{"id":18032,"date":"2021-09-12T05:13:41","date_gmt":"2021-09-12T05:13:41","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/?post_type=chapter&#038;p=18032"},"modified":"2021-09-12T05:49:38","modified_gmt":"2021-09-12T05:49:38","slug":"section-7-2-the-inverse-trigonometric-functions-continued","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/section-7-2-the-inverse-trigonometric-functions-continued\/","title":{"raw":"Section 7.2: The Inverse Trigonometric Functions (Continued)","rendered":"Section 7.2: The Inverse Trigonometric Functions (Continued)"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Find exact values of composite functions with inverse trigonometric functions.<\/li>\r\n \t<li>Write a trigonometric expression as an algebraic expression.<\/li>\r\n<\/ul>\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\nUse a right triangle to write [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{6}{x}\\right)\\right)[\/latex] as an algebraic expression. Assume that [latex]x[\/latex] is positive and in the domain of the inverse sine.\r\n\r\n[reveal-answer q=\"764834\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"764834\"]\r\n\r\nLet [latex]\\theta=\\sin^{-1}\\left(\\frac{6}{x}\\right)\\text{, so }\\sin\\theta=\\frac{6}{x}[\/latex].\u00a0 We can represent this on a triangle.\u00a0 Using the definition for sine, the opposite side is 6 and the hypotenuse is [latex]x[\/latex].\u00a0 Then we can find the adjacent side:\r\n<img class=\"alignright\" src=\"http:\/\/www.hutchmath.com\/Images\/righttriangleinverses.JPG\" alt=\"Right triangle with opposite side of 6 and hypotenuse of x\" \/>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;a^{2}+6^{2}=x^{2}&amp;&amp; \\text{Use the Pythagorean Theorem.} \\\\ &amp;a^{2}=x^{2}-36 &amp;&amp; \\text{Isolate a.} \\\\ &amp;a=\\pm\\sqrt{x^{2}-36} &amp;&amp;\\text{Take the square root of both sides}\\\\&amp;a=\\sqrt{x^{2}-36} &amp;&amp; \\text{Take the positive root. This is the adjacent side.} \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now that we have the triangle complete and\u00a0latex]\\theta=\\sin^{-1}\\left(\\frac{6}{x}\\right)[\/latex], we need to find the cosine.\u00a0 The definition of cosine is the adjacent side divided by the hypotenuse, so\u00a0[latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{6}{x}\\right)\\right)=\\frac{\\sqrt{x^{2}-36}}{x}[\/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\nUse a right triangle to write [latex]\\sin\\left(\\tan^{\u22121}\\left(4x\\right)\\right)[\/latex] as an algebraic expression.\u00a0Assume that [latex]x[\/latex] is positive and that the given inverse trigonometric function is defined for the expression in [latex]x[\/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>Key Concepts<\/h2>\r\n<ul>\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<\/ul>\r\n<h2 style=\"text-align: center;\">Section 7.2 Homework Exercises<\/h2>\r\n\r\n\r\nFor the following exercises, find the exact value, if possible, without a calculator. If it is not possible, explain why.\r\n\r\n1. [latex]\\cos^{\u22121}(\\sin(\\pi))[\/latex]\r\n\r\n2. [latex]\\sin^{\u22121}(\\cos(\\pi))[\/latex]\r\n\r\n3. [latex]\\tan^{\u22121}(\\sin(\\pi))[\/latex]\r\n\r\n4. [latex]\\cos^{\u22121}\\left(\\sin\\left(\\frac{\\pi}{3}\\right)\\right)[\/latex]\r\n\r\n5. [latex]\\tan^{\u22121}\\left(\\sin\\left(\\frac{\\pi}{3}\\right)\\right)[\/latex]\r\n\r\n6. [latex]\\sin^{\u22121}\\left(\\cos\\left(\\frac{\u2212\\pi}{2}\\right)\\right)[\/latex]\r\n\r\n7. [latex]\\tan^{\u22121}\\left(\\sin\\left(\\frac{4\\pi}{3}\\right)\\right)[\/latex]\r\n\r\n8. [latex]\\sin^{\u22121}\\left(\\sin\\left(\\frac{5\\pi}{6}\\right)\\right)[\/latex]\r\n\r\n9. [latex]\\tan^{\u22121}\\left(\\sin\\left(\\frac{\u22125\\pi}{2}\\right)\\right)[\/latex]\r\n\r\n10. [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{4}{5}\\right)\\right)[\/latex]\r\n\r\n11. [latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{3}{5}\\right)\\right)[\/latex]\r\n\r\n12. [latex]\\sin\\left(\\tan^{\u22121}\\left(\\frac{4}{3}\\right)\\right)[\/latex]\r\n\r\n13. [latex]\\cos\\left(\\tan^{\u22121}\\left(\\frac{12}{5}\\right)\\right)[\/latex]\r\n\r\n14. [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{1}{2}\\right)\\right)[\/latex]\r\n\r\nFor the following exercises, use a right triangle to write the expression as an algebraic expression.\u00a0 Assume that x is positive and in the domain of the given inverse trigonometric function.\r\n\r\n15. [latex]\\tan\\left(\\sin^{\u22121}\\left(x\\right)\\right)[\/latex]\r\n\r\n16. [latex]\\sin\\left(\\cos^{\u22121}\\left(x\\right)\\right)[\/latex]\r\n\r\n17. [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{1}{x}\\right)\\right)[\/latex]\r\n\r\n18. [latex]\\cos\\left(\\tan^{\u22121}\\left(3x\\right)\\right)[\/latex]\r\n\r\n19. [latex]\\sin\\left(\\tan^{\u22121}\\left(2x\\right)\\right)[\/latex]\r\n\r\n20. [latex]\\tan\\left(\\cos^{\u22121}\\left(4x\\right)\\right)[\/latex]\r\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li style=\"font-weight: 400;\">Find exact values of composite functions with inverse trigonometric functions.