{"id":162,"date":"2023-09-20T22:47:58","date_gmt":"2023-09-20T22:47:58","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/inverse-trigonometric-functions\/"},"modified":"2024-08-05T12:29:04","modified_gmt":"2024-08-05T12:29:04","slug":"inverse-functions-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/inverse-functions-learn-it-4\/","title":{"raw":"Inverse Functions: Learn It 4","rendered":"Inverse Functions: Learn It 4"},"content":{"raw":"<h2>Inverse Trigonometric Functions<\/h2>\r\n<p id=\"fs-id1170572478796\">The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse.<\/p>\r\n<p>Consider the sine function. The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}][\/latex]. By doing so, we define the inverse sine function on the domain [latex][-1,1][\/latex] such that for any [latex]x[\/latex] in the interval [latex][-1,1][\/latex], the inverse sine function tells us which angle [latex]\\theta [\/latex] in the interval [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}][\/latex] satisfies [latex] \\sin \\theta =x[\/latex].<\/p>\r\n<p>Similarly, we can restrict the domains of the other trigonometric functions to define <strong>inverse trigonometric functions<\/strong>, which are functions that tell us which angle in a certain interval has a specified trigonometric value.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>inverse trigonometric functions<\/h3>\r\n<p id=\"fs-id1170572547399\">The <strong>inverse sine function<\/strong>, denoted [latex] \\sin^{-1}[\/latex] or arcsin, and the inverse cosine function, denoted [latex]\\cos^{-1}[\/latex] or arccos, are defined on the domain [latex]D=\\{x|-1 \\le x \\le 1\\}[\/latex] as follows:<\/p>\r\n<div id=\"fs-id1170572547453\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{c}\\sin^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\sin (y)=x \\, \\text{and} \\, -\\frac{\\pi}{2} \\le y \\le \\frac{\\pi}{2};\\hfill \\\\ \\cos^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\cos (y)=x \\, \\text{and} \\, 0 \\le y \\le \\pi \\hfill \\end{array}[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1170572550974\">The <strong>inverse tangent function<\/strong>, denoted [latex]\\tan^{-1}[\/latex] or arctan, and inverse cotangent function, denoted [latex]\\cot^{-1}[\/latex] or arccot, are defined on the domain [latex]D=\\{x|-\\infty &lt; x &lt; \\infty \\}[\/latex] as follows:<\/p>\r\n<div id=\"fs-id1170572551028\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{c}\\tan^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\tan (y)=x \\, \\text{and} \\, -\\frac{\\pi}{2} \\le y \\le \\frac{\\pi}{2};\\hfill \\\\ \\cot^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\cot (y)=x \\, \\text{and} \\, 0 \\le y \\le \\pi \\hfill \\end{array}[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1170572434001\">The<strong> inverse cosecant function<\/strong>, denoted [latex]\\csc^{-1}[\/latex] or arccsc, and inverse secant function, denoted [latex]\\sec^{-1}[\/latex] or arcsec, are defined on the domain [latex]D=\\{x| \\, |x| \\ge 1\\}[\/latex] as follows:<\/p>\r\n<div id=\"fs-id1170572548730\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{c}\\csc^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\csc (y)=x \\, \\text{and} \\, -\\frac{\\pi}{2} \\le y \\le \\frac{\\pi}{2}, \\, y\\ne 0;\\hfill \\\\ \\sec^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\sec (y)=x \\, \\text{and} \\, 0 \\le y \\le \\pi, \\, y \\ne \\frac{\\pi}{2}\\hfill \\end{array}[\/latex]<\/div>\r\n<\/section>\r\n<h3>Graphs of Inverse Trigonometric Functions<\/h3>\r\n<p id=\"fs-id1170572551163\">To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line [latex]y=x[\/latex] (Figure 16).<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"851\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202603\/CNX_Calc_Figure_01_04_018.jpg\" alt=\"An image of six graphs. The first graph is of the function \u201cf(x) = sin inverse(x)\u201d, which is an increasing curve function. The function starts at the point (-1, -(pi\/2)) and increases until it ends at the point (1, (pi\/2)). The x intercept and y intercept are at the origin. The second graph is of the function \u201cf(x) = cos inverse (x)\u201d, which is a decreasing curved function. The function starts at the point (-1, pi) and decreases until it ends at the point (1, 0). The x intercept is at the point (1, 0). The y intercept is at the point (0, (pi\/2)). The third graph is of the function f(x) = tan inverse (x)\u201d, which is an increasing curve function. The function starts close to the horizontal line \u201cy = -(pi\/2)\u201d and increases until it comes close the \u201cy = (pi\/2)\u201d. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The x intercept and y intercept are both at the origin. The fourth graph is of the function \u201cf(x) = cot inverse (x)\u201d, which is a decreasing curved function. The function starts slightly below the horizontal line \u201cy = pi\u201d and decreases until it gets close the x axis. