{"id":1814,"date":"2025-07-28T20:27:46","date_gmt":"2025-07-28T20:27:46","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1814"},"modified":"2026-08-13T22:34:43","modified_gmt":"2026-08-13T22:34:43","slug":"the-other-trigonometric-functions-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/the-other-trigonometric-functions-learn-it-5\/","title":{"raw":"The Other Trigonometric Functions: Learn It 5","rendered":"The Other Trigonometric Functions: Learn It 5"},"content":{"raw":"<h2>Alternate Forms of the Pythagorean Identity<\/h2>\r\nWe can use these fundamental identities to derive alternative forms of the <strong>Pythagorean Identity<\/strong>, [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]. One form is obtained by dividing both sides by [latex]{\\cos }^{2}t:[\/latex]\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}\\frac{{\\cos }^{2}t}{{\\cos }^{2}t}+\\frac{{\\sin }^{2}t}{{\\cos }^{2}t}=\\frac{1}{{\\cos }^{2}t}\\\\ \\\\ 1+{\\tan }^{2}t={\\sec }^{2}t\\end{gathered}[\/latex]<\/div>\r\nThe other form is obtained by dividing both sides by [latex]{\\sin }^{2}t:[\/latex]\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}\\frac{{\\cos }^{2}t}{{\\sin }^{2}t}+\\frac{{\\sin }^{2}t}{{\\sin }^{2}t}=\\frac{1}{{\\sin }^{2}t}\\\\ \\\\ {\\cot }^{2}t+1={\\csc }^{2}t\\end{gathered}[\/latex]<\/div>\r\n<div><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>alternate Pythagorean identities<\/h3>\r\n<p style=\"text-align: center;\">[latex]1+{\\tan }^{2}t={\\sec }^{2}t[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\cot }^{2}t+1={\\csc }^{2}t[\/latex]<\/p>\r\n\r\n<\/section><\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">If [latex]\\text{cos}\\left(t\\right)=\\frac{12}{13}[\/latex] and [latex]t[\/latex] is in quadrant IV, find the values of the other five trigonometric functions.<span id=\"fs-id1165137444782\">\r\n<\/span><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003714\/CNX_Precalc_Figure_05_03_0082.jpg\" alt=\"Graph of circle with angle of t inscribed. Point of (12\/13, y) is at intersection of terminal side of angle and edge of circle.\" width=\"487\" height=\"383\" \/>[reveal-answer q=\"627612\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"627612\"]We can find the sine using the Pythagorean Identity, [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex], and the remaining functions by relating them to sine and cosine.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{\\left(\\frac{12}{13}\\right)}^{2}+{\\sin }^{2}t=1 \\\\ {\\sin }^{2}t=1-{\\left(\\frac{12}{13}\\right)}^{2} \\\\ {\\sin }^{2}t=1-\\frac{144}{169} \\\\ {\\sin }^{2}t=\\frac{25}{169} \\\\ \\sin t=\\pm \\sqrt{\\frac{25}{169}} \\\\ \\sin t=\\pm \\frac{\\sqrt{25}}{\\sqrt{169}} \\\\ \\sin t=\\pm \\frac{5}{13} \\end{gathered}[\/latex]<\/p>\r\nThe sign of the sine depends on the <em>y<\/em>-values in the quadrant where the angle is located. Since the angle is in quadrant IV, where the <em>y<\/em>-values are negative, its sine is negative, [latex]-\\frac{5}{13}[\/latex].\r\n\r\nThe remaining functions can be calculated using identities relating them to sine and cosine.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\tan t=\\frac{\\sin t}{\\cos t}=\\frac{-\\frac{5}{13}}{\\frac{12}{13}}=-\\frac{5}{12} \\\\ \\sec t=\\frac{1}{\\cos t}=\\frac{1}{\\frac{12}{13}}=\\frac{13}{12} \\\\ \\csc t=\\frac{1}{\\sin t}=\\frac{1}{-\\frac{5}{13}}=-\\frac{13}{5}\\\\ \\cot t=\\frac{1}{\\tan t}=\\frac{1}{-\\frac{5}{12}}=-\\frac{12}{5}\\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]322751[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]322752[\/ohm_question]<\/section><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>periodic functions<\/h3>\r\nA function that repeats its values in regular intervals is known as a <strong>periodic function<\/strong>. The trigonometric functions are periodic. For the four trigonometric functions, sine, cosine, cosecant