#set document(title: "8.2 Inverse Trigonometric Functions", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 8.2#h(0.6em)Inverse Trigonometric Functions We have been using the calculator keys #math.equation(block: false, alt: "sin to the power minus 1")[$sin^(− 1)$], #math.equation(block: false, alt: "cos to the power minus 1")[$cos^(− 1)$], and #math.equation(block: false, alt: "tan to the power minus 1")[$tan^(− 1)$] to find approximate values of #math.equation(block: false, alt: "θ")[$θ$] when we know either #math.equation(block: false, alt: "sin ⁡ θ , cos ⁡ θ")[$sin θ , " " cos θ$], or #math.equation(block: false, alt: "tan ⁡ θ")[$tan θ$]. For example, if we know that #math.equation(block: false, alt: "cos ⁡ θ equals 0.3")[$cos θ = 0.3$], then #math.equation(block: true, alt: "θ equals cos to the power minus 1 ⁡ open parenthesis 0.3 close parenthesis approximately equals 1.2661 radians")[$θ = cos^(− 1) ( 0.3 ) ≈ 1.2661 " " "radians"$] In other words, we use the #math.equation(block: false, alt: "sin to the power minus 1")[$sin^(− 1)$], #math.equation(block: false, alt: "cos to the power minus 1")[$cos^(− 1)$], and #math.equation(block: false, alt: "tan to the power minus 1")[$tan^(− 1)$] keys to solve trigonometric equations, just as we use square roots to solve quadratic equations. Using one of these keys performs the #strong[inverse operation] for computing a sine, cosine or tangent, just as extracting square roots is the inverse of squaring a number. Many functions can be described as an operation or as a sequence of operations on the input value, and this leads us to the notion of an inverse function. === Inverse of a Function Raising a number to the #math.equation(block: false, alt: "n to the power t h")[$n^(t h)$] power and taking #math.equation(block: false, alt: "n to the power t h")[$n^(t h)$] roots are an example of inverse operations. For example, if we first cube a number and then take the cube root of the result, we return to the original number. #figure(figph[inverse operations], alt: "inverse operations", caption: none) We say that the two functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed")[$f ( x ) = x^(3)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the cube root of x")[$g ( x ) = root(3, x)$] are #strong[inverse functions]. Each of the functions undoes the results of the other function. You can confirm this behavior by consulting the tables of values for the two functions. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed")[$f ( x ) = x^(3)$]]), [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 8")[$− 8$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus the fraction 1 over 2")[$− display(frac(1, 2))$]], [#math.equation(block: false, alt: "minus the fraction 1 over 8")[$− display(frac(1, 8))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "8")[$8$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the cube root of x")[$g ( x ) = root(3, x)$]]), [#math.equation(block: false, alt: "minus 8")[$− 8$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus the fraction 1 over 8")[$− display(frac(1, 8))$]], [#math.equation(block: false, alt: "minus the fraction 1 over 2")[$− display(frac(1, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "2")[$2$]], )) Observe that the table of values for #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the cube root of x")[$g ( x ) = root(3, x)$] can be obtained from the table for #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed")[$f ( x ) = x^(3)$] by interchanging the values of #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] in each ordered pair. In fact, we can often find a formula for the inverse function by interchanging the input and output variables in the formula for the function, and then solving for the new output variable. For our example, we start with the formula for the cubing function: #math.equation(block: true, alt: "y, equals x cubed, Interchange x and y .; x, equals y cubed, Solve for y .; y, equals the cube root of x, A formula for the inverse function.")[$y & = x^(3) & & "Interchange " bold(italic(x)) " and " bold(italic(y)) . \ x & = y^(3) & & "Solve for " bold(italic(y)) . \ y & = root(3, x) & & "A formula for the inverse function."$] We use the notation #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis")[$bold(italic(f))^(− 1) ( bold(italic(x)) )$] to denote the inverse function. Thus, we have just shown that the inverse function for #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed")[$f ( x ) = x^(3)$] is #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the cube root of x")[$f^(− 1) ( x ) = root(3, x)$]. #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The notation #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis")[$f^(− 1) ( x )$] does #math.equation(block: false, alt: "not")[$"not"$] mean #math.equation(block: false, alt: "the fraction 1 over f open parenthesis x close parenthesis")[$display(frac(1, f ( x )))$] ! For example, if #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed")[$f ( x ) = x^(3)$], we have #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the cube root of x")[$f^(− 1) ( x ) = root(3, x)$], but #math.equation(block: false, alt: "the fraction 1 over f open parenthesis x close parenthesis equals the fraction 1 over x cubed")[$display(frac(1, f ( x ))) = display(frac(1, x^(3)))$], and we know that #math.equation(block: false, alt: "the cube root of x not equal to the fraction 1 over x cubed")[$root(3, x) ≠ display(frac(1, x^(3)))$]. Although it is true that a negative exponent denotes a reciprocal, so that, for instance, #math.equation(block: false, alt: "a to the power minus 2 equals the fraction 1 over a squared")[$a^(− 2) = display(frac(1, a^(2)))$], the #math.equation(block: false, alt: "minus 1")[$− 1$] in #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis")[$f^(− 1) ( x )$] is #emph[not] an exponent; this is an entirely new use for the same symbol. ] #examplebox("Example 1")[][ Let #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x minus 3.")[$" " " " f ( x ) = display(frac(1, x − 3)) . " " " "$] Find a formula for the inverse function. #solutionbox[ We interchange the variables in the formula for #math.equation(block: false, alt: "f")[$f$] (that is, in #math.equation(block: false, alt: "y equals the fraction 1 over x minus 3")[$y = display(frac(1, x − 3))$]) to get #math.equation(block: false, alt: "x equals the fraction 1 over y minus 3.")[$" " " " x = display(frac(1, y − 3)) .$] Now we solve for #math.equation(block: false, alt: "y")[$y$] in terms of #math.equation(block: false, alt: "x")[$x$] by taking the reciprocal of both sides. #math.equation(block: false, alt: "y minus 3 equals the fraction 1 over x ,")[$" " " " y − 3 = display(frac(1, x)) , " " " "$] so #math.equation(block: false, alt: "y equals the fraction 1 over x plus 3.")[$" " " " y = display(frac(1, x)) + 3. " " " "$] The formula for the inverse function is #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the fraction 1 over x plus 3")[$f^(− 1) ( x ) = display(frac(1, x)) + 3$]. ] ] Let #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 x minus 6")[$g ( x ) = 2 x − 6$]. Find a formula for the inverse function. #math.equation(block: true, alt: "g to the power minus 1 open parenthesis x close parenthesis equals the fraction 1 over 2 x plus 3")[$g^(− 1) ( x ) = display(frac(1, 2)) x + 3$] In the examples above, note that #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis not equal to the fraction 1 over f open parenthesis x close parenthesis")[$f^(− 1) ( x ) ≠ display(frac(1, f ( x )))$], and #math.equation(block: false, alt: "g to the power minus 1 open parenthesis x close parenthesis not equal to the fraction 1 over g open parenthesis x close parenthesis")[$g^(− 1) ( x ) ≠ display(frac(1, g ( x )))$]. === The Graph of the Inverse If we graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed")[$f ( x ) = x^(3)$] and its inverse #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the cube root of x")[$f^(− 1) ( x ) = root(3, x)$] on the same set of axes, we see that the graphs are related in an interesting way, as shown below. The graphs are symmetric about the line #math.equation(block: false, alt: "y equals x")[$y = x$], which means that if we were to place a mirror along the line each graph would be the reflection of the other. This symmetry occurs because we interchanged the roles of #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] when we defined the inverse function. #figure(figph[graph of inverse function], alt: "graph of inverse function", caption: none) Note that, for this example, both graphs pass the vertical line test, so they are both graphs of functions. #examplebox("Example 2")[][ + Graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x minus 3")[$f ( x ) = display(frac(1, x − 3))$] and its inverse #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the fraction 1 over x plus 3")[$f^(− 1) ( x ) = display(frac(1, x)) + 3$] on the same grid, and sketch in the line #math.equation(block: false, alt: "y equals x")[$y = x$] to illustrate the symmetry. + Find the domain and range of #math.equation(block: false, alt: "f")[$f$], and the domain and range of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]. #solutionbox[ + The graph is shown at right. Note that #math.equation(block: false, alt: "f")[$f$] has a vertical asymptote at #math.equation(block: false, alt: "x equals 3")[$x = 3$], and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] has a horizontal asymptote at #math.equation(block: false, alt: "y equals 3")[$y = 3$]. #figure(figph[graph of inverse function], alt: "graph of inverse function", caption: none) + Note that #math.equation(block: false, alt: "f")[$f$] is undefined at #math.equation(block: false, alt: "x equals 3")[$x = 3$], and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is undefined at #math.equation(block: false, alt: "x equals 0")[$x = 0$]. Thus, the domain of #math.equation(block: false, alt: "f")[$f$] is all real numbers except 3, and the domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is all real numbers except 0. #linebreak() We can find the range of each function by examining its graph: - #math.equation(block: false, alt: "f")[$f$] has a horizontal asymptote at #math.equation(block: false, alt: "y equals 0")[$y = 0$], so there is no point on the graph with #math.equation(block: false, alt: "y equals 0")[$y = 0$]. Its range is all real numbers except 0. - Similarly, there is no point on the graph of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] with #math.equation(block: false, alt: "y equals 3")[$y = 3$], so its range is all real numbers except 3. ] ] + Graph #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 x minus 6")[$g ( x ) = 2 x − 6$] and its inverse function on the grid at right, and sketch in the line #math.equation(block: false, alt: "y equals x")[$y = x$] to show the symmetry. #figure(figph[grid], alt: "grid", caption: none) + Find the domain and range of #math.equation(block: false, alt: "g")[$g$], and the domain and range of #math.equation(block: false, alt: "g to the power minus 1")[$g^(− 1)$]. + #figure(figph[linear function and inverse], alt: "linear function and inverse", caption: none) + The domain and range of #math.equation(block: false, alt: "g")[$g$] and #math.equation(block: false, alt: "g to the power minus 1")[$g^(− 1)$] each include all real numbers From the examples above, we see that the domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the same as the range of #math.equation(block: false, alt: "f")[$f$], and the range of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the same as the domain of #math.equation(block: false, alt: "f")[$f$]. This should seem reasonable, because we obtain the inverse function by interchanging the values of the input and output variables. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Domain and Range of Inverse Function] Domain #math.equation(block: false, alt: "open parenthesis f close parenthesis equals")[$( f ) =$] Range #math.equation(block: false, alt: "open parenthesis f to the power minus 1 close parenthesis and")[$( f^(− 1) ) " " " " " " " " " " " " " " " " "and" " " " " " " " " " " " " " " " "$] Domain #math.equation(block: false, alt: "open parenthesis f to the power minus 1 close parenthesis equals")[$( f^(− 1) ) =$] Range #math.equation(block: false, alt: "open parenthesis f close parenthesis")[$( f )$] ] === Does Every Function Have an Inverse? In the Examples above, the inverse of the function turned out to be a function as well. But this is not always the case. Consider the function #math.equation(block: false, alt: "F open parenthesis x close parenthesis equals x squared minus 4")[$F ( x ) = x^(2) − 4$]. First we'll find a formula for the inverse. We interchange #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] and solve for #math.equation(block: false, alt: "y")[$y$]: #math.equation(block: true, alt: "x, equals y squared minus 4; y squared, equals x minus 4; y, equals plus or minus the square root of x minus 4")[$x & = y^(2) − 4 \ y^(2) & = x − 4 \ y & = ± sqrt(x − 4)$] The graphs of #math.equation(block: false, alt: "F")[$F$] and its inverse are shown at right. You can see that although #math.equation(block: false, alt: "F")[$F$] is a function, its inverse is not. #figure(figph[function and inverse], alt: "function and inverse", caption: none) Is there some way to predict whether the inverse of a function will be a function, too? Yes! In order for the inverse to be a function, its graph must pass the vertical line test. (Recall that if a graph passes the vertical line test, there is only one #math.equation(block: false, alt: "y")[$y$]-value for each value of #math.equation(block: false, alt: "x")[$x$].) Now, we obtain the inverse by interchanging #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] in the formula for a function, so the #emph[inverse] will be a function only if the #emph[original] function passes the #emph[horizontal] line test. A function that passes the horizontal line test is called #strong[one-to-one]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Horizontal Line Test] A function passes the #strong[Horizontal Line Test] if every horizontal line intersects the graph at most once. In that case, there is only one #math.equation(block: false, alt: "x")[$x$]-value for each #math.equation(block: false, alt: "y")[$y$]-value, and the function is called #strong[one-to-one]. ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ It is always the case that we can find the inverse of a function: we just interchange the domain and range values. However, the important question is whether the inverse is also a function. If the inverse is a function, we say that “#math.equation(block: false, alt: "f")[$f$] has an #strong[inverse function].” All of this leads us to the following result. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Inverse Function] A function #math.equation(block: false, alt: "f")[$f$] has an #strong[inverse function] if and only if #math.equation(block: false, alt: "f")[$f$] is one-to-one. ] #examplebox("Example 3")[][ + Which of the graphs below represent functions? + Which of the functions are one-to-one? + Which of these functions have inverse functions? #figure(figph[three graphs], alt: "three graphs", caption: none) #solutionbox[ + All three graphs pass the vertical line test, so all three represent functions. + Only the function represented by graph II passes the horizontal line test, so it is the only one-to-one function. + Only function II is one-to-one, so it is the only function that has an inverse function. ] ] Which of the functions below are one-to-one? #figure(figph[three graphs], alt: "three graphs", caption: none) I and III === Restricting the Domain: The Inverse Sine Function Sometimes it is so important that the inverse be a function that we are willing to sacrifice part of the original function to achieve this result. Look again at the graph of #math.equation(block: false, alt: "F open parenthesis x close parenthesis equals x squared minus 4")[$" " F ( x ) = x^(2) − 4$]. If we use only nonnegative #math.equation(block: false, alt: "x")[$x$]-values for the domain, we create a new function, #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x squared minus 4 , x greater than or equal to 0")[$f ( x ) = x^(2) − 4 , " " " " x ≥ 0$] The graph of this new function is shown as a dashed curve in the figure at right. #figure(figph[function and inverse function], alt: "function and inverse function", caption: none) The new function is one-to-one, and its inverse, #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the square root of x plus 4")[$f^(− 1) ( x ) = sqrt(x + 4)$], is also a function. (We could also have used only negative #math.equation(block: false, alt: "x")[$x$]-values for the domain, or some smaller interval, just as long as the resulting function is one-to-one.) We say that we have #emph[restricted the domain] of the original function, and we will use this technique to define inverse functions for the trigonometric functions. The sine function is not one-to-one; there are many angles that have the same sine value. In order to define its inverse function, we must restrict the domain of the sine to an interval on which the #math.equation(block: false, alt: "y")[$y$]-values do not repeat. But there are many candidates for such an interval; which one shall we choose? Why the restriction is needed, and why this interval. The dashed gray wave is the full sine; the solid blue piece is the restricted domain \[−π/2, π/2\]. Sweep the red horizontal line y = c up and down: on the full sine it crosses infinitely many times — many angles share the same sine — so “the angle whose sine is c” has no single answer. On the blue piece the line crosses exactly once for every c between −1 and 1: that one crossing is what sin⁻¹(c) means. The interval is the widest one through 0 you can take before heights start to repeat — slide c to ±1 and watch the crossing reach the ends of the blue piece exactly at its endpoints.{"functions":\[{"color":"\#6a6a6a","expression":"sin(x)","expression2":"","id":"full-sine","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#1f4e79","domain":\[-1.5708,1.5708\],"expression":"sin(x)","expression2":"","id":"restricted-branch","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#b8442a","expression":"c","expression2":"","id":"level-line","inequality":"lt","lineDash":\[\],"lineWidth":2,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-c","label":"Height c","max":1.5,"min":-1.5,"name":"c","step":0.05,"value":0.5}\],"title":"Horizontal line test: full sine vs the restricted branch","version":1,"viewport":{"centerX":0,"centerY":0,"scale":55}}It turns out that the most useful interval is found by starting at #math.equation(block: false, alt: "x equals 0")[$x = 0$] and moving as far as we can in either direction until the #math.equation(block: false, alt: "y")[$y$]-values begin to repeat. By doing so, we obtain the restricted domain #math.equation(block: false, alt: "the fraction minus π over 2 less than or equal to x less than or equal to the fraction π over 2")[$display(frac(− π, 2)) ≤ x ≤ display(frac(π, 2))$], as shown below. This piece of the function includes all of the original range values, from #math.equation(block: false, alt: "minus 1")[$− 1$] to #math.equation(block: false, alt: "1")[$1$]. #figure(figph[restricted sine], alt: "restricted sine", caption: none) Because the sine function is one-to-one on this domain, its inverse is a function. The graph of the inverse sine function, #math.equation(block: false, alt: "y equals sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$" " y = sin^(− 1) ( x )$], is shown below. Its domain is the same as the range of the sine function, namely #math.equation(block: false, alt: "minus 1 less than or equal to x less than or equal to 1")[$− 1 ≤ x ≤ 1$], and its range is our restricted domain for sine, #math.equation(block: false, alt: "the fraction minus π over 2 less than or equal to y less than or equal to the fraction π over 2")[$display(frac(− π, 2)) ≤ y ≤ display(frac(π, 2))$]. #figure(figph[inverse sine], alt: "inverse sine", caption: none) #notebox("Definition", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The inverse sine function] The function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$f ( x ) = sin^(− 1) ( x )$] is defined as follows: #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis equals θ if and only if sin ⁡ open parenthesis θ close parenthesis equals x and the fraction minus π over 2 less than or equal to θ less than or equal to the fraction π over 2")[$bold(sin)^(− 1) ( bold(italic(x)) ) = bold(italic(θ)) " " " " " " "if and only if" " " " " " " " " bold(sin) ( bold(italic(θ)) ) = bold(italic(x)) " " " " " " "and" " " " " " " display(frac(− bold(italic(π)), 2)) ≤ bold(italic(θ)) ≤ display(frac(bold(italic(π)), 2))$] ] In other words, #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$sin^(− 1) ( x )$] is the angle in radians, #emph[between #math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$] and #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], whose sine is #math.equation(block: false, alt: "x")[$x$]. There are many angles with a given sine value #math.equation(block: false, alt: "x")[$x$], but only one of these angles can be #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$sin^(− 1) ( x )$]. This is why the calculator's #math.equation(block: false, alt: "sin to the power minus 1")[$sin^(− 1)$] key only gives outputs between #math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$] and #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]. - If #math.equation(block: false, alt: "x equals sin ⁡ open parenthesis θ close parenthesis")[$x = sin ( θ )$] is positive, the inverse sine function delivers a first quadrant angle, #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to the fraction π over 2")[$0 ≤ θ ≤ display(frac(π, 2))$]. - If #math.equation(block: false, alt: "x equals sin ⁡ open parenthesis θ close parenthesis")[$x = sin ( θ )$] is negative, the inverse sine function delivers a fourth quadrant angle, #math.equation(block: false, alt: "the fraction minus π over 2 less than or equal to θ less than or equal to 0")[$display(frac(− π, 2)) ≤ θ ≤ 0$]. #examplebox("Example 4")[][ Simplify each expression without using a calculator. + #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis the fraction minus the square root of 3 over 2 close parenthesis")[$sin^(− 1) ( display(frac(− sqrt(3), 2)) )$] + #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis sin ⁡ open parenthesis π close parenthesis close parenthesis")[$sin^(− 1) ( sin ( π ) )$] #solutionbox[ + We know that #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis the fraction minus the square root of 3 over 2 close parenthesis")[$sin^(− 1) ( display(frac(− sqrt(3), 2)) )$] is an angle whose sine is #math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$], and that #math.equation(block: false, alt: "the fraction minus π over 2 less than or equal to θ less than or equal to the fraction π over 2")[$display(frac(− π, 2)) ≤ θ ≤ display(frac(π, 2))$]. Because #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] is negative, we must find an angle in the fourth quadrant, and because #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction π over 3 close parenthesis equals the fraction the square root of 3 over 2")[$sin ( display(frac(π, 3)) ) = display(frac(sqrt(3), 2))$], the reference angle for #math.equation(block: false, alt: "θ")[$θ$] is #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]. Thus, the angle we want is #math.equation(block: false, alt: "θ equals the fraction minus π over 3")[$θ = display(frac(− π, 3))$], and#math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis the fraction minus the square root of 3 over 2 close parenthesis equals the fraction minus π over 3")[$sin^(− 1) ( display(frac(− sqrt(3), 2)) ) = display(frac(− π, 3))$]. + First