{"id":14188,"date":"2018-09-27T16:40:26","date_gmt":"2018-09-27T16:40:26","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/inverse-trigonometric-functions\/"},"modified":"2021-02-04T22:40:18","modified_gmt":"2021-02-04T22:40:18","slug":"inverse-trigonometric-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/chapter\/inverse-trigonometric-functions\/","title":{"raw":"Walkthrough of Unit 5: Inverse Trig Functions and Solving Trig Equations","rendered":"Walkthrough of Unit 5: Inverse Trig Functions and Solving Trig Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li style=\"font-weight: 400\">Understand and use the inverse sine, cosine, and tangent functions.<\/li>\r\n \t<li style=\"font-weight: 400\">Find the exact value of expressions involving the inverse sine, cosine, and tangent functions.<\/li>\r\n \t<li style=\"font-weight: 400\">Use a calculator to evaluate inverse trigonometric functions.<\/li>\r\n \t<li>Use inverse trigonometric functions to solve right triangles.<\/li>\r\n \t<li style=\"font-weight: 400\">Find exact values of composite functions with inverse trigonometric functions.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Understanding and Using the Inverse Sine, Cosine, and Tangent Functions<\/h2>\r\nIn order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function \u201cundoes\u201d what the original trigonometric function \u201cdoes,\u201d as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in Figure 1.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27163959\/CNX_Precalc_Figure_06_03_013.jpg\" alt=\"A chart that says \u201cTrig Functinos\u201d, \u201cInverse Trig Functions\u201d, \u201cDomain: Measure of an angle\u201d, \u201cDomain: Ratio\u201d, \u201cRange: Ratio\u201d, and \u201cRange: Measure of an angle\u201d.\" width=\"731\" height=\"78\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\nFor example, if [latex]f(x)=\\sin x[\/latex], then we would write\u00a0[latex]f^{1}(x)={\\sin}^{-1}{x}[\/latex]. Be aware that [latex]{\\sin}^{-1}x[\/latex] does not mean [latex]\\frac{1}{\\sin{x}}[\/latex]. The following examples illustrate the inverse trigonometric functions:\r\n<ul>\r\n \t<li>Since [latex]\\sin\\left(\\frac{\\pi}{6}\\right)=\\frac{1}{2}[\/latex], then [latex]\\frac{\\pi}{6}=\\sin^{\u22121}(\\frac{1}{2})[\/latex].<\/li>\r\n \t<li>Since [latex]\\cos(\\pi)=\u22121[\/latex], then [latex]\\pi=\\cos^{\u22121}(\u22121)[\/latex].<\/li>\r\n \t<li>Since [latex]\\tan\\left(\\frac{\\pi}{4}\\right)=1[\/latex], then [latex]\\frac{\\pi}{4}=\\tan^{\u22121}(1)[\/latex].<\/li>\r\n<\/ul>\r\nIn previous sections, we evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that, for a <strong>one-to-one function<\/strong>, if [latex]f(a)=b[\/latex], then an inverse function would satisfy [latex]f^{\u22121}(b)=a[\/latex].\r\n\r\nBear in mind that the sine, cosine, and tangent functions are not one-to-one functions. The graph of each function would fail the horizontal line test. In fact, no periodic function can be one-to-one because each output in its range corresponds to at least one input in every period, and there are an infinite number of periods. As with other functions that are not one-to-one, we will need to restrict the <strong>domain<\/strong> of each function to yield a new function that is one-to-one. We choose a domain for each function that includes the number 0. Figure 2\u00a0shows the graph of the sine function limited to [latex]\\left[\\frac{\u2212\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex] and the graph of the cosine function limited to [0, \u03c0].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164001\/CNX_Precalc_Figure_06_03_001.jpg\" alt=\"Two side-by-side graphs. The first graph, graph A, shows half of a period of the function sine of x. The second graph, graph B, shows half a period of the function cosine of x.\" \/>\r\n<p style=\"text-align: center\"><strong>Figure 2.<\/strong> (a) Sine function on a restricted domain of [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex]; (b) Cosine function on a restricted domain of [0, \u03c0]<\/p>\r\nFigure 3\u00a0shows the graph of the tangent function limited to [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164003\/CNX_Precalc_Figure_06_03_003.jpg\" alt=\"A graph of one period of tangent of x, from -pi\/2 to pi\/2.\" \/>\r\n<p style=\"text-align: center\"><strong>Figure 3.\u00a0<\/strong>Tangent function on a restricted domain of [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex]<\/p>\r\nThese conventional choices for the restricted domain are somewhat arbitrary, but they have important, helpful characteristics. Each domain includes the origin and some positive values, and most importantly, each results in a one-to-one function that is invertible. The conventional choice for the restricted domain of the tangent function also has the useful property that it extends from one <strong>vertical asymptote<\/strong> to the next instead of being divided into two parts by an asymptote.\r\n\r\nOn these restricted domains, we can define the <strong>inverse trigonometric functions<\/strong>.\r\n<ul>\r\n \t<li>The <strong>inverse sine function<\/strong>\u00a0[latex]y=\\sin^{\u22121}x[\/latex] means [latex]x=\\sin y[\/latex]. The inverse sine function is sometimes called the <strong>arcsine<\/strong> function, and notated arcsin <em>x<\/em>.\r\n<div>\r\n<div style=\"text-align: center\">[latex]y=\\sin^{\u22121}x[\/latex] has\u00a0domain [\u22121, 1] and\u00a0range [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex]<\/div>\r\n<\/div><\/li>\r\n \t<li>The <strong>inverse cosine function<\/strong>\u00a0[latex]y=\\cos^{\u22121}x[\/latex] means [latex]x=\\cos y[\/latex]. The inverse cosine function is sometimes called the <strong>arccosine<\/strong> function, and notated arccos <em>x<\/em>.\r\n<div>\r\n<div style=\"text-align: center\">[latex]y=\\cos^{\u22121}x[\/latex] has\u00a0domain [\u22121, 1] and\u00a0range [0, \u03c0]<\/div>\r\n<\/div><\/li>\r\n \t<li>The <strong>inverse tangent function<\/strong>\u00a0[latex]y=\\tan^{\u22121}x[\/latex] means [latex]x=\\tan y[\/latex]. The inverse tangent function is sometimes called the <strong>arctangent<\/strong> function, and notated arctan <em>x<\/em>.\r\n<div>\r\n<div style=\"text-align: center\">[latex]y=\\tan^{\u22121}x[\/latex] has\u00a0domain (\u2212\u221e, \u221e) and\u00a0range [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex]<\/div>\r\n<\/div><\/li>\r\n<\/ul>\r\nThe graphs of the inverse functions are shown in Figure 4, Figure 5, and Figure 6. Notice that the output of each of these inverse functions is a <em>number, <\/em>an angle in radian measure. We see that [latex]\\sin^{\u22121}x[\/latex] has domain [\u22121, 1] and range [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], [latex]\\cos^{\u22121}x[\/latex] has domain [\u22121, 1] and range [0, \u03c0], and [latex]\\tan^{\u22121}x[\/latex] has domain of all real numbers and range [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex]. To find the <strong>domain<\/strong> and <strong>range<\/strong> of inverse trigonometric functions, switch the domain and range of the original functions. Each graph of the inverse trigonometric function is a reflection of the graph of the original function about the line [latex]y=x[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164005\/CNX_Precalc_Figure_06_03_004n.jpg\" alt=\"A graph of the functions of sine of x and arc sine of x. There is a dotted line y=x between the two graphs, to show inverse nature of the two functions\" width=\"731\" height=\"433\" \/> <b>Figure 4.<\/b>\u00a0The sine function and inverse sine (or arcsine) function[\/caption]\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164008\/CNX_Precalc_Figure_06_03_005n.jpg\" alt=\"A graph of the functions of cosine of x and arc cosine of x. There is a dotted line at y=x to show the inverse nature of the two functions.\" width=\"487\" height=\"343\" \/> <b>Figure 5.<\/b> The cosine function and inverse cosine (or arccosine) function[\/caption]\r\n\r\n&nbsp;\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164010\/CNX_Precalc_Figure_06_03_006n.jpg\" alt=\"A graph of the functions of tangent of x and arc tangent of x. There is a dotted line at y=x to show the inverse nature of the two functions.\" width=\"487\" height=\"433\" \/> <b>Figure 6.<\/b> The tangent function and inverse tangent (or arctangent) function[\/caption]\r\n\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>A General Note: Relations for Inverse Sine, Cosine, and Tangent Functions<\/h3>\r\nFor angles in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], if [latex]\\sin y=x[\/latex], then [latex]\\sin^{\u22121}x=y[\/latex].\r\n\r\nFor angles in the interval [0, \u03c0], if [latex]\\cos y=x[\/latex], then [latex]\\cos^{\u22121}x=y[\/latex].\r\n\r\nFor angles in the interval [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex], if [latex]\\tan y=x[\/latex], then [latex]\\tan^{\u22121}x=y[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Writing a Relation for an Inverse Function<\/h3>\r\nGiven [latex]\\sin\\left(\\frac{5\\pi}{12}\\right)\\approx 0.96593[\/latex], write a relation involving the inverse sine.\r\n\r\n[reveal-answer q=\"641490\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"641490\"]\r\n\r\nUse the relation for the inverse sine. If [latex]\\sin y=x[\/latex], then [latex]\\sin^{\u22121}x=y[\/latex].\r\n\r\nIn this problem, [latex]x=0.96593[\/latex], and [latex]y=\\frac{5\\pi}{12}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\sin^{\u22121}(0.96593)\\approx \\frac{5\\pi}{12}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]\\cos(0.5)\\approx 0.8776[\/latex], write a relation involving the inverse cosine.\r\n\r\n[reveal-answer q=\"359839\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"359839\"]\r\n\r\n[latex]\\arccos(0.8776)\\approx0.5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding the Exact Value of Expressions Involving the Inverse Sine, Cosine, and Tangent Functions<\/h2>\r\nNow that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the inverse functions when we are using the special angles, specifically [latex]\\frac{\\pi}{ 6} (30^\\circ)\\text{, }\\frac{\\pi}{ 4} (45^\\circ),\\text{ and } \\frac{\\pi}{ 3} (60^\\circ)[\/latex], and their reflections into other quadrants.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a \u201cspecial\u201d input value, evaluate an inverse trigonometric function.<\/h3>\r\n<ol>\r\n \t<li>Find angle\u00a0<em>x<\/em>\u00a0for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function.<\/li>\r\n \t<li>If\u00a0<em>x<\/em>\u00a0is not in the defined range of the inverse, find another angle\u00a0<em>y<\/em>\u00a0that is in the defined range and has the same sine, cosine, or tangent as\u00a0<em>x<\/em>, depending on which corresponds to the given inverse function.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Evaluating Inverse Trigonometric Functions for Special Input Values<\/h3>\r\nEvaluate each of the following.\r\n<p style=\"padding-left: 60px\">a. [latex]\\sin\u22121\\left(\\frac{1}{2}\\right)[\/latex]<\/p>\r\n<p style=\"padding-left: 60px\">b. [latex]\\sin\u22121\\left(\u2212\\frac{2}{\\sqrt{2}}\\right)[\/latex]<\/p>\r\n<p style=\"padding-left: 60px\">c. [latex]\\cos\u22121\\left(\u2212\\frac{3}{\\sqrt{2}}\\right)[\/latex]<\/p>\r\n<p style=\"padding-left: 60px\">d. [latex]\\tan^{\u2212 1}(1)[\/latex]<\/p>\r\n[reveal-answer q=\"666370\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"666370\"]\r\n<p style=\"padding-left: 60px\">a. Evaluating [latex]\\sin^{\u22121}(\\frac{1}{2})[\/latex] is the same as determining the angle that would have a sine value of [latex]\\frac{1}{2}[\/latex]. In other words, what angle <em>x<\/em> would satisfy [latex]\\sin(x)=\\frac{1}{2}[\/latex]? There are multiple values that would satisfy this relationship, such as [latex]\\frac{\\pi}{6}[\/latex] and [latex]\\frac{5\\pi}{6}[\/latex], but we know we need the angle in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], so the answer will be [latex]\\sin^{\u22121}(\\frac{1}{2})=\\frac{\\pi}{6}[\/latex]. Remember that the inverse is a function, so for each input, we will get exactly one output.<\/p>\r\n<p style=\"padding-left: 60px\">b. To evaluate [latex]\\sin^{\u22121}\\left(\u2212\\frac{\\sqrt{2}}{2}\\right)[\/latex], we know that [latex]\\frac{5\\pi}{4}[\/latex] and [latex]\\frac{7\\pi}{4}[\/latex] both have a sine value of [latex]\u2212\\frac{\\sqrt{2}}{2}[\/latex], but neither is in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex]. For that, we need the negative angle coterminal with [latex]\\frac{7\\pi}{4}:\\sin^{\u22121}\\left(\u2212\\frac{\\sqrt{2}}{2}\\right)=\u2212\\frac{\\pi}{4}[\/latex].<\/p>\r\n<p style=\"padding-left: 60px\">c. To evaluate [latex]\\cos^{\u22121}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)[\/latex], we are looking for an angle in the interval [0,\u03c0] with a cosine value of [latex]\u2212\\frac{\\sqrt{3}}{2}[\/latex]. The angle that satisfies this is [latex]\\cos^{\u22121}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)=\\frac{5\\pi}{6}[\/latex].<\/p>\r\n<p style=\"padding-left: 60px\">d. Evaluating [latex]\\tan^{\u22121}(1)[\/latex], we are looking for an angle in the interval [latex](\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2})[\/latex] with a tangent value of 1. The correct angle is [latex]\\tan^{\u22121}(1)=\\frac{\\pi}{4}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEvaluate each of the following.\r\n<ol>\r\n \t<li>[latex]\\sin^{\u22121}(\u22121)[\/latex]<\/li>\r\n \t<li>[latex]\\tan^{\u22121}(\u22121)[\/latex]<\/li>\r\n \t<li>[latex]\\cos^{\u22121}(\u22121)[\/latex]<\/li>\r\n \t<li>[latex]\\cos^{\u22121}(\\frac{1}{2})[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"333778\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"333778\"]\r\n\r\n1. [latex]\u2212\\frac{\\pi}{2}[\/latex];\r\n\r\n2. [latex]\u2212\\frac{\\pi}{4}[\/latex]\r\n\r\n3. [latex]\\pi[\/latex]\r\n\r\n4. [latex]\\frac{\\pi}{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173433[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Using a Calculator to Evaluate Inverse Trigonometric Functions<\/h2>\r\nTo evaluate <strong>inverse trigonometric functions<\/strong> that do not involve the special angles discussed previously, we will need to use a calculator or other type of technology. Most scientific calculators and calculator-emulating applications have specific keys or buttons for the inverse sine, cosine, and tangent functions. These may be labeled, for example, SIN-1, ARCSIN, or ASIN.