{"id":2056,"date":"2025-08-01T20:41:52","date_gmt":"2025-08-01T20:41:52","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=2056"},"modified":"2025-08-13T16:51:07","modified_gmt":"2025-08-13T16:51:07","slug":"right-triangle-trigonometry-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/right-triangle-trigonometry-learn-it-2\/","title":{"raw":"Right Triangle Trigonometry: Learn It 2","rendered":"Right Triangle Trigonometry: Learn It 2"},"content":{"raw":"<h2>Relating Angles and Their Functions<\/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.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.\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\/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\" \/> The side adjacent to one angle is opposite the other.[\/caption]\r\n\r\n<\/section>We will be asked to find all six trigonometric functions for a given angle in a triangle. Our 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\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.<\/strong>\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<\/section><section class=\"textbox example\" aria-label=\"Example\">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><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\" \/>[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<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<div class=\"bcc-box bcc-success\">\r\n\r\nEvaluate [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[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<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1 height=\"600\"]155312[\/ohm_question]<\/section><section aria-label=\"Try It\">\r\n<h2>Finding Trigonometric Functions of Special Angles Using Side Lengths<\/h2>\r\nWe have already discussed the trigonometric functions as they relate to the <strong>special angles<\/strong> on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of [latex]30^\\circ [\/latex], [latex]60^\\circ [\/latex], and [latex]45^\\circ[\/latex],\u00a0however, remember that when dealing with right triangles, we are limited to angles between [latex]0^\\circ \\text{ and } 90^\\circ[\/latex].\r\n\r\nSuppose we have a [latex]30^\\circ ,60^\\circ ,90^\\circ [\/latex] triangle, which can also be described as a [latex]\\frac{\\pi }{6}, \\frac{\\pi }{3},\\frac{\\pi }{2}[\/latex] triangle. The sides have lengths in the relation [latex]s,\\sqrt{3}s,2s[\/latex]. The sides of a [latex]45^\\circ ,45^\\circ ,90^\\circ [\/latex] triangle, which can also be described as a [latex]\\frac{\\pi }{4},\\frac{\\pi }{4},\\frac{\\pi }{2}[\/latex] triangle, have lengths in the relation [latex]s,s,\\sqrt{2}s[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003746\/CNX_Precalc_Figure_05_04_0082.jpg\" alt=\"Two side by side graphs of circles with inscribed angles. First circle has angle of pi\/3 inscribed. Second circle has angle of pi\/4 inscribed.\" width=\"975\" height=\"371\" \/> Side lengths of special triangles[\/caption]\r\n\r\nWe can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<div>\r\n<h3>ratios of special right triangles<\/h3>\r\nFor\u00a0<strong>every <\/strong>special right triangle:\r\n<ul>\r\n \t<li>For 45\u00ba or [latex]\\frac{\\pi}{4}[\/latex] angles\r\n<ul>\r\n \t<li>[latex]\\cos(x)=\\sin(x)=\\frac{\\sqrt{2}}{2}[\/latex]<\/li>\r\n \t<li>[latex]\\tan(x)=\\cot(x)=1[\/latex]<\/li>\r\n \t<li>[latex]\\sec(x)=\\csc(x)=\\sqrt{2}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>For 30\u00ba or [latex]\\frac{\\pi}{6}[\/latex] angles\r\n<ul>\r\n \t<li>[latex]\\cos(x)=\\frac{\\sqrt{3}}{2}[\/latex]<\/li>\r\n \t<li>[latex]\\sin(x)=\\frac{1}{2}[\/latex]<\/li>\r\n \t<li>[latex]\\tan(x)=\\frac{\\sqrt{3}}{3}[\/latex]<\/li>\r\n \t<li>[latex]\\cot(x)=\\sqrt{3}[\/latex]<\/li>\r\n \t<li>[latex]\\sec(x)=\\frac{2\\sqrt{3}}{3}[\/latex]<\/li>\r\n \t<li>[latex]\\csc(x)=2[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>For 60\u00ba or [latex]\\frac{\\pi}{3}[\/latex] angles\r\n<ul>\r\n \t<li>[latex]\\cos(x)=\\frac{1}{2}[\/latex]<\/li>\r\n \t<li>[latex]\\sin(x)=\\frac{\\sqrt{3}}{2}[\/latex]<\/li>\r\n \t<li>[latex]\\tan(x)=\\sqrt{3}[\/latex]<\/li>\r\n \t<li>[latex]\\cot(x)=\\frac{\\sqrt{3}}{3}[\/latex]<\/li>\r\n \t<li>[latex]\\sec(x)=2[\/latex]<\/li>\r\n \t<li>[latex]\\csc(x)=\\frac{2\\sqrt{3}}{3}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the exact value of the trigonometric functions of [latex]\\frac{\\pi }{3}[\/latex], using side lengths.