{"id":1985,"date":"2021-08-19T16:07:00","date_gmt":"2021-08-19T16:07:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-trigonometric-substitution\/"},"modified":"2021-11-19T03:02:42","modified_gmt":"2021-11-19T03:02:42","slug":"skills-review-for-trigonometric-substitution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-trigonometric-substitution\/","title":{"raw":"Skills Review for Trigonometric Substitution","rendered":"Skills Review for Trigonometric Substitution"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Complete the square<\/li>\r\n \t<li>Evaluate trigonometric functions using right triangles<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Trigonometric Substitution section, we will learn how to evaluate integrals that contain square roots with a strictly numeric quantity and variable quantity either added or subtracted. Each of these quantities can be rewritten as a quantity squared followed by applying a trigonometric substitution formula. Here we will review how to complete the square and evaluate trigonometric functions at an arbitrary angle [latex]\\theta[\/latex].\r\n<h2>Complete the Square<\/h2>\r\nCompleting the square can be used to create a perfect square trinomial which factors into a binomial squared. To complete the square, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign.\u00a0 Here is an example.\r\n\r\nWe will use the example [latex]{x}^{2}+4x+1[\/latex] to illustrate each step. To complete the square, the leading coefficient, <em>a<\/em>, must equal 1. If it does not, then divide the entire equation by <em>a<\/em>. Then, we can use the following procedures to complete the square and factor the resulting perfect square trinomial.\r\n<ol>\r\n \t<li>Multiply the <em>b <\/em>term by [latex]\\frac{1}{2}[\/latex] and square it.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(4\\right)=2\\hfill \\\\ {2}^{2}=4\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Add and subtract [latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex] to the expression. Simplify the original constant term and the subtracted perfect square trinomial constant. We have\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4+1-4\\hfill \\\\ = {x}^{2}+4x+4-3\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>The first three terms can now be factored as a perfect square trinomial.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4-3\\hfill \\\\ ={\\left(x+2\\right)}^{2}-3\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Create a perfect square trinomial using the method of complete the square<\/h3>\r\nComplete the square on: [latex]3\\left(x^2 - 10x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"107002\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"107002\"]\r\n\r\n&nbsp;\r\n\r\nAdd [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses and subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex]\r\n<p style=\"text-align: center;\">[latex]3\\left(x^2 - 10x+25\\right)-3\\cdot25[\/latex]<\/p>\r\n&nbsp;\r\n\r\nFactor the perfect square trinomial and simplify\r\n<p style=\"text-align: center;\">[latex]3\\left(x -5\\right)^2-75[\/latex].<\/p>\r\n.\r\n<h4><span style=\"text-decoration: underline;\">Analysis<\/span><\/h4>\r\nThe resulting expression is equivalent to the original expression. To test this, substitute a small value for [latex]x[\/latex], say [latex]x=3[\/latex].\r\n\r\n[latex]3\\left(3^2-10\\cdot3\\right) = \\quad -63 \\quad = 3(3-5)^2-75[\/latex]. True.\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]15002[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Evaluate Trigonometric Functions Using Right Triangles<\/h2>\r\nWhen working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle below.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003739\/CNX_Precalc_Figure_05_04_0052.jpg\" alt=\"Right triangle with angles alpha and beta. Sides are labeled hypotenuse, adjacent to alpha\/opposite to beta, and adjacent to beta\/opposite alpha.\" width=\"487\" height=\"181\" \/>\r\n\r\nWe will be asked to find all six trigonometric functions for a given angle in a triangle. Often, the strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.\r\n<div class=\"textbox\">\r\n<h3>How To: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.<\/h3>\r\n<ol>\r\n \t<li>If needed, draw the right triangle and label the angle provided.<\/li>\r\n \t<li>Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.