\r\n\r\nIn the previous chapter, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trigonometric functions, we can solve for the angles of a right triangle given two sides, and we can use a calculator to find the values to several decimal places.\r\n\r\nIn these examples and exercises, the answers will be interpreted as angles and we will use \u03b8 as the independent variable. The value displayed on the calculator may be in degrees or radians, so be sure to set the mode appropriate to the application.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Evaluating the Inverse Sine on a Calculator<\/h3>\r\nEvaluate [latex]\\sin^{\u22121}(0.97)[\/latex] using a calculator.\r\n\r\n[reveal-answer q=\"931769\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"931769\"]\r\n\r\nBecause the output of the inverse function is an angle, the calculator will give us a degree value if in degree mode and a radian value if in radian mode. Calculators also use the same domain restrictions on the angles as we are using.\r\n\r\nIn radian mode, [latex]\\sin^{\u22121}(0.97)\\approx1.3252[\/latex]. In degree mode, [latex]\\sin^{\u22121}(0.97)\\approx75.93^{\\circ}[\/latex]. Note that in calculus and beyond we will use radians in almost all cases.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\cos^{\u22121}(\u22120.4)[\/latex] using a calculator.\r\n\r\n[reveal-answer q=\"728477\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"728477\"]\r\n\r\n1.9823 or 113.578\u00b0\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173435[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given two sides of a right triangle like the one shown in Figure 7, find an angle.<\/h3>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164012\/CNX_Precalc_Figure_06_03_012.jpg\" alt=\"An illustration of a right triangle with an angle theta. Adjacent to theta is the side a, opposite theta is the side p, and the hypoteneuse is side h.\" width=\"487\" height=\"248\" \/> <b>Figure 7<\/b>[\/caption]\r\n<ol>\r\n \t<li>If one given side is the hypotenuse of length h and the side of length a adjacent to the desired angle is given, use the equation [latex]\\theta=\\cos^{\u22121}\\left(\\frac{a}{h}\\right)[\/latex].<\/li>\r\n \t<li>If one given side is the hypotenuse of length <em>h<\/em> and the side of length <em>p<\/em> opposite to the desired angle is given, use the equation [latex]\\theta=\\sin^{\u22121}\\left(\\frac{p}{h}\\right)[\/latex].<\/li>\r\n \t<li>If the two legs (the sides adjacent to the right angle) are given, then use the equation [latex]\\theta=\\tan^{\u22121}\\left(\\frac{p}{a}\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Applying the Inverse Cosine to a Right Triangle<\/h3>\r\nSolve the triangle in Figure 8\u00a0for the angle \u03b8.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164015\/CNX_Precalc_Figure_06_03_007.jpg\" alt=\"An illustration of a right triangle with the angle theta. Adjacent to the angle theta is a side with a length of 9 and a hypoteneuse of length 12.\" width=\"487\" height=\"200\" \/> <b>Figure 8<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"24088\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"24088\"]\r\n\r\nBecause we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;\\cos\\theta=\\frac{9}{12}\\\\ &amp;\\theta=\\cos^{\u22121}\\left(\\frac{9}{12}\\right) &amp;&amp; \\text{Apply definition of the inverse.} \\\\ &amp;\\theta\\approx0.7227\\text{ or about }41.4096^{\\circ} &amp;&amp; \\text{Evaluate.} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve the triangle in Figure 9\u00a0for the angle \u03b8.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164017\/CNX_Precalc_Figure_06_03_008.jpg\" alt=\"An illustration of a right triangle with the angle theta. Opposite to the angle theta is a side with a length of 6 and a hypoteneuse of length 10.\" width=\"487\" height=\"137\" \/> <b>Figure 9<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"230605\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"230605\"]\r\n\r\n[latex]\\sin^{\u22121}(0.6)=36.87^{\\circ}=0.6435[\/latex] radians\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]129737[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Finding Exact Values of Composite Functions with Inverse Trigonometric Functions<\/h2>\r\nThere are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let <em>f<\/em>(<em>x<\/em>) and <em>g<\/em>(<em>x<\/em>) be two different trigonometric functions belonging to the set {sin(<em>x<\/em>), cos(<em>x<\/em>), tan(<em>x<\/em>)} and let [latex]f^{\u22121}(y)[\/latex] and [latex]g^{\u22121}(y)[\/latex] be their inverses.\r\n<h3>Evaluating Compositions of the Form [latex]f\\left(f^{\u22121}(y)\\right)[\/latex] and [latex]f^{\u22121}(f(x))[\/latex]<\/h3>\r\nFor any trigonometric function, [latex]f(f^{\u22121}(y))=y[\/latex] for all <em>y<\/em> in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of <em>f<\/em> was defined to be identical to the domain of [latex]f^{\u22121}[\/latex]. However, we have to be a little more careful with expressions of the form [latex]f^{\u22121}(f(x))[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Compositions of a trigonometric function and its inverse<\/h3>\r\n<p style=\"text-align: center\">[latex]\\begin{align} &amp;\\sin(\\sin^{\u22121}x)=x\\text{ for }\u22121\\leq x\\leq1\\\\ &amp;\\cos(\\cos^{\u22121}x)=x\\text{ for }\u2212\\infty\\leq x\\leq1 \\\\ &amp;\\tan(\\tan^{\u22121}x)=x\\text{ for }\u2212\\infty\\text{ &lt; }x\\text{ &lt; }\\infty \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align} \\hfill &amp;\\sin^{\u22121}(\\sin x)=x\\text{ only for }\u2212\\frac{\\pi}{2} \\leq x \\leq \\frac{\\pi}{2} \\hfill \\\\ &amp;\\cos^{\u22121}(\\cos x)=x\\text{ only for }0\\leq x\\leq\\pi \\hfill \\\\ &amp;\\tan^{\u22121}(\\tan x)=x\\text{ only for }\u2212\\frac{\\pi}{2}\\text{ &lt; }x\\text{ &lt; }\\frac{\\pi}{2} \\end{align}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h3>Is it correct that [latex]\\sin^{\u22121}(\\sin x)=x[\/latex]?<\/h3>\r\n<em>No. This equation is correct if x belongs to the restricted domain [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex], but sine is defined for all real input values, and for x outside the restricted interval, the equation is not correct because its inverse always returns a value in\u00a0<em>[latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\\latex]<\/em>. The situation is similar for cosine and tangent and their inverses. For example, [latex]\\sin^{\u22121}\\left(\\sin\\left(\\frac{3\\pi}{4}\\right)\\right)=\\frac{\\pi}{4}[\/latex]. <\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To:<\/h3>\r\nGiven an expression of the form [latex]f^{\u22121}(f(\\theta))[\/latex] where [latex]f(\\theta)=\\sin\\theta\\text{, }\\cos\\theta\\text{, or }\\tan\\theta[\/latex], evaluate.\r\n<ol>\r\n \t<li>If \u03b8 is in the restricted domain of <em>f<\/em>,\u00a0then [latex]f^{\u22121}(f(\\theta))=\\theta[\/latex].<\/li>\r\n \t<li>If not, then find an angle \u03d5 within the restricted domain of <em>f<\/em> such that [latex]f(\\phi)=f(\\theta)[\/latex]. Then [latex]f^{\u22121}(f(\\theta))=\\phi[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Using Inverse Trigonometric Functions<\/h3>\r\nEvaluate the following:\r\n<ol>\r\n \t<li>[latex]\\sin^{\u22121}(\\sin(\\frac{\\pi}{3}))[\/latex]<\/li>\r\n \t<li>[latex]\\sin^{\u22121}(\\sin(\\frac{2\\pi}{3}))[\/latex]<\/li>\r\n \t<li>[latex]\\cos^{\u22121}(\\cos(\\frac{2\\pi}{3}))[\/latex]<\/li>\r\n \t<li>[latex]\\cos^{\u22121}(\\cos(\u2212\\frac{\\pi}{3}))[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"611200\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"611200\"]\r\n<ol>\r\n \t<li>[latex]\\frac{\\pi}{3}[\/latex] is in [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], so [latex]\\sin^{\u22121}(\\sin(\\frac{\\pi}{3}))=\\frac{\\pi}{3}[\/latex].<\/li>\r\n \t<li>[latex]\\frac{2\\pi}{3}[\/latex] is not in [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], but [latex]\\sin\\left(\\frac{2\\pi}{3}\\right)=\\sin\\left(\\frac{\\pi}{3}\\right)[\/latex], so [latex]\\sin^{\u22121}\\left(\\sin\\left(\\frac{2\\pi}{3}\\right)\\right)=\\frac{\\pi}{3}[\/latex].<\/li>\r\n \t<li>[latex]\\frac{2\\pi}{3}[\/latex] is in [0,\u03c0], so [latex]\\cos^{\u22121}\\left(\\cos\\left(\\frac{2\\pi}{3}\\right)\\right)=\\frac{2\\pi}{3}[\/latex].<\/li>\r\n \t<li>[latex]\u2212\\frac{\\pi}{3}[\/latex] is not in [0,\u03c0], but [latex]\\cos(\u2212\\frac{\\pi}{3})=\\cos\\left(\\frac{\\pi}{3}\\right)[\/latex] because cosine is an even function.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\tan^{\u22121}\\left(\\tan\\left(\\frac{\\pi}{8}\\right)\\right)[\/latex] and [latex]\\tan^{\u22121}\\left(\\tan\\left(\\frac{11\\pi}{9}\\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"356884\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"356884\"]\r\n\r\n[latex]\\frac{\\pi}{8}\\text{; }\\frac{2\\pi}{9}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Evaluating Compositions of the Form\u00a0[latex]f^{\u22121}(g(x))[\/latex]<\/h2>\r\nNow that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form [latex]f^{\u22121}(g(x))[\/latex]. For special values of <em>x<\/em>, we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is \u03b8, making the other [latex]\\frac{\\pi}{2}\u2212\\theta[\/latex]. Consider the sine and cosine of each angle of the right triangle in Figure 10.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164019\/CNX_Precalc_Figure_06_03_009.jpg\" alt=\"An illustration of a right triangle with angles theta and pi\/2 - theta. Opposite the angle theta and adjacent the angle pi\/2-theta is the side a. Adjacent the angle theta and opposite the angle pi\/2 - theta is the side b. The hypoteneuse is labeled c.\" width=\"487\" height=\"195\" \/> <b>Figure 10.<\/b> Right triangle illustrating the cofunction relationships[\/caption]\r\n\r\nBecause [latex]\\cos\\theta=\\frac{b}{c}=\\sin\\left(\\frac{\\pi}{2}\u2212\\theta\\right)[\/latex], we have [latex]\\sin^{\u22121}(\\cos\\theta)=\\frac{\\pi}{2}\u2212\\theta\\text{ if }0\\leq\\theta\\leq\\pi[\/latex]. If \u03b8 is not in this domain, then we need to find another angle that has the same cosine as \u03b8 and does belong to the restricted domain; we then subtract this angle from [latex]\\frac{\\pi}{2}[\/latex]. Similarly, [latex]\\sin\\theta=\\frac{a}{c}=\\cos\\left(\\frac{\\pi}{2}\u2212\\theta\\right)[\/latex], so [latex]\\cos^{\u22121}(\\sin\\theta)=\\frac{\\pi}{2}\u2212\\theta\\text{ if }\u2212\\frac{\\pi}{2}\\leq\\theta\\leq\\frac{\\pi}{2}[\/latex]. These are just the function-cofunction relationships presented in another way.\r\n<div class=\"textbox\">\r\n<h3>How To:\u00a0Given functions of the form [latex]\\sin^{\u22121}(\\cos x)\\text{ and }\\cos^{\u22121}(\\sin x)[\/latex], evaluate them.<\/h3>\r\n<ol>\r\n \t<li>If <em>x<\/em>\u00a0is\u00a0in\u00a0[0,\u03c0], then [latex]\\sin^{\u22121}(\\cos x)=\\frac{\\pi}{2}\u2212x[\/latex].<\/li>\r\n \t<li>If <em>x<\/em>\u00a0is\u00a0not\u00a0in\u00a0[0,\u03c0], then find another angle <em>y<\/em>\u00a0in\u00a0[0,\u03c0] such that [latex]\\cos y=\\cos x[\/latex].\r\n<div>\r\n<div style=\"text-align: center\">[latex]\\sin^{\u22121}(\\cos x)=\\frac{\\pi}{2}\u2212y[\/latex]<\/div>\r\n<\/div><\/li>\r\n \t<li>If <em>x<\/em>\u00a0is\u00a0in [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex], then [latex]\\cos^{\u22121}(\\sin x)=\\frac{\\pi}{2}\u2212x[\/latex].<\/li>\r\n \t<li>If <em>x<\/em>\u00a0is\u00a0not\u00a0in [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex], then find another angle <em>y<\/em>\u00a0in [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex] such that [latex]\\sin y=\\sin x[\/latex].\r\n<div>\r\n<div style=\"text-align: center\">[latex]\\cos^{\u22121}(\\sin x)=\\frac{\\pi}{2}\u2212y[\/latex]<\/div>\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Evaluating the Composition of an Inverse Sine with a Cosine<\/h3>\r\nEvaluate [latex]\\sin^{\u22121}(\\cos(\\frac{13\\pi}{6}))[\/latex]\r\n<ol>\r\n \t<li>by direct evaluation.<\/li>\r\n \t<li>by the method described previously.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"651517\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"651517\"]\r\n<ol>\r\n \t<li>Here, we can directly evaluate the inside of the composition.\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos\\left(\\frac{13\\pi}{6}\\right)&amp;=\\cos\\left(\\frac{\\pi}{6}+2\\pi\\right) \\\\ &amp;=\\cos\\left(\\frac{\\pi}{6}\\right) \\\\ &amp;=\\frac{\\sqrt{3}}{2} \\end{align}[\/latex]<\/div>\r\nNow, we can evaluate the inverse function as we did earlier.\r\n<div style=\"text-align: center\">[latex]\\sin^{\u22121}\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{\\pi}{3}[\/latex]<\/div><\/li>\r\n \t<li>We have [latex]x=\\frac{13\\pi}{6}[\/latex], [latex]y=\\frac{\\pi}{6}[\/latex], and\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin^{\u22121}\\left(\\cos\\left(\\frac{13\\pi}{6}\\right)\\right)=\\frac{\\pi}{2}\u2212\\frac{\\pi}{6} =\\frac{\\pi}{3} \\end{align}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\cos^{\u22121}(\\sin(\u2212\\frac{11\\pi}{4}))[\/latex].\r\n\r\n[reveal-answer q=\"325594\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"325594\"]\r\n\r\n[latex]\\frac{3\\pi}{4}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]129738[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Evaluating Compositions of the Form [latex]f(g^{\u22121}(x))[\/latex]<\/h2>\r\nTo evaluate compositions of the form [latex]f(g^{\u22121}(x))[\/latex], where <em>f<\/em> and <em>g<\/em> are any two of the functions sine, cosine, or tangent and <em>x<\/em> is any input in the domain of [latex]g\u22121[\/latex], we have exact formulas, such as [latex]\\sin\\left({\\cos}^{\u22121}x\\right)=\\sqrt{1\u2212{x}^{2}}[\/latex]. When we need to use them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagoras\u2019s relation between the lengths of the sides. We can use the Pythagorean identity, [latex]\\sin^{2}x+cos^{2}x=1[\/latex], to solve for one when given the other. We can also use the <strong>inverse trigonometric functions<\/strong> to find compositions involving algebraic expressions.