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The fifth graph is of the function \u201cf(x) = csc inverse (x)\u201d, a decreasing curved function. The function starts slightly below the x axis, then decreases until it hits a closed circle point at (-1, -(pi\/2)). The function then picks up again at the point (1, (pi\/2)), where is begins to decrease and approach the x axis, without ever touching the x axis. There is a horizontal asymptote at the x axis. The sixth graph is of the function \u201cf(x) = sec inverse (x)\u201d, an increasing curved function. The function starts slightly above the horizontal line \u201cy = (pi\/2)\u201d, then increases until it hits a closed circle point at (-1, pi). The function then picks up again at the point (1, 0), where is begins to increase and approach the horizontal line \u201cy = (pi\/2)\u201d, without ever touching the line. There is a horizontal asymptote at the \u201cy = (pi\/2)\u201d.\" width=\"851\" height=\"714\" \/> Figure 16. The graph of each of the inverse trigonometric functions is a reflection about the line [latex]y=x[\/latex] of the corresponding restricted trigonometric function.[\/caption]\r\n\r\n<p id=\"fs-id1170572366201\">When evaluating an inverse trigonometric function, the output is an angle.<\/p>\r\n<p>For example, to evaluate [latex]\\cos^{-1}(\\frac{1}{2})[\/latex], we need to find an angle [latex]\\theta [\/latex] such that [latex] \\cos \\theta =\\frac{1}{2}[\/latex]. Clearly, many angles have this property. However, given the definition of [latex]\\cos^{-1}[\/latex], we need the angle [latex]\\theta [\/latex] that not only solves this equation, but also lies in the interval [latex][0,\\pi][\/latex]. We conclude that [latex]\\cos^{-1}(\\frac{1}{2})=\\frac{\\pi}{3}[\/latex].<\/p>\r\n<section class=\"textbox proTip\">Knowing the common values of sine and cosine for key angles can simplify the process of evaluating inverse trigonometric functions, making it a smoother and quicker task.\r\n\r\n<table id=\"Table_05_02_01\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Angle<\/strong><\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{6}[\/latex], or [latex]30\u00b0[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{4}[\/latex], or [latex]45\u00b0[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{3}[\/latex], or [latex]60\u00b0[\/latex]<\/td>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex], or [latex]90\u00b0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cosine<\/strong><\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Sine<\/strong><\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section>\r\n<h3>Compositions of Inverse Trigonometric Functions<\/h3>\r\n<p id=\"fs-id1170572472189\">When working with inverse trigonometric functions, it's crucial to understand how composition works. For instance, when we compose [latex]\\sin[\/latex] and its inverse [latex]\\sin^{-1}[\/latex], such as in [latex]\\sin{(\\sin^{-1}(y))}[\/latex], we're essentially undoing the sine function, which should give us the original input, [latex]y[\/latex]. However, this holds true only if [latex]y[\/latex] falls within the range of [latex]\\sin^{-1}[\/latex], which is [latex][1,1][\/latex]. So [latex]\\sin{(\\sin^{-1}(y))}=y[\/latex] for [latex]-1 \\le y \\le 1[\/latex]. Conversely, when we consider [latex]\\sin^{-1}({\\sin(x)})[\/latex], the result is [latex]x[\/latex] only if [latex]x[\/latex] is within the restricted domain of [latex]\\sin^{-1}[\/latex], which is [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}][\/latex]. The same principle applies to [latex]\\cos[\/latex] and its inverse.<\/p>\r\n<section class=\"textbox example\">\r\n<p>For example, consider the two expressions [latex] \\sin (\\sin^{-1}(\\frac{\\sqrt{2}}{2}))[\/latex] and [latex]\\sin^{-1}(\\sin(\\pi))[\/latex]. For the first one, we simplify as follows:<\/p>\r\n<div id=\"fs-id1170572472267\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex] \\sin (\\sin^{-1}(\\frac{\\sqrt{2}}{2}))= \\sin (\\frac{\\pi}{4})=\\frac{\\sqrt{2}}{2}[\/latex]<\/div>\r\n<p>For the second one, we have<\/p>\r\n<center>[latex]\\sin^{-1}( \\sin (\\pi))=\\sin^{-1}(0)=0[\/latex]<\/center>This is because the value of [latex]\u03c0[\/latex] falls outside the restricted range of the inverse sine function, which is [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}][\/latex].<\/section>\r\n<p id=\"fs-id1170572089863\">To summarize,<\/p>\r\n<div id=\"fs-id1170572550786\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex] \\sin (\\sin^{-1}{y})=y \\, \\text{ if } \\, -1 \\le y \\le 1[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1170572550837\">and<\/p>\r\n<div id=\"fs-id1170572550840\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin^{-1}( \\sin {x})=x \\, \\text{ if } \\, -\\frac{\\pi}{2} \\le x \\le \\frac{\\pi}{2}[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1170572550900\">Similarly, for the cosine function,<\/p>\r\n<div id=\"fs-id1170572550903\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex] \\cos (\\cos^{-1}{y})=y \\, \\text{ if } \\, -1 \\le y \\le 1[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1170572549577\">and<\/p>\r\n<div id=\"fs-id1170572549580\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cos^{-1}( \\cos {x})=x \\, \\text{ if } 0 \\le x \\le \\pi[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1170572549631\">Similar properties hold for the other trigonometric functions and their inverses.