{"id":3239,"date":"2025-08-15T23:42:55","date_gmt":"2025-08-15T23:42:55","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3239"},"modified":"2025-10-14T18:31:32","modified_gmt":"2025-10-14T18:31:32","slug":"sum-and-difference-identities-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/sum-and-difference-identities-apply-it\/","title":{"raw":"Sum and Difference Identities: Apply It","rendered":"Sum and Difference Identities: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Use sum and difference formulas for sine, cosine, and tangent<\/li>\r\n \t<li style=\"font-weight: 400;\">Use sum and difference formulas to verify identities.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Use sum and difference formulas to verify identities<\/h2>\r\nVerifying an identity means demonstrating that the equation holds for all values of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems.\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given an identity, verify using sum and difference formulas.<\/strong>\r\n<ol>\r\n \t<li>Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient.<\/li>\r\n \t<li>Look for opportunities to use the sum and difference formulas.<\/li>\r\n \t<li>Rewrite sums or differences of quotients as single quotients.<\/li>\r\n \t<li>If the process becomes cumbersome, rewrite the expression in terms of sines and cosines.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<div><section class=\"textbox example\" aria-label=\"Example\">Verify the identity [latex]\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)=2\\sin \\alpha \\cos \\beta [\/latex].[reveal-answer q=\"464113\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"464113\"]We see that the left side of the equation includes the sines of the sum and the difference of angles.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta\\\\ \\sin \\left(\\alpha -\\beta \\right)=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\end{gathered}[\/latex]<\/p>\r\nWe can rewrite each using the sum and difference formulas.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)&amp;=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta +\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\\\ &amp;=2\\sin \\alpha \\cos \\beta \\end{align}[\/latex]<\/p>\r\nWe see that the identity is verified.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Verify the following identity.\r\n<p style=\"text-align: center;\">[latex]\\frac{\\sin \\left(\\alpha -\\beta \\right)}{\\cos \\alpha \\cos \\beta }=\\tan \\alpha -\\tan \\beta [\/latex]<\/p>\r\n[reveal-answer q=\"605610\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"605610\"]\r\n\r\nWe can begin by rewriting the numerator on the left side of the equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{\\sin \\left(\\alpha -\\beta \\right)}{\\cos \\alpha \\cos \\beta }&amp;=\\frac{\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta } \\\\ &amp;=\\frac{\\sin \\alpha \\cos \\beta}{\\cos \\alpha \\cos \\beta}-\\frac{\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta } &amp;&amp; \\text{Rewrite using a common denominator}. \\\\ &amp;=\\frac{\\sin \\alpha }{\\cos \\alpha }-\\frac{\\sin \\beta }{\\cos \\beta }&amp;&amp; \\text{Cancel}. \\\\ &amp;=\\tan \\alpha -\\tan \\beta &amp;&amp; \\text{Rewrite in terms of tangent}.\\end{align}[\/latex]<\/p>\r\nWe see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">Verify the identity: [latex]\\tan \\left(\\pi -\\theta \\right)=-\\tan \\theta [\/latex].[reveal-answer q=\"184004\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"184004\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(\\pi -\\theta \\right)&amp;=\\frac{\\tan \\left(\\pi \\right)-\\tan \\theta }{1+\\tan \\left(\\pi \\right)\\tan \\theta } \\\\ &amp;=\\frac{0-\\tan \\theta }{1+0\\cdot \\tan \\theta } \\\\ &amp;=-\\tan \\theta \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Let [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex] denote two non-vertical intersecting lines, and let [latex]\\theta [\/latex] denote the acute angle between [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex]. Show that\r\n<p style=\"text-align: center;\">[latex]\\tan \\theta =\\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}[\/latex]<\/p>\r\nwhere [latex]{m}_{1}[\/latex] and [latex]{m}_{2}[\/latex] are the slopes of [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex] respectively. (<strong>Hint:<\/strong> Use the fact that [latex]\\tan {\\theta }_{1}={m}_{1}[\/latex] and [latex]\\tan {\\theta }_{2}={m}_{2}[\/latex]. )\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164047\/CNX_Precalc_Figure_07_02_0052.jpg\" alt=\"Diagram of two non-vertical intersecting lines L1 and L2 also intersecting the x-axis. The acute angle formed by the intersection of L1 and L2 is theta. The acute angle formed by L2 and the x-axis is theta 1, and the acute angle formed by the x-axis and L1 is theta 2. \" width=\"487\" height=\"289\" \/>\r\n\r\n[reveal-answer q=\"429859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"429859\"]\r\n\r\nUsing the difference formula for tangent, this problem does not seem as daunting as it might.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\tan \\theta &amp;=\\tan \\left({\\theta }_{2}-{\\theta }_{1}\\right) \\\\ &amp;=\\frac{\\tan {\\theta }_{2}-\\tan {\\theta }_{1}}{1+\\tan {\\theta }_{1}\\tan {\\theta }_{2}} \\\\ &amp;=\\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]<span style=\"text-align: center; font-size: 0.9em;\">\u00a0<\/span>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164049\/CNX_Precalc_Figure_07_02_0062.jpg\" alt=\"Two right triangles. Both share the same base, 50 feet. The first has a height of 40 ft and hypotenuse S. The second has height 47 ft and hypotenuse R. The height sides of the triangles are overlapping. There is a B degree angle between R and the base, and an a degree angle between the two hypotenuses within the B degree angle. \" width=\"590\" height=\"322\" \/>For a climbing wall, a guy-wire [latex]R[\/latex] is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire [latex]S[\/latex] attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle [latex]\\alpha [\/latex] between the wires.