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Evaluating the Composition of a Sine with an Inverse Cosine<\/h3>\r\nFind an exact value for [latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{4}{5}\\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"530604\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"530604\"]\r\n\r\nBeginning with the inside, we can say there is some angle such that [latex]\\theta=\\cos^{\u22121}(\\frac{4}{5})[\/latex], which means [latex]\\cos\\theta=\\frac{4}{5}[\/latex], and we are looking for [latex]\\sin\\theta[\/latex]. We can use the Pythagorean identity to do this.\r\n<p style=\"text-align: center\">[latex]\\begin{align} &amp;\\sin^{2}\\theta+\\cos^{2}\\theta=1 &amp;&amp; \\text{Use our known value for cosine.} \\\\ &amp;\\sin^{2}\\theta+\\left(\\frac{4}{5}\\right)^{2}=1 &amp;&amp; \\text{Solve for sine.} \\\\ &amp;\\sin^{2}\\theta=1\u2212\\frac{16}{25} \\\\ &amp;\\sin\\theta=\\pm\\sqrt{\\frac{9}{25}}=\\pm\\frac{3}{5} \\end{align}[\/latex]<\/p>\r\nSince [latex]\\theta=\\cos^{\u22121}(\\frac{4}{5})[\/latex] is in quadrant I, [latex]\\sin{\\theta}[\/latex] must be positive, so the solution is [latex]\\frac{3}{5}[\/latex]. See Figure 11.\r\n<p style=\"text-align: center\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164021\/CNX_Precalc_Figure_06_03_010.jpg\" alt=\"An illustration of a right triangle with an angle theta. Oppostie the angle theta is a side with length 3. Adjacent the angle theta is a side with length 4. The hypoteneuse has angle of length 5.\" \/><\/p>\r\n<p style=\"text-align: center\"><strong>Figure 11.<\/strong> Right triangle illustrating that if [latex]\\cos\\theta=\\frac{4}{5}[\/latex], then [latex]\\sin\\theta=\\frac{3}{5}[\/latex]<\/p>\r\nWe know that the inverse cosine always gives an angle on the interval [0,\u00a0\u03c0], so we know that the sine of that angle must be positive; therefore [latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{4}{5}\\right)\\right)=\\sin\\theta=\\frac{3}{5}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\cos(\\tan^{\u22121}(\\frac{5}{12}))[\/latex].\r\n\r\n[reveal-answer q=\"754416\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"754416\"]\r\n\r\n[latex]\\frac{12}{13}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Evaluating the Composition of a Sine with an Inverse Tangent<\/h3>\r\nFind an exact value for [latex]\\sin\\left(\\tan^{\u22121}\\left(\\frac{7}{4}\\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"488532\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"488532\"]\r\n\r\nWhile we could use a similar technique as in Example 6, we will demonstrate a different technique here. From the inside, we know there is an angle such that [latex]\\tan\\theta=\\frac{7}{4}[\/latex]. We can envision this as the opposite and adjacent sides on a right triangle, as shown in Figure 12.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164023\/CNX_Precalc_Figure_06_03_011n.jpg\" alt=\"An illustration of a right triangle with angle theta. Adjacent the angle theta is a side with length 4. Opposite the angle theta is a side with length 7.\" width=\"487\" height=\"196\" \/> <b>Figure 12.<\/b> A right triangle with two sides known[\/caption]\r\n\r\nUsing the Pythagorean Theorem, we can find the hypotenuse of this triangle.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}4^{2}+7^{2}=\\text{hypotenuse}^{2} \\\\ \\text{hypotenuse}=\\sqrt{65} \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\cos(\\sin^{\u22121}(\\frac{7}{9}))[\/latex].\r\n\r\n[reveal-answer q=\"930711\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"930711\"]\r\n\r\n[latex]\\frac{4\\sqrt{2}}{9}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]129751[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Finding the Cosine of the Inverse Sine of an Algebraic Expression<\/h3>\r\nFind a simplified expression for [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{x}{3}\\right)\\right)[\/latex] for [latex]\u22123\\leq x\\leq3[\/latex].\r\n\r\n[reveal-answer q=\"764834\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"764834\"]\r\n\r\nWe know there is an angle\u00a0\u03b8 such that [latex]\\sin\\theta=\\frac{x}{3}\\\\[\/latex]\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;\\sin^{2}\\theta+\\cos^{2}\\theta=1 7&amp;&amp; \\text{Use the Pythagorean Theorem.} \\\\ &amp;\\left(\\frac{x}{3}\\right)^{2}+\\cos^{2}+\\cos^2\\theta=1 &amp;&amp; \\text{Solve for cosine.} \\\\ &amp;\\cos^{2}\\theta=1\u2212\\frac{x^{2}}{9} \\\\ &amp;\\cos\\theta=\\pm\\sqrt{\\frac{9\u2212x^{2}}{9}}=\\pm\\frac{\\sqrt{9\u2212x^{2}}}{3} \\end{align}[\/latex]<\/p>\r\nBecause we know that the inverse sine must give an angle on the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], we can deduce that the cosine of that angle must be positive.\r\n<p style=\"text-align: center\">[latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{x}{3}\\right)\\right)=\\frac{\\sqrt{9\u2212x^{2}}}{3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind a simplified expression for [latex]\\sin\\left(\\tan^{\u22121}\\left(4x\\right)\\right)\\\\[\/latex] for [latex]\u2212\\frac{1}{4}\\leq x \\leq\\frac{1}{4}[\/latex].\r\n\r\n[reveal-answer q=\"979016\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979016\"]\r\n\r\n[latex]\\frac{4x}{\\sqrt{16x^{2}+1}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]129755[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solving Linear Trigonometric Equations in Sine and Cosine<\/h2>\r\nTrigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The <strong>period<\/strong> of both the sine function and the cosine function is [latex]2\\pi [\/latex]. In other words, every [latex]2\\pi [\/latex] units, the <em>y-<\/em>values repeat. If we need to find all possible solutions, then we must add [latex]2\\pi k[\/latex], where [latex]k[\/latex] is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\\pi :[\/latex]\r\n<div style=\"text-align: center\">[latex]\\sin \\theta =\\sin \\left(\\theta \\pm 2k\\pi \\right)[\/latex]<\/div>\r\nThere are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 10: Solving a Linear Trigonometric Equation Involving the Cosine Function<\/h3>\r\nFind all possible exact solutions for the equation [latex]\\cos \\theta =\\frac{1}{2}[\/latex].\r\n\r\n[reveal-answer q=\"84784\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"84784\"]\r\n\r\nFrom the <strong>unit circle<\/strong>, we know that\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\cos \\theta =\\frac{1}{2} \\\\ \\theta =\\frac{\\pi }{3},\\frac{5\\pi }{3} \\end{gathered}[\/latex]<\/p>\r\nThese are the solutions in the interval [latex]\\left[0,2\\pi \\right][\/latex]. All possible solutions are given by\r\n<p style=\"text-align: center\">[latex]\\theta =\\frac{\\pi }{3}\\pm 2k\\pi \\text{ and }\\theta =\\frac{5\\pi }{3}\\pm 2k\\pi [\/latex]<\/p>\r\nwhere [latex]k[\/latex] is an integer.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 11: Solving a Linear Equation Involving the Sine Function<\/h3>\r\nFind all possible exact solutions for the equation [latex]\\sin t=\\frac{1}{2}[\/latex].\r\n\r\n[reveal-answer q=\"535703\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"535703\"]\r\n\r\nSolving for all possible values of <em>t<\/em> means that solutions include angles beyond the period of [latex]2\\pi [\/latex]. From the unit circle, we can see that the solutions are [latex]t=\\frac{\\pi }{6}[\/latex] and [latex]t=\\frac{5\\pi }{6}[\/latex]. But the problem is asking for all possible values that solve the equation. Therefore, the answer is\r\n<p style=\"text-align: center\">[latex]t=\\frac{\\pi }{6}\\pm 2\\pi k\\text{ and }t=\\frac{5\\pi }{6}\\pm 2\\pi k[\/latex]<\/p>\r\nwhere [latex]k[\/latex] is an integer.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a trigonometric equation, solve using algebra.<\/h3>\r\n<ul>\r\n \t<li>Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.<\/li>\r\n \t<li>Substitute the trigonometric expression with a single variable, such as [latex]x[\/latex] or [latex]u[\/latex].<\/li>\r\n \t<li>Solve the equation the same way an algebraic equation would be solved.<\/li>\r\n \t<li>Substitute the trigonometric expression back in for the variable in the resulting expressions.<\/li>\r\n \t<li>Solve for the angle.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 12: Solve the Trigonometric Equation in Linear Form<\/h3>\r\nSolve the equation exactly: [latex]2\\cos \\theta -3=-5,0\\le \\theta &lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"939405\"]Show Solution[\/reveal-answer]\r\n<p style=\"text-align: left\">[hidden-answer a=\"939405\"]<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}2\\cos \\theta -3=-5 \\\\ \\cos \\theta =-2 \\\\ \\cos \\theta =-1 \\\\ \\theta =\\pi \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve exactly the following linear equation on the interval [latex]\\left[0,2\\pi \\right):2\\sin x+1=0[\/latex].\r\n\r\n[reveal-answer q=\"500341\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"500341\"]\r\n\r\n[latex]x=\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149871[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solve Trigonometric Equations Using a Calculator<\/h2>\r\nNot all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 13: Using a Calculator to Solve a Trigonometric Equation Involving Sine<\/h3>\r\nUse a calculator to solve the equation [latex]\\sin \\theta =0.8[\/latex], where [latex]\\theta [\/latex] is in radians.\r\n\r\n[reveal-answer q=\"522194\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"522194\"]\r\n<h3><\/h3>\r\nMake sure mode is set to radians. To find [latex]\\theta [\/latex], use the inverse sine function. On most calculators, you will need to push the 2<sup>ND<\/sup> button and then the SIN button to bring up the [latex]{\\sin }^{-1}[\/latex] function. What is shown on the screen is [latex]{\\sin}^{-1}[\/latex](. The calculator is ready for the input within the parentheses. For this problem, we enter [latex]{\\sin }^{-1}\\left(0.8\\right)[\/latex], and press ENTER. Thus, to four decimals places,\r\n<p style=\"text-align: center\">[latex]{\\sin }^{-1}\\left(0.8\\right)\\approx 0.9273[\/latex]<\/p>\r\nThis is the solution in quadrant I. There is also a solution in quadrant II. To find this we subtract [latex]\/pi - 0.9273 \\approx 2.2143 [\/latex]\r\n\r\nThe general solution is\r\n<p style=\"text-align: center\">[latex]\\theta \\approx 0.9273\\pm 2\\pi k \\text{ and } \\theta \\approx 2.2143 \\pm 2\\pi k[\/latex]<\/p>\r\nThe angle measurement in degrees is\r\n<p style=\"text-align: center\">[latex]\\begin{align} \\theta &amp;\\approx {53.1}^{\\circ } \\\\ \\theta &amp;\\approx {180}^{\\circ }-{53.1}^{\\circ } \\\\ &amp;\\approx {126.9}^{\\circ } \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nNote that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using [latex]\\pi -\\theta [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 14: Using a Calculator to Solve a Trigonometric Equation Involving Secant<\/h3>\r\nUse a calculator to solve the equation [latex]\\sec \\theta =-4[\/latex], giving your answer in radians.\r\n\r\n[reveal-answer q=\"209133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"209133\"]\r\n\r\nWe can begin with some algebra.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sec \\theta =-4\\\\ \\frac{1}{\\cos \\theta }=-4\\\\ \\cos \\theta =-\\frac{1}{4}\\end{gathered}[\/latex]<\/p>\r\nCheck that the MODE is in radians. Now use the inverse cosine function.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}{\\cos }^{-1}\\left(-\\frac{1}{4}\\right)\\approx 1.8235 \\\\ \\theta \\approx 1.8235+2\\pi k \\end{gathered}[\/latex]<\/p>\r\nSince [latex]\\frac{\\pi }{2}\\approx 1.57[\/latex] and [latex]\\pi \\approx 3.14[\/latex], 1.8235 is between these two numbers, thus [latex]\\theta \\approx \\text{1}\\text{.8235}[\/latex] is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164111\/CNX_Precalc_Figure_07_05_0052.jpg\" alt=\"Graph of angles theta =approx 1.8235, theta prime =approx pi - 1.8235 = approx 1.3181, and then theta prime = pi + 1.3181 = approx 4.4597\" width=\"487\" height=\"380\" \/> <b>Figure 13.<\/b>[\/caption]\r\n\r\nSo, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi -\\text{1}\\text{.8235}\\approx \\text{1}\\text{.3181}\\text{.}[\/latex] The other solution in quadrant III is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi +\\text{1}\\text{.3181}\\approx \\text{4}\\text{.4597}\\text{.}[\/latex]\r\n\r\nThe solutions are [latex]\\theta \\approx 1.8235\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.4597\\pm 2\\pi k[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]\\cos \\theta =-0.2[\/latex].\r\n\r\n[reveal-answer q=\"145806\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"145806\"]\r\n\r\n[latex]\\theta \\approx 1.7722\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.5110\\pm 2\\pi k[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149873[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solving Equations Involving a Single Trigonometric Function<\/h2>\r\nWhen we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is [latex]\\pi [\/latex], not [latex]2\\pi [\/latex]. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of [latex]\\frac{\\pi }{2}[\/latex], unless, of course, a problem places its own restrictions on the domain.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 15: Solving a Problem Involving a Single Trigonometric Function<\/h3>\r\nSolve the problem exactly: [latex]2{\\sin }^{2}\\theta -1=0,0\\le \\theta &lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"467313\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"467313\"]\r\n\r\nAs this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\\sin \\theta [\/latex]. Then we will find the angles.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta -1=0 \\\\ 2{\\sin }^{2}\\theta =1 \\\\ {\\sin }^{2}\\theta =\\frac{1}{2} \\\\ \\sqrt{{\\sin }^{2}\\theta }=\\pm \\sqrt{\\frac{1}{2}} \\\\ \\sin \\theta =\\pm \\frac{1}{\\sqrt{2}}=\\pm \\frac{\\sqrt{2}}{2} \\\\ \\theta =\\frac{\\pi }{4},\\frac{3\\pi }{4},\\frac{5\\pi }{4},\\frac{7\\pi }{4} \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 16: Solving a Trigonometric Equation Involving Cosecant<\/h3>\r\nSolve the following equation exactly: [latex]\\csc \\theta =-2,0\\le \\theta &lt;4\\pi [\/latex].