{"id":3235,"date":"2025-08-15T23:42:11","date_gmt":"2025-08-15T23:42:11","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3235"},"modified":"2025-10-14T18:09:37","modified_gmt":"2025-10-14T18:09:37","slug":"simplifying-trigonometric-expressions-with-identities-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/simplifying-trigonometric-expressions-with-identities-apply-it\/","title":{"raw":"Simplifying Trigonometric Expressions with Identities: Apply It","rendered":"Simplifying Trigonometric Expressions with Identities: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Verify the fundamental trigonometric identities.<\/li>\r\n \t<li style=\"font-weight: 400;\">Simplify trigonometric expressions using algebra and the identities.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Pythagorean Identities<\/h2>\r\nThe second and third Pythagorean identities can be obtained by manipulating the first. The identity [latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta[\/latex] is found by rewriting the left side of the equation in terms of sine and cosine.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Prove: [latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta [\/latex]\r\n<div style=\"text-align: center;\">[latex]\\begin{align}1+{\\cot }^{2}\\theta&amp; =\\left(1+\\frac{{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)&amp;&amp; \\text{Rewrite the left side}. \\\\ &amp;=\\left(\\frac{{\\sin }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)+\\left(\\frac{{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)&amp;&amp; \\text{Write both terms with the common denominator}. \\\\ &amp;=\\frac{{\\sin }^{2}\\theta +{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta } \\\\ &amp;=\\frac{1}{{\\sin }^{2}\\theta } \\\\ &amp;={\\csc }^{2}\\theta \\end{align}[\/latex]<\/div>\r\n<\/section>\r\n<div>Similarly, [latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex] can be obtained by rewriting the left side of this identity in terms of sine and cosine.<\/div>\r\n<div><section class=\"textbox example\" aria-label=\"Example\">\r\n<div style=\"text-align: center;\">[latex]\\begin{align}1+{\\tan }^{2}\\theta &amp;=1+{\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)}^{2}&amp;&amp; \\text{Rewrite left side}. \\\\ &amp;={\\left(\\frac{\\cos \\theta }{\\cos \\theta }\\right)}^{2}+{\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)}^{2}&amp;&amp; \\text{Write both terms with the common denominator}. \\\\ &amp;=\\frac{{\\cos }^{2}\\theta +{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &amp;=\\frac{1}{{\\cos }^{2}\\theta } \\\\ &amp;={\\sec }^{2}\\theta \\end{align}[\/latex]<\/div>\r\n<\/section>\r\n<div><section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p class=\"whitespace-normal break-words\"><strong>How To: Using the Primary Pythagorean Identity<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Write the identity: [latex]\\sin^2\\theta + \\cos^2\\theta = 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Substitute the known value<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Solve for the unknown function squared<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Take the square root (remember: two possible values)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Use the quadrant to determine the correct sign<\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n<div><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]313599[\/ohm_question]<\/section><\/div>\r\n<div><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Always remember that taking a square root produces both positive and negative solutions. The quadrant information tells you which sign to use.<\/section><\/div>\r\n<section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]313600[\/ohm_question]<\/section>\r\n<div><section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">Rewrite [latex]\\sec^2\\theta[\/latex] in terms of sine only.<\/p>\r\n<p class=\"whitespace-normal break-words\">[reveal-answer q=\"616184\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"616184\"]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Start with the tangent-secant Pythagorean identity: [latex]1 + \\tan^2\\theta = \\sec^2\\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">So: [latex]\\sec^2\\theta = 1 + \\tan^2\\theta[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Now express [latex]\\tan^2\\theta[\/latex] in terms of sine: [latex]\\begin{aligned} \\tan^2\\theta &amp;= \\frac{\\sin^2\\theta}{\\cos^2\\theta} \\ &amp;= \\frac{\\sin^2\\theta}{1 - \\sin^2\\theta} &amp;&amp; \\text{Use } \\cos^2\\theta = 1 - \\sin^2\\theta \\end{aligned}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]\\begin{aligned} \\sec^2\\theta &amp;= 1 + \\frac{\\sin^2\\theta}{1 - \\sin^2\\theta} \\ &amp;= \\frac{1 - \\sin^2\\theta}{1 - \\sin^2\\theta} + \\frac{\\sin^2\\theta}{1 - \\sin^2\\theta} \\ &amp;= \\frac{1 - \\sin^2\\theta + \\sin^2\\theta}{1 - \\sin^2\\theta} \\ &amp;= \\frac{1}{1 - \\sin^2\\theta} \\end{aligned}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">[\/hidden-answer]<\/p>\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]313601[\/ohm_question]<\/section><\/div>\r\n<\/div>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li style=\"font-weight: 400;\">Verify the fundamental trigonometric identities.