<\/li>\r\n \t<li>Find the required function:\r\n<ul>\r\n \t<li>sine as the ratio of the opposite side to the hypotenuse<\/li>\r\n \t<li>cosine as the ratio of the adjacent side to the hypotenuse<\/li>\r\n \t<li>tangent as the ratio of the opposite side to the adjacent side<\/li>\r\n \t<li>secant as the ratio of the hypotenuse to the adjacent side<\/li>\r\n \t<li>cosecant as the ratio of the hypotenuse to the opposite side<\/li>\r\n \t<li>cotangent as the ratio of the adjacent side to the opposite side<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using a Right Triangle To EValuate All Six Trigonometric Functions<\/h3>\r\nUsing the triangle shown below, evaluate [latex]\\sin \\alpha[\/latex], [latex]\\cos \\alpha[\/latex], [latex]\\tan \\alpha[\/latex], [latex]\\sec \\alpha[\/latex], [latex]\\csc \\alpha [\/latex], and [latex]\\cot \\alpha[\/latex].<span id=\"fs-id1165137542988\">\r\n<\/span>\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003741\/CNX_Precalc_Figure_05_04_0062.jpg\" alt=\"Right triangle with sides of 3, 4, and 5. Angle alpha is also labeled.\" width=\"487\" height=\"162\" \/>\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"626810\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"626810\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\sin \\alpha =\\frac{\\text{opposite }\\alpha }{\\text{hypotenuse}}=\\frac{4}{5}\\\\ &amp;\\cos \\alpha =\\frac{\\text{adjacent to }\\alpha }{\\text{hypotenuse}}=\\frac{3}{5} \\\\ &amp;\\tan \\alpha =\\frac{\\text{opposite }\\alpha }{\\text{adjacent to }\\alpha }=\\frac{4}{3} \\\\ &amp;\\sec \\alpha =\\frac{\\text{hypotenuse}}{\\text{adjacent to }\\alpha }=\\frac{5}{3} \\\\ &amp;\\csc \\alpha =\\frac{\\text{hypotenuse}}{\\text{opposite }\\alpha }=\\frac{5}{4} \\\\ &amp;\\cot \\alpha =\\frac{\\text{adjacent to }\\alpha }{\\text{opposite }\\alpha }=\\frac{3}{4} \\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\nUsing the triangle shown below, evaluate [latex]\\sin t[\/latex], [latex]\\cos t[\/latex], [latex]\\tan t[\/latex], [latex]\\sec t[\/latex], [latex]\\csc t[\/latex], and [latex]\\cot t[\/latex].\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003744\/CNX_Precalc_Figure_05_04_0072.jpg\" alt=\"Right triangle with sides 33, 56, and 65. Angle t is also labeled.\" width=\"487\" height=\"204\" \/>\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"252611\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"252611\"]\r\n\r\n[latex]\\begin{align}&amp;\\sin t=\\frac{33}{65},\\cos t=\\frac{56}{65},\\tan t=\\frac{33}{56}, \\\\ &amp;\\sec t=\\frac{65}{56},\\csc t=\\frac{65}{33},\\cot t=\\frac{56}{33} \\end{align}[\/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]155312[\/ohm_question]\r\n\r\nHint: The Pythagorean Theorem can be used to find the length of the missing side.\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Complete the square<\/li>\n<li>Evaluate trigonometric functions using right triangles<\/li>\n<\/ul>\n<\/div>\n<p>In the Trigonometric Substitution section, we will learn how to evaluate integrals that contain square roots with a strictly numeric quantity and variable quantity either added or subtracted. Each of these quantities can be rewritten as a quantity squared followed by applying a trigonometric substitution formula. Here we will review how to complete the square and evaluate trigonometric functions at an arbitrary angle [latex]\\theta[\/latex].<\/p>\n<h2>Complete the Square<\/h2>\n<p>Completing the square can be used to create a perfect square trinomial which factors into a binomial squared. To complete the square, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign.\u00a0 Here is an example.<\/p>\n<p>We will use the example [latex]{x}^{2}+4x+1[\/latex] to illustrate each step. To complete the square, the leading coefficient, <em>a<\/em>, must equal 1. If it does not, then divide the entire equation by <em>a<\/em>. Then, we can use the following procedures to complete the square and factor the resulting perfect square trinomial.<\/p>\n<ol>\n<li>Multiply the <em>b <\/em>term by [latex]\\frac{1}{2}[\/latex] and square it.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(4\\right)=2\\hfill \\\\ {2}^{2}=4\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Add and subtract [latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex] to the expression. Simplify the original constant term and the subtracted perfect square trinomial constant. We have\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4+1-4\\hfill \\\\ = {x}^{2}+4x+4-3\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>The first three terms can now be factored as a perfect square trinomial.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4-3\\hfill \\\\ ={\\left(x+2\\right)}^{2}-3\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<div class=\"textbox exercises\">\n<h3>Example: Create a perfect square trinomial using the method of complete the square<\/h3>\n<p>Complete the square on: [latex]3\\left(x^2 - 10x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q107002\">Show Solution<\/span><\/p>\n<div id=\"q107002\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<p>Add [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses and subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]3\\left(x^2 - 10x+25\\right)-3\\cdot25[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Factor the perfect square trinomial and simplify<\/p>\n<p style=\"text-align: center;\">[latex]3\\left(x -5\\right)^2-75[\/latex].