\r\n\r\n[reveal-answer q=\"605306\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"605306\"]\r\n\r\nWe want all values of [latex]\\theta [\/latex] for which [latex]\\csc \\theta =-2[\/latex] over the interval [latex]0\\le \\theta &lt;4\\pi [\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\csc \\theta =-2 \\\\ \\frac{1}{\\sin \\theta }=-2 \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6},\\frac{19\\pi }{6},\\frac{23\\pi }{6} \\end{gathered}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nAs [latex]\\sin \\theta =-\\frac{1}{2}[\/latex], notice that all four solutions are in the third and fourth quadrants.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 17: Solving an Equation Involving Tangent<\/h3>\r\nSolve the equation exactly: [latex]\\tan \\left(\\theta -\\frac{\\pi }{2}\\right)=1,0\\le \\theta &lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"484899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"484899\"]\r\n\r\nRecall that the tangent function has a period of [latex]\\pi [\/latex]. On the interval [latex]\\left[0,\\pi \\right)[\/latex], and at the angle of [latex]\\frac{\\pi }{4}[\/latex], the tangent has a value of 1. However, the angle we want is [latex]\\left(\\theta -\\frac{\\pi }{2}\\right)[\/latex]. Thus, if [latex]\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex], then\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\theta -\\frac{\\pi }{2}=\\frac{\\pi }{4}\\\\ \\theta =\\frac{3\\pi }{4}\\pm k\\pi \\end{gathered}[\/latex]<\/p>\r\nOver the interval [latex]\\left[0,2\\pi \\right)[\/latex], we have two solutions:\r\n<p style=\"text-align: center\">[latex]\\theta =\\frac{3\\pi }{4}\\text{ and }\\theta =\\frac{3\\pi }{4}+\\pi =\\frac{7\\pi }{4}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind all solutions for [latex]\\tan x=\\sqrt{3}[\/latex].\r\n\r\n[reveal-answer q=\"629684\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"629684\"]\r\n\r\n[latex]\\frac{\\pi }{3}\\pm \\pi k[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173739[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 18: Identify all Solutions to the Equation Involving Tangent<\/h3>\r\nIdentify all exact solutions to the equation [latex]2\\left(\\tan x+3\\right)=5+\\tan x,0\\le x&lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"949694\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"949694\"]\r\n\r\nWe can solve this equation using only algebra. Isolate the expression [latex]\\tan x[\/latex] on the left side of the equals sign.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} 2\\left(\\tan x\\right)+2\\left(3\\right) =5+\\tan x \\\\ 2\\tan x+6 =5+\\tan x \\\\ 2\\tan x-\\tan x =5 - 6 \\\\ \\tan x =-1\\end{gathered}[\/latex]<\/p>\r\nThere are two angles on the unit circle that have a tangent value of [latex]-1:\\theta =\\frac{3\\pi }{4}[\/latex] and [latex]\\theta =\\frac{7\\pi }{4}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solving Trigonometric Equations in Quadratic Form<\/h2>\r\nSolving a <strong>quadratic equation<\/strong> may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as [latex]x[\/latex] or [latex]u[\/latex]. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 19: Solving a Trigonometric Equation in Quadratic Form<\/h3>\r\nSolve the equation exactly: [latex]{\\cos }^{2}\\theta +3\\cos \\theta -1=0,0\\le \\theta &lt;2\\pi[\/latex].\r\n\r\n[reveal-answer q=\"563509\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"563509\"]\r\n\r\nWe begin by using substitution and replacing cos [latex]\\theta[\/latex] with [latex]x[\/latex]. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\\cos \\theta =x[\/latex]. We have\r\n<p style=\"text-align: center\">[latex]{x}^{2}+3x - 1=0[\/latex]<\/p>\r\nThe equation cannot be factored, so we will use the <strong>quadratic formula<\/strong> [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{align} x&amp;=\\frac{-3\\pm \\sqrt{{\\left(-3\\right)}^{2}-4\\left(1\\right)\\left(-1\\right)}}{2}&amp;=\\frac{-3\\pm \\sqrt{13}}{2} \\end{align}[\/latex]<\/p>\r\nReplace [latex]x[\/latex] with [latex]\\cos \\theta[\/latex], and solve. Thus,\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\cos \\theta =\\frac{-3\\pm \\sqrt{13}}{2}\\theta ={\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right)\\end{gathered}[\/latex]<\/p>\r\nNote that only the + sign is used. This is because we get an error when we solve [latex]\\theta ={\\cos }^{-1}\\left(\\frac{-3-\\sqrt{13}}{2}\\right)[\/latex] on a calculator, since the domain of the inverse cosine function is [latex]\\left[-1,1\\right][\/latex]. However, there is a second solution:\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\theta &amp;={\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\\\ &amp;\\approx 1.26 \\end{align}[\/latex]<\/p>\r\nThis terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\theta &amp;=2\\pi -{\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\\\ &amp;\\approx 5.02 \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 20: Solving a Trigonometric Equation in Quadratic Form by Factoring<\/h3>\r\nSolve the equation exactly: [latex]2{\\sin }^{2}\\theta -5\\sin \\theta +3=0,0\\le \\theta \\le 2\\pi[\/latex].\r\n\r\n[reveal-answer q=\"693370\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"693370\"]\r\n\r\nUsing grouping, this quadratic can be factored. Either make the real substitution, [latex]\\sin \\theta =u[\/latex], or imagine it, as we factor:\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta -5\\sin \\theta +3=0 \\\\ \\left(2\\sin \\theta -3\\right)\\left(\\sin \\theta -1\\right)=0 \\end{gathered}[\/latex]<\/p>\r\nNow set each factor equal to zero.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}2\\sin \\theta -3=0 \\\\ 2\\sin \\theta =3 \\\\ \\sin \\theta =\\frac{3}{2} \\\\ \\text{ } \\\\ \\sin \\theta -1=0 \\\\ \\sin \\theta =1 \\end{gathered}[\/latex]<\/p>\r\nNext solve for [latex]\\theta :\\sin \\theta \\ne \\frac{3}{2}[\/latex], as the range of the sine function is [latex]\\left[-1,1\\right][\/latex]. However, [latex]\\sin \\theta =1[\/latex], giving the solution [latex]\\theta =\\frac{\\pi }{2}[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nMake sure to check all solutions on the given domain as some factors have no solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]{\\sin }^{2}\\theta =2\\cos \\theta +2,0\\le \\theta \\le 2\\pi[\/latex]. [Hint: Make a substitution to express the equation only in terms of cosine.]\r\n\r\n[reveal-answer q=\"608693\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"608693\"]\r\n\r\n[latex]\\cos \\theta =-1,\\theta =\\pi [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149010[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 21: Solving a Trigonometric Equation Using Algebra<\/h3>\r\nSolve exactly:\r\n<p style=\"text-align: center\">[latex]2{\\sin }^{2}\\theta +\\sin \\theta =0;0\\le \\theta &lt;2\\pi[\/latex]<\/p>\r\n[reveal-answer q=\"204542\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"204542\"]\r\n\r\nThis problem should appear familiar as it is similar to a quadratic. Let [latex]\\sin \\theta =x[\/latex]. The equation becomes [latex]2{x}^{2}+x=0[\/latex]. We begin by factoring:\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}2{x}^{2}+x=0\\\\ x\\left(2x+1\\right)=0\\end{gathered}[\/latex]<\/p>\r\nSet each factor equal to zero.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}x=0 \\\\ 2x+1=0 \\\\ x=-\\frac{1}{2} \\end{gathered}[\/latex]<\/p>\r\nThen, substitute back into the equation the original expression [latex]\\sin \\theta[\/latex] for [latex]x[\/latex]. Thus,\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sin \\theta =0 \\\\ \\theta =0,\\pi \\\\ \\text{ } \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6} \\end{gathered}[\/latex]<\/p>\r\nThe solutions within the domain [latex]0\\le \\theta &lt;2\\pi[\/latex] are [latex]\\theta =0,\\pi ,\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex].\r\n\r\nIf we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta +\\sin \\theta =0 \\\\ \\sin \\theta \\left(2\\sin \\theta +1\\right)=0 \\\\ \\sin \\theta =0 \\\\ \\theta =0,\\pi \\\\ \\text{ } \\\\ 2\\sin \\theta +1=0 \\\\ 2\\sin \\theta =-1 \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6} \\end{gathered}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can see the solutions on the graph in Figure 3. On the interval [latex]0\\le \\theta &lt;2\\pi[\/latex], the graph crosses the <em>x-<\/em>axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164113\/CNX_Precalc_Figure_07_05_0042.jpg\" alt=\"Graph of 2*(sin(theta))^2 + sin(theta) from 0 to 2pi. Zeros are at 0, pi, 7pi\/6, and 11pi\/6.\" width=\"731\" height=\"256\" \/> <b>Figure 14<\/b>[\/caption]\r\n\r\nWe can verify the solutions on the <a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/chapter\/use-sum-and-difference-formulas-for-cosine\/\" target=\"_blank\" rel=\"noopener\">unit circle<\/a> in Sum and Difference Identities\u00a0as well.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 22: Solving a Trigonometric Equation Quadratic in Form<\/h3>\r\nSolve the equation quadratic in form exactly: [latex]2{\\sin }^{2}\\theta -3\\sin \\theta +1=0,0\\le \\theta &lt;2\\pi[\/latex].\r\n\r\n[reveal-answer q=\"699307\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"699307\"]\r\n\r\nWe can factor using grouping. Solution values of [latex]\\theta[\/latex] can be found on the unit circle:\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\left(2\\sin \\theta -1\\right)\\left(\\sin \\theta -1\\right)=0 \\\\ 2\\sin \\theta -1=0 \\\\ \\sin \\theta =\\frac{1}{2} \\\\ \\theta =\\frac{\\pi }{6},\\frac{5\\pi }{6} \\\\ \\text{ } \\\\ \\sin \\theta =1 \\\\ \\theta =\\frac{\\pi }{2} \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve the quadratic equation [latex]2{\\cos }^{2}\\theta +\\cos \\theta =0[\/latex].\r\n\r\n[reveal-answer q=\"736189\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"736189\"]\r\n\r\n[latex]\\frac{\\pi }{2},\\frac{2\\pi }{3},\\frac{4\\pi }{3},\\frac{3\\pi }{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>An inverse function is one that \u201cundoes\u201d another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function.<\/li>\r\n \t<li>Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains.<\/li>\r\n \t<li>For any trigonometric function [latex]f(x)[\/latex], if [latex]x=f^{\u22121}(y)[\/latex], then [latex]f(x)=y[\/latex]. However, [latex]f(x)=y[\/latex] only implies [latex]x=f^{\u22121}(y)[\/latex] if <em>x<\/em> is in the restricted domain of <em>f<\/em>.<\/li>\r\n \t<li>Special angles are the outputs of inverse trigonometric functions for special input values; for example, [latex]\\frac{\\pi}{4}=\\tan^{\u22121}( 1 )\\text{ and }\\frac{\\pi}{6}=\\sin^{\u22121}(\\frac{1}{2})[\/latex].<\/li>\r\n \t<li>A calculator will return an angle within the restricted domain of the original trigonometric function.<\/li>\r\n \t<li>Inverse functions allow us to find an angle when given two sides of a right triangle.<\/li>\r\n \t<li>In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example, [latex]\\sin\\left(\\cos^{\u22121}\\left(x\\right)\\right)=\\sqrt{1\u2212x^{2}}[\/latex].<\/li>\r\n \t<li>If the inside function is a trigonometric function, then the only possible combinations are [latex]\\sin^{\u22121}\\left(\\cos x\\right)=\\frac{\\pi}{2}\u2212x[\/latex] if [latex]0\\leq x\\leq\\pi[\/latex] and [latex]\\cos^{\u22121}\\left(\\sin x\\right)=\\frac{\\pi}{2}\u2212x[\/latex] if [latex]\u2212\\frac{\\pi}{2}\\leq x \\leq\\frac{\\pi}{2}[\/latex].<\/li>\r\n \t<li>When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function.<\/li>\r\n \t<li>When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides.<\/li>\r\n \t<li>When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution.<\/li>\r\n \t<li>Equations involving a single trigonometric function can be solved or verified using the unit circle.<\/li>\r\n \t<li>We can also solve trigonometric equations using a graphing calculator.<\/li>\r\n \t<li>Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc.<\/li>\r\n<\/ul>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl class=\"definition\">\r\n \t<dt>arccosine<\/dt>\r\n \t<dd>another name for the inverse cosine; [latex]\\arccos x=\\cos^{\u22121}x[\/latex]<\/dd>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dt>arcsine<\/dt>\r\n \t<dd>another name for the inverse sine; [latex]\\arcsin x=\\sin^{\u22121}x[\/latex]<\/dd>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dt>arctangent<\/dt>\r\n \t<dd>another name for the inverse tangent; [latex]\\arctan x=\\tan^{\u22121}x[\/latex]<\/dd>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dt>inverse cosine function<\/dt>\r\n \t<dd>the function [latex]\\cos^{\u22121}x[\/latex], which is the inverse of the cosine function and the angle that has a cosine equal to a given number<\/dd>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dt>inverse sine function<\/dt>\r\n \t<dd>the function [latex]\\sin^{\u22121}x[\/latex], which is the inverse of the sine function and the angle that has a sine equal to a given number<\/dd>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dt>inverse tangent function<\/dt>\r\n \t<dd>the function [latex]\\tan^{\u22121}x[\/latex], which is the inverse of the tangent function and the angle that has a tangent equal to a given number<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li style=\"font-weight: 400\">Understand and use the inverse sine, cosine, and tangent functions.<\/li>\n<li style=\"font-weight: 400\">Find the exact value of expressions involving the inverse sine, cosine, and tangent functions.<\/li>\n<li style=\"font-weight: 400\">Use a calculator to evaluate inverse trigonometric functions.<\/li>\n<li>Use inverse trigonometric functions to solve right triangles.<\/li>\n<li style=\"font-weight: 400\">Find exact values of composite functions with inverse trigonometric functions.<\/li>\n<\/ul>\n<\/div>\n<h2>Understanding and Using the Inverse Sine, Cosine, and Tangent Functions<\/h2>\n<p>In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function \u201cundoes\u201d what the original trigonometric function \u201cdoes,\u201d as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in Figure 1.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27163959\/CNX_Precalc_Figure_06_03_013.jpg\" alt=\"A chart that says \u201cTrig Functinos\u201d, \u201cInverse Trig Functions\u201d, \u201cDomain: Measure of an angle\u201d, \u201cDomain: Ratio\u201d, \u201cRange: Ratio\u201d, and \u201cRange: Measure of an angle\u201d.