<\/p>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How to: <\/strong><strong>Composing Inverse Trig Functions<\/strong><\/p>\r\n<ol>\r\n\t<li><strong>Check the Range<\/strong>: Ensure the value inside the inverse function is within the inverse function's range. For [latex]\\sin^{-1}[\/latex], the value must be between [latex]-1[\/latex] and [latex]1[\/latex].<\/li>\r\n\t<li><strong>Apply the Function<\/strong>: Perform the composition by applying the inverse function first.<\/li>\r\n\t<li><strong>Reverse the Process<\/strong>: Apply the original trigonometric function to the result.<\/li>\r\n\t<li><strong>Restrict the Range<\/strong>: Remember that for [latex]\\sin{(\\sin^{-1}({x}))}[\/latex] and [latex]\\cos{(\\cos^{-1}({x}))}[\/latex], the original [latex]x[\/latex] is retrieved only if it's in the principal range of the inverse function.<\/li>\r\n\t<li><strong>Verify<\/strong>: Plug the result back into the original function to confirm the outcome.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170572549644\">Evaluate each of the following expressions.<\/p>\r\n<ol id=\"fs-id1170572549647\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>[latex]\\sin^{-1}\\left(-\\frac{\\sqrt{3}}{2}\\right)[\/latex]<\/li>\r\n\t<li>[latex] \\tan \\left(\\tan^{-1}\\left(-\\frac{1}{\\sqrt{3}}\\right)\\right)[\/latex]<\/li>\r\n\t<li>[latex]\\cos^{-1}\\left( \\cos \\left(\\frac{5\\pi}{4}\\right)\\right)[\/latex]<\/li>\r\n\t<li>[latex]\\sin^{-1}\\left( \\cos \\left(\\frac{2\\pi}{3}\\right)\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"fs-id1170572549361\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170572549361\"]<\/p>\r\n<ol id=\"fs-id1170572549361\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>Evaluating [latex]\\sin^{-1}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)[\/latex] is equivalent to finding the angle [latex]\\theta [\/latex] such that [latex] \\sin \\theta =\u2212\\frac{\\sqrt{3}}{2}[\/latex] and [latex]\u2212\\frac{\\pi}{2} \\le \\theta \\le \\frac{\\pi}{2}[\/latex]. <br \/>\r\n<br \/>\r\nThe angle [latex]\\theta =\u2212\\frac{\\pi}{3}[\/latex] satisfies these two conditions. Therefore, [latex]\\sin^{-1}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)=\u2212\\frac{\\pi}{3}[\/latex].<\/li>\r\n\t<li>First we use the fact that [latex]\\tan^{-1}\\left(-\\frac{1}{\\sqrt{3}}\\right)=\u2212\\frac{\\pi}{6}[\/latex]. Then [latex] \\tan \\left(\\frac{\\pi}{6}\\right)=-\\frac{1}{\\sqrt{3}}[\/latex]. <br \/>\r\n<br \/>\r\nTherefore, [latex] \\tan \\left(\\tan^{-1}\\left(-\\frac{1}{\\sqrt{3}}\\right)\\right)=-\\frac{1}{\\sqrt{3}}[\/latex].<\/li>\r\n\t<li>To evaluate [latex]\\cos^{-1}\\left( \\cos \\left(\\frac{5\\pi}{4}\\right)\\right)[\/latex], first use the fact that [latex] \\cos \\left(\\frac{5\\pi}{4}\\right)=\u2212\\frac{\\sqrt{2}}{2}[\/latex]. Then we need to find the angle [latex]\\theta [\/latex] such that [latex] \\cos (\\theta )=\u2212\\frac{\\sqrt{2}}{2}[\/latex] and [latex]0 \\le \\theta \\le \\pi[\/latex]. <br \/>\r\n<br \/>\r\nSince [latex]\\frac{3\\pi}{4}[\/latex] satisfies both these conditions, we have [latex] \\cos \\left(\\cos^{-1}\\left(\\frac{5\\pi}{4}\\right)\\right)= \\cos \\left(\\cos^{-1}\\left(\u2212\\frac{\\sqrt{2}}{2}\\right)\\right)=\\frac{3\\pi}{4}[\/latex].<\/li>\r\n\t<li>Since [latex] \\cos \\left(\\frac{2\\pi}{3}\\right)=-\\frac{1}{2}[\/latex], we need to evaluate [latex]\\sin^{-1}\\left(-\\frac{1}{2}\\right)[\/latex]. That is, we need to find the angle [latex]\\theta [\/latex] such that [latex] \\sin (\\theta )=-\\frac{1}{2}[\/latex] and [latex]\u2212\\frac{\\pi}{2} \\le \\theta \\le \\frac{\\pi}{2}[\/latex]. <br \/>\r\n<br \/>\r\nSince [latex]\u2212\\frac{\\pi}{6}[\/latex] satisfies both these conditions, we can conclude that [latex]\\sin^{-1}\\left( \\cos \\left(\\frac{2\\pi}{3}\\right)\\right)=\\sin^{-1}\\left(-\\frac{1}{2}\\right)=\u2212\\frac{\\pi}{6}[\/latex].<\/li>\r\n<\/ol>\r\n\r\nWatch the following video to see the worked solution to this example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/JFQ8maupdT8?controls=0&amp;start=934&amp;end=1240&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\n<p>You can view the transcript for this video using <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.4InverseFunctions934to1240_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]197034[\/ohm_question]<\/p>\r\n<\/section>\r\n<div id=\"fs-id1170572169395\" class=\"textbox tryit\">\r\n<h3>Activity: The Maximum Value of a Function<\/h3>\r\n<p id=\"fs-id1170572169402\">In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don\u2019t know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.