[reveal-answer q=\"547484\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"547484\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\sin \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{opp}}{\\text{hyp}}=\\frac{\\sqrt{3}s}{2s}=\\frac{\\sqrt{3}}{2}\\\\ &amp;\\cos \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{adj}}{\\text{hyp}}=\\frac{s}{2s}=\\frac{1}{2}\\\\ &amp;\\tan \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{opp}}{\\text{adj}}=\\frac{\\sqrt{3}s}{s}=\\sqrt{3}\\\\ &amp;\\sec \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{hyp}}{\\text{adj}}=\\frac{2s}{s}=2\\\\ &amp;\\csc \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{hyp}}{\\text{opp}}=\\frac{2s}{\\sqrt{3}s}=\\frac{2}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3} \\\\ &amp;\\cot \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{adj}}{\\text{opp}}=\\frac{s}{\\sqrt{3}s}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">Find the exact value of the trigonometric functions of [latex]\\frac{\\pi }{4}[\/latex], using side lengths.[reveal-answer q=\"676808\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"676808\"][latex]\\sin \\left(\\frac{\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\cos \\left(\\frac{\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex],\r\n[latex]\\sec \\left(\\frac{\\pi }{4}\\right)=\\sqrt{2},\\csc\\left(\\frac{\\pi }{4}\\right)=\\sqrt{2},\\cot \\left(\\frac{\\pi }{4}\\right)=1[\/latex][\/hidden-answer]<\/section><\/section>","rendered":"<h2>Relating Angles and Their Functions<\/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.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.<\/p>\n<figure style=\"width: 487px\" 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\/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\" \/><figcaption class=\"wp-caption-text\">The side adjacent to one angle is opposite the other.<\/figcaption><\/figure>\n<\/section>\n<p>We will be asked to find all six trigonometric functions for a given angle in a triangle. Our 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<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.<\/strong><\/p>\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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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><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<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q626810\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<div class=\"bcc-box bcc-success\">\n<p>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<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q252611\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm155312\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=155312&theme=lumen&iframe_resize_id=ohm155312&source=tnh\" width=\"100%\" height=\"600\"><\/iframe><\/section>\n<section aria-label=\"Try It\">\n<h2>Finding Trigonometric Functions of Special Angles Using Side Lengths<\/h2>\n<p>We have already discussed the trigonometric functions as they relate to the <strong>special angles<\/strong> on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of [latex]30^\\circ[\/latex], [latex]60^\\circ[\/latex], and [latex]45^\\circ[\/latex],\u00a0however, remember that when dealing with right triangles, we are limited to angles between [latex]0^\\circ \\text{ and } 90^\\circ[\/latex].<\/p>\n<p>Suppose we have a [latex]30^\\circ ,60^\\circ ,90^\\circ[\/latex] triangle, which can also be described as a [latex]\\frac{\\pi }{6}, \\frac{\\pi }{3},\\frac{\\pi }{2}[\/latex] triangle. The sides have lengths in the relation [latex]s,\\sqrt{3}s,2s[\/latex]. The sides of a [latex]45^\\circ ,45^\\circ ,90^\\circ[\/latex] triangle, which can also be described as a [latex]\\frac{\\pi }{4},\\frac{\\pi }{4},\\frac{\\pi }{2}[\/latex] triangle, have lengths in the relation [latex]s,s,\\sqrt{2}s[\/latex].