\" width=\"731\" height=\"78\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p>For example, if [latex]f(x)=\\sin x[\/latex], then we would write\u00a0[latex]f^{1}(x)={\\sin}^{-1}{x}[\/latex]. Be aware that [latex]{\\sin}^{-1}x[\/latex] does not mean [latex]\\frac{1}{\\sin{x}}[\/latex]. The following examples illustrate the inverse trigonometric functions:<\/p>\n<ul>\n<li>Since [latex]\\sin\\left(\\frac{\\pi}{6}\\right)=\\frac{1}{2}[\/latex], then [latex]\\frac{\\pi}{6}=\\sin^{\u22121}(\\frac{1}{2})[\/latex].<\/li>\n<li>Since [latex]\\cos(\\pi)=\u22121[\/latex], then [latex]\\pi=\\cos^{\u22121}(\u22121)[\/latex].<\/li>\n<li>Since [latex]\\tan\\left(\\frac{\\pi}{4}\\right)=1[\/latex], then [latex]\\frac{\\pi}{4}=\\tan^{\u22121}(1)[\/latex].<\/li>\n<\/ul>\n<p>In previous sections, we evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that, for a <strong>one-to-one function<\/strong>, if [latex]f(a)=b[\/latex], then an inverse function would satisfy [latex]f^{\u22121}(b)=a[\/latex].<\/p>\n<p>Bear in mind that the sine, cosine, and tangent functions are not one-to-one functions. The graph of each function would fail the horizontal line test. In fact, no periodic function can be one-to-one because each output in its range corresponds to at least one input in every period, and there are an infinite number of periods. As with other functions that are not one-to-one, we will need to restrict the <strong>domain<\/strong> of each function to yield a new function that is one-to-one. We choose a domain for each function that includes the number 0. Figure 2\u00a0shows the graph of the sine function limited to [latex]\\left[\\frac{\u2212\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex] and the graph of the cosine function limited to [0, \u03c0].<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164001\/CNX_Precalc_Figure_06_03_001.jpg\" alt=\"Two side-by-side graphs. The first graph, graph A, shows half of a period of the function sine of x. The second graph, graph B, shows half a period of the function cosine of x.\" \/><\/p>\n<p style=\"text-align: center\"><strong>Figure 2.<\/strong> (a) Sine function on a restricted domain of [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex]; (b) Cosine function on a restricted domain of [0, \u03c0]<\/p>\n<p>Figure 3\u00a0shows the graph of the tangent function limited to [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex].<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164003\/CNX_Precalc_Figure_06_03_003.jpg\" alt=\"A graph of one period of tangent of x, from -pi\/2 to pi\/2.\" \/><\/p>\n<p style=\"text-align: center\"><strong>Figure 3.\u00a0<\/strong>Tangent function on a restricted domain of [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex]<\/p>\n<p>These conventional choices for the restricted domain are somewhat arbitrary, but they have important, helpful characteristics. Each domain includes the origin and some positive values, and most importantly, each results in a one-to-one function that is invertible. The conventional choice for the restricted domain of the tangent function also has the useful property that it extends from one <strong>vertical asymptote<\/strong> to the next instead of being divided into two parts by an asymptote.<\/p>\n<p>On these restricted domains, we can define the <strong>inverse trigonometric functions<\/strong>.<\/p>\n<ul>\n<li>The <strong>inverse sine function<\/strong>\u00a0[latex]y=\\sin^{\u22121}x[\/latex] means [latex]x=\\sin y[\/latex]. The inverse sine function is sometimes called the <strong>arcsine<\/strong> function, and notated arcsin <em>x<\/em>.\n<div>\n<div style=\"text-align: center\">[latex]y=\\sin^{\u22121}x[\/latex] has\u00a0domain [\u22121, 1] and\u00a0range [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex]<\/div>\n<\/div>\n<\/li>\n<li>The <strong>inverse cosine function<\/strong>\u00a0[latex]y=\\cos^{\u22121}x[\/latex] means [latex]x=\\cos y[\/latex]. The inverse cosine function is sometimes called the <strong>arccosine<\/strong> function, and notated arccos <em>x<\/em>.\n<div>\n<div style=\"text-align: center\">[latex]y=\\cos^{\u22121}x[\/latex] has\u00a0domain [\u22121, 1] and\u00a0range [0, \u03c0]<\/div>\n<\/div>\n<\/li>\n<li>The <strong>inverse tangent function<\/strong>\u00a0[latex]y=\\tan^{\u22121}x[\/latex] means [latex]x=\\tan y[\/latex]. The inverse tangent function is sometimes called the <strong>arctangent<\/strong> function, and notated arctan <em>x<\/em>.\n<div>\n<div style=\"text-align: center\">[latex]y=\\tan^{\u22121}x[\/latex] has\u00a0domain (\u2212\u221e, \u221e) and\u00a0range [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/li>\n<\/ul>\n<p>The graphs of the inverse functions are shown in Figure 4, Figure 5, and Figure 6. Notice that the output of each of these inverse functions is a <em>number, <\/em>an angle in radian measure. We see that [latex]\\sin^{\u22121}x[\/latex] has domain [\u22121, 1] and range [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], [latex]\\cos^{\u22121}x[\/latex] has domain [\u22121, 1] and range [0, \u03c0], and [latex]\\tan^{\u22121}x[\/latex] has domain of all real numbers and range [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex]. To find the <strong>domain<\/strong> and <strong>range<\/strong> of inverse trigonometric functions, switch the domain and range of the original functions. Each graph of the inverse trigonometric function is a reflection of the graph of the original function about the line [latex]y=x[\/latex].<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164005\/CNX_Precalc_Figure_06_03_004n.jpg\" alt=\"A graph of the functions of sine of x and arc sine of x. There is a dotted line y=x between the two graphs, to show inverse nature of the two functions\" width=\"731\" height=\"433\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4.<\/b>\u00a0The sine function and inverse sine (or arcsine) function<\/p>\n<\/div>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164008\/CNX_Precalc_Figure_06_03_005n.jpg\" alt=\"A graph of the functions of cosine of x and arc cosine of x. There is a dotted line at y=x to show the inverse nature of the two functions.\" width=\"487\" height=\"343\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5.<\/b> The cosine function and inverse cosine (or arccosine) function<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164010\/CNX_Precalc_Figure_06_03_006n.jpg\" alt=\"A graph of the functions of tangent of x and arc tangent of x. There is a dotted line at y=x to show the inverse nature of the two functions.\" width=\"487\" height=\"433\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6.<\/b> The tangent function and inverse tangent (or arctangent) function<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Relations for Inverse Sine, Cosine, and Tangent Functions<\/h3>\n<p>For angles in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], if [latex]\\sin y=x[\/latex], then [latex]\\sin^{\u22121}x=y[\/latex].<\/p>\n<p>For angles in the interval [0, \u03c0], if [latex]\\cos y=x[\/latex], then [latex]\\cos^{\u22121}x=y[\/latex].<\/p>\n<p>For angles in the interval [latex]\\left(\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right)[\/latex], if [latex]\\tan y=x[\/latex], then [latex]\\tan^{\u22121}x=y[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Writing a Relation for an Inverse Function<\/h3>\n<p>Given [latex]\\sin\\left(\\frac{5\\pi}{12}\\right)\\approx 0.96593[\/latex], write a relation involving the inverse sine.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q641490\">Show Solution<\/span><\/p>\n<div id=\"q641490\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the relation for the inverse sine. If [latex]\\sin y=x[\/latex], then [latex]\\sin^{\u22121}x=y[\/latex].<\/p>\n<p>In this problem, [latex]x=0.96593[\/latex], and [latex]y=\\frac{5\\pi}{12}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\sin^{\u22121}(0.96593)\\approx \\frac{5\\pi}{12}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given [latex]\\cos(0.5)\\approx 0.8776[\/latex], write a relation involving the inverse cosine.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q359839\">Show Solution<\/span><\/p>\n<div id=\"q359839\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\arccos(0.8776)\\approx0.5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding the Exact Value of Expressions Involving the Inverse Sine, Cosine, and Tangent Functions<\/h2>\n<p>Now that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the inverse functions when we are using the special angles, specifically [latex]\\frac{\\pi}{ 6} (30^\\circ)\\text{, }\\frac{\\pi}{ 4} (45^\\circ),\\text{ and } \\frac{\\pi}{ 3} (60^\\circ)[\/latex], and their reflections into other quadrants.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a \u201cspecial\u201d input value, evaluate an inverse trigonometric function.<\/h3>\n<ol>\n<li>Find angle\u00a0<em>x<\/em>\u00a0for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function.<\/li>\n<li>If\u00a0<em>x<\/em>\u00a0is not in the defined range of the inverse, find another angle\u00a0<em>y<\/em>\u00a0that is in the defined range and has the same sine, cosine, or tangent as\u00a0<em>x<\/em>, depending on which corresponds to the given inverse function.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Evaluating Inverse Trigonometric Functions for Special Input Values<\/h3>\n<p>Evaluate each of the following.<\/p>\n<p style=\"padding-left: 60px\">a. [latex]\\sin\u22121\\left(\\frac{1}{2}\\right)[\/latex]<\/p>\n<p style=\"padding-left: 60px\">b. [latex]\\sin\u22121\\left(\u2212\\frac{2}{\\sqrt{2}}\\right)[\/latex]<\/p>\n<p style=\"padding-left: 60px\">c. [latex]\\cos\u22121\\left(\u2212\\frac{3}{\\sqrt{2}}\\right)[\/latex]<\/p>\n<p style=\"padding-left: 60px\">d. [latex]\\tan^{\u2212 1}(1)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q666370\">Show Solution<\/span><\/p>\n<div id=\"q666370\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"padding-left: 60px\">a. Evaluating [latex]\\sin^{\u22121}(\\frac{1}{2})[\/latex] is the same as determining the angle that would have a sine value of [latex]\\frac{1}{2}[\/latex]. In other words, what angle <em>x<\/em> would satisfy [latex]\\sin(x)=\\frac{1}{2}[\/latex]? There are multiple values that would satisfy this relationship, such as [latex]\\frac{\\pi}{6}[\/latex] and [latex]\\frac{5\\pi}{6}[\/latex], but we know we need the angle in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], so the answer will be [latex]\\sin^{\u22121}(\\frac{1}{2})=\\frac{\\pi}{6}[\/latex]. Remember that the inverse is a function, so for each input, we will get exactly one output.<\/p>\n<p style=\"padding-left: 60px\">b. To evaluate [latex]\\sin^{\u22121}\\left(\u2212\\frac{\\sqrt{2}}{2}\\right)[\/latex], we know that [latex]\\frac{5\\pi}{4}[\/latex] and [latex]\\frac{7\\pi}{4}[\/latex] both have a sine value of [latex]\u2212\\frac{\\sqrt{2}}{2}[\/latex], but neither is in the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex]. For that, we need the negative angle coterminal with [latex]\\frac{7\\pi}{4}:\\sin^{\u22121}\\left(\u2212\\frac{\\sqrt{2}}{2}\\right)=\u2212\\frac{\\pi}{4}[\/latex].<\/p>\n<p style=\"padding-left: 60px\">c. To evaluate [latex]\\cos^{\u22121}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)[\/latex], we are looking for an angle in the interval [0,\u03c0] with a cosine value of [latex]\u2212\\frac{\\sqrt{3}}{2}[\/latex]. The angle that satisfies this is [latex]\\cos^{\u22121}\\left(\u2212\\frac{\\sqrt{3}}{2}\\right)=\\frac{5\\pi}{6}[\/latex].<\/p>\n<p style=\"padding-left: 60px\">d. Evaluating [latex]\\tan^{\u22121}(1)[\/latex], we are looking for an angle in the interval [latex](\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2})[\/latex] with a tangent value of 1. The correct angle is [latex]\\tan^{\u22121}(1)=\\frac{\\pi}{4}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Evaluate each of the following.<\/p>\n<ol>\n<li>[latex]\\sin^{\u22121}(\u22121)[\/latex]<\/li>\n<li>[latex]\\tan^{\u22121}(\u22121)[\/latex]<\/li>\n<li>[latex]\\cos^{\u22121}(\u22121)[\/latex]<\/li>\n<li>[latex]\\cos^{\u22121}(\\frac{1}{2})[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q333778\">Show Solution<\/span><\/p>\n<div id=\"q333778\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. [latex]\u2212\\frac{\\pi}{2}[\/latex];<\/p>\n<p>2. [latex]\u2212\\frac{\\pi}{4}[\/latex]<\/p>\n<p>3. [latex]\\pi[\/latex]<\/p>\n<p>4. [latex]\\frac{\\pi}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173433\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173433&theme=oea&iframe_resize_id=ohm173433\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using a Calculator to Evaluate Inverse Trigonometric Functions<\/h2>\n<p>To evaluate <strong>inverse trigonometric functions<\/strong> that do not involve the special angles discussed previously, we will need to use a calculator or other type of technology. Most scientific calculators and calculator-emulating applications have specific keys or buttons for the inverse sine, cosine, and tangent functions. These may be labeled, for example, SIN-1, ARCSIN, or ASIN.<\/p>\n<p>In the previous chapter, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trigonometric functions, we can solve for the angles of a right triangle given two sides, and we can use a calculator to find the values to several decimal places.<\/p>\n<p>In these examples and exercises, the answers will be interpreted as angles and we will use \u03b8 as the independent variable. The value displayed on the calculator may be in degrees or radians, so be sure to set the mode appropriate to the application.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 3: Evaluating the Inverse Sine on a Calculator<\/h3>\n<p>Evaluate [latex]\\sin^{\u22121}(0.97)[\/latex] using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q931769\">Show Solution<\/span><\/p>\n<div id=\"q931769\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because the output of the inverse function is an angle, the calculator will give us a degree value if in degree mode and a radian value if in radian mode. Calculators also use the same domain restrictions on the angles as we are using.