note that #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis sin ⁡ open parenthesis π close parenthesis close parenthesis not equal to π")[$sin^(− 1) ( sin ( π ) ) ≠ π$] ! We start by simplifying the expression inside parentheses: #math.equation(block: false, alt: "sin ⁡ open parenthesis π close parenthesis equals 0")[$bold(sin) ( bold(italic(π)) ) = 0$], so #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis sin ⁡ open parenthesis π close parenthesis close parenthesis equals sin to the power minus 1 ⁡ open parenthesis 0 close parenthesis")[$sin^(− 1) ( bold(sin) ( bold(italic(π)) ) ) = sin^(− 1) ( 0 )$]. We want an angle whose sine is 0, and which lies in the interval #math.equation(block: false, alt: "the fraction minus π over 2 less than or equal to θ less than or equal to the fraction π over 2")[$display(frac(− π, 2)) ≤ θ ≤ display(frac(π, 2))$]. This angle is 0, so #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis sin ⁡ open parenthesis π close parenthesis close parenthesis equals 0")[$sin^(− 1) ( sin ( π ) ) = 0$]. (It is also true that #math.equation(block: false, alt: "sin ⁡ open parenthesis π close parenthesis equals 0")[$sin ( π ) = 0$], but #math.equation(block: false, alt: "π")[$π$] is not in the interval #math.equation(block: false, alt: "the fraction minus π over 2 less than or equal to θ less than or equal to the fraction π over 2")[$display(frac(− π, 2)) ≤ θ ≤ display(frac(π, 2))$].) ] ] Simplify each expression without using a calculator. + #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open bracket sin ⁡ open parenthesis the fraction π over 3 close parenthesis close bracket")[$sin^(− 1) [ sin ( display(frac(π, 3)) ) ]$] + #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open bracket sin ⁡ open parenthesis the fraction 2 π over 3 close parenthesis close bracket")[$sin^(− 1) [ sin ( display(frac(2 π, 3)) ) ]$] + #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] + #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] === The Inverse Cosine and Inverse Tangent Functions The cosine and tangent functions are also periodic, so, just as with the sine function, to define their inverse functions we must restrict their domains to intervals where they are one-to-one. The graph of cosine is shown below. #figure(figph[restricted cosine], alt: "restricted cosine", caption: none) Once again the choice of these intervals is arbitrary. If we start at #math.equation(block: false, alt: "θ equals 0")[$θ = 0$] on the cosine graph, we can move in only one direction, either right or left, without encountering repeated #math.equation(block: false, alt: "y")[$y$]-values. We choose to move in the positive direction, to obtain the interval #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to π")[$0 ≤ θ ≤ π$], as shown above. On this domain, the inverse of cosine is a function. Its graph is shown below. #figure(figph[inverse cosine], alt: "inverse cosine", caption: none) #notebox("Definition", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The inverse cosine function] The function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals cos to the power minus 1 ⁡ open parenthesis x close parenthesis")[$f ( x ) = cos^(− 1) ( x )$] is defined as follows: #math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis x close parenthesis equals θ if and only if cos ⁡ open parenthesis θ close parenthesis equals x and 0 less than or equal to θ less than or equal to π")[$bold(cos)^(− 1) ( bold(italic(x)) ) = bold(italic(θ)) " " " " " " "if and only if" " " " " " " " " bold(cos) ( bold(italic(θ)) ) = bold(italic(x)) " " " " " " "and" " " " " " " 0 ≤ bold(italic(θ)) ≤ bold(italic(π))$] ] The range of the inverse cosine function is #math.equation(block: false, alt: "0 less than or equal to y less than or equal to π")[$0 ≤ y ≤ π$], so it delivers angles in the first and second quadrants. (Compare to the inverse sine, whose outputs are angles in the first or fourth quadrants.) #examplebox("Example 5")[][ Simplify each expression without using a calculator. + #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis the fraction minus the square root of 3 over 2 close parenthesis")[$cos^(− 1) ( display(frac(− sqrt(3), 2)) )$] + #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open bracket cos ⁡ open parenthesis the fraction 7 π over 4 close parenthesis close bracket")[$cos^(− 1) [ cos ( display(frac(7 π, 4)) ) ]$] #solutionbox[ + The expression#math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis the fraction minus the square root of 3 over 2 close parenthesis")[$cos^(− 1) ( display(frac(− sqrt(3), 2)) )$] represents an angle in the second quadrant whose cosine is #math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]. Because #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction π over 6 close parenthesis equals the fraction the square root of 3 over 2")[$cos ( display(frac(π, 6)) ) = display(frac(sqrt(3), 2))$], the reference angle for #math.equation(block: false, alt: "θ")[$θ$] is #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]. The angle in the second quadrant with this reference angle is #math.equation(block: false, alt: "θ equals the fraction 5 π over 6")[$θ = display(frac(5 π, 6))$], so #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis the fraction minus the square root of 3 over 2 close parenthesis equals the fraction 5 π over 6")[$cos^(− 1) ( display(frac(− sqrt(3), 2)) ) = display(frac(5 π, 6))$]. + Because #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction 7 π over 4 close parenthesis")[$cos ( display(frac(7 π, 4)) )$] is positive, the angle #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open bracket cos ⁡ open parenthesis the fraction 7 π over 4 close parenthesis close bracket")[$cos^(− 1) [ cos ( display(frac(7 π, 4)) ) ]$] is a first quadrant angle whose cosine is the same as #math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction 7 π over 4 close parenthesis")[$cos ( display(frac(7 π, 4)) )$]. This angle is #math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$], so #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open bracket cos ⁡ open parenthesis the fraction 7 π over 4 close parenthesis close bracket equals the fraction π over 4")[$cos^(− 1) [ cos ( display(frac(7 π, 4)) ) ] = display(frac(π, 4))$]. ] ] Simplify #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis minus 1 close parenthesis")[$cos^(− 1) ( − 1 )$] without using a calculator. #math.equation(block: true, alt: "π")[$π$] Finally, consider the graph of the tangent function. To choose a convenient interval on which the tangent is one-to-one, we start at #math.equation(block: false, alt: "x equals 0")[$x = 0$] and move as far as we can in either direction along the #math.equation(block: false, alt: "x")[$x$]-axis. In this way we obtain one cycle of the graph, on the interval #math.equation(block: false, alt: "the fraction minus π over 2 less than θ less than the fraction π over 2")[$display(frac(− π, 2)) < θ < display(frac(π, 2))$], as shown at right. #figure(figph[tan], alt: "tan", caption: none) The range of the tangent on that interval includes all real numbers. Consequently, the domain of the inverse tangent function includes all real numbers, and its range is the interval #math.equation(block: false, alt: "the fraction minus π over 2 less than y less than the fraction π over 2")[$display(frac(− π, 2)) < y < display(frac(π, 2))$]. The graph of the inverse tangent function is shown below. Its outputs are angles in the first and fourth quadrants. #figure(figph[triangle], alt: "triangle", caption: none) #notebox("Definition", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The inverse tangent function] The function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals tan to the power minus 1 ⁡ open parenthesis x close parenthesis")[$f ( x ) = tan^(− 1) ( x )$] is defined as follows: #math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis x close parenthesis equals θ if and only if tan ⁡ open parenthesis θ close parenthesis equals x and the fraction minus π over 2 less than θ less than the fraction π over 2")[$bold(tan)^(− 1) ( bold(italic(x)) ) = bold(italic(θ)) " " " " " " "if and only if" " " " " " " " " bold(tan) ( bold(italic(θ)) ) = bold(italic(x)) " " " " " " "and" " " " " " " display(frac(− bold(italic(π)), 2)) < bold(italic(θ)) < display(frac(bold(italic(π)), 2))$] ] #examplebox("Example 6")[][ Simplify each expression without using a calculator. + #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis minus the square root of 3 close parenthesis")[$tan^(− 1) ( − sqrt(3) )$] + #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open bracket tan ⁡ open parenthesis the fraction 7 π over 4 close parenthesis close bracket")[$tan^(− 1) [ tan ( display(frac(7 π, 4)) ) ]$] #solutionbox[ + The expression #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis minus the square root of 3 close parenthesis")[$tan^(− 1) ( − sqrt(3) )$] represents an angle between #math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$] and #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$] whose tangent is #math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$]. Now, #math.equation(block: false, alt: "tan ⁡ open parenthesis the fraction π over 3 close parenthesis equals the square root of 3")[$tan ( display(frac(π, 3)) ) = sqrt(3)$], and #math.equation(block: false, alt: "tan ⁡ open parenthesis minus θ close parenthesis equals minus tan ⁡ open parenthesis θ close parenthesis")[$tan ( − θ ) = − tan ( θ )$], so #math.equation(block: false, alt: "tan ⁡ open parenthesis the fraction minus π over 3 close parenthesis equals the fraction minus π over 3 minus the square root of 3")[$tan ( display(frac(− π, 3)) ) = display(frac(− π, 3)) − sqrt(3)$]. Thus, #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis minus the square root of 3 close parenthesis equals the fraction minus π over 3")[$tan^(− 1) ( − sqrt(3) ) = display(frac(− π, 3))$]. + Because #math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$] is in the fourth quadrant, #math.equation(block: false, alt: "tan ⁡ open parenthesis the fraction 7 π over 4 close parenthesis")[$tan ( display(frac(7 π, 4)) )$] is negative. The angle #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open bracket tan ⁡ open parenthesis the fraction 7 π over 4 close parenthesis close bracket")[$tan^(− 1) [ tan ( display(frac(7 π, 4)) ) ]$] is also in the fourth quadrant, but it must be between #math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$] and 0. The angle #math.equation(block: false, alt: "the fraction minus π over 4")[$display(frac(− π, 4))$] is coterminal with #math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$], so #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open bracket tan ⁡ open parenthesis the fraction 7 π over 4 close parenthesis close bracket equals the fraction minus π over 4")[$tan^(− 1) [ tan ( display(frac(7 π, 4)) ) ] = display(frac(− π, 4))$] ] ] Simplify #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis minus 1 close parenthesis")[$tan^(− 1) ( − 1 )$] without using a calculator. #math.equation(block: true, alt: "the fraction minus π over 4")[$display(frac(− π, 4))$] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the Examples above, perhaps you noticed that #math.equation(block: false, alt: "sin ⁡ open parenthesis sin to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis equals x")[$sin ( sin^(− 1) ( x ) ) = x$] for any value of #math.equation(block: false, alt: "x")[$x$] in the domain of the inverse sine. In other words, if we find the inverse sine of a number between #math.equation(block: false, alt: "minus 1")[$− 1$] and #math.equation(block: false, alt: "1")[$1$], and then take the sine of the resulting angle, we get the original number back again. The same is true for the cosine and tangent, so that #math.equation(block: false, alt: "cos ⁡ open parenthesis cos to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis equals x")[$cos ( cos^(− 1) ( x ) ) = x$] and #math.equation(block: false, alt: "tan ⁡ open parenthesis tan to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis equals x")[$tan ( tan^(− 1) ( x ) ) = x$]. However, if we apply the functions in the opposite order, we may not get back the original number. For example, #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open bracket sin ⁡ open parenthesis the fraction 3 π over 4 close parenthesis close bracket equals sin to the power minus 1 ⁡ open parenthesis the fraction minus 1 over the square root of 2 close parenthesis equals the fraction π over 4")[$sin^(− 1) [ sin ( display(frac(3 π, 4)) ) ] = sin^(− 1) ( display(frac(− 1, sqrt(2))) ) = display(frac(π, 4))$] This inequality occurs because the inverse sine function returns just one of the many angles whose sine is #math.equation(block: false, alt: "the fraction minus 1 over the square root of 2")[$display(frac(− 1, sqrt(2)))$], and that angle may not be the angle we started with. ] We summarize the content of the Caution above as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Inverse Trigonometric Functions] #math.equation(block: true, alt: "sin ⁡ open parenthesis sin to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis, equals x for minus 1 less than or equal to x less than or equal to 1; cos ⁡ open parenthesis cos to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis, equals x for minus 1 less than or equal to x less than or equal to 1; tan ⁡ open parenthesis tan to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis, equals x for all x")[$sin ( sin^(− 1) ( x ) ) & = x " " " " " " " " "for" " " − 1 ≤ x ≤ 1 \ cos ( cos^(− 1) ( x ) ) & = x " " " " " " " " "for" " " − 1 ≤ x ≤ 1 \ tan ( tan^(− 1) ( x ) ) & = x " " " " " " " " "for all" " " x$] However, #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis sin ⁡ open parenthesis x close parenthesis close parenthesis may not be equal to x")[$sin^(− 1) ( sin ( x ) ) " " " " "may not be equal to" " " x$] #math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis cos ⁡ open parenthesis x close parenthesis close parenthesis may not be equal to x")[$cos^(− 1) ( cos ( x ) ) " " " " "may not be equal to" " " x$] #math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis tan ⁡ open parenthesis x close parenthesis close parenthesis may not be equal to x")[$tan^(− 1) ( tan ( x ) ) " " " " "may not be equal to" " " x$] ] === Modeling with Inverse Functions The inverse trig functions are used to model situations in which an angle is described in terms of one of its trigonometric ratios. #examplebox("Example 7")[][ The bottom of a 3-meter tall tapestry on a chateau wall is at your eye level. The angle #math.equation(block: false, alt: "θ")[$θ$] subtended vertically by the tapestry changes as you approach the wall. + Express your distance from the wall, #math.equation(block: false, alt: "d")[$d$], as a function of #math.equation(block: false, alt: "θ")[$θ$]. + Express #math.equation(block: false, alt: "θ")[$θ$] as a function of #math.equation(block: false, alt: "d")[$d$]. #solutionbox[ + We draw a sketch of the triangle formed by the tapestry and the lines of sight to its bottom and top, as shown at right. From the triangle we see that #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 3 over d")[$tan ( θ ) = display(frac(3, d))$], so #math.equation(block: false, alt: "d equals the fraction 3 over tan ⁡ open parenthesis θ close parenthesis")[$d = display(frac(3, tan ( θ )))$]. #figure(figph[triangle], alt: "triangle", caption: none) + Because #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 3 over d")[$tan ( θ ) = display(frac(3, d))$], #math.equation(block: false, alt: "θ equals tan to the power minus 1 ⁡ open parenthesis the fraction 3 over d close parenthesis")[$θ = tan^(− 1) ( display(frac(3, d)) )$]. ] ] The tapestry from the previous example includes a 2-meter tall unicorn with its feet at the bottom of the tapestry. + Express #math.equation(block: false, alt: "α")[$α$], the angle subtended vertically by the unicorn, as a function of #math.equation(block: false, alt: "d")[$d$], your distance to the tapestry. + Express #math.equation(block: false, alt: "β")[$β$], the angle subtended by the portion of the tapestry above the unicorn, as a function of #math.equation(block: false, alt: "d")[$d$]. (See the figure at right). #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "α equals tan to the power minus 1 ⁡ open parenthesis the fraction 2 over d close parenthesis")[$α = tan^(− 1) ( display(frac(2, d)) )$] + #math.equation(block: false, alt: "β equals tan to the power minus 1 ⁡ open parenthesis the fraction 3 over d close parenthesis minus tan to the power minus 1 ⁡ open parenthesis the fraction 2 over d close parenthesis")[$β = tan^(− 1) ( display(frac(3, d)) ) − tan^(− 1) ( display(frac(2, d)) )$] === Alternate Notations The inverse sine function, #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$sin^(− 1) ( x )$], is also called the #strong[arcsine] function and denoted by #strong[arcsin #math.equation(block: false, alt: "open parenthesis x close parenthesis")[$( x )$]]. (This terminology reminds us that the output of the inverse sine function is an angle, or the arc on a unit circle determined by that angle, as shown at right.) #figure(figph[unit circle], alt: "unit circle", caption: none) Similarly, the inverse cosine function is sometimes denoted by #strong[arccos #math.equation(block: false, alt: "open parenthesis x close parenthesis")[$( x )$]], and the inverse tangent function by #strong[arctan #math.equation(block: false, alt: "open parenthesis x close parenthesis")[$( x )$]]. Some computer programs use the notation #strong[asin #math.equation(block: false, alt: "open parenthesis x close parenthesis")[$( x )$]], #strong[accos #math.equation(block: false, alt: "open parenthesis x close parenthesis")[$( x )$]], and #strong[atan #math.equation(block: false, alt: "open parenthesis x close parenthesis")[$( x )$]]. #examplebox("Example 8")[][ Simplify each expression. + #math.equation(block: false, alt: "cos ⁡ open bracket arccos ⁡ open parenthesis the fraction 1 over 3 close parenthesis close bracket")[$cos [ arccos ( display(frac(1, 3)) ) ]$] + #math.equation(block: false, alt: "arcsin ⁡ open bracket sin ⁡ open parenthesis the fraction 3 π over 2 close parenthesis close bracket")[$arcsin [ sin ( display(frac(3 π, 2)) ) ]$] #solutionbox[ + Keep in mind that #math.equation(block: false, alt: "arccos ⁡ open parenthesis the fraction 1 over 3 close parenthesis")[$arccos ( display(frac(1, 3)) )$] is an angle. We'll call it #math.equation(block: false, alt: "θ")[$θ$], so that #math.equation(block: false, alt: "θ equals arccos ⁡ open parenthesis the fraction 1 over 3 close parenthesis")[$θ = arccos ( display(frac(1, 3)) )$], and hence #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 3")[$cos ( θ ) = display(frac(1, 3))$]. Then substituting #math.equation(block: false, alt: "θ")[$θ$] for #math.equation(block: false, alt: "arccos ⁡ open parenthesis the fraction 1 over 3 close parenthesis")[$arccos ( display(frac(1, 3)) )$] in the original expression gives us #math.equation(block: true, alt: "cos ⁡ open bracket arccos ⁡ open parenthesis the fraction 1 over 3 close parenthesis close bracket equals cos ⁡ open parenthesis θ close parenthesis equals the fraction 1 over 3")[$cos [ arccos ( display(frac(1, 3)) ) ] = cos ( θ ) = display(frac(1, 3))$] + We start inside the parentheses: #math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction 3 π over 2 close parenthesis equals minus 1")[$sin ( display(frac(3 π, 2)) ) = − 1$]. So #math.equation(block: false, alt: "arcsin ⁡ open bracket sin ⁡ open parenthesis the fraction 3 π over 2 close parenthesis close bracket equals arcsin ⁡ open parenthesis minus 1 close parenthesis")[$arcsin [ sin ( display(frac(3 π, 2)) ) ] = arcsin ( − 1 )$]. Now #math.equation(block: false, alt: "arcsin ⁡ open parenthesis minus 1 close parenthesis")[$arcsin ( − 1 )$] is the angle between #math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$] and 0 whose sine is #math.equation(block: false, alt: "minus 1")[$− 1$], and that angle is #math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$]. Thus, #math.equation(block: false, alt: "arcsin ⁡ open bracket sin ⁡ open parenthesis the fraction 3 π over 2 close parenthesis close bracket equals the fraction minus π over 2")[$arcsin [ sin ( display(frac(3 π, 2)) ) ] = display(frac(− π, 2))$] ] ] Simplify each expression. + #math.equation(block: false, alt: "tan ⁡ open bracket arctan ⁡ open parenthesis the fraction minus 5 over 2 close parenthesis close bracket")[$tan [ arctan ( display(frac(− 5, 2)) ) ]$] + #math.equation(block: false, alt: "arctan ⁡ open bracket tan ⁡ open parenthesis the fraction minus 5 π over 6 close parenthesis close bracket")[$arctan [ tan ( display(frac(− 5 π, 6)) ) ]$] + #math.equation(block: false, alt: "the fraction minus 5 over 2")[$display(frac(− 5, 2))$] + #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$] === Simplifying Expressions The key to simplifying expressions involving inverse trigonometric functions is to remember that the inverse sine, cosine, or tangent of a number can be treated as an angle. If we assign a name such as #math.equation(block: false, alt: "θ")[$θ$] or #math.equation(block: false, alt: "ϕ")[$ϕ$] to the inverse trig value, it can often clarify the computations. #examplebox("Example 9")[][ Evaluate #math.equation(block: false, alt: "sin ⁡ open bracket cos to the power minus 1 ⁡ open parenthesis the fraction 3 over 5 close parenthesis close bracket")[$sin [ cos^(− 1) ( display(frac(3, 5)) ) ]$]. #solutionbox[ Let #math.equation(block: false, alt: "θ")[$θ$] represent the angle #math.equation(block: false, alt: "cos to the power minus 1 ⁡ the fraction 3 over 5")[$cos^(− 1) display(frac(3, 5))$], so that #math.equation(block: false, alt: "θ equals cos to the power minus 1 ⁡ open parenthesis the fraction 3 over 5 close parenthesis")[$θ = cos^(− 1) ( display(frac(3, 5)) )$], and consequently #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 5")[$cos ( θ ) = display(frac(3, 5))$]. Then #math.equation(block: false, alt: "sin ⁡ open bracket cos to the power minus 1 ⁡ open parenthesis the fraction 3 over 5 close parenthesis close bracket")[$sin [ cos^(− 1) ( display(frac(3, 5)) ) ]$] simplifies to #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$], and we know that #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction 3 over 5")[$cos ( θ ) = display(frac(3, 5))$]. If we know one of the trigonometric ratios of an angle, we can find all the others, either by using a diagram or by using identities. In this case we'll use the 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 have #math.equation(block: true, alt: "sin squared ⁡ open parenthesis θ close parenthesis plus open parenthesis the fraction 3 over 5 close parenthesis squared equals 1")[$sin^(2) ( θ ) + attach(( display(frac(3, 5)) ), t: 2) = 1$] and solving for #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] gives us #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction plus or minus 4 over 5")[$sin ( θ ) = display(frac(± 4, 5))$]. But #math.equation(block: false, alt: "θ equals cos to the power minus 1 ⁡ open parenthesis the fraction 3 over 5 close parenthesis")[$θ = cos^(− 1) ( display(frac(3, 5)) )$] is a first quadrant angle, so its sine is positive. Thus, #math.equation(block: false, alt: "sin ⁡ open bracket cos to the power minus 1 ⁡ open parenthesis the fraction 3 over 5 close parenthesis close bracket equals the fraction 4 over 5")[$sin [ cos^(− 1) ( display(frac(3, 5)) ) ] = display(frac(4, 5))$]. ] ] Evaluate #math.equation(block: false, alt: "cos ⁡ open bracket tan to the power minus 1 ⁡ open parenthesis the fraction 2 over 3 close parenthesis close bracket")[$cos [ tan^(− 1) ( display(frac(2, 3)) ) ]$]. #math.equation(block: true, alt: "the fraction 3 over the square root of 13")[$display(frac(3, sqrt(13)))$] We can verify the results of the previous example using a calculator, but the same