<\/p>\n<p>.<\/p>\n<h4><span style=\"text-decoration: underline;\">Analysis<\/span><\/h4>\n<p>The resulting expression is equivalent to the original expression. To test this, substitute a small value for [latex]x[\/latex], say [latex]x=3[\/latex].<\/p>\n<p>[latex]3\\left(3^2-10\\cdot3\\right) = \\quad -63 \\quad = 3(3-5)^2-75[\/latex]. True.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm15002\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15002&theme=oea&iframe_resize_id=ohm15002&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Evaluate Trigonometric Functions Using Right Triangles<\/h2>\n<p>When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003739\/CNX_Precalc_Figure_05_04_0052.jpg\" alt=\"Right triangle with angles alpha and beta. Sides are labeled hypotenuse, adjacent to alpha\/opposite to beta, and adjacent to beta\/opposite alpha.\" width=\"487\" height=\"181\" \/><\/p>\n<p>We will be asked to find all six trigonometric functions for a given angle in a triangle. Often, the strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.<\/h3>\n<ol>\n<li>If needed, draw the right triangle and label the angle provided.<\/li>\n<li>Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.<\/li>\n<li>Find the required function:\n<ul>\n<li>sine as the ratio of the opposite side to the hypotenuse<\/li>\n<li>cosine as the ratio of the adjacent side to the hypotenuse<\/li>\n<li>tangent as the ratio of the opposite side to the adjacent side<\/li>\n<li>secant as the ratio of the hypotenuse to the adjacent side<\/li>\n<li>cosecant as the ratio of the hypotenuse to the opposite side<\/li>\n<li>cotangent as the ratio of the adjacent side to the opposite side<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using a Right Triangle To EValuate All Six Trigonometric Functions<\/h3>\n<p>Using the triangle shown below, evaluate [latex]\\sin \\alpha[\/latex], [latex]\\cos \\alpha[\/latex], [latex]\\tan \\alpha[\/latex], [latex]\\sec \\alpha[\/latex], [latex]\\csc \\alpha[\/latex], and [latex]\\cot \\alpha[\/latex].<span id=\"fs-id1165137542988\"><br \/>\n<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003741\/CNX_Precalc_Figure_05_04_0062.jpg\" alt=\"Right triangle with sides of 3, 4, and 5. Angle alpha is also labeled.\" width=\"487\" height=\"162\" \/><\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q626810\">Show Solution<\/span><\/p>\n<div id=\"q626810\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\sin \\alpha =\\frac{\\text{opposite }\\alpha }{\\text{hypotenuse}}=\\frac{4}{5}\\\\ &\\cos \\alpha =\\frac{\\text{adjacent to }\\alpha }{\\text{hypotenuse}}=\\frac{3}{5} \\\\ &\\tan \\alpha =\\frac{\\text{opposite }\\alpha }{\\text{adjacent to }\\alpha }=\\frac{4}{3} \\\\ &\\sec \\alpha =\\frac{\\text{hypotenuse}}{\\text{adjacent to }\\alpha }=\\frac{5}{3} \\\\ &\\csc \\alpha =\\frac{\\text{hypotenuse}}{\\text{opposite }\\alpha }=\\frac{5}{4} \\\\ &\\cot \\alpha =\\frac{\\text{adjacent to }\\alpha }{\\text{opposite }\\alpha }=\\frac{3}{4} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Using the triangle shown below, evaluate [latex]\\sin t[\/latex], [latex]\\cos t[\/latex], [latex]\\tan t[\/latex], [latex]\\sec t[\/latex], [latex]\\csc t[\/latex], and [latex]\\cot t[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003744\/CNX_Precalc_Figure_05_04_0072.jpg\" alt=\"Right triangle with sides 33, 56, and 65. Angle t is also labeled.\" width=\"487\" height=\"204\" \/><\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q252611\">Show Solution<\/span><\/p>\n<div id=\"q252611\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{align}&\\sin t=\\frac{33}{65},\\cos t=\\frac{56}{65},\\tan t=\\frac{33}{56}, \\\\ &\\sec t=\\frac{65}{56},\\csc t=\\frac{65}{33},\\cot t=\\frac{56}{33} \\end{align}[\/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=\"ohm155312\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=155312&theme=oea&iframe_resize_id=ohm155312\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p>Hint: The Pythagorean Theorem can be used to find the length of the missing side.<\/p>\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-1985\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Modification and Revision. <strong>Provided by<\/strong>: Lumen Learning. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/.\">https:\/\/courses.lumenlearning.com\/precalculus\/.<\/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":349141,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen 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