<\/p>\n<p>In radian mode, [latex]\\sin^{\u22121}(0.97)\\approx1.3252[\/latex]. In degree mode, [latex]\\sin^{\u22121}(0.97)\\approx75.93^{\\circ}[\/latex]. Note that in calculus and beyond we will use radians in almost all cases.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\cos^{\u22121}(\u22120.4)[\/latex] using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q728477\">Show Solution<\/span><\/p>\n<div id=\"q728477\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.9823 or 113.578\u00b0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173435\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173435&theme=oea&iframe_resize_id=ohm173435\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given two sides of a right triangle like the one shown in Figure 7, find an angle.<\/h3>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164012\/CNX_Precalc_Figure_06_03_012.jpg\" alt=\"An illustration of a right triangle with an angle theta. Adjacent to theta is the side a, opposite theta is the side p, and the hypoteneuse is side h.\" width=\"487\" height=\"248\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<ol>\n<li>If one given side is the hypotenuse of length h and the side of length a adjacent to the desired angle is given, use the equation [latex]\\theta=\\cos^{\u22121}\\left(\\frac{a}{h}\\right)[\/latex].<\/li>\n<li>If one given side is the hypotenuse of length <em>h<\/em> and the side of length <em>p<\/em> opposite to the desired angle is given, use the equation [latex]\\theta=\\sin^{\u22121}\\left(\\frac{p}{h}\\right)[\/latex].<\/li>\n<li>If the two legs (the sides adjacent to the right angle) are given, then use the equation [latex]\\theta=\\tan^{\u22121}\\left(\\frac{p}{a}\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Applying the Inverse Cosine to a Right Triangle<\/h3>\n<p>Solve the triangle in Figure 8\u00a0for the angle \u03b8.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164015\/CNX_Precalc_Figure_06_03_007.jpg\" alt=\"An illustration of a right triangle with the angle theta. Adjacent to the angle theta is a side with a length of 9 and a hypoteneuse of length 12.\" width=\"487\" height=\"200\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q24088\">Show Solution<\/span><\/p>\n<div id=\"q24088\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&\\cos\\theta=\\frac{9}{12}\\\\ &\\theta=\\cos^{\u22121}\\left(\\frac{9}{12}\\right) && \\text{Apply definition of the inverse.} \\\\ &\\theta\\approx0.7227\\text{ or about }41.4096^{\\circ} && \\text{Evaluate.} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve the triangle in Figure 9\u00a0for the angle \u03b8.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164017\/CNX_Precalc_Figure_06_03_008.jpg\" alt=\"An illustration of a right triangle with the angle theta. Opposite to the angle theta is a side with a length of 6 and a hypoteneuse of length 10.\" width=\"487\" height=\"137\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q230605\">Show Solution<\/span><\/p>\n<div id=\"q230605\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\sin^{\u22121}(0.6)=36.87^{\\circ}=0.6435[\/latex] radians<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm129737\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129737&theme=oea&iframe_resize_id=ohm129737\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Finding Exact Values of Composite Functions with Inverse Trigonometric Functions<\/h2>\n<p>There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let <em>f<\/em>(<em>x<\/em>) and <em>g<\/em>(<em>x<\/em>) be two different trigonometric functions belonging to the set {sin(<em>x<\/em>), cos(<em>x<\/em>), tan(<em>x<\/em>)} and let [latex]f^{\u22121}(y)[\/latex] and [latex]g^{\u22121}(y)[\/latex] be their inverses.<\/p>\n<h3>Evaluating Compositions of the Form [latex]f\\left(f^{\u22121}(y)\\right)[\/latex] and [latex]f^{\u22121}(f(x))[\/latex]<\/h3>\n<p>For any trigonometric function, [latex]f(f^{\u22121}(y))=y[\/latex] for all <em>y<\/em> in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of <em>f<\/em> was defined to be identical to the domain of [latex]f^{\u22121}[\/latex]. However, we have to be a little more careful with expressions of the form [latex]f^{\u22121}(f(x))[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Compositions of a trigonometric function and its inverse<\/h3>\n<p style=\"text-align: center\">[latex]\\begin{align} &\\sin(\\sin^{\u22121}x)=x\\text{ for }\u22121\\leq x\\leq1\\\\ &\\cos(\\cos^{\u22121}x)=x\\text{ for }\u2212\\infty\\leq x\\leq1 \\\\ &\\tan(\\tan^{\u22121}x)=x\\text{ for }\u2212\\infty\\text{ < }x\\text{ < }\\infty \\end{align}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} \\hfill &\\sin^{\u22121}(\\sin x)=x\\text{ only for }\u2212\\frac{\\pi}{2} \\leq x \\leq \\frac{\\pi}{2} \\hfill \\\\ &\\cos^{\u22121}(\\cos x)=x\\text{ only for }0\\leq x\\leq\\pi \\hfill \\\\ &\\tan^{\u22121}(\\tan x)=x\\text{ only for }\u2212\\frac{\\pi}{2}\\text{ < }x\\text{ < }\\frac{\\pi}{2} \\end{align}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>Is it correct that [latex]\\sin^{\u22121}(\\sin x)=x[\/latex]?<\/h3>\n<p><em>No. This equation is correct if x belongs to the restricted domain [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex], but sine is defined for all real input values, and for x outside the restricted interval, the equation is not correct because its inverse always returns a value in\u00a0<em>[latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\\latex]<\/em>. The situation is similar for cosine and tangent and their inverses. For example, [latex]\\sin^{\u22121}\\left(\\sin\\left(\\frac{3\\pi}{4}\\right)\\right)=\\frac{\\pi}{4}[\/latex]. <\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To:<\/h3>\n<p>Given an expression of the form [latex]f^{\u22121}(f(\\theta))[\/latex] where [latex]f(\\theta)=\\sin\\theta\\text{, }\\cos\\theta\\text{, or }\\tan\\theta[\/latex], evaluate.<\/p>\n<ol>\n<li>If \u03b8 is in the restricted domain of <em>f<\/em>,\u00a0then [latex]f^{\u22121}(f(\\theta))=\\theta[\/latex].<\/li>\n<li>If not, then find an angle \u03d5 within the restricted domain of <em>f<\/em> such that [latex]f(\\phi)=f(\\theta)[\/latex]. Then [latex]f^{\u22121}(f(\\theta))=\\phi[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Using Inverse Trigonometric Functions<\/h3>\n<p>Evaluate the following:<\/p>\n<ol>\n<li>[latex]\\sin^{\u22121}(\\sin(\\frac{\\pi}{3}))[\/latex]<\/li>\n<li>[latex]\\sin^{\u22121}(\\sin(\\frac{2\\pi}{3}))[\/latex]<\/li>\n<li>[latex]\\cos^{\u22121}(\\cos(\\frac{2\\pi}{3}))[\/latex]<\/li>\n<li>[latex]\\cos^{\u22121}(\\cos(\u2212\\frac{\\pi}{3}))[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q611200\">Show Solution<\/span><\/p>\n<div id=\"q611200\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\frac{\\pi}{3}[\/latex] is in [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], so [latex]\\sin^{\u22121}(\\sin(\\frac{\\pi}{3}))=\\frac{\\pi}{3}[\/latex].<\/li>\n<li>[latex]\\frac{2\\pi}{3}[\/latex] is not in [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], but [latex]\\sin\\left(\\frac{2\\pi}{3}\\right)=\\sin\\left(\\frac{\\pi}{3}\\right)[\/latex], so [latex]\\sin^{\u22121}\\left(\\sin\\left(\\frac{2\\pi}{3}\\right)\\right)=\\frac{\\pi}{3}[\/latex].<\/li>\n<li>[latex]\\frac{2\\pi}{3}[\/latex] is in [0,\u03c0], so [latex]\\cos^{\u22121}\\left(\\cos\\left(\\frac{2\\pi}{3}\\right)\\right)=\\frac{2\\pi}{3}[\/latex].<\/li>\n<li>[latex]\u2212\\frac{\\pi}{3}[\/latex] is not in [0,\u03c0], but [latex]\\cos(\u2212\\frac{\\pi}{3})=\\cos\\left(\\frac{\\pi}{3}\\right)[\/latex] because cosine is an even function.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\tan^{\u22121}\\left(\\tan\\left(\\frac{\\pi}{8}\\right)\\right)[\/latex] and [latex]\\tan^{\u22121}\\left(\\tan\\left(\\frac{11\\pi}{9}\\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q356884\">Show Solution<\/span><\/p>\n<div id=\"q356884\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\pi}{8}\\text{; }\\frac{2\\pi}{9}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Evaluating Compositions of the Form\u00a0[latex]f^{\u22121}(g(x))[\/latex]<\/h2>\n<p>Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form [latex]f^{\u22121}(g(x))[\/latex]. For special values of <em>x<\/em>, we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is \u03b8, making the other [latex]\\frac{\\pi}{2}\u2212\\theta[\/latex]. Consider the sine and cosine of each angle of the right triangle in Figure 10.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164019\/CNX_Precalc_Figure_06_03_009.jpg\" alt=\"An illustration of a right triangle with angles theta and pi\/2 - theta. Opposite the angle theta and adjacent the angle pi\/2-theta is the side a. Adjacent the angle theta and opposite the angle pi\/2 - theta is the side b. The hypoteneuse is labeled c.\" width=\"487\" height=\"195\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10.<\/b> Right triangle illustrating the cofunction relationships<\/p>\n<\/div>\n<p>Because [latex]\\cos\\theta=\\frac{b}{c}=\\sin\\left(\\frac{\\pi}{2}\u2212\\theta\\right)[\/latex], we have [latex]\\sin^{\u22121}(\\cos\\theta)=\\frac{\\pi}{2}\u2212\\theta\\text{ if }0\\leq\\theta\\leq\\pi[\/latex]. If \u03b8 is not in this domain, then we need to find another angle that has the same cosine as \u03b8 and does belong to the restricted domain; we then subtract this angle from [latex]\\frac{\\pi}{2}[\/latex]. Similarly, [latex]\\sin\\theta=\\frac{a}{c}=\\cos\\left(\\frac{\\pi}{2}\u2212\\theta\\right)[\/latex], so [latex]\\cos^{\u22121}(\\sin\\theta)=\\frac{\\pi}{2}\u2212\\theta\\text{ if }\u2212\\frac{\\pi}{2}\\leq\\theta\\leq\\frac{\\pi}{2}[\/latex]. These are just the function-cofunction relationships presented in another way.<\/p>\n<div class=\"textbox\">\n<h3>How To:\u00a0Given functions of the form [latex]\\sin^{\u22121}(\\cos x)\\text{ and }\\cos^{\u22121}(\\sin x)[\/latex], evaluate them.<\/h3>\n<ol>\n<li>If <em>x<\/em>\u00a0is\u00a0in\u00a0[0,\u03c0], then [latex]\\sin^{\u22121}(\\cos x)=\\frac{\\pi}{2}\u2212x[\/latex].<\/li>\n<li>If <em>x<\/em>\u00a0is\u00a0not\u00a0in\u00a0[0,\u03c0], then find another angle <em>y<\/em>\u00a0in\u00a0[0,\u03c0] such that [latex]\\cos y=\\cos x[\/latex].\n<div>\n<div style=\"text-align: center\">[latex]\\sin^{\u22121}(\\cos x)=\\frac{\\pi}{2}\u2212y[\/latex]<\/div>\n<\/div>\n<\/li>\n<li>If <em>x<\/em>\u00a0is\u00a0in [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex], then [latex]\\cos^{\u22121}(\\sin x)=\\frac{\\pi}{2}\u2212x[\/latex].<\/li>\n<li>If <em>x<\/em>\u00a0is\u00a0not\u00a0in [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex], then find another angle <em>y<\/em>\u00a0in [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex] such that [latex]\\sin y=\\sin x[\/latex].\n<div>\n<div style=\"text-align: center\">[latex]\\cos^{\u22121}(\\sin x)=\\frac{\\pi}{2}\u2212y[\/latex]<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Evaluating the Composition of an Inverse Sine with a Cosine<\/h3>\n<p>Evaluate [latex]\\sin^{\u22121}(\\cos(\\frac{13\\pi}{6}))[\/latex]<\/p>\n<ol>\n<li>by direct evaluation.<\/li>\n<li>by the method described previously.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q651517\">Show Solution<\/span><\/p>\n<div id=\"q651517\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Here, we can directly evaluate the inside of the composition.\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos\\left(\\frac{13\\pi}{6}\\right)&=\\cos\\left(\\frac{\\pi}{6}+2\\pi\\right) \\\\ &=\\cos\\left(\\frac{\\pi}{6}\\right) \\\\ &=\\frac{\\sqrt{3}}{2} \\end{align}[\/latex]<\/div>\n<p>Now, we can evaluate the inverse function as we did earlier.<\/p>\n<div style=\"text-align: center\">[latex]\\sin^{\u22121}\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{\\pi}{3}[\/latex]<\/div>\n<\/li>\n<li>We have [latex]x=\\frac{13\\pi}{6}[\/latex], [latex]y=\\frac{\\pi}{6}[\/latex], and\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin^{\u22121}\\left(\\cos\\left(\\frac{13\\pi}{6}\\right)\\right)=\\frac{\\pi}{2}\u2212\\frac{\\pi}{6} =\\frac{\\pi}{3} \\end{align}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\cos^{\u22121}(\\sin(\u2212\\frac{11\\pi}{4}))[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q325594\">Show Solution<\/span><\/p>\n<div id=\"q325594\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{3\\pi}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm129738\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129738&theme=oea&iframe_resize_id=ohm129738\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Evaluating Compositions of the Form [latex]f(g^{\u22121}(x))[\/latex]<\/h2>\n<p>To evaluate compositions of the form [latex]f(g^{\u22121}(x))[\/latex], where <em>f<\/em> and <em>g<\/em> are any two of the functions sine, cosine, or tangent and <em>x<\/em> is any input in the domain of [latex]g\u22121[\/latex], we have exact formulas, such as [latex]\\sin\\left({\\cos}^{\u22121}x\\right)=\\sqrt{1\u2212{x}^{2}}[\/latex]. When we need to use them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagoras\u2019s relation between the lengths of the sides. We can use the Pythagorean identity, [latex]\\sin^{2}x+cos^{2}x=1[\/latex], to solve for one when given the other. We can also use the <strong>inverse trigonometric functions<\/strong> to find compositions involving algebraic expressions.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 7: Evaluating the Composition of a Sine with an Inverse Cosine<\/h3>\n<p>Find an exact value for [latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{4}{5}\\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q530604\">Show Solution<\/span><\/p>\n<div id=\"q530604\" class=\"hidden-answer\" style=\"display: none\">\n<p>Beginning with the inside, we can say there is some angle such that [latex]\\theta=\\cos^{\u22121}(\\frac{4}{5})[\/latex], which means [latex]\\cos\\theta=\\frac{4}{5}[\/latex], and we are looking for [latex]\\sin\\theta[\/latex]. We can use the Pythagorean identity to do this.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} &\\sin^{2}\\theta+\\cos^{2}\\theta=1 && \\text{Use our known value for cosine.} \\\\ &\\sin^{2}\\theta+\\left(\\frac{4}{5}\\right)^{2}=1 && \\text{Solve for sine.} \\\\ &\\sin^{2}\\theta=1\u2212\\frac{16}{25} \\\\ &\\sin\\theta=\\pm\\sqrt{\\frac{9}{25}}=\\pm\\frac{3}{5} \\end{align}[\/latex]<\/p>\n<p>Since [latex]\\theta=\\cos^{\u22121}(\\frac{4}{5})[\/latex] is in quadrant I, [latex]\\sin{\\theta}[\/latex] must be positive, so the solution is [latex]\\frac{3}{5}[\/latex]. See Figure 11.