technique can be applied to simplify similar expressions involving variables. #examplebox("Example 10")[][ Simplify #math.equation(block: false, alt: "tan ⁡ open bracket sin to the power minus 1 ⁡ open parenthesis x close parenthesis close bracket")[$" " tan [ sin^(− 1) ( x ) ]$], assuming that #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 1")[$0 ≤ x ≤ 1$]. #solutionbox[ Let #math.equation(block: false, alt: "θ equals sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$" " θ = sin^(− 1) ( x )$], so that #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals x")[$sin ( θ ) = x$], and we can rewrite the expression #math.equation(block: false, alt: "tan ⁡ open bracket sin to the power minus 1 ⁡ open parenthesis x close parenthesis close bracket")[$" " tan [ sin^(− 1) ( x ) ]$] as #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$]. Now we use identities to write #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] in terms of #math.equation(block: false, alt: "x")[$x$]. First, from the Pythagorean identity we know that #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the square root of 1 minus sin squared ⁡ open parenthesis θ close parenthesis equals the square root of 1 minus x squared")[$cos ( θ ) = sqrt(1 − sin^(2) ( θ )) = sqrt(1 − x^(2))$] and then #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction sin ⁡ open parenthesis θ close parenthesis over cos ⁡ open parenthesis θ close parenthesis equals the fraction x over the square root of 1 minus x squared")[$tan ( θ ) = display(frac(sin ( θ ), cos ( θ ))) = display(frac(x, sqrt(1 − x^(2))))$] ] ] Simplify #math.equation(block: false, alt: "sin ⁡ open bracket tan to the power minus 1 ⁡ open parenthesis z close parenthesis close bracket")[$sin [ tan^(− 1) ( z ) ]$], assuming that #math.equation(block: false, alt: "z greater than or equal to 0")[$z ≥ 0$]. #math.equation(block: true, alt: "the fraction z over the square root of 1 plus z squared")[$display(frac(z, sqrt(1 + z^(2))))$] Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Find a formula for the inverse function. + State the domain and range of the inverse function. + Graph the function and its inverse on the same grid. + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over 2 x minus 4")[$f ( x ) = display(frac(1, 2)) x − 4$] + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 x plus 6")[$g ( x ) = 3 x + 6$] + #math.equation(block: false, alt: "F open parenthesis x close parenthesis equals 2 plus the fraction 1 over x")[$F ( x ) = 2 + display(frac(1, x))$] + #math.equation(block: false, alt: "G open parenthesis x close parenthesis equals the fraction 1 over x plus 5")[$G ( x ) = display(frac(1, x + 5))$] + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals the square root of x plus 2")[$h ( x ) = sqrt(x + 2)$] + #math.equation(block: false, alt: "H open parenthesis x close parenthesis equals 3 plus the cube root of x")[$H ( x ) = 3 + root(3, x)$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals 2 x plus 8")[$f^(− 1) ( x ) = 2 x + 8$] + Dom: #math.equation(block: false, alt: "open parenthesis minus ∞ , ∞ close parenthesis")[$( − upright(∞) , upright(∞) )$] Rge: #math.equation(block: false, alt: "open parenthesis minus ∞ , ∞ close parenthesis")[$( − upright(∞) , upright(∞) )$] + #figure(figph[linear graph and inverse], alt: "linear graph and inverse", caption: none) + + #math.equation(block: false, alt: "g to the power minus 1 open parenthesis x close parenthesis equals the fraction 1 over 3 x minus 2")[$g^(− 1) ( x ) = display(frac(1, 3)) x − 2$] + Dom: #math.equation(block: false, alt: "open parenthesis minus ∞ , ∞ close parenthesis")[$( − upright(∞) , upright(∞) )$] Rge: #math.equation(block: false, alt: "open parenthesis minus ∞ , ∞ close parenthesis")[$( − upright(∞) , upright(∞) )$] + #figure(figph[linear function and inverse], alt: "linear function and inverse", caption: none) + + #math.equation(block: false, alt: "F to the power minus 1 open parenthesis x close parenthesis equals the fraction 1 over x minus 2")[$F^(− 1) ( x ) = display(frac(1, x − 2))$] + Dom: #math.equation(block: false, alt: "x not equal to 0")[$x ≠ 0$] Rge: #math.equation(block: false, alt: "y not equal to 2")[$y ≠ 2$] + #figure(figph[transformations of reciprocal function], alt: "transformations of reciprocal function", caption: none) + + #math.equation(block: false, alt: "G to the power minus 1 open parenthesis x close parenthesis equals the fraction 1 over x minus 5")[$G^(− 1) ( x ) = display(frac(1, x)) − 5$] + Dom: #math.equation(block: false, alt: "x not equal to minus 5")[$x ≠ − 5$] Rge: #math.equation(block: false, alt: "y not equal to 0")[$y ≠ 0$] + #figure(figph[transformations of reciprocal function], alt: "transformations of reciprocal function", caption: none) + + #math.equation(block: false, alt: "h to the power minus 1 open parenthesis x close parenthesis equals x squared minus 2")[$h^(− 1) ( x ) = x^(2) − 2$] + Dom: #math.equation(block: false, alt: "x greater than or equal to 0")[$x ≥ 0$] Rge: #math.equation(block: false, alt: "y greater than or equal to minus 2")[$y ≥ − 2$] + #figure(figph[transformations of square root], alt: "transformations of square root", caption: none) + + #math.equation(block: false, alt: "H to the power minus 1 open parenthesis x close parenthesis equals open parenthesis x minus 3 close parenthesis cubed")[$H^(− 1) ( x ) = ( x − 3 )^(3)$] + Dom: #math.equation(block: false, alt: "open parenthesis minus ∞ , ∞ close parenthesis")[$( − upright(∞) , upright(∞) )$] Rge: #math.equation(block: false, alt: "open parenthesis minus ∞ , ∞ close parenthesis")[$( − upright(∞) , upright(∞) )$] + #figure(figph[transformations of cube root], alt: "transformations of cube root", caption: none) ] === Section 8.2 Summary ==== Vocabulary - Inverse function - One-to-one - Subtend ==== Concepts + Using one of the calculator keys #math.equation(block: false, alt: "sin to the power minus 1 , cos to the power minus 1")[$sin^(− 1) , " " cos^(− 1)$], or #math.equation(block: false, alt: "tan to the power minus 1")[$tan^(− 1)$] performs the inverse operation for computing a sine, cosine or tangent. + Two functions are called inverse functions if each "undoes" the results of the other function. + If #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] is a function, we can often find a formula for the inverse function by interchanging #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] in the formula for the function, and then solving for #math.equation(block: false, alt: "y")[$y$]. + The graphs of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] and #math.equation(block: false, alt: "y equals f to the power minus 1 open parenthesis x close parenthesis")[$y = f^(− 1) ( x )$] are symmetric about the line #math.equation(block: false, alt: "y equals x")[$y = x$]. + The domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the same as the range of #math.equation(block: false, alt: "f")[$f$], and the range of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the same as the domain of #math.equation(block: false, alt: "f")[$f$]. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Horizontal Line Test] A function passes the #strong[Horizontal Line Test] if every horizontal line intersects the graph at most once. In that case, there is only one #math.equation(block: false, alt: "x")[$x$]-value for each #math.equation(block: false, alt: "y")[$y$]-value, and the function is called #strong[one-to-one]. ] + A function #math.equation(block: false, alt: "f")[$f$] has an inverse function if and only if #math.equation(block: false, alt: "f")[$f$] is one-to-one. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Definitions of the inverse trig functions] #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis equals θ if and only if sin ⁡ open parenthesis θ close parenthesis equals x and the fraction minus π over 2 less than or equal to θ less than or equal to the fraction π over 2")[$sin^(− 1) ( x ) = θ " " " " " " "if and only if" " " " " " " " " sin ( θ ) = x " " " " " " "and" " " " " " " display(frac(− π, 2)) ≤ θ ≤ display(frac(π, 2))$]#math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis x close parenthesis equals θ if and only if cos ⁡ open parenthesis θ close parenthesis equals x and 0 less than or equal to θ less than or equal to π")[$cos^(− 1) ( x ) = θ " " " " " " "if and only if" " " " " " " " " cos ( θ ) = x " " " " " " "and" " " " " " " 0 ≤ θ ≤ π$]#math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis x close parenthesis equals θ if and only if tan ⁡ open parenthesis θ close parenthesis equals x and the fraction minus π over 2 less than θ less than the fraction π over 2")[$tan^(− 1) ( x ) = θ " " " " " " "if and only if" " " " " " " " " tan ( θ ) = x " " " " " " "and" " " " " " " display(frac(− π, 2)) < θ < display(frac(π, 2))$] ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Inverse Trigonometric Functions] #math.equation(block: true, alt: "sin ⁡ open parenthesis sin to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis equals x for minus 1 less than or equal to x less than or equal to 1")[$sin ( sin^(− 1) ( x ) ) = x " " " " " " " " "for" " " − 1 ≤ x ≤ 1$]#math.equation(block: true, alt: "cos ⁡ open parenthesis cos to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis equals x for minus 1 less than or equal to x less than or equal to 1")[$cos ( cos^(− 1) ( x ) ) = x " " " " " " " " "for" " " − 1 ≤ x ≤ 1$]#math.equation(block: true, alt: "tan ⁡ open parenthesis tan to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis equals x for all x")[$tan ( tan^(− 1) ( x ) ) = x " " " " " " " " "for all" " " x$]However, #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis sin ⁡ open parenthesis x close parenthesis close parenthesis may not be equal to x")[$sin^(− 1) ( sin ( x ) ) " " " " "may not be equal to" " " x$]#math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis cos ⁡ open parenthesis x close parenthesis close parenthesis may not be equal to x")[$cos^(− 1) ( cos ( x ) ) " " " " "may not be equal to" " " x$]#math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis tan ⁡ open parenthesis x close parenthesis close parenthesis may not be equal to x")[$tan^(− 1) ( tan ( x ) ) " " " " "may not be equal to" " " x$] ] + The inverse sine function is also called the arcsine function and denoted by #math.equation(block: false, alt: "arcsin ⁡ open parenthesis x close parenthesis")[$arcsin ( x )$]. Similarly, the inverse cosine function is sometimes denoted by #math.equation(block: false, alt: "arccos ⁡ open parenthesis x close parenthesis")[$arccos ( x )$], and the inverse tangent function by #math.equation(block: false, alt: "arctan ⁡ open parenthesis x close parenthesis")[$arctan ( x )$]. + When simplifying expressions involving inverse trigonometric functions, it can often clarify the computations if we assign a name such as #math.equation(block: false, alt: "θ")[$θ$] or #math.equation(block: false, alt: "ϕ")[$ϕ$] to the inverse trig value. ==== Study Questions + Here is a table of values defining a function #math.equation(block: false, alt: "f")[$f$]. Make a table of values for #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$].#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], )) + What does it mean for a function to be one-to-one? Give an example. + Why do we restrict the domains of the trig functions when we define their inverse functions? + Which of the following expressions is undefined? Why?