The secant function : sec ⁡ open parenthesis θ close parenthesis equals the fraction 1 over cos ⁡ open parenthesis θ close parenthesis")[$"The" " " " " "secant" " " "function" : " " " " " " " " " " " " sec ( θ ) = display(frac(1, cos ( θ )))$]#math.equation(block: true, alt: "The cosecant function : csc ⁡ open parenthesis θ close parenthesis equals the fraction 1 over sin ⁡ open parenthesis θ close parenthesis")[$"The" " " " " "cosecant" " " "function" : " " " " " " " " csc ( θ ) = display(frac(1, sin ( θ )))$]#math.equation(block: true, alt: "The cotangent function : cot ⁡ open parenthesis θ close parenthesis equals the fraction 1 over tan ⁡ open parenthesis θ close parenthesis")[$"The" " " " " "cotangent" " " "function" : " " " " " " cot ( θ ) = display(frac(1, tan ( θ )))$] ] + We can obtain graphs of the secant, cosecant, and cotangent functions as the reciprocals of the three basic functions. + We can solve equations of the form #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis equals k")[$sec ( θ ) = k$], #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis equals k")[$csc ( θ ) = k$], and #math.equation(block: false, alt: "cot ⁡ open parenthesis θ close parenthesis equals k")[$cot ( θ ) = k$] by taking the reciprocal of both sides. + If we know one of the trigonometric ratios for an angle, we can use identities to find any of the others. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Cotangent Identity] #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction 1 over tan ⁡ open parenthesis θ close parenthesis equals the fraction cos ⁡ open parenthesis θ close parenthesis over sin ⁡ open parenthesis θ close parenthesis , sin ⁡ open parenthesis θ close parenthesis not equal to 0")[$cot ( θ ) = display(frac(1, tan ( θ ))) = display(frac(cos ( θ ), sin ( θ ))) , " " " " " " " " sin ( θ ) ≠ 0$] ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Two More Pythagorean Identities] #math.equation(block: true, alt: "1 plus tan squared ⁡ open parenthesis θ close parenthesis equals sec squared ⁡ open parenthesis θ close parenthesis 1 plus cot squared ⁡ open parenthesis θ close parenthesis equals csc squared ⁡ open parenthesis θ close parenthesis")[$1 + tan^(2) ( θ ) = sec^(2) ( θ ) " " " " " " " " " " " " " " " " " " " " 1 + cot^(2) ( θ ) = csc^(2) ( θ )$] ] + We can often simplify trigonometric expressions by first converting all the trig ratios to sines and cosines. ==== Study Questions + Delbert says that #math.equation(block: false, alt: "sec ⁡ open parenthesis x close parenthesis")[$sec ( x )$] is just another way of writing #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis x close parenthesis")[$cos^(− 1) ( x )$], because #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis x close parenthesis equals the fraction 1 over cos ⁡ open parenthesis x close parenthesis")[$cos^(− 1) ( x ) = display(frac(1, cos ( x )))$]. Is he correct? Explain your reasoning. + Each of the following functions is related to the sine function in a different way. Explain how. #math.equation(block: true, alt: "cos ⁡ open parenthesis x close parenthesis , csc ⁡ open parenthesis x close parenthesis , and sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$cos ( x ) , " " " " csc ( x ) , " " " " "and" " " " " sin^(− 1) ( x )$] + Using Study Question \#2 as an example, name three functions related to the tangent function, and explain how they are related. + Why do the graphs of #math.equation(block: false, alt: "y equals csc ⁡ open parenthesis x close parenthesis")[$y = csc ( x )$] and #math.equation(block: false, alt: "y equals cot ⁡ open parenthesis x close parenthesis")[$y = cot ( x )$] have vertical asymptotes at the same #math.equation(block: false, alt: "x")[$x$]-values? ==== Skills + Evaluate the reciprocal trig functions for angles in degrees or radians \#1–20 + Find values or expressions for the six trig ratios \#21–28 + Evaluate the reciprocal trig functions in applications \#29–32 + Given one trig ratio, find the others \#33–46, 71–80 + Evaluate expressions exactly \#47–52 + Graph the secant, cosecant, and cotangent functions \#53–58 + Identify graphs of the reciprocal trig functions \#59–64 + Solve equations in secant, cosecant, and cotangent \#65–70 + Use identities to simplify or evaluate expressions \#81–94 === Homework 8-3 For Problems 1–8, evaluate. Round answers to 3 decimal places. #math.equation(block: true, alt: "csc ⁡ open parenthesis 27 degrees close parenthesis")[$csc ( 27^(∘) )$] #math.equation(block: true, alt: "2.203")[$2.203$] #math.equation(block: true, alt: "sec ⁡ open parenthesis 8 degrees close parenthesis")[$sec ( 8^(∘) )$] #math.equation(block: true, alt: "cot ⁡ open parenthesis 65 degrees close parenthesis")[$cot ( 65^(∘) )$] #math.equation(block: true, alt: "0.466")[$0.466$] #math.equation(block: true, alt: "csc ⁡ open parenthesis 11 degrees close parenthesis")[$csc ( 11^(∘) )$] #math.equation(block: true, alt: "sec ⁡ open parenthesis 1.4 close parenthesis")[$sec ( 1.4 )$] #math.equation(block: true, alt: "5.883")[$5.883$] #math.equation(block: true, alt: "cot ⁡ open parenthesis 4.3 close parenthesis")[$cot ( 4.3 )$] #math.equation(block: true, alt: "csc ⁡ open parenthesis the fraction 5 π over 16 close parenthesis")[$csc ( display(frac(5 π, 16)) )$] #math.equation(block: true, alt: "1.203")[$1.203$] #math.equation(block: true, alt: "sec ⁡ open parenthesis the fraction 7 π over 20 close parenthesis")[$sec ( display(frac(7 π, 20)) )$] For Problems 9–16, evaluate. Give exact values. #math.equation(block: true, alt: "csc ⁡ open parenthesis 30 degrees close parenthesis")[$csc ( 30^(∘) )$] #math.equation(block: true, alt: "2")[$2$] #math.equation(block: true, alt: "sec ⁡ open parenthesis 0 degrees close parenthesis")[$sec ( 0^(∘) )$] #math.equation(block: true, alt: "cot ⁡ open parenthesis 45 degrees close parenthesis")[$cot ( 45^(∘) )$] #math.equation(block: true, alt: "1")[$1$] #math.equation(block: true, alt: "csc ⁡ open parenthesis 60 degrees close parenthesis")[$csc ( 60^(∘) )$] #math.equation(block: true, alt: "sec ⁡ open parenthesis 150 degrees close parenthesis")[$sec ( 150^(∘) )$] #math.equation(block: true, alt: "the fraction minus 2 the square root of 3 over 3")[$display(frac(− 2 sqrt(3), 3))$] #math.equation(block: true, alt: "cot ⁡ open parenthesis 120 degrees close parenthesis")[$cot ( 120^(∘) )$] #math.equation(block: true, alt: "csc ⁡ open parenthesis 135 degrees close parenthesis")[$csc ( 135^(∘) )$] #math.equation(block: true, alt: "the square root of 2")[$sqrt(2)$] #math.equation(block: true, alt: "sec ⁡ open parenthesis 270 degrees close parenthesis")[$sec ( 270^(∘) )$] For Problems 17–18, complete the tables with exact values. #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#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")[$π$]]), [#math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis")[$sec ( θ )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis")[$csc ( θ )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "cot ⁡ open parenthesis θ close parenthesis")[$cot ( θ )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#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")[$π$]]), [#math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis")[$sec ( θ )$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction 2 the square root of 3 over 3")[$display(frac(2 sqrt(3), 3))$]], [#math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$]], [#math.equation(block: false, alt: "2")[$2$]], [undefined], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus the square root of 2")[$− sqrt(2)$]], [#math.equation(block: false, alt: "minus the fraction 2 the square root of 3 over 3")[$− display(frac(2 sqrt(3), 3))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis")[$csc ( θ )$]], [undefined], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$]], [#math.equation(block: false, alt: "the fraction 2 the square root of 3 over 3")[$display(frac(2 sqrt(3), 3))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction 2 the square root of 3 over 3")[$display(frac(2 sqrt(3), 3))$]], [#math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$]], [#math.equation(block: false, alt: "2")[$2$]], [undefined], [#math.equation(block: false, alt: "cot ⁡ open parenthesis θ close parenthesis")[$cot ( θ )$]], [undefined], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 3")[$display(frac(sqrt(3), 3))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 3")[$display(frac(− sqrt(3), 3))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$]], [undefined], )) #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#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")[$2 π$]]), [#math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis")[$sec ( θ )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis")[$csc ( θ )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "cot ⁡ open parenthesis θ close parenthesis")[$cot ( θ )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) Evaluate. Round answers to three decimal places. + #math.equation(block: false, alt: "cos ⁡ open parenthesis 0.2 close parenthesis")[$cos ( 0.2 )$] + #math.equation(block: false, alt: "open parenthesis cos ⁡ open parenthesis 0.2 close parenthesis close parenthesis to the power minus 1")[$( cos ( 0.2 ) )^(− 1)$] + #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis 0.2 close parenthesis")[$cos^(− 1) ( 0.2 )$] + #math.equation(block: false, alt: "the fraction 1 over cos ⁡ open parenthesis 0.2 close parenthesis")[$display(frac(1, cos ( 0.2 )))$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction 1 over 0.2 close parenthesis")[$cos ( display(frac(1, 0.2)) )$] + #math.equation(block: false, alt: "sec ⁡ open parenthesis 0.2 close parenthesis")[$sec ( 0.2 )$] + #math.equation(block: false, alt: "0.980")[$0.980$] + #math.equation(block: false, alt: "1.020")[$1.020$] + #math.equation(block: false, alt: "1.369")[$1.369$] + #math.equation(block: false, alt: "1.020")[$1.020$] + #math.equation(block: false, alt: "0.284")[$0.284$] + #math.equation(block: false, alt: "1.020")[$1.020$] Evaluate. Round answers to three decimal places. + #math.equation(block: false, alt: "tan ⁡ open parenthesis 3.2 close parenthesis")[$tan ( 3.2 )$] + #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis 3.2 close parenthesis")[$tan^(− 1) ( 3.2 )$] + #math.equation(block: false, alt: "cot ⁡ open parenthesis 3.2 close parenthesis")[$cot ( 3.2 )$] + #math.equation(block: false, alt: "the fraction 1 over tan ⁡ open parenthesis 3.2 close parenthesis")[$display(frac(1, tan ( 3.2 )))$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis the fraction 1 over 3.2 close parenthesis")[$tan ( display(frac(1, 3.2)) )$] + #math.equation(block: false, alt: "open parenthesis tan ⁡ open parenthesis 3.2 close parenthesis close parenthesis to the power minus 1")[$( tan ( 3.2 ) )^(− 1)$] For Problems 21–28, find exact values for the six trigonometric ratios of the angle #math.equation(block: false, alt: "θ")[$θ$]. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 4 over 5")[$sin ( θ ) = display(frac(4, 5))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 5")[$" " cos ( θ ) = display(frac(3, 5))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 4 over 3")[$" " tan ( θ ) = display(frac(4, 3))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction 5 over 3")[$" " sec ( θ ) = display(frac(5, 3))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction 5 over 4")[$" " csc ( θ ) = display(frac(5, 4))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 4")[$" " cot ( θ ) = display(frac(3, 4))$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 4 over the square root of 41")[$sin ( θ ) = display(frac(4, sqrt(41)))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 5 over the square root of 41")[$" " cos ( θ ) = display(frac(5, sqrt(41)))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 4 over 5")[$" " tan ( θ ) = display(frac(4, 5))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 41 over 5")[$" " sec ( θ ) = display(frac(sqrt(41), 5))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 41 over 4")[$" " csc ( θ ) = display(frac(sqrt(41), 4))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction 5 over 4")[$" " cot ( θ ) = display(frac(5, 4))$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[angle], alt: "angle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 5 over the square root of 74")[$sin ( θ ) = display(frac(5, sqrt(74)))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 7 over the square root of 74")[$" " cos ( θ ) = display(frac(− 7, sqrt(74)))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus 5 over 7")[$" " tan ( θ ) = display(frac(− 5, 7))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 74 over 7")[$" " sec ( θ ) = display(frac(− sqrt(74), 7))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 74 over 5")[$" " csc ( θ ) = display(frac(sqrt(74), 5))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction minus 7 over 5")[$" " cot ( θ ) = display(frac(− 7, 5))$] #figure(figph[angle], alt: "angle", caption: none) #figure(figph[angle], alt: "angle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus 5 over 8")[$sin ( θ ) = display(frac(− 5, 8))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 39 over 8")[$" " cos ( θ ) = display(frac(sqrt(39), 8))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 5 over the square root of 39")[$" " tan ( θ ) = display(frac(5, sqrt(39)))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction minus 8 over the square root of 39")[$" " sec ( θ ) = display(frac(− 8, sqrt(39)))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction minus 8 over 5")[$" " csc ( θ ) = display(frac(− 8, 5))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 39 over 5")[$" " cot ( θ ) = display(frac(sqrt(39), 5))$] #figure(figph[angle], alt: "angle", caption: none) The distance that sunlight must travel to pass through a layer of Earth's atmosphere depends on both the thickness of the atmosphere and the angle of the sun. #figure(figph[sunlight], alt: "sunlight", caption: none) + Write an expression for the distance, #math.equation(block: false, alt: "d")[$d$], that sunlight travels through a layer of atmosphere of thickness #math.equation(block: false, alt: "h")[$h$]. + Find the distance (to the nearest mile) that sunlight travels through a 100-mile layer of atmosphere when the sun is #math.equation(block: false, alt: "40 degrees")[$40^(∘)$] above the horizon. + #math.equation(block: false, alt: "d equals h csc ⁡ open parenthesis θ close parenthesis")[$d = h csc ( θ )$] + 155.572 miles In railroad design, the degree of curvature of a section of track is the angle subtended by a chord 100 feet long. #figure(figph[railroad track], alt: "railroad track", caption: none) + Use the figure to write an expression for the radius, #math.equation(block: false, alt: "r")[$r$], of a curve whose degree of curvature is #math.equation(block: false, alt: "θ")[$θ$]. (Hint: The bisector of the angle #math.equation(block: false, alt: "θ")[$θ$] is perpendicular to the chord.) + Find the radius of a curve whose degree of curvature is #math.equation(block: false, alt: "43 degrees")[$43^(∘)$]. When a plane is tilted by an angle #math.equation(block: false, alt: "θ")[$θ$] from the horizontal, the time required for a ball starting from rest to roll a horizontal distance of #math.equation(block: false, alt: "l")[$l$] feet on the plane is #math.equation(block: true, alt: "t equals the square root of the fraction l over 8 csc ⁡ open parenthesis 2 θ close parenthesis seconds")[$t = sqrt(display(frac(l, 8)) csc ( 2 θ )) " " " " "seconds"$] #figure(figph[inclined plane], alt: "inclined plane", caption: none) + How long, to the nearest 0.01 second, will it take the ball to roll 2 feet horizontally on a plane tilted by #math.equation(block: false, alt: "12 degrees")[$12^(∘)$]? + Solve the formula for #math.equation(block: false, alt: "l")[$l$] in terms of #math.equation(block: false, alt: "t")[$t$] and #math.equation(block: false, alt: "θ")[$θ$]. + 0.78 sec + #math.equation(block: false, alt: "l equals 8 t squared sin ⁡ open parenthesis 2 θ close parenthesis")[$l = 8 t^(2) sin ( 2 θ )$] After a heavy rainfall, the depth, #math.equation(block: false, alt: "D")[$D$], of the runoff flow at a distance #math.equation(block: false, alt: "x")[$x$] feet from the watershed down a slope at angle #math.equation(block: false, alt: "α")[$α$] is given by #math.equation(block: true, alt: "D equals open parenthesis k x close parenthesis to the power 0.6 open parenthesis cot ⁡ open parenthesis α close parenthesis close parenthesis to the power 0.3 inches")[$D = ( k x )^(0.6) ( cot ( α ) )^(0.3) " " " " "inches"$] where #math.equation(block: false, alt: "k")[$k$] is a constant determined by the surface roughness and the intensity of the runoff. + How deep, to the nearest 0.01 inch, is the runoff 100 feet down a slope of #math.equation(block: false, alt: "10 degrees")[$10^(∘)$] if #math.equation(block: false, alt: "k equals 0.0006")[$k = 0.0006$]? + Solve the formula for #math.equation(block: false, alt: "x")[$x$] in terms of #math.equation(block: false, alt: "D")[$D$] and #math.equation(block: false, alt: "α")[$α$]. For Problems 33–38, write algebraic expressions for the six trigonometric ratios of the angle #math.equation(block: false, alt: "θ")[$θ$]. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 7 over the square root of x squared plus 49")[$sin ( θ ) = display(frac(7, sqrt(x^(2) + 49)))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction x over the square root of x squared plus 49")[$" " cos ( θ ) = display(frac(x, sqrt(x^(2) + 49)))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 7 over x")[$" " tan ( θ ) = display(frac(7, x))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction the square root of x squared plus 49 over x")[$" " sec ( θ ) = display(frac(sqrt(x^(2) + 49), x))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction the square root of x squared plus 49 over 7")[$" " csc ( θ ) = display(frac(sqrt(x^(2) + 49), 7))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction x over 7")[$" " cot ( θ ) = display(frac(x, 7))$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals S")[$sin ( θ ) = S$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the square root of 1 minus S squared")[$" " cos ( θ ) = sqrt(1 − S^(2))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction S over the square root of 1 minus S squared")[$" " tan ( θ ) = display(frac(S, sqrt(1 − S^(2))))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction 1 over the square root of 1 minus S squared")[$" " sec ( θ ) = display(frac(1, sqrt(1 − S^(2))))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction 1 over S")[$" " csc ( θ ) = display(frac(1, S))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 1 minus S squared over S")[$" " cot ( θ ) = display(frac(sqrt(1 − S^(2)), S))$] #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[angle], alt: "angle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 9 minus a squared over 3")[$sin ( θ ) = display(frac(− sqrt(9 − a^(2)), 3))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction a over 3")[$" " cos ( θ ) = display(frac(a, 3))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 9 minus a squared over a")[$" " tan ( θ ) = display(frac(− sqrt(9 − a^(2)), a))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction 3 over a")[$" " sec ( θ ) = display(frac(3, a))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction minus 3 over the square root of 9 minus a squared")[$" " csc ( θ ) = display(frac(− 3, sqrt(9 − a^(2))))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction minus a over the square root of 9 minus a squared")[$" " cot ( θ ) = display(frac(− a, sqrt(9 − a^(2))))$] #figure(figph[angle], alt: "angle", caption: none) The diagram shows a unit circle. Find six line segments whose lengths are, respectively, #math.equation(block: false, alt: "sin ⁡ open parenthesis t close parenthesis , cos ⁡ open parenthesis t close parenthesis , tan ⁡ open parenthesis t close parenthesis , sec ⁡ open parenthesis t close parenthesis , csc ⁡ open parenthesis t close parenthesis ,")[$sin ( t ) , " " cos ( t ) , " " tan ( t ) , " " sec ( t ) , " " csc ( t ) ,$] and #math.equation(block: false, alt: "cot ⁡ open parenthesis t close parenthesis")[$cot ( t )$]. #figure(figph[unit circle], alt: "unit circle", caption: none) #math.equation(block: true, alt: "A C , O A , B D , O D , O E , E F")[$A C , " " O A , " " B D , " " O D , " " O E , " " E F$] Use the figure in Problem 39 to find each area in terms of the angle #math.equation(block: false, alt: "t")[$t$]. + #math.equation(block: false, alt: "△ O A C")[$upright(△) O A C$] + #math.equation(block: false, alt: "△ O B D")[$upright(△) O B D$] + sector #math.equation(block: false, alt: "O B C")[$O B C$] + #math.equation(block: false, alt: "△ O F E")[$upright(△) O F E$] For Problems 41–46, sketch the reference angle, and find exact values for all six trigonometric functions of the angle. #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis equals 2 , θ")[$sec ( θ ) = 2 , " " " " θ$] in Quadrant IV #figure(figph[angle], alt: "angle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 3 over 2")[$" " sin ( θ ) = display(frac(− sqrt(3), 2))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 2")[$" " cos ( θ ) = display(frac(1, 2))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus the square root of 3")[$" " tan ( θ ) = − sqrt(3)$], #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis equals 2")[$" " sec ( θ ) = 2$], #math.equation(block: true, alt: "csc ⁡ open parenthesis θ close parenthesis equals the fraction minus 2 the square root of 3 over 3")[$" " csc ( θ ) = display(frac(− 2 sqrt(3), 3))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 3 over 3")[$" " cot ( θ ) = display(frac(− sqrt(3), 3))$] #math.equation(block: false, alt: "csc ⁡ open parenthesis ϕ close parenthesis equals 4 , ϕ")[$csc ( ϕ ) = 4 , " " " " ϕ$] in Quadrant II #math.equation(block: false, alt: "csc ⁡ open parenthesis α close parenthesis equals 3 , α")[$csc ( α ) = 3 , " " " " α$] in Quadrant I #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis α close parenthesis equals the fraction 1 over 3")[$sin ( α ) = display(frac(1, 3))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis α close parenthesis equals the fraction 2 the square root of 2 over 3")[$" " cos ( α ) = display(frac(2 sqrt(2), 3))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis α close parenthesis equals the fraction the square root of 2 over 4")[$" " tan ( α ) = display(frac(sqrt(2), 4))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis α close parenthesis equals the fraction 3 the square root of 2 over 4")[$" " sec ( α ) = display(frac(3 sqrt(2), 4))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis α close parenthesis equals 3")[$" " csc ( α ) = 3$], #math.equation(block: true, alt: "cot ⁡ open parenthesis α close parenthesis equals 2 the square root of 2")[$" " cot ( α ) = 2 sqrt(2)$] #math.equation(block: false, alt: "sec ⁡ open parenthesis β close parenthesis equals 4 , β")[$sec ( β ) = 4 , " " " " β$] in Quadrant IV #math.equation(block: false, alt: "cot ⁡ open parenthesis γ close parenthesis equals the fraction 1 over 4 , γ")[$cot ( γ ) = display(frac(1, 4)) , " " " " γ$] in Quadrant III #figure(figph[angle], alt: "angle", caption: none) #math.equation(block: true, alt: "sin ⁡ open parenthesis γ close parenthesis equals the fraction minus 4 over the square root of 17")[$sin ( γ ) = display(frac(− 4, sqrt(17)))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis γ close parenthesis equals the fraction minus 1 over the square root of 17")[$cos ( γ ) = display(frac(− 1, sqrt(17)))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis γ close parenthesis equals 4")[$tan ( γ ) = 4$], #math.equation(block: true, alt: "sec ⁡ open parenthesis γ close parenthesis equals minus the square root of 17")[$sec ( γ ) = − sqrt(17)$], #math.equation(block: true, alt: "csc ⁡ open parenthesis γ close parenthesis equals the fraction minus the square root of 17 over 4")[$csc ( γ ) = display(frac(− sqrt(17), 4))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis γ close parenthesis equals the fraction 1 over 4")[$cot ( γ ) = display(frac(1, 4))$] #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals 6 , θ")[$tan ( θ ) = 6 , " " " " θ$] in Quadrant I For Problems 47–52, evaluate. #math.equation(block: true, alt: "4 cot ⁡ open parenthesis the fraction π over 3 close parenthesis plus 2 sec ⁡ open parenthesis the fraction π over 4 close parenthesis")[$4 cot ( display(frac(π, 3)) ) + 2 sec ( display(frac(π, 4)) )$] #math.equation(block: true, alt: "the fraction 4 the square root of 3 over 3 plus 2 the square root of 2")[$display(frac(4 sqrt(3), 3)) + 2 sqrt(2)$] #math.equation(block: true, alt: "the fraction 1 over 2 csc ⁡ open parenthesis the fraction π over 6 close parenthesis minus the fraction 1 over 4 cot ⁡ open parenthesis the fraction π over 6 close parenthesis")[$display(frac(1, 2)) csc ( display(frac(π, 6)) ) − display(frac(1, 4)) cot ( display(frac(π, 6)) )$] #math.equation(block: true, alt: "the fraction 1 over 2 csc ⁡ open parenthesis the fraction 5 π over 3 close parenthesis cot ⁡ open parenthesis the fraction 3 π over 4 close parenthesis")[$display(frac(1, 2)) csc ( display(frac(5 π, 3)) ) cot ( display(frac(3 π, 4)) )$] #math.equation(block: true, alt: "the fraction the square root of 3 over 3")[$display(frac(sqrt(3), 3))$] #math.equation(block: true, alt: "6 cot ⁡ open parenthesis the fraction 7 π over 6 close parenthesis sec ⁡ open parenthesis the fraction 5 π over 4 close parenthesis")[$6 cot ( display(frac(7 π, 6)) ) sec ( display(frac(5 π, 4)) )$] #math.equation(block: true, alt: "open bracket csc ⁡ open parenthesis the fraction 2 π over 3 close parenthesis minus sec ⁡ open parenthesis the fraction 3 π over 4 close parenthesis close bracket squared")[$attach([ csc ( display(frac(2 π, 3)) ) − sec ( display(frac(3 π, 4)) ) ], t: 2)$] #math.equation(block: true, alt: "the fraction 4 the square root of 6 over 3 plus the fraction 10 over 3")[$display(frac(4 sqrt(6), 3)) + display(frac(10, 3))$] #math.equation(block: true, alt: "sec squared ⁡ open parenthesis the fraction 5 π over 6 close parenthesis csc squared ⁡ open parenthesis the fraction 4 π over 3 close parenthesis")[$sec^(2) ( display(frac(5 π, 6)) ) csc^(2) ( display(frac(4 π, 3)) )$] Complete the table and sketch a graph of #math.equation(block: false, alt: "y equals sec ⁡ open parenthesis x close parenthesis")[$y = sec ( x )$]. #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "sec ⁡ open parenthesis x close parenthesis")[$sec ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(figph[grid], alt: "grid", caption: none) #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "sec ⁡ open parenthesis x close parenthesis")[$sec ( x )$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$]], [undefined], [#math.equation(block: false, alt: "minus the square root of 2")[$− sqrt(2)$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus the square root of 2")[$− sqrt(2)$]], [undefined], [#math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$]], [#math.equation(block: false, alt: "1")[$1$]], )) #figure(figph[secant], alt: "secant", caption: none) Complete the table and sketch a graph of #math.equation(block: false, alt: "y equals csc ⁡ open parenthesis x close parenthesis")[$y = csc ( x )$]. #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "csc ⁡ open parenthesis x close parenthesis")[$csc ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(figph[grid], alt: "grid", caption: none) Use the graph of #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis x close parenthesis")[$y = sin ( x )$] to sketch a graph of its reciprocal, #math.equation(block: false, alt: "y equals csc ⁡ open parenthesis x close parenthesis")[$y = csc ( x )$]. #figure(figph[sine], alt: "sine", caption: none) #figure(figph[sine and cosecant], alt: "sine and cosecant", caption: none) Use the graph of #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$y = cos ( x )$] to sketch a graph of its reciprocal, #math.equation(block: false, alt: "y equals sec ⁡ open parenthesis x close parenthesis")[$y = sec ( x )$]. #figure(figph[cosine], alt: "cosine", caption: none) Complete the table and sketch a graph of #math.equation(block: false, alt: "y equals cot ⁡ open parenthesis x close parenthesis")[$y = cot ( x )$]. #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "cot ⁡ open parenthesis x close parenthesis")[$cot ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(figph[grid], alt: "grid", caption: none) #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "cot ⁡ open parenthesis x close parenthesis")[$cot ( x )$]], [undefined], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [undefined], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [undefined], )) #figure(figph[cotangent], alt: "cotangent", caption: none) Use the graphs of #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$y = cos ( x )$] and #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis x close parenthesis")[$y = sin ( x )$] to sketch a graph of #math.equation(block: false, alt: "y equals cot ⁡ open parenthesis x close parenthesis equals the fraction cos ⁡ open parenthesis x close parenthesis over sin ⁡ open parenthesis x close parenthesis")[$y = cot ( x ) = display(frac(cos ( x ), sin ( x )))$]. #figure(figph[cos and sin], alt: "cos and sin", caption: none) For