<\/p>\n<p style=\"text-align: center\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164021\/CNX_Precalc_Figure_06_03_010.jpg\" alt=\"An illustration of a right triangle with an angle theta. Oppostie the angle theta is a side with length 3. Adjacent the angle theta is a side with length 4. The hypoteneuse has angle of length 5.\" \/><\/p>\n<p style=\"text-align: center\"><strong>Figure 11.<\/strong> Right triangle illustrating that if [latex]\\cos\\theta=\\frac{4}{5}[\/latex], then [latex]\\sin\\theta=\\frac{3}{5}[\/latex]<\/p>\n<p>We know that the inverse cosine always gives an angle on the interval [0,\u00a0\u03c0], so we know that the sine of that angle must be positive; therefore [latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{4}{5}\\right)\\right)=\\sin\\theta=\\frac{3}{5}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\cos(\\tan^{\u22121}(\\frac{5}{12}))[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q754416\">Show Solution<\/span><\/p>\n<div id=\"q754416\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{12}{13}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Evaluating the Composition of a Sine with an Inverse Tangent<\/h3>\n<p>Find an exact value for [latex]\\sin\\left(\\tan^{\u22121}\\left(\\frac{7}{4}\\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q488532\">Show Solution<\/span><\/p>\n<div id=\"q488532\" class=\"hidden-answer\" style=\"display: none\">\n<p>While we could use a similar technique as in Example 6, we will demonstrate a different technique here. From the inside, we know there is an angle such that [latex]\\tan\\theta=\\frac{7}{4}[\/latex]. We can envision this as the opposite and adjacent sides on a right triangle, as shown in Figure 12.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164023\/CNX_Precalc_Figure_06_03_011n.jpg\" alt=\"An illustration of a right triangle with angle theta. Adjacent the angle theta is a side with length 4. Opposite the angle theta is a side with length 7.\" width=\"487\" height=\"196\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 12.<\/b> A right triangle with two sides known<\/p>\n<\/div>\n<p>Using the Pythagorean Theorem, we can find the hypotenuse of this triangle.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}4^{2}+7^{2}=\\text{hypotenuse}^{2} \\\\ \\text{hypotenuse}=\\sqrt{65} \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\cos(\\sin^{\u22121}(\\frac{7}{9}))[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q930711\">Show Solution<\/span><\/p>\n<div id=\"q930711\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{4\\sqrt{2}}{9}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm129751\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129751&theme=oea&iframe_resize_id=ohm129751\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 9: Finding the Cosine of the Inverse Sine of an Algebraic Expression<\/h3>\n<p>Find a simplified expression for [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{x}{3}\\right)\\right)[\/latex] for [latex]\u22123\\leq x\\leq3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q764834\">Show Solution<\/span><\/p>\n<div id=\"q764834\" class=\"hidden-answer\" style=\"display: none\">\n<p>We know there is an angle\u00a0\u03b8 such that [latex]\\sin\\theta=\\frac{x}{3}\\\\[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&\\sin^{2}\\theta+\\cos^{2}\\theta=1 7&& \\text{Use the Pythagorean Theorem.} \\\\ &\\left(\\frac{x}{3}\\right)^{2}+\\cos^{2}+\\cos^2\\theta=1 && \\text{Solve for cosine.} \\\\ &\\cos^{2}\\theta=1\u2212\\frac{x^{2}}{9} \\\\ &\\cos\\theta=\\pm\\sqrt{\\frac{9\u2212x^{2}}{9}}=\\pm\\frac{\\sqrt{9\u2212x^{2}}}{3} \\end{align}[\/latex]<\/p>\n<p>Because we know that the inverse sine must give an angle on the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], we can deduce that the cosine of that angle must be positive.<\/p>\n<p style=\"text-align: center\">[latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{x}{3}\\right)\\right)=\\frac{\\sqrt{9\u2212x^{2}}}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find a simplified expression for [latex]\\sin\\left(\\tan^{\u22121}\\left(4x\\right)\\right)\\\\[\/latex] for [latex]\u2212\\frac{1}{4}\\leq x \\leq\\frac{1}{4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q979016\">Show Solution<\/span><\/p>\n<div id=\"q979016\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{4x}{\\sqrt{16x^{2}+1}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm129755\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129755&theme=oea&iframe_resize_id=ohm129755\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solving Linear Trigonometric Equations in Sine and Cosine<\/h2>\n<p>Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The <strong>period<\/strong> of both the sine function and the cosine function is [latex]2\\pi[\/latex]. In other words, every [latex]2\\pi[\/latex] units, the <em>y-<\/em>values repeat. If we need to find all possible solutions, then we must add [latex]2\\pi k[\/latex], where [latex]k[\/latex] is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\\pi :[\/latex]<\/p>\n<div style=\"text-align: center\">[latex]\\sin \\theta =\\sin \\left(\\theta \\pm 2k\\pi \\right)[\/latex]<\/div>\n<p>There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 10: Solving a Linear Trigonometric Equation Involving the Cosine Function<\/h3>\n<p>Find all possible exact solutions for the equation [latex]\\cos \\theta =\\frac{1}{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q84784\">Show Solution<\/span><\/p>\n<div id=\"q84784\" class=\"hidden-answer\" style=\"display: none\">\n<p>From the <strong>unit circle<\/strong>, we know that<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\cos \\theta =\\frac{1}{2} \\\\ \\theta =\\frac{\\pi }{3},\\frac{5\\pi }{3} \\end{gathered}[\/latex]<\/p>\n<p>These are the solutions in the interval [latex]\\left[0,2\\pi \\right][\/latex]. All possible solutions are given by<\/p>\n<p style=\"text-align: center\">[latex]\\theta =\\frac{\\pi }{3}\\pm 2k\\pi \\text{ and }\\theta =\\frac{5\\pi }{3}\\pm 2k\\pi[\/latex]<\/p>\n<p>where [latex]k[\/latex] is an integer.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 11: Solving a Linear Equation Involving the Sine Function<\/h3>\n<p>Find all possible exact solutions for the equation [latex]\\sin t=\\frac{1}{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q535703\">Show Solution<\/span><\/p>\n<div id=\"q535703\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solving for all possible values of <em>t<\/em> means that solutions include angles beyond the period of [latex]2\\pi[\/latex]. From the unit circle, we can see that the solutions are [latex]t=\\frac{\\pi }{6}[\/latex] and [latex]t=\\frac{5\\pi }{6}[\/latex]. But the problem is asking for all possible values that solve the equation. Therefore, the answer is<\/p>\n<p style=\"text-align: center\">[latex]t=\\frac{\\pi }{6}\\pm 2\\pi k\\text{ and }t=\\frac{5\\pi }{6}\\pm 2\\pi k[\/latex]<\/p>\n<p>where [latex]k[\/latex] is an integer.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a trigonometric equation, solve using algebra.<\/h3>\n<ul>\n<li>Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.<\/li>\n<li>Substitute the trigonometric expression with a single variable, such as [latex]x[\/latex] or [latex]u[\/latex].<\/li>\n<li>Solve the equation the same way an algebraic equation would be solved.<\/li>\n<li>Substitute the trigonometric expression back in for the variable in the resulting expressions.<\/li>\n<li>Solve for the angle.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 12: Solve the Trigonometric Equation in Linear Form<\/h3>\n<p>Solve the equation exactly: [latex]2\\cos \\theta -3=-5,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q939405\">Show Solution<\/span><\/p>\n<p style=\"text-align: left\">\n<div id=\"q939405\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{gathered}2\\cos \\theta -3=-5 \\\\ \\cos \\theta =-2 \\\\ \\cos \\theta =-1 \\\\ \\theta =\\pi \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve exactly the following linear equation on the interval [latex]\\left[0,2\\pi \\right):2\\sin x+1=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q500341\">Show Solution<\/span><\/p>\n<div id=\"q500341\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149871\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149871&theme=oea&iframe_resize_id=ohm149871\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solve Trigonometric Equations Using a Calculator<\/h2>\n<p>Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 13: Using a Calculator to Solve a Trigonometric Equation Involving Sine<\/h3>\n<p>Use a calculator to solve the equation [latex]\\sin \\theta =0.8[\/latex], where [latex]\\theta[\/latex] is in radians.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q522194\">Show Solution<\/span><\/p>\n<div id=\"q522194\" class=\"hidden-answer\" style=\"display: none\">\n<h3><\/h3>\n<p>Make sure mode is set to radians. To find [latex]\\theta[\/latex], use the inverse sine function. On most calculators, you will need to push the 2<sup>ND<\/sup> button and then the SIN button to bring up the [latex]{\\sin }^{-1}[\/latex] function. What is shown on the screen is [latex]{\\sin}^{-1}[\/latex](. The calculator is ready for the input within the parentheses. For this problem, we enter [latex]{\\sin }^{-1}\\left(0.8\\right)[\/latex], and press ENTER. Thus, to four decimals places,<\/p>\n<p style=\"text-align: center\">[latex]{\\sin }^{-1}\\left(0.8\\right)\\approx 0.9273[\/latex]<\/p>\n<p>This is the solution in quadrant I. There is also a solution in quadrant II. To find this we subtract [latex]\/pi - 0.9273 \\approx 2.2143[\/latex]<\/p>\n<p>The general solution is<\/p>\n<p style=\"text-align: center\">[latex]\\theta \\approx 0.9273\\pm 2\\pi k \\text{ and } \\theta \\approx 2.2143 \\pm 2\\pi k[\/latex]<\/p>\n<p>The angle measurement in degrees is<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} \\theta &\\approx {53.1}^{\\circ } \\\\ \\theta &\\approx {180}^{\\circ }-{53.1}^{\\circ } \\\\ &\\approx {126.9}^{\\circ } \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using [latex]\\pi -\\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 14: Using a Calculator to Solve a Trigonometric Equation Involving Secant<\/h3>\n<p>Use a calculator to solve the equation [latex]\\sec \\theta =-4[\/latex], giving your answer in radians.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q209133\">Show Solution<\/span><\/p>\n<div id=\"q209133\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can begin with some algebra.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sec \\theta =-4\\\\ \\frac{1}{\\cos \\theta }=-4\\\\ \\cos \\theta =-\\frac{1}{4}\\end{gathered}[\/latex]<\/p>\n<p>Check that the MODE is in radians. Now use the inverse cosine function.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}{\\cos }^{-1}\\left(-\\frac{1}{4}\\right)\\approx 1.8235 \\\\ \\theta \\approx 1.8235+2\\pi k \\end{gathered}[\/latex]<\/p>\n<p>Since [latex]\\frac{\\pi }{2}\\approx 1.57[\/latex] and [latex]\\pi \\approx 3.14[\/latex], 1.8235 is between these two numbers, thus [latex]\\theta \\approx \\text{1}\\text{.8235}[\/latex] is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164111\/CNX_Precalc_Figure_07_05_0052.jpg\" alt=\"Graph of angles theta =approx 1.8235, theta prime =approx pi - 1.8235 = approx 1.3181, and then theta prime = pi + 1.3181 = approx 4.4597\" width=\"487\" height=\"380\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 13.<\/b><\/p>\n<\/div>\n<p>So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi -\\text{1}\\text{.8235}\\approx \\text{1}\\text{.3181}\\text{.}[\/latex] The other solution in quadrant III is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi +\\text{1}\\text{.3181}\\approx \\text{4}\\text{.4597}\\text{.}[\/latex]<\/p>\n<p>The solutions are [latex]\\theta \\approx 1.8235\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.4597\\pm 2\\pi k[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve [latex]\\cos \\theta =-0.2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q145806\">Show Solution<\/span><\/p>\n<div id=\"q145806\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\theta \\approx 1.7722\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.5110\\pm 2\\pi k[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149873\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149873&theme=oea&iframe_resize_id=ohm149873\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solving Equations Involving a Single Trigonometric Function<\/h2>\n<p>When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is [latex]\\pi[\/latex], not [latex]2\\pi[\/latex]. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of [latex]\\frac{\\pi }{2}[\/latex], unless, of course, a problem places its own restrictions on the domain.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 15: Solving a Problem Involving a Single Trigonometric Function<\/h3>\n<p>Solve the problem exactly: [latex]2{\\sin }^{2}\\theta -1=0,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q467313\">Show Solution<\/span><\/p>\n<div id=\"q467313\" class=\"hidden-answer\" style=\"display: none\">\n<p>As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\\sin \\theta[\/latex]. Then we will find the angles.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta -1=0 \\\\ 2{\\sin }^{2}\\theta =1 \\\\ {\\sin }^{2}\\theta =\\frac{1}{2} \\\\ \\sqrt{{\\sin }^{2}\\theta }=\\pm \\sqrt{\\frac{1}{2}} \\\\ \\sin \\theta =\\pm \\frac{1}{\\sqrt{2}}=\\pm \\frac{\\sqrt{2}}{2} \\\\ \\theta =\\frac{\\pi }{4},\\frac{3\\pi }{4},\\frac{5\\pi }{4},\\frac{7\\pi }{4} \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 16: Solving a Trigonometric Equation Involving Cosecant<\/h3>\n<p>Solve the following equation exactly: [latex]\\csc \\theta =-2,0\\le \\theta <4\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q605306\">Show Solution<\/span><\/p>\n<div id=\"q605306\" class=\"hidden-answer\" style=\"display: none\">\n<p>We want all values of [latex]\\theta[\/latex] for which [latex]\\csc \\theta =-2[\/latex] over the interval [latex]0\\le \\theta <4\\pi[\/latex].