<\/li>\n<li style=\"font-weight: 400;\">Simplify trigonometric expressions using algebra and the identities.<\/li>\n<\/ul>\n<\/section>\n<h2>Pythagorean Identities<\/h2>\n<p>The second and third Pythagorean identities can be obtained by manipulating the first. The identity [latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta[\/latex] is found by rewriting the left side of the equation in terms of sine and cosine.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Prove: [latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}1+{\\cot }^{2}\\theta& =\\left(1+\\frac{{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)&& \\text{Rewrite the left side}. \\\\ &=\\left(\\frac{{\\sin }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)+\\left(\\frac{{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)&& \\text{Write both terms with the common denominator}. \\\\ &=\\frac{{\\sin }^{2}\\theta +{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta } \\\\ &=\\frac{1}{{\\sin }^{2}\\theta } \\\\ &={\\csc }^{2}\\theta \\end{align}[\/latex]<\/div>\n<\/section>\n<div>Similarly, [latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex] can be obtained by rewriting the left side of this identity in terms of sine and cosine.<\/div>\n<div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div style=\"text-align: center;\">[latex]\\begin{align}1+{\\tan }^{2}\\theta &=1+{\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)}^{2}&& \\text{Rewrite left side}. \\\\ &={\\left(\\frac{\\cos \\theta }{\\cos \\theta }\\right)}^{2}+{\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)}^{2}&& \\text{Write both terms with the common denominator}. \\\\ &=\\frac{{\\cos }^{2}\\theta +{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &=\\frac{1}{{\\cos }^{2}\\theta } \\\\ &={\\sec }^{2}\\theta \\end{align}[\/latex]<\/div>\n<\/section>\n<div>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p class=\"whitespace-normal break-words\"><strong>How To: Using the Primary Pythagorean Identity<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Write the identity: [latex]\\sin^2\\theta + \\cos^2\\theta = 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Substitute the known value<\/li>\n<li class=\"whitespace-normal break-words\">Solve for the unknown function squared<\/li>\n<li class=\"whitespace-normal break-words\">Take the square root (remember: two possible values)<\/li>\n<li class=\"whitespace-normal break-words\">Use the quadrant to determine the correct sign<\/li>\n<\/ol>\n<\/section>\n<\/div>\n<div>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm313599\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=313599&theme=lumen&iframe_resize_id=ohm313599&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n<div>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Always remember that taking a square root produces both positive and negative solutions. The quadrant information tells you which sign to use.<\/section>\n<\/div>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm313600\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=313600&theme=lumen&iframe_resize_id=ohm313600&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">Rewrite [latex]\\sec^2\\theta[\/latex] in terms of sine only.<\/p>\n<p class=\"whitespace-normal break-words\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q616184\">Show Solution<\/button><\/p>\n<div id=\"q616184\" class=\"hidden-answer\" style=\"display: none\">\n<p class=\"whitespace-pre-wrap break-words\">Start with the tangent-secant Pythagorean identity: [latex]1 + \\tan^2\\theta = \\sec^2\\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">So: [latex]\\sec^2\\theta = 1 + \\tan^2\\theta[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Now express [latex]\\tan^2\\theta[\/latex] in terms of sine: [latex]\\begin{aligned} \\tan^2\\theta &= \\frac{\\sin^2\\theta}{\\cos^2\\theta} \\ &= \\frac{\\sin^2\\theta}{1 - \\sin^2\\theta} && \\text{Use } \\cos^2\\theta = 1 - \\sin^2\\theta \\end{aligned}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]\\begin{aligned} \\sec^2\\theta &= 1 + \\frac{\\sin^2\\theta}{1 - \\sin^2\\theta} \\ &= \\frac{1 - \\sin^2\\theta}{1 - \\sin^2\\theta} + \\frac{\\sin^2\\theta}{1 - \\sin^2\\theta} \\ &= \\frac{1 - \\sin^2\\theta + \\sin^2\\theta}{1 - \\sin^2\\theta} \\ &= \\frac{1}{1 - \\sin^2\\theta} \\end{aligned}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm313601\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=313601&theme=lumen&iframe_resize_id=ohm313601&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n<\/div>\n","protected":false},"author":67,"menu_order":9,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":201,"module-header":"apply_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3235"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/67"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3235\/revisions"}],"predecessor-version":[{"id":4664,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3235\/revisions\/4664"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/201"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3235\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3235"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3235"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3235"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3235"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}