<\/p>\r\n<p id=\"fs-id1170572169416\">This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable [latex]x[\/latex].<\/p>\r\n<ol id=\"fs-id1170572169426\">\r\n\t<li>Consider the graph in Figure 17 of the function [latex]y= \\sin x + \\cos x[\/latex]. Describe its overall shape. Is it periodic? How do you know?\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202607\/CNX_Calc_Figure_01_04_016.jpg\" alt=\"An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of the function \u201cy = sin(x) + cos(x)\u201d, a curved wave function. The graph of the function decreases until it hits the approximate point (-(3pi\/4), -1.4), where it increases until the approximate point ((pi\/4), 1.4), where it begins to decrease again. The x intercepts shown on this graph of the function are at (-(5pi\/4), 0), (-(pi\/4), 0), and ((3pi\/4), 0). The y intercept is at (0, 1).\" width=\"325\" height=\"308\" \/> Figure 17. The graph of [latex]y= \\sin x + \\cos x[\/latex].[\/caption]\r\n<br \/>\r\nUsing a graphing calculator or other graphing device, estimate the [latex]x[\/latex]- and [latex]y[\/latex]-values of the maximum point for the graph (the first such point where [latex]x&gt;0[\/latex]). It may be helpful to express the [latex]x[\/latex]-value as a multiple of [latex]\\pi[\/latex].<\/li>\r\n\t<li>Now consider other graphs of the form [latex]y=A \\sin x + B \\cos x[\/latex] for various values of [latex]A[\/latex] and [latex]B[\/latex]. Sketch the graph when [latex]A = 2[\/latex] and [latex]B = 1[\/latex], and find the [latex]x[\/latex]- and [latex]y[\/latex]-values for the maximum point. (Remember to express the [latex]x[\/latex]-value as a multiple of [latex]\\pi[\/latex], if possible.) Has it moved?<\/li>\r\n\t<li>Repeat for [latex]A = 1, \\, B = 2[\/latex]. Is there any relationship to what you found in part (2)?<\/li>\r\n\t<li>Complete the following table, adding a few choices of your own for [latex]A[\/latex] and [latex]B[\/latex]:<br \/>\r\n<table id=\"fs-id1170572554057\" class=\"unnumbered\" summary=\"A table containing 8 columns and 9 rows is shown. The first column is labeled \u201cA\u201d and contains the values \u201c0,1,1,1,2,2,3, and 4.\u201d The second column is labeled \u201cB\u201d and contains the values \u201c1,0,1,2,1,2,4, and 3.\u201d The third column is labeled \u201cx\u201d and has no values for any of the rows. The fourth column is labeled \u201cy\u201d and contains no values for any of the rows. The fifth column is separated from the fourth column by a gutter, is labeled \u201cA\u201d and contains the values \u201cthe square root of 3,1,12, and 5.\u201d The sixth column is labeled \u201cB\u201d and contains the values 1, the square root of 3,5, and 12.\u201d The seventh column is labeled \u201cx\u201d and contains no values for any of the rows. The eighth column is labeled \u201cy\u201d and contains no values for any of the rows.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]A[\/latex]<\/th>\r\n<th>[latex]B[\/latex]<\/th>\r\n<th>[latex]x[\/latex]<\/th>\r\n<th>[latex]y[\/latex]<\/th>\r\n<th>\u00a0<\/th>\r\n<th>[latex]A[\/latex]<\/th>\r\n<th>[latex]B[\/latex]<\/th>\r\n<th>[latex]x[\/latex]<\/th>\r\n<th>[latex]y[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td rowspan=\"8\">\u00a0<\/td>\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1<\/td>\r\n<td>0<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1<\/td>\r\n<td>1<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>12<\/td>\r\n<td>5<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>5<\/td>\r\n<td>12<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2<\/td>\r\n<td>1<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2<\/td>\r\n<td>2<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4<\/td>\r\n<td>3<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n\t<li>Try to figure out the formula for the [latex]y[\/latex]-values.<\/li>\r\n\t<li>The formula for the [latex]x[\/latex]-values is a little harder. The most helpful points from the table are [latex](1,1), \\, (1,\\sqrt{3}), \\, (\\sqrt{3},1)[\/latex]. (<em>Hint<\/em>: <em>Consider inverse trigonometric functions.)<\/em><\/li>\r\n\t<li>If you found formulas for parts (5) and (6), show that they work together. That is, substitute the [latex]x[\/latex]-value formula you found into [latex]y=A \\sin x + B \\cos x[\/latex] and simplify it to arrive at the [latex]y[\/latex]-value formula you found.<\/li>\r\n<\/ol>\r\n<\/div>","rendered":"<h2>Inverse Trigonometric Functions<\/h2>\n<p id=\"fs-id1170572478796\">The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse.<\/p>\n<p>Consider the sine function. The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}][\/latex]. By doing so, we define the inverse sine function on the domain [latex][-1,1][\/latex] such that for any [latex]x[\/latex] in the interval [latex][-1,1][\/latex], the inverse sine function tells us which angle [latex]\\theta[\/latex] in the interval [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}][\/latex] satisfies [latex]\\sin \\theta =x[\/latex].<\/p>\n<p>Similarly, we can restrict the domains of the other trigonometric functions to define <strong>inverse trigonometric functions<\/strong>, which are functions that tell us which angle in a certain interval has a specified trigonometric value.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>inverse trigonometric functions<\/h3>\n<p id=\"fs-id1170572547399\">The <strong>inverse sine function<\/strong>, denoted [latex]\\sin^{-1}[\/latex] or arcsin, and the inverse cosine function, denoted [latex]\\cos^{-1}[\/latex] or arccos, are defined on the domain [latex]D=\\{x|-1 \\le x \\le 1\\}[\/latex] as follows:<\/p>\n<div id=\"fs-id1170572547453\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{c}\\sin^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\sin (y)=x \\, \\text{and} \\, -\\frac{\\pi}{2} \\le y \\le \\frac{\\pi}{2};\\hfill \\\\ \\cos^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\cos (y)=x \\, \\text{and} \\, 0 \\le y \\le \\pi \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572550974\">The <strong>inverse tangent function<\/strong>, denoted [latex]\\tan^{-1}[\/latex] or arctan, and inverse cotangent function, denoted [latex]\\cot^{-1}[\/latex] or arccot, are defined on the domain [latex]D=\\{x|-\\infty < x < \\infty \\}[\/latex] as follows:<\/p>\n<div id=\"fs-id1170572551028\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{c}\\tan^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\tan (y)=x \\, \\text{and} \\, -\\frac{\\pi}{2} \\le y \\le \\frac{\\pi}{2};\\hfill \\\\ \\cot^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\cot (y)=x \\, \\text{and} \\, 0 \\le y \\le \\pi \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572434001\">The<strong> inverse cosecant function<\/strong>, denoted [latex]\\csc^{-1}[\/latex] or arccsc, and inverse secant function, denoted [latex]\\sec^{-1}[\/latex] or arcsec, are defined on the domain [latex]D=\\{x| \\, |x| \\ge 1\\}[\/latex] as follows:<\/p>\n<div id=\"fs-id1170572548730\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{c}\\csc^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\csc (y)=x \\, \\text{and} \\, -\\frac{\\pi}{2} \\le y \\le \\frac{\\pi}{2}, \\, y\\ne 0;\\hfill \\\\ \\sec^{-1}(x)=y \\,\\, \\text{if and only if} \\, \\sec (y)=x \\, \\text{and} \\, 0 \\le y \\le \\pi, \\, y \\ne \\frac{\\pi}{2}\\hfill \\end{array}[\/latex]<\/div>\n<\/section>\n<h3>Graphs of Inverse Trigonometric Functions<\/h3>\n<p id=\"fs-id1170572551163\">To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line [latex]y=x[\/latex] (Figure 16).<\/p>\n<figure style=\"width: 851px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202603\/CNX_Calc_Figure_01_04_018.jpg\" alt=\"An image of six graphs. The first graph is of the function \u201cf(x) = sin inverse(x)\u201d, which is an increasing curve function. The function starts at the point (-1, -(pi\/2)) and increases until it ends at the point (1, (pi\/2)). The x intercept and y intercept are at the origin. The second graph is of the function \u201cf(x) = cos inverse (x)\u201d, which is a decreasing curved function. The function starts at the point (-1, pi) and decreases until it ends at the point (1, 0). The x intercept is at the point (1, 0). The y intercept is at the point (0, (pi\/2)). The third graph is of the function f(x) = tan inverse (x)\u201d, which is an increasing curve function. The function starts close to the horizontal line \u201cy = -(pi\/2)\u201d and increases until it comes close the \u201cy = (pi\/2)\u201d. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The x intercept and y intercept are both at the origin. The fourth graph is of the function \u201cf(x) = cot inverse (x)\u201d, which is a decreasing curved function. The function starts slightly below the horizontal line \u201cy = pi\u201d and decreases until it gets close the x axis. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The fifth graph is of the function \u201cf(x) = csc inverse (x)\u201d, a decreasing curved function. The function starts slightly below the x axis, then decreases until it hits a closed circle point at (-1, -(pi\/2)). The function then picks up again at the point (1, (pi\/2)), where is begins to decrease and approach the x axis, without ever touching the x axis. There is a horizontal asymptote at the x axis. The sixth graph is of the function \u201cf(x) = sec inverse (x)\u201d, an increasing curved function. The function starts slightly above the horizontal line \u201cy = (pi\/2)\u201d, then increases until it hits a closed circle point at (-1, pi). The function then picks up again at the point (1, 0), where is begins to increase and approach the horizontal line \u201cy = (pi\/2)\u201d, without ever touching the line. There is a horizontal asymptote at the \u201cy = (pi\/2)\u201d.