\n\n\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\csc \\theta =-2 \\\\ \\frac{1}{\\sin \\theta }=-2 \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6},\\frac{19\\pi }{6},\\frac{23\\pi }{6} \\end{gathered}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>As [latex]\\sin \\theta =-\\frac{1}{2}[\/latex], notice that all four solutions are in the third and fourth quadrants.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 17: Solving an Equation Involving Tangent<\/h3>\n<p>Solve the equation exactly: [latex]\\tan \\left(\\theta -\\frac{\\pi }{2}\\right)=1,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q484899\">Show Solution<\/span><\/p>\n<div id=\"q484899\" class=\"hidden-answer\" style=\"display: none\">\n<p>Recall that the tangent function has a period of [latex]\\pi[\/latex]. On the interval [latex]\\left[0,\\pi \\right)[\/latex], and at the angle of [latex]\\frac{\\pi }{4}[\/latex], the tangent has a value of 1. However, the angle we want is [latex]\\left(\\theta -\\frac{\\pi }{2}\\right)[\/latex]. Thus, if [latex]\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex], then<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\theta -\\frac{\\pi }{2}=\\frac{\\pi }{4}\\\\ \\theta =\\frac{3\\pi }{4}\\pm k\\pi \\end{gathered}[\/latex]<\/p>\n<p>Over the interval [latex]\\left[0,2\\pi \\right)[\/latex], we have two solutions:<\/p>\n<p style=\"text-align: center\">[latex]\\theta =\\frac{3\\pi }{4}\\text{ and }\\theta =\\frac{3\\pi }{4}+\\pi =\\frac{7\\pi }{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find all solutions for [latex]\\tan x=\\sqrt{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q629684\">Show Solution<\/span><\/p>\n<div id=\"q629684\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\pi }{3}\\pm \\pi k[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173739\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173739&theme=oea&iframe_resize_id=ohm173739\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 18: Identify all Solutions to the Equation Involving Tangent<\/h3>\n<p>Identify all exact solutions to the equation [latex]2\\left(\\tan x+3\\right)=5+\\tan x,0\\le x<2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q949694\">Show Solution<\/span><\/p>\n<div id=\"q949694\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can solve this equation using only algebra. Isolate the expression [latex]\\tan x[\/latex] on the left side of the equals sign.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} 2\\left(\\tan x\\right)+2\\left(3\\right) =5+\\tan x \\\\ 2\\tan x+6 =5+\\tan x \\\\ 2\\tan x-\\tan x =5 - 6 \\\\ \\tan x =-1\\end{gathered}[\/latex]<\/p>\n<p>There are two angles on the unit circle that have a tangent value of [latex]-1:\\theta =\\frac{3\\pi }{4}[\/latex] and [latex]\\theta =\\frac{7\\pi }{4}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solving Trigonometric Equations in Quadratic Form<\/h2>\n<p>Solving a <strong>quadratic equation<\/strong> may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as [latex]x[\/latex] or [latex]u[\/latex]. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 19: Solving a Trigonometric Equation in Quadratic Form<\/h3>\n<p>Solve the equation exactly: [latex]{\\cos }^{2}\\theta +3\\cos \\theta -1=0,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q563509\">Show Solution<\/span><\/p>\n<div id=\"q563509\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by using substitution and replacing cos [latex]\\theta[\/latex] with [latex]x[\/latex]. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\\cos \\theta =x[\/latex]. We have<\/p>\n<p style=\"text-align: center\">[latex]{x}^{2}+3x - 1=0[\/latex]<\/p>\n<p>The equation cannot be factored, so we will use the <strong>quadratic formula<\/strong> [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} x&=\\frac{-3\\pm \\sqrt{{\\left(-3\\right)}^{2}-4\\left(1\\right)\\left(-1\\right)}}{2}&=\\frac{-3\\pm \\sqrt{13}}{2} \\end{align}[\/latex]<\/p>\n<p>Replace [latex]x[\/latex] with [latex]\\cos \\theta[\/latex], and solve. Thus,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\cos \\theta =\\frac{-3\\pm \\sqrt{13}}{2}\\theta ={\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right)\\end{gathered}[\/latex]<\/p>\n<p>Note that only the + sign is used. This is because we get an error when we solve [latex]\\theta ={\\cos }^{-1}\\left(\\frac{-3-\\sqrt{13}}{2}\\right)[\/latex] on a calculator, since the domain of the inverse cosine function is [latex]\\left[-1,1\\right][\/latex]. However, there is a second solution:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\theta &={\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\\\ &\\approx 1.26 \\end{align}[\/latex]<\/p>\n<p>This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\theta &=2\\pi -{\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\\\ &\\approx 5.02 \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 20: Solving a Trigonometric Equation in Quadratic Form by Factoring<\/h3>\n<p>Solve the equation exactly: [latex]2{\\sin }^{2}\\theta -5\\sin \\theta +3=0,0\\le \\theta \\le 2\\pi[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q693370\">Show Solution<\/span><\/p>\n<div id=\"q693370\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using grouping, this quadratic can be factored. Either make the real substitution, [latex]\\sin \\theta =u[\/latex], or imagine it, as we factor:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta -5\\sin \\theta +3=0 \\\\ \\left(2\\sin \\theta -3\\right)\\left(\\sin \\theta -1\\right)=0 \\end{gathered}[\/latex]<\/p>\n<p>Now set each factor equal to zero.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}2\\sin \\theta -3=0 \\\\ 2\\sin \\theta =3 \\\\ \\sin \\theta =\\frac{3}{2} \\\\ \\text{ } \\\\ \\sin \\theta -1=0 \\\\ \\sin \\theta =1 \\end{gathered}[\/latex]<\/p>\n<p>Next solve for [latex]\\theta :\\sin \\theta \\ne \\frac{3}{2}[\/latex], as the range of the sine function is [latex]\\left[-1,1\\right][\/latex]. However, [latex]\\sin \\theta =1[\/latex], giving the solution [latex]\\theta =\\frac{\\pi }{2}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Make sure to check all solutions on the given domain as some factors have no solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve [latex]{\\sin }^{2}\\theta =2\\cos \\theta +2,0\\le \\theta \\le 2\\pi[\/latex]. [Hint: Make a substitution to express the equation only in terms of cosine.]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q608693\">Show Solution<\/span><\/p>\n<div id=\"q608693\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\cos \\theta =-1,\\theta =\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149010\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149010&theme=oea&iframe_resize_id=ohm149010\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 21: Solving a Trigonometric Equation Using Algebra<\/h3>\n<p>Solve exactly:<\/p>\n<p style=\"text-align: center\">[latex]2{\\sin }^{2}\\theta +\\sin \\theta =0;0\\le \\theta <2\\pi[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q204542\">Show Solution<\/span><\/p>\n<div id=\"q204542\" class=\"hidden-answer\" style=\"display: none\">\n<p>This problem should appear familiar as it is similar to a quadratic. Let [latex]\\sin \\theta =x[\/latex]. The equation becomes [latex]2{x}^{2}+x=0[\/latex]. We begin by factoring:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}2{x}^{2}+x=0\\\\ x\\left(2x+1\\right)=0\\end{gathered}[\/latex]<\/p>\n<p>Set each factor equal to zero.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}x=0 \\\\ 2x+1=0 \\\\ x=-\\frac{1}{2} \\end{gathered}[\/latex]<\/p>\n<p>Then, substitute back into the equation the original expression [latex]\\sin \\theta[\/latex] for [latex]x[\/latex]. Thus,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sin \\theta =0 \\\\ \\theta =0,\\pi \\\\ \\text{ } \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6} \\end{gathered}[\/latex]<\/p>\n<p>The solutions within the domain [latex]0\\le \\theta <2\\pi[\/latex] are [latex]\\theta =0,\\pi ,\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex].\n\nIf we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.\n\n\n<p style=\"text-align: center\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta +\\sin \\theta =0 \\\\ \\sin \\theta \\left(2\\sin \\theta +1\\right)=0 \\\\ \\sin \\theta =0 \\\\ \\theta =0,\\pi \\\\ \\text{ } \\\\ 2\\sin \\theta +1=0 \\\\ 2\\sin \\theta =-1 \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6} \\end{gathered}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can see the solutions on the graph in Figure 3. On the interval [latex]0\\le \\theta <2\\pi[\/latex], the graph crosses the <em>x-<\/em>axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164113\/CNX_Precalc_Figure_07_05_0042.jpg\" alt=\"Graph of 2*(sin(theta))^2 + sin(theta) from 0 to 2pi. Zeros are at 0, pi, 7pi\/6, and 11pi\/6.\" width=\"731\" height=\"256\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 14<\/b><\/p>\n<\/div>\n<p>We can verify the solutions on the <a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/chapter\/use-sum-and-difference-formulas-for-cosine\/\" target=\"_blank\" rel=\"noopener\">unit circle<\/a> in Sum and Difference Identities\u00a0as well.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 22: Solving a Trigonometric Equation Quadratic in Form<\/h3>\n<p>Solve the equation quadratic in form exactly: [latex]2{\\sin }^{2}\\theta -3\\sin \\theta +1=0,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q699307\">Show Solution<\/span><\/p>\n<div id=\"q699307\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can factor using grouping. Solution values of [latex]\\theta[\/latex] can be found on the unit circle:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\left(2\\sin \\theta -1\\right)\\left(\\sin \\theta -1\\right)=0 \\\\ 2\\sin \\theta -1=0 \\\\ \\sin \\theta =\\frac{1}{2} \\\\ \\theta =\\frac{\\pi }{6},\\frac{5\\pi }{6} \\\\ \\text{ } \\\\ \\sin \\theta =1 \\\\ \\theta =\\frac{\\pi }{2} \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve the quadratic equation [latex]2{\\cos }^{2}\\theta +\\cos \\theta =0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q736189\">Show Solution<\/span><\/p>\n<div id=\"q736189\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\pi }{2},\\frac{2\\pi }{3},\\frac{4\\pi }{3},\\frac{3\\pi }{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>An inverse function is one that \u201cundoes\u201d another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function.<\/li>\n<li>Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains.<\/li>\n<li>For any trigonometric function [latex]f(x)[\/latex], if [latex]x=f^{\u22121}(y)[\/latex], then [latex]f(x)=y[\/latex]. However, [latex]f(x)=y[\/latex] only implies [latex]x=f^{\u22121}(y)[\/latex] if <em>x<\/em> is in the restricted domain of <em>f<\/em>.<\/li>\n<li>Special angles are the outputs of inverse trigonometric functions for special input values; for example, [latex]\\frac{\\pi}{4}=\\tan^{\u22121}( 1 )\\text{ and }\\frac{\\pi}{6}=\\sin^{\u22121}(\\frac{1}{2})[\/latex].<\/li>\n<li>A calculator will return an angle within the restricted domain of the original trigonometric function.<\/li>\n<li>Inverse functions allow us to find an angle when given two sides of a right triangle.<\/li>\n<li>In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example, [latex]\\sin\\left(\\cos^{\u22121}\\left(x\\right)\\right)=\\sqrt{1\u2212x^{2}}[\/latex].<\/li>\n<li>If the inside function is a trigonometric function, then the only possible combinations are [latex]\\sin^{\u22121}\\left(\\cos x\\right)=\\frac{\\pi}{2}\u2212x[\/latex] if [latex]0\\leq x\\leq\\pi[\/latex] and [latex]\\cos^{\u22121}\\left(\\sin x\\right)=\\frac{\\pi}{2}\u2212x[\/latex] if [latex]\u2212\\frac{\\pi}{2}\\leq x \\leq\\frac{\\pi}{2}[\/latex].<\/li>\n<li>When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function.<\/li>\n<li>When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides.<\/li>\n<li>When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution.<\/li>\n<li>Equations involving a single trigonometric function can be solved or verified using the unit circle.<\/li>\n<li>We can also solve trigonometric equations using a graphing calculator.<\/li>\n<li>Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc.<\/li>\n<\/ul>\n<div>\n<h2>Glossary<\/h2>\n<dl class=\"definition\">\n<dt>arccosine<\/dt>\n<dd>another name for the inverse cosine; [latex]\\arccos x=\\cos^{\u22121}x[\/latex]<\/dd>\n<\/dl>\n<dl class=\"definition\">\n<dt>arcsine<\/dt>\n<dd>another name for the inverse sine; [latex]\\arcsin x=\\sin^{\u22121}x[\/latex]<\/dd>\n<\/dl>\n<dl class=\"definition\">\n<dt>arctangent<\/dt>\n<dd>another name for the inverse tangent; [latex]\\arctan x=\\tan^{\u22121}x[\/latex]<\/dd>\n<\/dl>\n<dl class=\"definition\">\n<dt>inverse cosine function<\/dt>\n<dd>the function [latex]\\cos^{\u22121}x[\/latex], which is the inverse of the cosine function and the angle that has a cosine equal to a given number<\/dd>\n<\/dl>\n<dl class=\"definition\">\n<dt>inverse sine function<\/dt>\n<dd>the function [latex]\\sin^{\u22121}x[\/latex], which is the inverse of the sine function and the angle that has a sine equal to a given number<\/dd>\n<\/dl>\n<dl class=\"definition\">\n<dt>inverse tangent function<\/dt>\n<dd>the function [latex]\\tan^{\u22121}x[\/latex], which is the inverse of the tangent function and the angle that has a tangent equal to a given number<\/dd>\n<\/dl>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-14188\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-14188","chapter","type-chapter","status-publish","hentry"],"part":16037,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14188","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":13,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14188\/revisions"}],"predecessor-version":[{"id":16082,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14188\/revisions\/16082"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/parts\/16037"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14188\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/media?parent=14188"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=14188"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/contributor?post=14188"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/license?post=14188"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}