Problems 59–64, + Graph each function for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$], and write a simpler expression for the function. + Show algebraically that your new expression is equivalent to the original one. #math.equation(block: true, alt: "y equals the fraction csc ⁡ open parenthesis x close parenthesis over cot ⁡ open parenthesis x close parenthesis")[$y = display(frac(csc ( x ), cot ( x )))$] #math.equation(block: true, alt: "the fraction csc ⁡ open parenthesis x close parenthesis over cot ⁡ open parenthesis x close parenthesis, equals the fraction the fraction 1 over sin ⁡ open parenthesis x close parenthesis over the fraction cos ⁡ open parenthesis x close parenthesis over sin ⁡ open parenthesis x close parenthesis; equals the fraction 1 over sin ⁡ open parenthesis x close parenthesis divided by the fraction cos ⁡ open parenthesis x close parenthesis over sin ⁡ open parenthesis x close parenthesis; equals the fraction 1 over sin ⁡ open parenthesis x close parenthesis times the fraction sin ⁡ open parenthesis x close parenthesis over cos ⁡ open parenthesis x close parenthesis; equals the fraction 1 over cos ⁡ open parenthesis x close parenthesis; equals sec ⁡ open parenthesis x close parenthesis")[$display(frac(csc ( x ), cot ( x ))) & = display(frac(display(frac(1, sin ( x ))), display(frac(cos ( x ), sin ( x ))))) \ & = display(frac(1, sin ( x ))) ÷ display(frac(cos ( x ), sin ( x ))) \ & = display(frac(1, sin ( x ))) ⋅ display(frac(sin ( x ), cos ( x ))) \ & = display(frac(1, cos ( x ))) \ & = sec ( x )$] #math.equation(block: true, alt: "y equals the fraction sec ⁡ open parenthesis x close parenthesis over tan ⁡ open parenthesis x close parenthesis")[$y = display(frac(sec ( x ), tan ( x )))$] #math.equation(block: true, alt: "y equals the fraction sec ⁡ open parenthesis x close parenthesis cot ⁡ open parenthesis x close parenthesis over csc ⁡ open parenthesis x close parenthesis")[$y = display(frac(sec ( x ) cot ( x ), csc ( x )))$] #math.equation(block: true, alt: "the fraction sec ⁡ open parenthesis x close parenthesis cot ⁡ open parenthesis x close parenthesis over csc ⁡ open parenthesis x close parenthesis equals the fraction the fraction 1 over cos ⁡ open parenthesis x close parenthesis times the fraction cos ⁡ open parenthesis x close parenthesis over sin ⁡ open parenthesis x close parenthesis over the fraction 1 over sin ⁡ open parenthesis x close parenthesis equals the fraction the fraction 1 over sin ⁡ open parenthesis x close parenthesis over the fraction 1 over sin ⁡ open parenthesis x close parenthesis equals 1")[$display(frac(sec ( x ) cot ( x ), csc ( x ))) = display(frac(display(frac(1, cos ( x ))) ⋅ display(frac(cos ( x ), sin ( x ))), display(frac(1, sin ( x ))))) = display(frac(display(frac(1, sin ( x ))), display(frac(1, sin ( x ))))) = 1$] #math.equation(block: true, alt: "y equals the fraction csc ⁡ open parenthesis x close parenthesis tan ⁡ x over sec ⁡ open parenthesis x close parenthesis")[$y = display(frac(csc ( x ) tan x, sec ( x )))$] #math.equation(block: true, alt: "y equals tan ⁡ open parenthesis x close parenthesis csc ⁡ open parenthesis x close parenthesis")[$y = tan ( x ) csc ( x )$] #math.equation(block: true, alt: "tan ⁡ open parenthesis x close parenthesis csc ⁡ open parenthesis x close parenthesis equals the fraction sin ⁡ open parenthesis x close parenthesis over cos ⁡ open parenthesis x close parenthesis times the fraction 1 over sin ⁡ open parenthesis x close parenthesis equals the fraction 1 over cos ⁡ open parenthesis x close parenthesis equals sec ⁡ open parenthesis x close parenthesis")[$tan ( x ) csc ( x ) = display(frac(sin ( x ), cos ( x ))) ⋅ display(frac(1, sin ( x ))) = display(frac(1, cos ( x ))) = sec ( x )$] #math.equation(block: true, alt: "y equals sin ⁡ open parenthesis x close parenthesis sec ⁡ open parenthesis x close parenthesis")[$y = sin ( x ) sec ( x )$] For Problems 65–70, find all solutions between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$]. #math.equation(block: true, alt: "3 csc ⁡ open parenthesis θ close parenthesis plus 2 equals 8")[$3 csc ( θ ) + 2 = 8$] #math.equation(block: true, alt: "the fraction π over 6 , the fraction 5 π over 6")[$display(frac(π, 6)) , " " display(frac(5 π, 6))$] #math.equation(block: true, alt: "minus 2 sec ⁡ open parenthesis θ close parenthesis plus 7 equals 3")[$− 2 sec ( θ ) + 7 = 3$] #math.equation(block: true, alt: "the square root of 2 sec ⁡ open parenthesis θ close parenthesis equals minus 2")[$sqrt(2) sec ( θ ) = − 2$] #math.equation(block: true, alt: "the fraction 3 π over 4 , the fraction 5 π over 4")[$display(frac(3 π, 4)) , " " display(frac(5 π, 4))$] #math.equation(block: true, alt: "8 plus csc ⁡ open parenthesis θ close parenthesis equals 6")[$8 + csc ( θ ) = 6$] #math.equation(block: true, alt: "2 cot ⁡ open parenthesis θ close parenthesis equals minus the square root of 12")[$2 cot ( θ ) = − sqrt(12)$] #math.equation(block: true, alt: "the fraction 5 π over 6 , the fraction 11 π over 6")[$display(frac(5 π, 6)) , " " display(frac(11 π, 6))$] #math.equation(block: true, alt: "the square root of 3 cot ⁡ open parenthesis θ close parenthesis equals 1")[$sqrt(3) cot ( θ ) = 1$] For Problems 71–76, use identities to find exact values or to write algebraic expressions. If #math.equation(block: false, alt: "tan ⁡ open parenthesis α close parenthesis equals minus 2")[$tan ( α ) = − 2$] and #math.equation(block: false, alt: "the fraction π over 2 less than α less than π")[$display(frac(π, 2)) < α < π$], find #math.equation(block: false, alt: "cos ⁡ open parenthesis α close parenthesis")[$cos ( α )$]. #math.equation(block: true, alt: "the fraction minus the square root of 5 over 5")[$display(frac(− sqrt(5), 5))$] If #math.equation(block: false, alt: "cot ⁡ open parenthesis β close parenthesis equals the fraction 5 over 4")[$cot ( β ) = display(frac(5, 4))$] and #math.equation(block: false, alt: "π less than β less than the fraction 3 π over 2")[$π < β < display(frac(3 π, 2))$], find #math.equation(block: false, alt: "sin ⁡ open parenthesis β close parenthesis")[$sin ( β )$]. If #math.equation(block: false, alt: "sec ⁡ open parenthesis x close parenthesis equals the fraction a over 2")[$sec ( x ) = display(frac(a, 2))$] and #math.equation(block: false, alt: "0 less than x less than the fraction π over 2")[$0 < x < display(frac(π, 2))$], find #math.equation(block: false, alt: "tan ⁡ open parenthesis x close parenthesis")[$tan ( x )$]. #math.equation(block: true, alt: "the fraction the square root of a squared minus 4 over 2")[$display(frac(sqrt(a^(2) − 4), 2))$] If #math.equation(block: false, alt: "csc ⁡ open parenthesis y close parenthesis equals the fraction 1 over b")[$csc ( y ) = display(frac(1, b))$] and #math.equation(block: false, alt: "the fraction π over 2 less than y less than π")[$display(frac(π, 2)) < y < π$], find #math.equation(block: false, alt: "cot ⁡ open parenthesis y close parenthesis")[$cot ( y )$]. If #math.equation(block: false, alt: "csc ⁡ open parenthesis ϕ close parenthesis equals w")[$csc ( ϕ ) = w$] and #math.equation(block: false, alt: "the fraction 3 π over 2 less than ϕ less than 2 π")[$display(frac(3 π, 2)) < ϕ < 2 π$], find #math.equation(block: false, alt: "cos ⁡ ϕ")[$cos ϕ$]. #math.equation(block: true, alt: "the fraction the square root of w squared minus 1 over minus w")[$display(frac(sqrt(w^(2) − 1), − w))$] If #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis equals the fraction 3 over z")[$sec ( θ ) = display(frac(3, z))$] and #math.equation(block: false, alt: "π less than θ less than the fraction 3 π over 2")[$π < θ < display(frac(3 π, 2))$], find #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$]. For Problems 77–80, find exact values for #math.equation(block: false, alt: "sec ⁡ open parenthesis s close parenthesis , csc ⁡ open parenthesis s close parenthesis ,")[$sec ( s ) , " " csc ( s ) ,$] and #math.equation(block: false, alt: "cot ⁡ open parenthesis s close parenthesis")[$cot ( s )$]. #figure(figph[unit circle], alt: "unit circle", caption: none) #math.equation(block: true, alt: "sec ⁡ open parenthesis s close parenthesis equals the fraction minus 5 over 4")[$sec ( s ) = display(frac(− 5, 4))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis s close parenthesis equals the fraction 5 over 3")[$" " csc ( s ) = display(frac(5, 3))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis s close parenthesis equals the fraction minus 4 over 3")[$" " cot ( s ) = display(frac(− 4, 3))$] #figure(figph[unit circle], alt: "unit circle", caption: none) #figure(figph[unit circle], alt: "unit circle", caption: none) #math.equation(block: true, alt: "sec ⁡ open parenthesis s close parenthesis equals the fraction 1 over the square root of 1 minus w squared")[$sec ( s ) = display(frac(1, sqrt(1 − w^(2))))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis s close parenthesis equals the fraction 1 over w")[$" " csc ( s ) = display(frac(1, w))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis s close parenthesis equals the fraction the square root of 1 minus w squared over w")[$" " cot ( s ) = display(frac(sqrt(1 − w^(2)), w))$] #figure(figph[unit circle], alt: "unit circle", caption: none) For Problems 81–88, write the expression in terms of sine and cosine, and simplify. #math.equation(block: true, alt: "sec ⁡ open parenthesis θ close parenthesis tan ⁡ open parenthesis θ close parenthesis")[$sec ( θ ) tan ( θ )$] #math.equation(block: true, alt: "the fraction sin ⁡ open parenthesis θ close parenthesis over cos squared ⁡ open parenthesis θ close parenthesis")[$display(frac(sin ( θ ), cos^(2) ( θ )))$] #math.equation(block: true, alt: "csc ⁡ open parenthesis ϕ close parenthesis cot ⁡ open parenthesis ϕ close parenthesis")[$csc ( ϕ ) cot ( ϕ )$] #math.equation(block: true, alt: "the fraction csc ⁡ open parenthesis t close parenthesis over cot ⁡ open parenthesis t close parenthesis")[$display(frac(csc ( t ), cot ( t )))$] #math.equation(block: true, alt: "sec ⁡ open parenthesis t close parenthesis")[$sec ( t )$] #math.equation(block: true, alt: "the fraction tan ⁡ open parenthesis v close parenthesis over sec ⁡ open parenthesis v close parenthesis")[$display(frac(tan ( v ), sec ( v )))$] #math.equation(block: true, alt: "sec ⁡ open parenthesis β close parenthesis minus tan ⁡ open parenthesis β close parenthesis")[$sec ( β ) − tan ( β )$] #math.equation(block: true, alt: "the fraction 1 minus sin ⁡ open parenthesis β close parenthesis over cos ⁡ open parenthesis β close parenthesis")[$display(frac(1 − sin ( β ), cos ( β )))$] #math.equation(block: true, alt: "cot ⁡ open parenthesis α close parenthesis plus csc ⁡ open parenthesis α close parenthesis")[$cot ( α ) + csc ( α )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis x close parenthesis tan ⁡ open parenthesis x close parenthesis minus sec ⁡ open parenthesis x close parenthesis")[$sin ( x ) tan ( x ) − sec ( x )$] #math.equation(block: true, alt: "minus cos ⁡ open parenthesis x close parenthesis")[$− cos ( x )$] #math.equation(block: true, alt: "csc ⁡ y minus cos ⁡ y cot ⁡ y")[$csc y − cos y cot y$] Prove the Pythagorean identity #math.equation(block: false, alt: "1 plus tan squared ⁡ open parenthesis θ close parenthesis equals sec squared ⁡ open parenthesis θ close parenthesis")[$1 + tan^(2) ( θ ) = sec^(2) ( θ )$]. (Hint: Start with the identity #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$cos^(2) ( θ ) + sin^(2) ( θ ) = 1$] and divide both sides of the equation by #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis")[$cos^(2) ( θ )$].) #math.equation(block: true, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis, equals 1; the fraction cos squared ⁡ open parenthesis θ close parenthesis over cos squared ⁡ open parenthesis θ close parenthesis plus the fraction sin squared ⁡ open parenthesis θ close parenthesis over cos squared ⁡ open parenthesis θ close parenthesis, equals the fraction 1 over cos squared ⁡ open parenthesis θ close parenthesis; 1 plus tan squared ⁡ open parenthesis θ close parenthesis, equals sec squared ⁡ open parenthesis θ close parenthesis")[$cos^(2) ( θ ) + sin^(2) ( θ ) & = 1 \ display(frac(cos^(2) ( θ ), cos^(2) ( θ ))) + display(frac(sin^(2) ( θ ), cos^(2) ( θ ))) & = display(frac(1, cos^(2) ( θ ))) \ 1 + tan^(2) ( θ ) & = sec^(2) ( θ )$] Prove the Pythagorean identity #math.equation(block: false, alt: "1 plus cot squared ⁡ open parenthesis θ close parenthesis equals csc squared ⁡ open parenthesis θ close parenthesis")[$1 + cot^(2) ( θ ) = csc^(2) ( θ )$]. (Hint: Start with the identity #math.equation(block: false, alt: "cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis equals 1")[$cos^(2) ( θ ) + sin^(2) ( θ ) = 1$] and divide both sides of the equation by #math.equation(block: false, alt: "sin squared ⁡ open parenthesis θ close parenthesis")[$sin^(2) ( θ )$].) Suppose that #math.equation(block: false, alt: "cot ⁡ open parenthesis θ close parenthesis equals 5")[$cot ( θ ) = 5$] and #math.equation(block: false, alt: "θ")[$θ$] lies in the third quadrant. + Use the Pythagorean identity to find the value of #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis")[$csc ( θ )$]. + Use identities to find the values of the other four trig functions of #math.equation(block: false, alt: "θ")[$θ$]. + #math.equation(block: false, alt: "csc ⁡ open parenthesis θ close parenthesis equals minus the square root of 26")[$csc ( θ ) = − sqrt(26)$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 26 over 26 , cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 5 the square root of 26 over 26 , tan ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 5 , sec ⁡ open parenthesis θ close parenthesis equals the fraction minus the square root of 26 over 5")[$sin ( θ ) = display(frac(− sqrt(26), 26)) , " " cos ( θ ) = display(frac(− 5 sqrt(26), 26)) , " " tan ( θ ) = display(frac(1, 5)) , " " sec ( θ ) = display(frac(− sqrt(26), 5))$] Suppose that #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus 2")[$tan ( θ ) = − 2$] and #math.equation(block: false, alt: "θ")[$θ$] lies in the second quadrant. + Use the Pythagorean identity to find the value of #math.equation(block: false, alt: "sec ⁡ open parenthesis θ close parenthesis")[$sec ( θ )$]. + Use identities to find the values of the other four trig functions of #math.equation(block: false, alt: "θ")[$θ$]. Write each of the other five trig functions in terms of #math.equation(block: false, alt: "sin ⁡ open parenthesis t close parenthesis")[$sin ( t )$] only. #math.equation(block: true, alt: "cos ⁡ open parenthesis t close parenthesis equals plus or minus the square root of 1 minus sin squared ⁡ open parenthesis t close parenthesis")[$cos ( t ) = ± sqrt(1 − sin^(2) ( t ))$], #math.equation(block: true, alt: "tan ⁡ open parenthesis t close parenthesis equals the fraction plus or minus sin ⁡ open parenthesis t close parenthesis over the square root of 1 minus sin squared ⁡ open parenthesis t close parenthesis")[$" " tan ( t ) = display(frac(± sin ( t ), sqrt(1 − sin^(2) ( t ))))$], #math.equation(block: true, alt: "sec ⁡ open parenthesis t close parenthesis equals the fraction plus or minus 1 over the square root of 1 minus sin squared ⁡ open parenthesis t close parenthesis")[$" " sec ( t ) = display(frac(± 1, sqrt(1 − sin^(2) ( t ))))$], #math.equation(block: true, alt: "csc ⁡ open parenthesis t close parenthesis equals the fraction 1 over sin ⁡ open parenthesis t close parenthesis")[$" " csc ( t ) = display(frac(1, sin ( t )))$], #math.equation(block: true, alt: "cot ⁡ open parenthesis t close parenthesis equals the fraction plus or minus the square root of 1 minus sin squared ⁡ open parenthesis t close parenthesis over sin ⁡ open parenthesis t close parenthesis")[$" " cot ( t ) = display(frac(± sqrt(1 − sin^(2) ( t )), sin ( t )))$] Write each of the other five trig functions in terms of #math.equation(block: false, alt: "cos ⁡ open parenthesis t close parenthesis")[$cos ( t )$] only. Show that if the angles of a triangle are #math.equation(block: false, alt: "A , B ,")[$A , " " B ,$] and #math.equation(block: false, alt: "C")[$C$] and the opposite sides are respectively #math.equation(block: false, alt: "a , b ,")[$a , " " b ,$] and #math.equation(block: false, alt: "c ,")[$c ,$] then #math.equation(block: true, alt: "a csc ⁡ open parenthesis A close parenthesis equals b csc ⁡ open parenthesis B close parenthesis equals c csc ⁡ open parenthesis C close parenthesis")[$a csc ( A ) = b csc ( B ) = c csc ( C )$] #math.equation(block: true, alt: "the fraction a over sin ⁡ open parenthesis A close parenthesis, equals the fraction b over sin ⁡ open parenthesis B close parenthesis equals the fraction c over sin ⁡ open parenthesis C close parenthesis; a times the fraction 1 over sin ⁡ open parenthesis A close parenthesis, equals b times the fraction 1 over sin ⁡ open parenthesis B close parenthesis equals c times the fraction 1 over sin ⁡ open parenthesis C close parenthesis; a csc ⁡ open parenthesis A close parenthesis, equals b csc ⁡ open parenthesis B close parenthesis equals c csc ⁡ open parenthesis C close parenthesis")[$display(frac(a, sin ( A ))) & = display(frac(b, sin ( B ))) = display(frac(c, sin ( C ))) \ a ⋅ display(frac(1, sin ( A ))) & = b ⋅ display(frac(1, sin ( B ))) = c ⋅ display(frac(1, sin ( C ))) \ a csc ( A ) & = b csc ( B ) = c csc ( C )$] The figure shows a unit circle and an angle #math.equation(block: false, alt: "θ")[$θ$] in standard position. Each of the six trigonometric ratios for #math.equation(block: false, alt: "θ")[$θ$] is represented by the length of a line segment in the figure. Find the line segment for each ratio, and explain your choice. #figure(figph[triangle], alt: "triangle", caption: none)