and secant, a revolution of one circle, or [latex]2\\pi [\/latex], will result in the same outputs for these functions. And for tangent and cotangent, only a half a revolution will result in the same outputs.swer]\r\n[hidden-answer a=\"99481\"]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin t=y=-\\frac{\\sqrt{3}}{2}\\\\ \\cos t=x=-\\frac{1}{2}\\\\ \\tan t=\\frac{\\sin t}{\\cos t}=\\frac{-\\frac{\\sqrt{3}}{2}}{-\\frac{1}{2}}=\\sqrt{3}\\\\ \\sec t=\\frac{1}{\\cos t}=\\frac{1}{-\\frac{1}{2}}=-2\\\\ \\csc t=\\frac{1}{\\sin t}=\\frac{1}{-\\frac{\\sqrt{3}}{2}}=-\\frac{2\\sqrt{3}}{3}\\\\ \\cot t=\\frac{1}{\\tan t}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}\\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">If [latex]\\sin \\left(t\\right)=-\\frac{\\sqrt{3}}{2}[\/latex] and [latex]\\text{cos}\\left(t\\right)=\\frac{1}{2}[\/latex], find [latex]\\text{sec}\\left(t\\right),\\text{csc}\\left(t\\right),\\text{tan}\\left(t\\right),\\text{ cot}\\left(t\\right)[\/latex].[reveal-answer q=\"940244\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"940244\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} \\sec t=\\frac{1}{\\cos t}=\\frac{1}{\\frac{1}{2}}=2 \\\\ \\csc t=\\frac{1}{\\sin t}=\\frac{1}{-\\frac{\\sqrt{3}}{2}}-\\frac{2\\sqrt{3}}{3} \\\\ \\tan t=\\frac{\\sin t}{\\cos t}=\\frac{-\\frac{\\sqrt{3}}{2}}{\\frac{1}{2}}=-\\sqrt{3} \\\\ \\cot t=\\frac{1}{\\tan t}=\\frac{1}{-\\sqrt{3}}=-\\frac{\\sqrt{3}}{3} \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]322766[\/ohm_question]<\/section>","rendered":"<h2>Alternate Forms of the Pythagorean Identity<\/h2>\n<p>We can use these fundamental identities to derive alternative forms of the <strong>Pythagorean Identity<\/strong>, [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]. One form is obtained by dividing both sides by [latex]{\\cos }^{2}t:[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}\\frac{{\\cos }^{2}t}{{\\cos }^{2}t}+\\frac{{\\sin }^{2}t}{{\\cos }^{2}t}=\\frac{1}{{\\cos }^{2}t}\\\\ \\\\ 1+{\\tan }^{2}t={\\sec }^{2}t\\end{gathered}[\/latex]<\/div>\n<p>The other form is obtained by dividing both sides by [latex]{\\sin }^{2}t:[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}\\frac{{\\cos }^{2}t}{{\\sin }^{2}t}+\\frac{{\\sin }^{2}t}{{\\sin }^{2}t}=\\frac{1}{{\\sin }^{2}t}\\\\ \\\\ {\\cot }^{2}t+1={\\csc }^{2}t\\end{gathered}[\/latex]<\/div>\n<div>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>alternate Pythagorean identities<\/h3>\n<p style=\"text-align: center;\">[latex]1+{\\tan }^{2}t={\\sec }^{2}t[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{\\cot }^{2}t+1={\\csc }^{2}t[\/latex]<\/p>\n<\/section>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">If [latex]\\text{cos}\\left(t\\right)=\\frac{12}{13}[\/latex] and [latex]t[\/latex] is in quadrant IV, find the values of the other five trigonometric functions.<span id=\"fs-id1165137444782\"><br \/>\n<\/span><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003714\/CNX_Precalc_Figure_05_03_0082.jpg\" alt=\"Graph of circle with angle of t inscribed. Point of (12\/13, y) is at intersection of terminal side of angle and edge of circle.\" width=\"487\" height=\"383\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q627612\">Show Solution<\/button><\/p>\n<div id=\"q627612\" class=\"hidden-answer\" style=\"display: none\">We can find the sine using the Pythagorean Identity, [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex], and the remaining functions by relating them to sine and cosine.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{\\left(\\frac{12}{13}\\right)}^{2}+{\\sin }^{2}t=1 \\\\ {\\sin }^{2}t=1-{\\left(\\frac{12}{13}\\right)}^{2} \\\\ {\\sin }^{2}t=1-\\frac{144}{169} \\\\ {\\sin }^{2}t=\\frac{25}{169} \\\\ \\sin t=\\pm \\sqrt{\\frac{25}{169}} \\\\ \\sin t=\\pm \\frac{\\sqrt{25}}{\\sqrt{169}} \\\\ \\sin t=\\pm \\frac{5}{13} \\end{gathered}[\/latex]<\/p>\n<p>The sign of the sine depends on the <em>y<\/em>-values in the quadrant where the angle is located. Since the angle is in quadrant IV, where the <em>y<\/em>-values are negative, its sine is negative, [latex]-\\frac{5}{13}[\/latex].