<\/li>\n<li>Write a trigonometric expression as an algebraic expression.<\/li>\n<\/ul>\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>Use a right triangle to write [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{6}{x}\\right)\\right)[\/latex] as an algebraic expression. Assume that [latex]x[\/latex] is positive and in the domain of the inverse sine.<\/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>Let [latex]\\theta=\\sin^{-1}\\left(\\frac{6}{x}\\right)\\text{, so }\\sin\\theta=\\frac{6}{x}[\/latex].\u00a0 We can represent this on a triangle.\u00a0 Using the definition for sine, the opposite side is 6 and the hypotenuse is [latex]x[\/latex].\u00a0 Then we can find the adjacent side:<br \/>\n<img decoding=\"async\" class=\"alignright\" src=\"http:\/\/www.hutchmath.com\/Images\/righttriangleinverses.JPG\" alt=\"Right triangle with opposite side of 6 and hypotenuse of x\" \/><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&a^{2}+6^{2}=x^{2}&& \\text{Use the Pythagorean Theorem.} \\\\ &a^{2}=x^{2}-36 && \\text{Isolate a.} \\\\ &a=\\pm\\sqrt{x^{2}-36} &&\\text{Take the square root of both sides}\\\\&a=\\sqrt{x^{2}-36} && \\text{Take the positive root. This is the adjacent side.} \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now that we have the triangle complete and\u00a0latex]\\theta=\\sin^{-1}\\left(\\frac{6}{x}\\right)[\/latex], we need to find the cosine.\u00a0 The definition of cosine is the adjacent side divided by the hypotenuse, so\u00a0[latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{6}{x}\\right)\\right)=\\frac{\\sqrt{x^{2}-36}}{x}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Use a right triangle to write [latex]\\sin\\left(\\tan^{\u22121}\\left(4x\\right)\\right)[\/latex] as an algebraic expression.\u00a0Assume that [latex]x[\/latex] is positive and that the given inverse trigonometric function is defined for the expression in [latex]x[\/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>Key Concepts<\/h2>\n<ul>\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<\/ul>\n<h2 style=\"text-align: center;\">Section 7.2 Homework Exercises<\/h2>\n<p>For the following exercises, find the exact value, if possible, without a calculator. If it is not possible, explain why.<\/p>\n<p>1. [latex]\\cos^{\u22121}(\\sin(\\pi))[\/latex]<\/p>\n<p>2. [latex]\\sin^{\u22121}(\\cos(\\pi))[\/latex]<\/p>\n<p>3. [latex]\\tan^{\u22121}(\\sin(\\pi))[\/latex]<\/p>\n<p>4. [latex]\\cos^{\u22121}\\left(\\sin\\left(\\frac{\\pi}{3}\\right)\\right)[\/latex]<\/p>\n<p>5. [latex]\\tan^{\u22121}\\left(\\sin\\left(\\frac{\\pi}{3}\\right)\\right)[\/latex]<\/p>\n<p>6. [latex]\\sin^{\u22121}\\left(\\cos\\left(\\frac{\u2212\\pi}{2}\\right)\\right)[\/latex]<\/p>\n<p>7. [latex]\\tan^{\u22121}\\left(\\sin\\left(\\frac{4\\pi}{3}\\right)\\right)[\/latex]<\/p>\n<p>8. [latex]\\sin^{\u22121}\\left(\\sin\\left(\\frac{5\\pi}{6}\\right)\\right)[\/latex]<\/p>\n<p>9. [latex]\\tan^{\u22121}\\left(\\sin\\left(\\frac{\u22125\\pi}{2}\\right)\\right)[\/latex]<\/p>\n<p>10. [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{4}{5}\\right)\\right)[\/latex]<\/p>\n<p>11. [latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{3}{5}\\right)\\right)[\/latex]<\/p>\n<p>12. [latex]\\sin\\left(\\tan^{\u22121}\\left(\\frac{4}{3}\\right)\\right)[\/latex]<\/p>\n<p>13. [latex]\\cos\\left(\\tan^{\u22121}\\left(\\frac{12}{5}\\right)\\right)[\/latex]<\/p>\n<p>14. [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{1}{2}\\right)\\right)[\/latex]<\/p>\n<p>For the following exercises, use a right triangle to write the expression as an algebraic expression.\u00a0 Assume that x is positive and in the domain of the given inverse trigonometric function.<\/p>\n<p>15. [latex]\\tan\\left(\\sin^{\u22121}\\left(x\\right)\\right)[\/latex]<\/p>\n<p>16. [latex]\\sin\\left(\\cos^{\u22121}\\left(x\\right)\\right)[\/latex]<\/p>\n<p>17. [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{1}{x}\\right)\\right)[\/latex]<\/p>\n<p>18. [latex]\\cos\\left(\\tan^{\u22121}\\left(3x\\right)\\right)[\/latex]<\/p>\n<p>19. [latex]\\sin\\left(\\tan^{\u22121}\\left(2x\\right)\\right)[\/latex]<\/p>\n<p>20. [latex]\\tan\\left(\\cos^{\u22121}\\left(4x\\right)\\right)[\/latex]<\/p>\n","protected":false},"author":264444,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-18032","chapter","type-chapter","status-publish","hentry"],"part":14191,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18032","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/users\/264444"}],"version-history":[{"count":18,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18032\/revisions"}],"predecessor-version":[{"id":18055,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18032\/revisions\/18055"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/parts\/14191"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18032\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=18032"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=18032"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=18032"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=18032"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}