\" width=\"851\" height=\"714\" \/><figcaption class=\"wp-caption-text\">Figure 16. The graph of each of the inverse trigonometric functions is a reflection about the line [latex]y=x[\/latex] of the corresponding restricted trigonometric function.<\/figcaption><\/figure>\n<p id=\"fs-id1170572366201\">When evaluating an inverse trigonometric function, the output is an angle.<\/p>\n<p>For example, to evaluate [latex]\\cos^{-1}(\\frac{1}{2})[\/latex], we need to find an angle [latex]\\theta[\/latex] such that [latex]\\cos \\theta =\\frac{1}{2}[\/latex]. Clearly, many angles have this property. However, given the definition of [latex]\\cos^{-1}[\/latex], we need the angle [latex]\\theta[\/latex] that not only solves this equation, but also lies in the interval [latex][0,\\pi][\/latex]. We conclude that [latex]\\cos^{-1}(\\frac{1}{2})=\\frac{\\pi}{3}[\/latex].<\/p>\n<section class=\"textbox proTip\">Knowing the common values of sine and cosine for key angles can simplify the process of evaluating inverse trigonometric functions, making it a smoother and quicker task.<\/p>\n<table id=\"Table_05_02_01\" summary=\"..\">\n<tbody>\n<tr>\n<td><strong>Angle<\/strong><\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{6}[\/latex], or [latex]30\u00b0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{4}[\/latex], or [latex]45\u00b0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{3}[\/latex], or [latex]60\u00b0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex], or [latex]90\u00b0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Cosine<\/strong><\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Sine<\/strong><\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<h3>Compositions of Inverse Trigonometric Functions<\/h3>\n<p id=\"fs-id1170572472189\">When working with inverse trigonometric functions, it&#8217;s crucial to understand how composition works. For instance, when we compose [latex]\\sin[\/latex] and its inverse [latex]\\sin^{-1}[\/latex], such as in [latex]\\sin{(\\sin^{-1}(y))}[\/latex], we&#8217;re essentially undoing the sine function, which should give us the original input, [latex]y[\/latex]. However, this holds true only if [latex]y[\/latex] falls within the range of [latex]\\sin^{-1}[\/latex], which is [latex][1,1][\/latex]. So [latex]\\sin{(\\sin^{-1}(y))}=y[\/latex] for [latex]-1 \\le y \\le 1[\/latex]. Conversely, when we consider [latex]\\sin^{-1}({\\sin(x)})[\/latex], the result is [latex]x[\/latex] only if [latex]x[\/latex] is within the restricted domain of [latex]\\sin^{-1}[\/latex], which is [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}][\/latex]. The same principle applies to [latex]\\cos[\/latex] and its inverse.<\/p>\n<section class=\"textbox example\">\n<p>For example, consider the two expressions [latex]\\sin (\\sin^{-1}(\\frac{\\sqrt{2}}{2}))[\/latex] and [latex]\\sin^{-1}(\\sin(\\pi))[\/latex]. For the first one, we simplify as follows:<\/p>\n<div id=\"fs-id1170572472267\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin (\\sin^{-1}(\\frac{\\sqrt{2}}{2}))= \\sin (\\frac{\\pi}{4})=\\frac{\\sqrt{2}}{2}[\/latex]<\/div>\n<p>For the second one, we have<\/p>\n<div style=\"text-align: center;\">[latex]\\sin^{-1}( \\sin (\\pi))=\\sin^{-1}(0)=0[\/latex]<\/div>\n<p>This is because the value of [latex]\u03c0[\/latex] falls outside the restricted range of the inverse sine function, which is [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}][\/latex].<\/section>\n<p id=\"fs-id1170572089863\">To summarize,<\/p>\n<div id=\"fs-id1170572550786\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin (\\sin^{-1}{y})=y \\, \\text{ if } \\, -1 \\le y \\le 1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572550837\">and<\/p>\n<div id=\"fs-id1170572550840\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin^{-1}( \\sin {x})=x \\, \\text{ if } \\, -\\frac{\\pi}{2} \\le x \\le \\frac{\\pi}{2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572550900\">Similarly, for the cosine function,<\/p>\n<div id=\"fs-id1170572550903\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cos (\\cos^{-1}{y})=y \\, \\text{ if } \\, -1 \\le y \\le 1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572549577\">and<\/p>\n<div id=\"fs-id1170572549580\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cos^{-1}( \\cos {x})=x \\, \\text{ if } 0 \\le x \\le \\pi[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572549631\">Similar properties hold for the other trigonometric functions and their inverses.<\/p>\n<section class=\"textbox questionHelp\">\n<p><strong>How to: <\/strong><strong>Composing Inverse Trig Functions<\/strong><\/p>\n<ol>\n<li><strong>Check the Range<\/strong>: Ensure the value inside the inverse function is within the inverse function&#8217;s range. For [latex]\\sin^{-1}[\/latex], the value must be between [latex]-1[\/latex] and [latex]1[\/latex].