{"id":202,"date":"2025-02-13T22:44:58","date_gmt":"2025-02-13T22:44:58","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-trigonometric-equations-with-identities\/"},"modified":"2025-10-14T16:09:36","modified_gmt":"2025-10-14T16:09:36","slug":"solving-trigonometric-equations-with-identities","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-trigonometric-equations-with-identities\/","title":{"raw":"Simplifying Trigonometric Expressions With Identities: Learn It 1","rendered":"Simplifying Trigonometric Expressions With Identities: Learn It 1"},"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>Verify the fundamental trigonometric identities<\/h2>\r\nIdentities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways.\r\n\r\nTo verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the <strong>Pythagorean identities<\/strong>, the even-odd identities, the reciprocal identities, and the quotient identities.\r\n\r\nWe will begin with the <strong>Pythagorean identities<\/strong>, which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities.\r\n<table id=\"Table_07_01_01\" summary=\"&quot;Pythagorean Identities&quot; with three cells. First: sin(theta)^2 + cos(theta)^2 = 1. Second: 1 + cot(theta)^2 = csc(theta)^2. Third: 1 + tan(theta)^2 = sec(theta)^2.\"><colgroup> <col \/> <col \/> <col \/><\/colgroup>\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center;\" colspan=\"3\">Pythagorean Identities<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]{\\sin }^{2}\\theta +{\\cos }^{2}\\theta =1[\/latex]<\/td>\r\n<td>[latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta[\/latex]<\/td>\r\n<td>[latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe second and third 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><\/div>\r\n<div><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a trigonometric identity, verify that it is true.\r\n<\/strong>\r\n<ol id=\"fs-id2191946\">\r\n \t<li>Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.<\/li>\r\n \t<li>Look for opportunities to factor expressions, square a binomial, or add fractions.<\/li>\r\n \t<li>Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.<\/li>\r\n \t<li>If these steps do not yield the desired result, try converting all terms to sines and cosines.<\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n<div>\r\n<div class=\"bcc-box bcc-success\"><section class=\"textbox example\" aria-label=\"Example\">Verify the identity [latex]\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }={\\sin }^{2}\\theta[\/latex][reveal-answer q=\"565941\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"565941\"]As the left side is more complicated, let\u2019s begin there.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }&amp;=\\frac{\\left({\\tan }^{2}\\theta +1\\right)-1}{{\\sec }^{2}\\theta }&amp;&amp; {\\sec}^{2}\\theta ={\\tan }^{2}\\theta +1 \\\\ &amp;=\\frac{{\\tan }^{2}\\theta }{{\\sec }^{2}\\theta } \\\\ &amp;={\\tan }^{2}\\theta \\left(\\frac{1}{{\\sec }^{2}\\theta }\\right) \\\\ &amp;={\\tan }^{2}\\theta \\left({\\cos }^{2}\\theta \\right)&amp;&amp; {\\cos }^{2}\\theta =\\frac{1}{{\\sec }^{2}\\theta } \\\\ &amp;=\\left(\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta }\\right)\\left({\\cos }^{2}\\theta \\right)&amp;&amp; {\\tan}^{2}\\theta =\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &amp;=\\left(\\frac{{\\sin }^{2}\\theta }{\\cancel{{\\cos }^{2}\\theta}}\\right)\\left(\\cancel{{\\cos }^{2}\\theta} \\right) \\\\ &amp;={\\sin }^{2}\\theta \\end{align}[\/latex]<\/p>\r\nThere is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }&amp;=\\frac{{\\sec }^{2}\\theta }{{\\sec }^{2}\\theta }-\\frac{1}{{\\sec }^{2}\\theta } \\\\ &amp;=1-{\\cos }^{2}\\theta \\\\ &amp;={\\sin }^{2}\\theta \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/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>Verify the fundamental trigonometric identities<\/h2>\n<p>Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways.<\/p>\n<p>To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the <strong>Pythagorean identities<\/strong>, the even-odd identities, the reciprocal identities, and the quotient identities.<\/p>\n<p>We will begin with the <strong>Pythagorean identities<\/strong>, which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities.<\/p>\n<table id=\"Table_07_01_01\" summary=\"&quot;Pythagorean Identities&quot; with three cells. First: sin(theta)^2 + cos(theta)^2 = 1. Second: 1 + cot(theta)^2 = csc(theta)^2. Third: 1 + tan(theta)^2 = sec(theta)^2.\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<thead>\n<tr>\n<th style=\"text-align: center;\" colspan=\"3\">Pythagorean Identities<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]{\\sin }^{2}\\theta +{\\cos }^{2}\\theta =1[\/latex]<\/td>\n<td>[latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta[\/latex]<\/td>\n<td>[latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The second and third 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<div>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a trigonometric identity, verify that it is true.<br \/>\n<\/strong><\/p>\n<ol id=\"fs-id2191946\">\n<li>Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.<\/li>\n<li>Look for opportunities to factor expressions, square a binomial, or add fractions.<\/li>\n<li>Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.<\/li>\n<li>If these steps do not yield the desired result, try converting all terms to sines and cosines.<\/li>\n<\/ol>\n<\/section>\n<\/div>\n<div>\n<div class=\"bcc-box bcc-success\">\n<section class=\"textbox example\" aria-label=\"Example\">Verify the identity [latex]\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }={\\sin }^{2}\\theta[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q565941\">Show Solution<\/button><\/p>\n<div id=\"q565941\" class=\"hidden-answer\" style=\"display: none\">As the left side is more complicated, let\u2019s begin there.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }&=\\frac{\\left({\\tan }^{2}\\theta +1\\right)-1}{{\\sec }^{2}\\theta }&& {\\sec}^{2}\\theta ={\\tan }^{2}\\theta +1 \\\\ &=\\frac{{\\tan }^{2}\\theta }{{\\sec }^{2}\\theta } \\\\ &={\\tan }^{2}\\theta \\left(\\frac{1}{{\\sec }^{2}\\theta }\\right) \\\\ &={\\tan }^{2}\\theta \\left({\\cos }^{2}\\theta \\right)&& {\\cos }^{2}\\theta =\\frac{1}{{\\sec }^{2}\\theta } \\\\ &=\\left(\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta }\\right)\\left({\\cos }^{2}\\theta \\right)&& {\\tan}^{2}\\theta =\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &=\\left(\\frac{{\\sin }^{2}\\theta }{\\cancel{{\\cos }^{2}\\theta}}\\right)\\left(\\cancel{{\\cos }^{2}\\theta} \\right) \\\\ &={\\sin }^{2}\\theta \\end{align}[\/latex]<\/p>\n<p>There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }&=\\frac{{\\sec }^{2}\\theta }{{\\sec }^{2}\\theta }-\\frac{1}{{\\sec }^{2}\\theta } \\\\ &=1-{\\cos }^{2}\\theta \\\\ &={\\sin }^{2}\\theta \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n","protected":false},"author":6,"menu_order":6,"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\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":201,"module-header":"learn_it","content_attributions":[{"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":""}],"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\/202"}],"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\/6"}],"version-history":[{"count":14,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/202\/revisions"}],"predecessor-version":[{"id":1969,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/202\/revisions\/1969"}],"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\/202\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=202"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=202"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=202"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=202"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}