<\/p>\n<figure style=\"width: 975px\" 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\/27003746\/CNX_Precalc_Figure_05_04_0082.jpg\" alt=\"Two side by side graphs of circles with inscribed angles. First circle has angle of pi\/3 inscribed. Second circle has angle of pi\/4 inscribed.\" width=\"975\" height=\"371\" \/><figcaption class=\"wp-caption-text\">Side lengths of special triangles<\/figcaption><\/figure>\n<p>We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<div>\n<h3>ratios of special right triangles<\/h3>\n<p>For\u00a0<strong>every <\/strong>special right triangle:<\/p>\n<ul>\n<li>For 45\u00ba or [latex]\\frac{\\pi}{4}[\/latex] angles\n<ul>\n<li>[latex]\\cos(x)=\\sin(x)=\\frac{\\sqrt{2}}{2}[\/latex]<\/li>\n<li>[latex]\\tan(x)=\\cot(x)=1[\/latex]<\/li>\n<li>[latex]\\sec(x)=\\csc(x)=\\sqrt{2}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>For 30\u00ba or [latex]\\frac{\\pi}{6}[\/latex] angles\n<ul>\n<li>[latex]\\cos(x)=\\frac{\\sqrt{3}}{2}[\/latex]<\/li>\n<li>[latex]\\sin(x)=\\frac{1}{2}[\/latex]<\/li>\n<li>[latex]\\tan(x)=\\frac{\\sqrt{3}}{3}[\/latex]<\/li>\n<li>[latex]\\cot(x)=\\sqrt{3}[\/latex]<\/li>\n<li>[latex]\\sec(x)=\\frac{2\\sqrt{3}}{3}[\/latex]<\/li>\n<li>[latex]\\csc(x)=2[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>For 60\u00ba or [latex]\\frac{\\pi}{3}[\/latex] angles\n<ul>\n<li>[latex]\\cos(x)=\\frac{1}{2}[\/latex]<\/li>\n<li>[latex]\\sin(x)=\\frac{\\sqrt{3}}{2}[\/latex]<\/li>\n<li>[latex]\\tan(x)=\\sqrt{3}[\/latex]<\/li>\n<li>[latex]\\cot(x)=\\frac{\\sqrt{3}}{3}[\/latex]<\/li>\n<li>[latex]\\sec(x)=2[\/latex]<\/li>\n<li>[latex]\\csc(x)=\\frac{2\\sqrt{3}}{3}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the exact value of the trigonometric functions of [latex]\\frac{\\pi }{3}[\/latex], using side lengths.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q547484\">Show Solution<\/button><\/p>\n<div id=\"q547484\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\sin \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{opp}}{\\text{hyp}}=\\frac{\\sqrt{3}s}{2s}=\\frac{\\sqrt{3}}{2}\\\\ &\\cos \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{adj}}{\\text{hyp}}=\\frac{s}{2s}=\\frac{1}{2}\\\\ &\\tan \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{opp}}{\\text{adj}}=\\frac{\\sqrt{3}s}{s}=\\sqrt{3}\\\\ &\\sec \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{hyp}}{\\text{adj}}=\\frac{2s}{s}=2\\\\ &\\csc \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{hyp}}{\\text{opp}}=\\frac{2s}{\\sqrt{3}s}=\\frac{2}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3} \\\\ &\\cot \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{adj}}{\\text{opp}}=\\frac{s}{\\sqrt{3}s}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">Find the exact value of the trigonometric functions of [latex]\\frac{\\pi }{4}[\/latex], using side lengths.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q676808\">Show Solution<\/button><\/p>\n<div id=\"q676808\" class=\"hidden-answer\" style=\"display: none\">[latex]\\sin \\left(\\frac{\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\cos \\left(\\frac{\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex],<br \/>\n[latex]\\sec \\left(\\frac{\\pi }{4}\\right)=\\sqrt{2},\\csc\\left(\\frac{\\pi }{4}\\right)=\\sqrt{2},\\cot \\left(\\frac{\\pi }{4}\\right)=1[\/latex]<\/div>\n<\/div>\n<\/section>\n<\/section>\n","protected":false},"author":13,"menu_order":6,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":221,"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\/2056"}],"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":5,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2056\/revisions"}],"predecessor-version":[{"id":2068,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2056\/revisions\/2068"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/221"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2056\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=2056"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=2056"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=2056"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=2056"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}