<\/li>\n<li><strong>Apply the Function<\/strong>: Perform the composition by applying the inverse function first.<\/li>\n<li><strong>Reverse the Process<\/strong>: Apply the original trigonometric function to the result.<\/li>\n<li><strong>Restrict the Range<\/strong>: Remember that for [latex]\\sin{(\\sin^{-1}({x}))}[\/latex] and [latex]\\cos{(\\cos^{-1}({x}))}[\/latex], the original [latex]x[\/latex] is retrieved only if it&#8217;s in the principal range of the inverse function.<\/li>\n<li><strong>Verify<\/strong>: Plug the result back into the original function to confirm the outcome.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572549644\">Evaluate each of the following expressions.<\/p>\n<ol id=\"fs-id1170572549647\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\sin^{-1}\\left(-\\frac{\\sqrt{3}}{2}\\right)[\/latex]<\/li>\n<li>[latex]\\tan \\left(\\tan^{-1}\\left(-\\frac{1}{\\sqrt{3}}\\right)\\right)[\/latex]<\/li>\n<li>[latex]\\cos^{-1}\\left( \\cos \\left(\\frac{5\\pi}{4}\\right)\\right)[\/latex]<\/li>\n<li>[latex]\\sin^{-1}\\left( \\cos \\left(\\frac{2\\pi}{3}\\right)\\right)[\/latex]<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572549361\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572549361\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572549361\" style=\"list-style-type: lower-alpha;\">\n<li>Evaluating [latex]\\sin^{-1}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)[\/latex] is equivalent to finding the angle [latex]\\theta[\/latex] such that [latex]\\sin \\theta =\u2212\\frac{\\sqrt{3}}{2}[\/latex] and [latex]\u2212\\frac{\\pi}{2} \\le \\theta \\le \\frac{\\pi}{2}[\/latex].\n<p>The angle [latex]\\theta =\u2212\\frac{\\pi}{3}[\/latex] satisfies these two conditions. Therefore, [latex]\\sin^{-1}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)=\u2212\\frac{\\pi}{3}[\/latex].<\/li>\n<li>First we use the fact that [latex]\\tan^{-1}\\left(-\\frac{1}{\\sqrt{3}}\\right)=\u2212\\frac{\\pi}{6}[\/latex]. Then [latex]\\tan \\left(\\frac{\\pi}{6}\\right)=-\\frac{1}{\\sqrt{3}}[\/latex].\n<p>Therefore, [latex]\\tan \\left(\\tan^{-1}\\left(-\\frac{1}{\\sqrt{3}}\\right)\\right)=-\\frac{1}{\\sqrt{3}}[\/latex].<\/li>\n<li>To evaluate [latex]\\cos^{-1}\\left( \\cos \\left(\\frac{5\\pi}{4}\\right)\\right)[\/latex], first use the fact that [latex]\\cos \\left(\\frac{5\\pi}{4}\\right)=\u2212\\frac{\\sqrt{2}}{2}[\/latex]. Then we need to find the angle [latex]\\theta[\/latex] such that [latex]\\cos (\\theta )=\u2212\\frac{\\sqrt{2}}{2}[\/latex] and [latex]0 \\le \\theta \\le \\pi[\/latex].\n<p>Since [latex]\\frac{3\\pi}{4}[\/latex] satisfies both these conditions, we have [latex]\\cos \\left(\\cos^{-1}\\left(\\frac{5\\pi}{4}\\right)\\right)= \\cos \\left(\\cos^{-1}\\left(\u2212\\frac{\\sqrt{2}}{2}\\right)\\right)=\\frac{3\\pi}{4}[\/latex].<\/li>\n<li>Since [latex]\\cos \\left(\\frac{2\\pi}{3}\\right)=-\\frac{1}{2}[\/latex], we need to evaluate [latex]\\sin^{-1}\\left(-\\frac{1}{2}\\right)[\/latex]. That is, we need to find the angle [latex]\\theta[\/latex] such that [latex]\\sin (\\theta )=-\\frac{1}{2}[\/latex] and [latex]\u2212\\frac{\\pi}{2} \\le \\theta \\le \\frac{\\pi}{2}[\/latex].\n<p>Since [latex]\u2212\\frac{\\pi}{6}[\/latex] satisfies both these conditions, we can conclude that [latex]\\sin^{-1}\\left( \\cos \\left(\\frac{2\\pi}{3}\\right)\\right)=\\sin^{-1}\\left(-\\frac{1}{2}\\right)=\u2212\\frac{\\pi}{6}[\/latex].<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/JFQ8maupdT8?controls=0&amp;start=934&amp;end=1240&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the transcript for this video using <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.4InverseFunctions934to1240_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm197034\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=197034&theme=lumen&iframe_resize_id=ohm197034&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<div id=\"fs-id1170572169395\" class=\"textbox tryit\">\n<h3>Activity: The Maximum Value of a Function<\/h3>\n<p id=\"fs-id1170572169402\">In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don\u2019t know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.<\/p>\n<p id=\"fs-id1170572169416\">This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable [latex]x[\/latex].<\/p>\n<ol id=\"fs-id1170572169426\">\n<li>Consider the graph in Figure 17 of the function [latex]y= \\sin x + \\cos x[\/latex]. Describe its overall shape. Is it periodic? How do you know?<br \/>\n<figure style=\"width: 325px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202607\/CNX_Calc_Figure_01_04_016.jpg\" alt=\"An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of the function \u201cy = sin(x) + cos(x)\u201d, a curved wave function. The graph of the function decreases until it hits the approximate point (-(3pi\/4), -1.4), where it increases until the approximate point ((pi\/4), 1.4), where it begins to decrease again. The x intercepts shown on this graph of the function are at (-(5pi\/4), 0), (-(pi\/4), 0), and ((3pi\/4), 0). The y intercept is at (0, 1).