<\/p>\n<p>The remaining functions can be calculated using identities relating them to sine and cosine.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\tan t=\\frac{\\sin t}{\\cos t}=\\frac{-\\frac{5}{13}}{\\frac{12}{13}}=-\\frac{5}{12} \\\\ \\sec t=\\frac{1}{\\cos t}=\\frac{1}{\\frac{12}{13}}=\\frac{13}{12} \\\\ \\csc t=\\frac{1}{\\sin t}=\\frac{1}{-\\frac{5}{13}}=-\\frac{13}{5}\\\\ \\cot t=\\frac{1}{\\tan t}=\\frac{1}{-\\frac{5}{12}}=-\\frac{12}{5}\\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm322751\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=322751&theme=lumen&iframe_resize_id=ohm322751&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm322752\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=322752&theme=lumen&iframe_resize_id=ohm322752&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>periodic functions<\/h3>\n<p>A function that repeats its values in regular intervals is known as a <strong>periodic function<\/strong>. The trigonometric functions are periodic. For the four trigonometric functions, sine, cosine, cosecant and secant, a revolution of one circle, or [latex]2\\pi[\/latex], will result in the same outputs for these functions. And for tangent and cotangent, only a half a revolution will result in the same outputs.swer]<\/p>\n<div id=\"q99481\" class=\"hidden-answer\" style=\"display: none\">\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin t=y=-\\frac{\\sqrt{3}}{2}\\\\ \\cos t=x=-\\frac{1}{2}\\\\ \\tan t=\\frac{\\sin t}{\\cos t}=\\frac{-\\frac{\\sqrt{3}}{2}}{-\\frac{1}{2}}=\\sqrt{3}\\\\ \\sec t=\\frac{1}{\\cos t}=\\frac{1}{-\\frac{1}{2}}=-2\\\\ \\csc t=\\frac{1}{\\sin t}=\\frac{1}{-\\frac{\\sqrt{3}}{2}}=-\\frac{2\\sqrt{3}}{3}\\\\ \\cot t=\\frac{1}{\\tan t}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}\\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">If [latex]\\sin \\left(t\\right)=-\\frac{\\sqrt{3}}{2}[\/latex] and [latex]\\text{cos}\\left(t\\right)=\\frac{1}{2}[\/latex], find [latex]\\text{sec}\\left(t\\right),\\text{csc}\\left(t\\right),\\text{tan}\\left(t\\right),\\text{ cot}\\left(t\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q940244\">Show Solution<\/button><\/p>\n<div id=\"q940244\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{gathered} \\sec t=\\frac{1}{\\cos t}=\\frac{1}{\\frac{1}{2}}=2 \\\\ \\csc t=\\frac{1}{\\sin t}=\\frac{1}{-\\frac{\\sqrt{3}}{2}}-\\frac{2\\sqrt{3}}{3} \\\\ \\tan t=\\frac{\\sin t}{\\cos t}=\\frac{-\\frac{\\sqrt{3}}{2}}{\\frac{1}{2}}=-\\sqrt{3} \\\\ \\cot t=\\frac{1}{\\tan t}=\\frac{1}{-\\sqrt{3}}=-\\frac{\\sqrt{3}}{3} \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm322766\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=322766&theme=lumen&iframe_resize_id=ohm322766&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":13,"menu_order":29,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1814","chapter","type-chapter","status-publish","hentry"],"part":178,"module-header":"learn_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\/1814","targetHints":{"allow":["GET"]}}],"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\/13"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1814\/revisions"}],"predecessor-version":[{"id":6367,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1814\/revisions\/6367"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/178"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1814\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1814"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1814"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1814"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1814"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}