+ #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis 0 close parenthesis")[$cos^(− 1) ( 0 )$] + #math.equation(block: false, alt: "arctan ⁡ open parenthesis minus 2 close parenthesis")[$arctan ( − 2 )$] + #math.equation(block: false, alt: "arcsin ⁡ open parenthesis minus 2 close parenthesis")[$arcsin ( − 2 )$] + + Write #math.equation(block: false, alt: "x")[$x$] as a function of #math.equation(block: false, alt: "θ")[$θ$]. + Write #math.equation(block: false, alt: "θ")[$θ$] as a function of #math.equation(block: false, alt: "x")[$x$]. #figure(figph[triangle], alt: "triangle", caption: none) + + Write #math.equation(block: false, alt: "x")[$x$] as a function of #math.equation(block: false, alt: "θ")[$θ$]. + Write #math.equation(block: false, alt: "θ")[$θ$] as a function of #math.equation(block: false, alt: "x")[$x$]. #figure(figph[triangle], alt: "triangle", caption: none) ==== Skills + Decide whether a function has an inverse function \#1–8 + Evaluate the inverse trig functions \#9–20 + Model problems with inverse trig functions \#21–24 + Solve formulas \#25–30 + Simplify expressions involving the inverse trig functions \#31–42, 51–68 + Graph the inverse trig functions \#43–50, 69 and 70 === Homework 8-2 In Problems 1–4, which functions have an inverse function? Explain your answer. #figure(figph[graph], alt: "graph", caption: none) No inverse: Some horizontal lines intersect the curve in more than one point. #figure(figph[transformed reciprocal function], alt: "transformed reciprocal function", caption: none) #figure(figph[graph], alt: "graph", caption: none) Inverse exists: The function is 1-1. #figure(figph[quartic graph], alt: "quartic graph", caption: none) For Problems 5–8, graph the function and decide if it has an inverse function. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals sin ⁡ open parenthesis 2 x close parenthesis minus cos ⁡ open parenthesis x close parenthesis")[$f ( x ) = sin ( 2 x ) − cos ( x )$] #figure(figph[graph], alt: "graph", caption: none) No inverse #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals 4 e to the power minus open parenthesis x / 4 close parenthesis squared")[$g ( x ) = 4 e^(− ( x / 4 )^(2))$] #math.equation(block: true, alt: "G open parenthesis x close parenthesis equals the square root of 25 minus x squared")[$G ( x ) = sqrt(25 − x^(2))$] #figure(figph[simicircle], alt: "simicircle", caption: none) \> No inverse #math.equation(block: true, alt: "F open parenthesis x close parenthesis equals ln ⁡ open parenthesis x cubed plus 8 close parenthesis")[$F ( x ) = ln ( x^(3) + 8 )$] For Problems 9–14, use a calculator to evaluate. Round your answers to the nearest tenth of a degree. #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis 0.2838 close parenthesis")[$sin^(− 1) ( 0.2838 )$] #math.equation(block: true, alt: "16.5 degrees")[$16.5^(∘)$] #math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis 4.8972 close parenthesis")[$tan^(− 1) ( 4.8972 )$] #math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis 0.6894 close parenthesis")[$cos^(− 1) ( 0.6894 )$] #math.equation(block: true, alt: "46.4 degrees")[$46.4^(∘)$] #math.equation(block: true, alt: "arccos ⁡ open parenthesis minus 0.8134 close parenthesis")[$arccos ( − 0.8134 )$] #math.equation(block: true, alt: "arctan ⁡ open parenthesis minus 1.2765 close parenthesis")[$arctan ( − 1.2765 )$] #math.equation(block: true, alt: "equals 51.9 degrees")[$= 51.9^(∘)$] #math.equation(block: true, alt: "arcsin ⁡ open parenthesis minus 0.7493 close parenthesis")[$arcsin ( − 0.7493 )$] For Problems 15–20, give exact values in radians. #math.equation(block: true, alt: "cos to the power minus 1 ⁡ open parenthesis the fraction minus 1 over the square root of 2 close parenthesis")[$cos^(− 1) ( display(frac(− 1, sqrt(2))) )$] #math.equation(block: true, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$] #math.equation(block: true, alt: "tan to the power minus 1 ⁡ open parenthesis minus 1 close parenthesis")[$tan^(− 1) ( − 1 )$] #math.equation(block: true, alt: "sin to the power minus 1 ⁡ open parenthesis the fraction minus 1 over 2 close parenthesis")[$sin^(− 1) ( display(frac(− 1, 2)) )$] #math.equation(block: true, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$] #math.equation(block: true, alt: "arccos ⁡ open parenthesis the fraction the square root of 3 over 2 close parenthesis")[$arccos ( display(frac(sqrt(3), 2)) )$] #math.equation(block: true, alt: "arctan ⁡ open parenthesis the fraction 1 over the square root of 3 close parenthesis")[$arctan ( display(frac(1, sqrt(3))) )$] #math.equation(block: true, alt: "the fraction π over 6")[$display(frac(π, 6))$] #math.equation(block: true, alt: "arcsin ⁡ open parenthesis minus 1 close parenthesis")[$arcsin ( − 1 )$] For Problems 21–26, sketch a figure to help you model each problem. Delbert is watching the launch of a satellite at Cape Canaveral. The viewing area is 500 yards from the launch site. The angle of elevation, #math.equation(block: false, alt: "θ")[$θ$], of Delbert's line of sight increases as the booster rocket rises. + Write a formula for the height, #math.equation(block: false, alt: "h")[$h$], of the rocket as a function of #math.equation(block: false, alt: "θ")[$θ$]. + Write a formula for #math.equation(block: false, alt: "θ")[$θ$] as a function of #math.equation(block: false, alt: "h")[$h$]. + Evaluate the formula in part (b) for #math.equation(block: false, alt: "h equals 1000")[$h = 1000$], and interpret the result. #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "h equals 500 tan ⁡ open parenthesis θ close parenthesis")[$h = 500 tan ( θ )$] + #math.equation(block: false, alt: "θ equals tan to the power minus 1 ⁡ open parenthesis the fraction h over 500 close parenthesis")[$θ = tan^(− 1) ( display(frac(h, 500)) )$] + #math.equation(block: false, alt: "θ equals tan to the power minus 1 ⁡ open parenthesis 2 close parenthesis")[$θ = tan^(− 1) ( 2 )$], so the angle of elevation is #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis 2 close parenthesis approximately equals 63.4 degrees")[$tan^(− 1) ( 2 ) ≈ 63.4^(∘)$] when the rocket is 1000 yd high. Francine's house lies under the flight path from the city airport, and commercial airliners pass overhead at an altitude of 35,000 feet. As Francine watches an airplane recede, its angle of elevation, #math.equation(block: false, alt: "θ")[$θ$], decreases. + Write a formula for the horizontal distance, #math.equation(block: false, alt: "d")[$d$], to the airplane as a function of #math.equation(block: false, alt: "θ")[$θ$]. + Write a formula for #math.equation(block: false, alt: "θ")[$θ$] as a function of #math.equation(block: false, alt: "d")[$d$]. + Evaluate the formula in part (b) for #math.equation(block: false, alt: "d equals 20 , 000")[$d = 20 , 000$], and interpret the result. While driving along the interstate, you approach an enormous 50-foot-wide billboard that sits just beside the road. Your viewing angle, #math.equation(block: false, alt: "θ")[$θ$], increases as you get closer to the billboard. + Write a formula for your distance, #math.equation(block: false, alt: "d")[$d$], from the billboard as a function of #math.equation(block: false, alt: "θ")[$θ$]. + Write a formula for #math.equation(block: false, alt: "θ")[$θ$] as a function of #math.equation(block: false, alt: "d")[$d$]. + Evaluate the formula in part (b) for #math.equation(block: false, alt: "d equals 200")[$d = 200$], and interpret the result. #figure(figph[triangle], alt: "triangle", caption: none) + #math.equation(block: false, alt: "d equals the fraction 50 over tan ⁡ θ")[$d = display(frac(50, tan θ))$] + #math.equation(block: false, alt: "θ equals tan to the power minus 1 ⁡ open parenthesis the fraction 50 over d close parenthesis")[$θ = tan^(− 1) ( display(frac(50, d)) )$] + #math.equation(block: false, alt: "θ equals tan to the power minus 1 ⁡ open parenthesis 0.25 close parenthesis")[$θ = tan^(− 1) ( 0.25 )$]; the bilboard subtends an angle of #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis 0.25 close parenthesis approximately equals 14 degrees")[$tan^(− 1) ( 0.25 ) ≈ 14^(∘)$] at a distance of 200 ft. Emma is walking along the bank of a straight river toward a 20-meter long bridge over the river. Let #math.equation(block: false, alt: "θ")[$θ$] be the angle subtended horizontally by Emma's view of the bridge. + Write a formula for Emma's distance from the bridge, #math.equation(block: false, alt: "d")[$d$], as a function of #math.equation(block: false, alt: "θ")[$θ$]. + Write a formula for #math.equation(block: false, alt: "θ")[$θ$] as a function of #math.equation(block: false, alt: "d")[$d$]. + Evaluate the formula in part (b) for #math.equation(block: false, alt: "d equals 500")[$d = 500$], and interpret the result. Martin is viewing a 4-meter tall painting whose base is 1 meter above his eye level. + Write a formula for #math.equation(block: false, alt: "α")[$α$], the angle subtended from Martin's eye level to the bottom of the painting, when he stands #math.equation(block: false, alt: "x")[$x$] meters from the wall. + Write a formula for #math.equation(block: false, alt: "β")[$β$], the angle subtended by the painting, in terms of #math.equation(block: false, alt: "x")[$x$]. + Evaluate the formula in part (b) for #math.equation(block: false, alt: "x equals 5")[$x = 5$], and interpret the result. #figure(figph[triangles], alt: "triangles", caption: none) + #math.equation(block: false, alt: "α equals tan to the power minus 1 ⁡ open parenthesis the fraction 1 over x close parenthesis")[$α = tan^(− 1) ( display(frac(1, x)) )$] + #math.equation(block: false, alt: "β equals tan to the power minus 1 ⁡ open parenthesis the fraction 5 over x close parenthesis minus tan to the power minus 1 ⁡ open parenthesis the fraction 1 over x close parenthesis")[$β = tan^(− 1) ( display(frac(5, x)) ) − tan^(− 1) ( display(frac(1, x)) )$] + #math.equation(block: false, alt: "β equals 45 degrees minus tan to the power minus 1 ⁡ open parenthesis the fraction 1 over 5 close parenthesis")[$β = 45^(∘) − tan^(− 1) ( display(frac(1, 5)) )$], so the painting subtends an angle of #math.equation(block: false, alt: "45 degrees minus tan to the power minus 1 ⁡ open parenthesis the fraction 1 over 5 close parenthesis approximately equals 33.7 degrees")[$45^(∘) − tan^(− 1) ( display(frac(1, 5)) ) ≈ 33.7^(∘)$] when Martin is 5 meters from the wall. A 5-foot mirror is positioned so that its bottom is 1.5 feet below Jane's eye level. + Write a formula for #math.equation(block: false, alt: "α")[$α$], the angle subtended by the section of mirror below Jane's eye level, when she stands #math.equation(block: false, alt: "x")[$x$] feet from the mirror. + Write a formula for #math.equation(block: false, alt: "θ")[$θ$], the angle subtended by the entire mirror, in terms of #math.equation(block: false, alt: "x")[$x$]. + Evaluate the formula in part (b) for #math.equation(block: false, alt: "x equals 10")[$x = 10$], and interpret the result. For Problems 27–32, solve the formula for the given variable. #math.equation(block: false, alt: "V equals V sub 0 sin ⁡ open parenthesis 2 π ω t plus ϕ close parenthesis")[$V = V_(0) sin ( 2 π ω t + ϕ )$], for #math.equation(block: false, alt: "t")[$t$] #math.equation(block: true, alt: "t equals the fraction 1 over 2 π ω open parenthesis sin to the power minus 1 ⁡ the fraction V over V sub 0 minus ϕ close parenthesis")[$t = display(frac(1, 2 π ω)) ( sin^(− 1) display(frac(V, V_(0))) − ϕ )$] #math.equation(block: false, alt: "R equals the fraction 1 over 32 v sub 0 squared sin ⁡ open parenthesis 2 θ close parenthesis")[$R = display(frac(1, 32)) v_(0)^(2) sin ( 2 θ )$], for #math.equation(block: false, alt: "θ")[$θ$] #math.equation(block: false, alt: "the fraction a over sin ⁡ open parenthesis A close parenthesis equals the fraction b over sin ⁡ open parenthesis B close parenthesis")[$display(frac(a, sin ( A ))) = display(frac(b, sin ( B )))$], for #math.equation(block: false, alt: "A")[$A$] #math.equation(block: true, alt: "A equals sin to the power minus 1 ⁡ open parenthesis the fraction a sin ⁡ open parenthesis B close parenthesis over b close parenthesis")[$A = sin^(− 1) ( display(frac(a sin ( B ), b)) )$] #math.equation(block: false, alt: "c squared equals a squared plus b squared minus 2 a b cos ⁡ open parenthesis C close parenthesis")[$c^(2) = a^(2) + b^(2) − 2 a b cos ( C )$], for #math.equation(block: false, alt: "C")[$C$] #math.equation(block: false, alt: "P equals the fraction k over R to the power 4 cos ⁡ open parenthesis θ close parenthesis")[$P = display(frac(k, R^(4) cos ( θ )))$] for #math.equation(block: false, alt: "θ")[$θ$] #math.equation(block: true, alt: "θ equals plus or minus cos to the power minus 1 ⁡ open parenthesis the fraction k over P R to the power 4 close parenthesis")[$θ = ± cos^(− 1) ( display(frac(k, P R^(4))) )$] #math.equation(block: false, alt: "the fraction r over z equals the fraction 1 over tan ⁡ open parenthesis α plus β close parenthesis")[$display(frac(r, z)) = display(frac(1, tan ( α + β )))$], for #math.equation(block: false, alt: "α")[$α$] For Problems 33–38, find exact values without using a calculator. #math.equation(block: true, alt: "tan ⁡ open bracket sin to the power minus 1 ⁡ open parenthesis the fraction 2 over 3 close parenthesis close bracket")[$tan [ sin^(− 1) ( display(frac(2, 3)) ) ]$] #math.equation(block: true, alt: "the fraction 2 over the square root of 5")[$display(frac(2, sqrt(5)))$] #math.equation(block: true, alt: "tan ⁡ open bracket cos to the power minus 1 ⁡ open parenthesis the fraction 3 over 4 close parenthesis close bracket")[$tan [ cos^(− 1) ( display(frac(3, 4)) ) ]$] #math.equation(block: true, alt: "cos ⁡ open bracket tan to the power minus 1 ⁡ open parenthesis minus 2 close parenthesis close bracket")[$cos [ tan^(− 1) ( − 2 ) ]$] #math.equation(block: true, alt: "the fraction 1 over the square root of 5")[$display(frac(1, sqrt(5)))$] #math.equation(block: true, alt: "sin ⁡ open bracket tan to the power minus 1 ⁡ open parenthesis the fraction minus 3 over the square root of 5 close parenthesis close bracket")[$sin [ tan^(− 1) ( display(frac(− 3, sqrt(5))) ) ]$] #math.equation(block: true, alt: "sin ⁡ open bracket cos to the power minus 1 ⁡ open parenthesis the fraction minus 2 the square root of 6 over 7 close parenthesis close bracket")[$sin [ cos^(− 1) ( display(frac(− 2 sqrt(6), 7)) ) ]$] #math.equation(block: true, alt: "the fraction 5 over 7")[$display(frac(5, 7))$] #math.equation(block: true, alt: "cos ⁡ open bracket sin to the power minus 1 ⁡ open parenthesis the fraction minus 2 over 7 close parenthesis close bracket")[$cos [ sin^(− 1) ( display(frac(− 2, 7)) ) ]$] For Problems 39–44, simplify the expression. #math.equation(block: true, alt: "tan ⁡ open parenthesis cos to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis")[$tan ( cos^(− 1) ( x ) )$] #math.equation(block: true, alt: "the fraction the square root of 1 minus x squared over x")[$display(frac(sqrt(1 − x^(2)), x))$] #math.equation(block: true, alt: "cos ⁡ open parenthesis tan to the power minus 1 ⁡ open parenthesis a close parenthesis close parenthesis")[$cos ( tan^(− 1) ( a ) )$] #math.equation(block: true, alt: "cos ⁡ open parenthesis sin to the power minus 1 ⁡ open parenthesis h close parenthesis close parenthesis")[$cos ( sin^(− 1) ( h ) )$] #math.equation(block: true, alt: "the square root of 1 minus h squared")[$sqrt(1 − h^(2))$] #math.equation(block: true, alt: "sin ⁡ open parenthesis cos to the power minus 1 ⁡ open parenthesis v close parenthesis close parenthesis")[$sin ( cos^(− 1) ( v ) )$] #math.equation(block: true, alt: "sin ⁡ open parenthesis tan to the power minus 1 ⁡ open parenthesis 2 t close parenthesis close parenthesis")[$sin ( tan^(− 1) ( 2 t ) )$] #math.equation(block: true, alt: "the fraction 2 t over the square root of 4 t squared plus 1")[$display(frac(2 t, sqrt(4 t^(2) + 1)))$] #math.equation(block: true, alt: "tan ⁡ open parenthesis sin to the power minus 1 ⁡ open parenthesis 3 b close parenthesis close parenthesis")[$tan ( sin^(− 1) ( 3 b ) )$] For Problems 45–47, complete the table of values and sketch the function. #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$display(frac(− sqrt(2), 2))$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "1")[$1$]]), [#math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis x close parenthesis")[$cos^(− 1) ( x )$]], [$#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$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$frac(− sqrt(3), 2)$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$frac(− sqrt(2), 2)$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$frac(− 1, 2)$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$frac(sqrt(2), 2)$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$frac(sqrt(3), 2)$]], [#math.equation(block: false, alt: "1")[$1$]]), [#math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis x close parenthesis")[$cos^(− 1) ( x )$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 6")[$frac(5 π, 6)$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$frac(3 π, 4)$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$frac(2 π, 3)$]], [#math.equation(block: false, alt: "the fraction π over 2")[$frac(π, 2)$]], [#math.equation(block: false, alt: "the fraction π over 3")[$frac(π, 3)$]], [#math.equation(block: false, alt: "the fraction π over 4")[$frac(π, 4)$]], [#math.equation(block: false, alt: "the fraction π over 6")[$frac(π, 6)$]], [#math.equation(block: false, alt: "0")[$0$]], )) #figure(figph[arccosine], alt: "arccosine", caption: none) #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$display(frac(− sqrt(2), 2))$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "1")[$1$]]), [#math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$sin^(− 1) ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(figph[grid], alt: "grid", caption: none) #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus 1 over the square root of 3")[$display(frac(− 1, sqrt(3)))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 3")[$display(frac(1, sqrt(3)))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]]), [#math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis x close parenthesis")[$tan^(− 1) ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(figph[grid], alt: "grid", caption: none) #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus 1 over the square root of 3")[$frac(− 1, sqrt(3))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over the square root of 3")[$frac(1, sqrt(3))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]]), [#math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis x close parenthesis")[$cos^(− 1) ( x )$]], [#math.equation(block: false, alt: "the fraction minus π over 2")[$frac(− π, 2)$]], [#math.equation(block: false, alt: "the fraction minus π over 3")[$frac(− π, 3)$]], [#math.equation(block: false, alt: "the fraction minus π over 6")[$frac(− π, 6)$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 6")[$frac(π, 6)$]], [#math.equation(block: false, alt: "the fraction π over 4")[$frac(π, 4)$]], [#math.equation(block: false, alt: "the fraction π over 3")[$frac(π, 3)$]], )) #figure(figph[arctangent], alt: "arctangent", caption: none) Use a graphing calculator to answer each of the following questions. Then explain the results. + Does #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")[$cos^(− 1) ( x ) = display(frac(1, cos ( x )))$]? + Does #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis equals the fraction 1 over sin ⁡ open parenthesis x close parenthesis")[$sin^(− 1) ( x ) = display(frac(1, sin ( x )))$]? + Does #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis x close parenthesis equals the fraction 1 over tan ⁡ open parenthesis x close parenthesis")[$tan^(− 1) ( x ) = display(frac(1, tan ( x )))$]? + Sketch a graph of #math.equation(block: false, alt: "y equals cos to the power minus 1 ⁡ open parenthesis x close parenthesis")[$y = cos^(− 1) ( x )$], and label the scales on the axes. + Use transformations to sketch graphs of #math.equation(block: false, alt: "y equals 2 cos to the power minus 1 ⁡ open parenthesis x close parenthesis")[$y = 2 cos^(− 1) ( x )$] and #math.equation(block: false, alt: "y equals cos to the power minus 1 ⁡ open parenthesis 2 x close parenthesis")[$y = cos^(− 1) ( 2 x )$]. + Does #math.equation(block: false, alt: "2 cos to the power minus 1 ⁡ open parenthesis x close parenthesis equals cos to the power minus 1 ⁡ open parenthesis 2 x close parenthesis")[$2 cos^(− 1) ( x ) = cos^(− 1) ( 2 x )$]? a–b. #figure(figph[transformations of arccos], alt: "transformations of arccos", caption: none) c. No + Sketch a graph of #math.equation(block: false, alt: "y equals sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$y = sin^(− 1) ( x )$], and label the scales on the axes. + Use transformations to sketch graphs of #math.equation(block: false, alt: "y equals the fraction 1 over 2 sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$y = display(frac(1, 2)) sin^(− 1) ( x )$] and #math.equation(block: false, alt: "y equals sin to the power minus 1 ⁡ open parenthesis the fraction 1 over 2 x close parenthesis")[$y = sin^(− 1) ( display(frac(1, 2)) x )$]. + Does #math.equation(block: false, alt: "the fraction 1 over 2 sin to the power minus 1 ⁡ open parenthesis x close parenthesis equals sin to the power minus 1 ⁡ open parenthesis the fraction 1 over 2 x close parenthesis")[$display(frac(1, 2)) sin^(− 1) ( x ) = sin^(− 1) ( display(frac(1, 2)) x )$]? + Sketch a graph of #math.equation(block: false, alt: "y equals tan to the power minus 1 ⁡ open parenthesis x close parenthesis")[$y = tan^(− 1) ( x )$], and label the scales on the axes. + Use technology to graph #math.equation(block: false, alt: "y equals the fraction sin to the power minus 1 ⁡ open parenthesis x close parenthesis over cos to the power minus 1 ⁡ open parenthesis x close parenthesis")[$y = display(frac(sin^(− 1) ( x ), cos^(− 1) ( x )))$] on a suitable domain. + Does #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis x close parenthesis equals the fraction sin to the power minus 1 ⁡ open parenthesis x close parenthesis over cos to the power minus 1 ⁡ open parenthesis x close parenthesis")[$tan^(− 1) ( x ) = display(frac(sin^(− 1) ( x ), cos^(− 1) ( x )))$]? a. #figure(figph[arctangent], alt: "arctangent", caption: none) c. No + Use technology to graph #math.equation(block: false, alt: "y equals the cube root of x")[$" " y = root(3, x) " "$] and #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis x close parenthesis")[$" " tan^(− 1) ( x ) " "$] for #math.equation(block: false, alt: "minus 10 less than or equal to x less than or equal to 10")[$− 10 ≤ x ≤ 10$]. + Describe the similarities and differences in the two graphs. Use the identities from Section 8.1 to help you find exact values for the expressions in Problems 53–58. #math.equation(block: true, alt: "sin ⁡ open parenthesis open parenthesis 2 tan to the power minus 1 ⁡ open parenthesis 4 close parenthesis close parenthesis")[$sin \( \( 2 tan^(− 1) \( 4 \) \)$] #math.equation(block: true, alt: "the fraction 8 over 17")[$display(frac(8, 17))$] #math.equation(block: true, alt: "cos ⁡ open bracket 2 sin to the power minus 1 ⁡ open parenthesis the fraction 5 over 13 close parenthesis close bracket")[$cos [ 2 sin^(− 1) ( display(frac(5, 13)) ) ]$] #math.equation(block: true, alt: "tan ⁡ open bracket 2 cos to the power minus 1 ⁡ open parenthesis the fraction 3 over 4 close parenthesis close bracket")[$tan [ 2 cos^(− 1) ( display(frac(3, 4)) ) ]$] #math.equation(block: true, alt: "the fraction 16 over 65")[$display(frac(16, 65))$] #math.equation(block: true, alt: "sin ⁡ open bracket 2 cos to the power minus 1 ⁡ open parenthesis the fraction minus 4 over 5 close parenthesis close bracket")[$sin [ 2 cos^(− 1) ( display(frac(− 4, 5)) ) ]$] #math.equation(block: true, alt: "tan ⁡ open bracket 2 sin to the power minus 1 ⁡ open parenthesis the fraction 1 over 3 close parenthesis close bracket")[$tan [ 2 sin^(− 1) ( display(frac(1, 3)) ) ]$] #math.equation(block: true, alt: "the fraction 4 the square root of 2 over 7")[$display(frac(4 sqrt(2), 7))$] #math.equation(block: true, alt: "cos ⁡ open bracket 2 tan to the power minus 1 ⁡ open parenthesis the fraction 3 over 2 close parenthesis close bracket")[$cos [ 2 tan^(− 1) ( display(frac(3, 2)) ) ]$] Let #math.equation(block: false, alt: "α equals cos to the power minus 1 ⁡ open parenthesis the fraction minus 4 over 5 close parenthesis , β equals sin to the power minus 1 ⁡ open parenthesis the fraction 5 over 13 close parenthesis")[$α = cos^(− 1) ( display(frac(− 4, 5)) ) , " " β = sin^(− 1) ( display(frac(5, 13)) )$]. Find exact values for the following. + #math.equation(block: false, alt: "cos ⁡ open parenthesis α plus β close parenthesis")[$cos ( α + β )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis α plus β close parenthesis")[$sin ( α + β )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis α minus β close parenthesis")[$cos ( α − β )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis α minus β close parenthesis")[$sin ( α − β )$] + #math.equation(block: false, alt: "the fraction minus 63 over 65")[$display(frac(− 63, 65))$] + #math.equation(block: false, alt: "the fraction 16 over 65")[$display(frac(16, 65))$] + #math.equation(block: false, alt: "the fraction minus 33 over 65")[$display(frac(− 33, 65))$] + #math.equation(block: false, alt: "the fraction 56 over 65")[$display(frac(56, 65))$] Let #math.equation(block: false, alt: "α equals sin to the power minus 1 ⁡ open parenthesis the fraction minus 15 over 17 close parenthesis , β equals tan to the power minus 1 ⁡ open parenthesis the fraction 4 over 3 close parenthesis")[$α = sin^(− 1) ( display(frac(− 15, 17)) ) , " " β = tan^(− 1) ( display(frac(4, 3)) )$]. Find exact values for the following. + #math.equation(block: false, alt: "cos ⁡ open parenthesis α plus β close parenthesis")[$cos ( α + β )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis α plus β close parenthesis")[$sin ( α + β )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis α minus β close parenthesis")[$cos ( α − β )$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis α minus β close parenthesis")[$sin ( α − β )$] Find an exact value for #math.equation(block: false, alt: "sin ⁡ open bracket tan to the power minus 1 ⁡ open parenthesis the fraction 3 over 4 close parenthesis minus sin to the power minus 1 ⁡ open parenthesis the fraction minus 4 over 5 close parenthesis close bracket")[$sin [ tan^(− 1) ( display(frac(3, 4)) ) − sin^(− 1) ( display(frac(− 4, 5)) ) ]$]. #math.equation(block: true, alt: "1")[$1$] Find an exact value for #math.equation(block: false, alt: "cos ⁡ open bracket tan to the power minus 1 ⁡ open parenthesis the fraction 5 over 12 close parenthesis plus sin to the power minus 1 ⁡ open parenthesis the fraction minus 3 over 5 close parenthesis close bracket")[$cos [ tan^(− 1) ( display(frac(5, 12)) ) + sin^(− 1) ( display(frac(− 3, 5)) ) ]$]. Express in terms of #math.equation(block: false, alt: "x")[$x$] without trigonometric functions. + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 tan to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis")[$sin ( 2 tan^(− 1) ( x ) )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 sin to the power minus 1 ⁡ open parenthesis x close parenthesis close parenthesis")[$cos ( 2 sin^(− 1) ( x ) )$] + #math.equation(block: false, alt: "the fraction 2 x over x squared plus 1")[$display(frac(2 x, x^(2) + 1))$] + #math.equation(block: false, alt: "1 minus x squared")[$1 − x^(2)$] Express in terms of #math.equation(block: false, alt: "w")[$w$] without trigonometric functions. + #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 cos to the power minus 1 ⁡ open parenthesis w close parenthesis close parenthesis")[$sin ( 2 cos^(− 1) ( w ) )$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 tan to the power minus 1 ⁡ open parenthesis w close parenthesis close parenthesis")[$cos ( 2 tan^(− 1) ( w ) )$] If #math.equation(block: false, alt: "x equals 5 sin ⁡ open parenthesis θ close parenthesis , 0 degrees less than θ less than 90 degrees")[$x = 5 sin ( θ ) , " " 0^(∘) < θ < 90^(∘)$], express #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis")[$sin ( 2 θ )$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis")[$cos ( 2 θ )$] in terms of #math.equation(block: false, alt: "x")[$x$]. #math.equation(block: true, alt: "sin ⁡ open parenthesis 2 θ close parenthesis equals the fraction 2 x the square root of 25 minus x squared over 25")[$sin ( 2 θ ) = display(frac(2 x sqrt(25 − x^(2)), 25))$], #math.equation(block: true, alt: "cos ⁡ open parenthesis 2 θ close parenthesis equals the fraction 25 minus 2 x squared over 25")[$cos ( 2 θ ) = display(frac(25 − 2 x^(2), 25))$] If #math.equation(block: false, alt: "x minus 1 equals 2 cos ⁡ open parenthesis θ close parenthesis , 0 degrees less than θ less than 90 degrees")[$" " x − 1 = 2 cos ( θ ) , " " 0^(∘) < θ < 90^(∘)$], express #math.equation(block: false, alt: "sin ⁡ open parenthesis 2 θ close parenthesis")[$sin ( 2 θ )$] and #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 θ close parenthesis")[$cos ( 2 θ )$] in terms of #math.equation(block: false, alt: "x")[$x$]. If #math.equation(block: false, alt: "x equals 3 tan ⁡ open parenthesis θ close parenthesis")[$" " x = 3 tan ( θ )$], write #math.equation(block: false, alt: "θ plus the fraction 1 over 4 sin ⁡ open parenthesis 2 θ close parenthesis")[$θ + display(frac(1, 4)) sin ( 2 θ )$] in terms of #math.equation(block: false, alt: "x")[$x$]. #math.equation(block: true, alt: "arctan ⁡ open parenthesis the fraction x over 3 plus the fraction 3 x over 2 open parenthesis x squared plus 9 close parenthesis close parenthesis")[$arctan ( display(frac(x, 3)) + display(frac(3 x, 2 ( x^(2) + 9 ))) )$] If #math.equation(block: false, alt: "x equals 5 cos ⁡ open parenthesis θ close parenthesis")[$" " x = 5 cos ( θ )$], write #math.equation(block: false, alt: "the fraction θ over 2 minus cos ⁡ open parenthesis 2 θ close parenthesis")[$display(frac(θ, 2)) − cos ( 2 θ )$] in terms of #math.equation(block: false, alt: "x")[$x$]. + For what values of #math.equation(block: false, alt: "x")[$x$] is the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sin ⁡ open parenthesis arcsin ⁡ open parenthesis x close parenthesis close parenthesis")[$" " f ( x ) = sin ( arcsin ( x ) ) " "$] defined? + Is #math.equation(block: false, alt: "sin ⁡ open parenthesis arcsin ⁡ open parenthesis x close parenthesis close parenthesis equals x")[$" " sin ( arcsin ( x ) ) = x " "$] for all #math.equation(block: false, alt: "x")[$x$] where it is defined? If not, for what values of #math.equation(block: false, alt: "x")[$x$] is the equation false? + For what values of #math.equation(block: false, alt: "x")[$x$] is the function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals arcsin ⁡ open parenthesis sin ⁡ open parenthesis x close parenthesis close parenthesis")[$" " g ( x ) = arcsin ( sin ( x ) ) " "$] defined? + Is #math.equation(block: false, alt: "arcsin ⁡ open parenthesis sin ⁡ open parenthesis x close parenthesis close parenthesis equals x")[$" " arcsin ( sin ( x ) ) = x " "$] for all #math.equation(block: false, alt: "x")[$x$] where it is defined? If not, for what values of #math.equation(block: false, alt: "x")[$x$] is the equation false? + #math.equation(block: false, alt: "minus 1 less than or equal to x less than or equal to 1")[$− 1 ≤ x ≤ 1$] + Yes. + All + #math.equation(block: false, alt: "x less than the fraction minus π over 2")[$x < display(frac(− π, 2))$] or #math.equation(block: false, alt: "x greater than the fraction π over 2")[$x > display(frac(π, 2))$] + For what values of #math.equation(block: false, alt: "x")[$x$] is the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals cos ⁡ open parenthesis arccos ⁡ open parenthesis x close parenthesis close parenthesis")[$" " f ( x ) = cos ( arccos ( x ) ) " "$] defined? + Is #math.equation(block: false, alt: "cos ⁡ open parenthesis arccos ⁡ open parenthesis x close parenthesis close parenthesis equals x")[$" " cos ( arccos ( x ) ) = x " "$] for all #math.equation(block: false, alt: "x")[$x$] where it is defined? If not, for what values of #math.equation(block: false, alt: "x")[$x$] is the equation false? + For what values of #math.equation(block: false, alt: "x")[$x$] is the function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals arccos ⁡ open parenthesis cos ⁡ open parenthesis x close parenthesis close parenthesis")[$" " g ( x ) = arccos ( cos ( x ) ) " "$] defined? + Is #math.equation(block: false, alt: "arccos ⁡ open parenthesis cos ⁡ open parenthesis x close parenthesis close parenthesis equals x")[$" " arccos ( cos ( x ) ) = x " "$] for all #math.equation(block: false, alt: "x")[$x$] where it is defined? If not, for what values of #math.equation(block: false, alt: "x")[$x$] is the equation false? Use your calculator to graph #math.equation(block: false, alt: "y equals sin to the power minus 1 ⁡ open parenthesis x close parenthesis plus cos to the power minus 1 ⁡ open parenthesis x close parenthesis")[$" " y = sin^(− 1) ( x ) + cos^(− 1) ( x )$]. + State the domain and range of the graph. + Explain why the graph looks as it does. + Domain: #math.equation(block: false, alt: "minus 1 less than or equal to x less than or equal to 1")[$− 1 ≤ x ≤ 1$], range: #math.equation(block: false, alt: "open brace the fraction π over 2 close brace")[$\{ display(frac(π, 2)) \}$] + Let #math.equation(block: false, alt: "θ equals sin to the power minus 1 ⁡ open parenthesis x close parenthesis")[$θ = sin^(− 1) ( x )$]. Then #math.equation(block: false, alt: "x equals sin ⁡ open parenthesis θ close parenthesis equals cos ⁡ open parenthesis the fraction π over 2 minus θ close parenthesis")[$x = sin ( θ ) = cos ( display(frac(π, 2)) − θ )$] and #math.equation(block: false, alt: "cos to the power minus 1 ⁡ open parenthesis x close parenthesis equals the fraction π over 2 minus θ")[$cos^(− 1) ( x ) = display(frac(π, 2)) − θ$]. So #math.equation(block: false, alt: "sin to the power minus 1 ⁡ open parenthesis x close parenthesis plus cos to the power minus 1 ⁡ open parenthesis x close parenthesis equals θ plus open parenthesis the fraction π over 2 minus θ close parenthesis equals the fraction π over 2")[$" " sin^(− 1) ( x ) + cos^(− 1) ( x ) = θ + ( display(frac(π, 2)) − θ ) = display(frac(π, 2))$] . Use your calculator to graph #math.equation(block: false, alt: "y equals tan to the power minus 1 ⁡ open parenthesis x close parenthesis plus tan to the power minus 1 ⁡ open parenthesis the fraction 1 over x close parenthesis")[$" " y = tan^(− 1) ( x ) + tan^(− 1) ( display(frac(1, x)) )$]. + State the domain and range of the graph. + Explain why the graph looks as it does. In Problems 73–74, we find a formula for the area under part of a semicircle. Use the figure of a unit circle to answer the following. + Write an expression for the area of the shaded sector in terms of #math.equation(block: false, alt: "θ")[$θ$]. + How are #math.equation(block: false, alt: "θ")[$θ$] and #math.equation(block: false, alt: "t")[$t$] related in the figure? (Hint: Write an expression for #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$].) + Combine your answers to (a) and (b) to write an expression for the area of the sector in terms of #math.equation(block: false, alt: "t")[$t$]. #figure(figph[unit circle], alt: "unit circle", caption: none) + #math.equation(block: false, alt: "the fraction θ over 2")[$display(frac(θ, 2))$] + #math.equation(block: false, alt: "t equals sin ⁡ open parenthesis θ close parenthesis")[$t = sin ( θ )$] + #math.equation(block: false, alt: "the fraction 1 over 2 arcsin ⁡ open parenthesis t close parenthesis")[$frac(1, 2) arcsin ( t )$] Use the figure of a unit circle to answer the following. + Write an expression for the height of the shaded triangle in terms of #math.equation(block: false, alt: "t")[$t$]. (Hint: Use the Pythagorean theorem.) + Write an expression for the area of the triangle in terms of #math.equation(block: false, alt: "t")[$t$]. + Combine your answers to (b) and to Problem 73 to write an expression for the area bounded above by the upper semicircle, below by the #math.equation(block: false, alt: "x")[$x$]-axis, on the left by the #math.equation(block: false, alt: "y")[$y$]-axis, and on the right by #math.equation(block: false, alt: "x equals t")[$x = t$], when #math.equation(block: false, alt: "0 less than or equal to t less than or equal to 1")[$0 ≤ t ≤ 1$]. #figure(figph[unit circle], alt: "unit circle", caption: none)