\" width=\"325\" height=\"308\" \/><figcaption class=\"wp-caption-text\">Figure 17. The graph of [latex]y= \\sin x + \\cos x[\/latex].<\/figcaption><\/figure>\n<p>\nUsing a graphing calculator or other graphing device, estimate the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-values of the maximum point for the graph (the first such point where [latex]x>0[\/latex]). It may be helpful to express the [latex]x[\/latex]-value as a multiple of [latex]\\pi[\/latex].<\/li>\n<li>Now consider other graphs of the form [latex]y=A \\sin x + B \\cos x[\/latex] for various values of [latex]A[\/latex] and [latex]B[\/latex]. Sketch the graph when [latex]A = 2[\/latex] and [latex]B = 1[\/latex], and find the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-values for the maximum point. (Remember to express the [latex]x[\/latex]-value as a multiple of [latex]\\pi[\/latex], if possible.) Has it moved?<\/li>\n<li>Repeat for [latex]A = 1, \\, B = 2[\/latex]. Is there any relationship to what you found in part (2)?<\/li>\n<li>Complete the following table, adding a few choices of your own for [latex]A[\/latex] and [latex]B[\/latex]:<br \/>\n<table id=\"fs-id1170572554057\" class=\"unnumbered\" summary=\"A table containing 8 columns and 9 rows is shown. The first column is labeled \u201cA\u201d and contains the values \u201c0,1,1,1,2,2,3, and 4.\u201d The second column is labeled \u201cB\u201d and contains the values \u201c1,0,1,2,1,2,4, and 3.\u201d The third column is labeled \u201cx\u201d and has no values for any of the rows. The fourth column is labeled \u201cy\u201d and contains no values for any of the rows. The fifth column is separated from the fourth column by a gutter, is labeled \u201cA\u201d and contains the values \u201cthe square root of 3,1,12, and 5.\u201d The sixth column is labeled \u201cB\u201d and contains the values 1, the square root of 3,5, and 12.\u201d The seventh column is labeled \u201cx\u201d and contains no values for any of the rows. The eighth column is labeled \u201cy\u201d and contains no values for any of the rows.\">\n<thead>\n<tr valign=\"top\">\n<th>[latex]A[\/latex]<\/th>\n<th>[latex]B[\/latex]<\/th>\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]y[\/latex]<\/th>\n<th>\u00a0<\/th>\n<th>[latex]A[\/latex]<\/th>\n<th>[latex]B[\/latex]<\/th>\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]y[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>0<\/td>\n<td>1<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td rowspan=\"8\">\u00a0<\/td>\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<td>1<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1<\/td>\n<td>0<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>1<\/td>\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1<\/td>\n<td>1<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>12<\/td>\n<td>5<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1<\/td>\n<td>2<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>5<\/td>\n<td>12<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2<\/td>\n<td>1<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2<\/td>\n<td>2<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3<\/td>\n<td>4<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4<\/td>\n<td>3<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Try to figure out the formula for the [latex]y[\/latex]-values.<\/li>\n<li>The formula for the [latex]x[\/latex]-values is a little harder. The most helpful points from the table are [latex](1,1), \\, (1,\\sqrt{3}), \\, (\\sqrt{3},1)[\/latex]. (<em>Hint<\/em>: <em>Consider inverse trigonometric functions.)<\/em><\/li>\n<li>If you found formulas for parts (5) and (6), show that they work together. That is, substitute the [latex]x[\/latex]-value formula you found into [latex]y=A \\sin x + B \\cos x[\/latex] and simplify it to arrive at the [latex]y[\/latex]-value formula you found.<\/li>\n<\/ol>\n<\/div>\n","protected":false},"author":6,"menu_order":14,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"1.4 Inverse Functions\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":759,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"1.4 Inverse Functions","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/162"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":21,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/162\/revisions"}],"predecessor-version":[{"id":4477,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/162\/revisions\/4477"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/759"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/162\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=162"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=162"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=162"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=162"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}