\u00a0<span id=\"fs-id1386435\">\r\n<\/span>[reveal-answer q=\"48078\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"48078\"]\r\n\r\nLet\u2019s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that [latex]\\tan \\beta =\\frac{47}{50}[\/latex], and [latex]\\tan \\left(\\beta -\\alpha \\right)=\\frac{40}{50}=\\frac{4}{5}[\/latex]. We can then use difference formula for tangent.\r\n<p style=\"text-align: center;\">[latex]\\tan \\left(\\beta -\\alpha \\right)=\\frac{\\tan \\beta -\\tan \\alpha }{1+\\tan \\beta \\tan \\alpha }[\/latex]<\/p>\r\nNow, substituting the values we know into the formula, we have\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\frac{4}{5}=\\frac{\\frac{47}{50}-\\tan \\alpha }{1+\\frac{47}{50}\\tan \\alpha } \\\\ 4\\left(1+\\frac{47}{50}\\tan \\alpha \\right)=5\\left(\\frac{47}{50}-\\tan \\alpha \\right) \\end{gathered}[\/latex]<\/p>\r\nUse the distributive property, and then simplify the functions.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}4\\left(1\\right)+4\\left(\\frac{47}{50}\\right)\\tan \\alpha =5\\left(\\frac{47}{50}\\right)-5\\tan \\alpha\\\\ 4+3.76\\tan \\alpha =4.7 - 5\\tan \\alpha \\\\ 5\\tan \\alpha +3.76\\tan \\alpha =0.7\\\\ 8.76\\tan \\alpha =0.7\\\\ \\tan \\alpha \\approx 0.07991 \\\\ {\\tan }^{-1}\\left(0.07991\\right)\\approx .079741 \\end{gathered}[\/latex]<\/p>\r\nNow we can calculate the angle in degrees.\r\n<p style=\"text-align: center;\">[latex]\\alpha \\approx 0.079741\\left(\\frac{180}{\\pi }\\right)\\approx {4.57}^{\\circ }[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nOccasionally, when an application appears that includes a right triangle, we may think that solving is a matter of applying the Pythagorean Theorem. That may be partially true, but it depends on what the problem is asking and what information is given.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li style=\"font-weight: 400;\">Use sum and difference formulas for sine, cosine, and tangent<\/li>\n<li style=\"font-weight: 400;\">Use sum and difference formulas to verify identities.<\/li>\n<\/ul>\n<\/section>\n<h2>Use sum and difference formulas to verify identities<\/h2>\n<p>Verifying an identity means demonstrating that the equation holds for all values of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given an identity, verify using sum and difference formulas.<\/strong><\/p>\n<ol>\n<li>Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient.<\/li>\n<li>Look for opportunities to use the sum and difference formulas.<\/li>\n<li>Rewrite sums or differences of quotients as single quotients.<\/li>\n<li>If the process becomes cumbersome, rewrite the expression in terms of sines and cosines.<\/li>\n<\/ol>\n<\/section>\n<div>\n<section class=\"textbox example\" aria-label=\"Example\">Verify the identity [latex]\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)=2\\sin \\alpha \\cos \\beta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q464113\">Show Solution<\/button><\/p>\n<div id=\"q464113\" class=\"hidden-answer\" style=\"display: none\">We see that the left side of the equation includes the sines of the sum and the difference of angles.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta\\\\ \\sin \\left(\\alpha -\\beta \\right)=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\end{gathered}[\/latex]<\/p>\n<p>We can rewrite each using the sum and difference formulas.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)&=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta +\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\\\ &=2\\sin \\alpha \\cos \\beta \\end{align}[\/latex]<\/p>\n<p>We see that the identity is verified.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Verify the following identity.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\sin \\left(\\alpha -\\beta \\right)}{\\cos \\alpha \\cos \\beta }=\\tan \\alpha -\\tan \\beta[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q605610\">Show Solution<\/button><\/p>\n<div id=\"q605610\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can begin by rewriting the numerator on the left side of the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{\\sin \\left(\\alpha -\\beta \\right)}{\\cos \\alpha \\cos \\beta }&=\\frac{\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta } \\\\ &=\\frac{\\sin \\alpha \\cos \\beta}{\\cos \\alpha \\cos \\beta}-\\frac{\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta } && \\text{Rewrite using a common denominator}. \\\\ &=\\frac{\\sin \\alpha }{\\cos \\alpha }-\\frac{\\sin \\beta }{\\cos \\beta }&& \\text{Cancel}. \\\\ &=\\tan \\alpha -\\tan \\beta && \\text{Rewrite in terms of tangent}.\\end{align}[\/latex]<\/p>\n<p>We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">Verify the identity: [latex]\\tan \\left(\\pi -\\theta \\right)=-\\tan \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q184004\">Show Solution<\/button><\/p>\n<div id=\"q184004\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(\\pi -\\theta \\right)&=\\frac{\\tan \\left(\\pi \\right)-\\tan \\theta }{1+\\tan \\left(\\pi \\right)\\tan \\theta } \\\\ &=\\frac{0-\\tan \\theta }{1+0\\cdot \\tan \\theta } \\\\ &=-\\tan \\theta \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Let [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex] denote two non-vertical intersecting lines, and let [latex]\\theta[\/latex] denote the acute angle between [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex]. Show that<\/p>\n<p style=\"text-align: center;\">[latex]\\tan \\theta =\\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}[\/latex]<\/p>\n<p>where [latex]{m}_{1}[\/latex] and [latex]{m}_{2}[\/latex] are the slopes of [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex] respectively. (<strong>Hint:<\/strong> Use the fact that [latex]\\tan {\\theta }_{1}={m}_{1}[\/latex] and [latex]\\tan {\\theta }_{2}={m}_{2}[\/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\/27164047\/CNX_Precalc_Figure_07_02_0052.jpg\" alt=\"Diagram of two non-vertical intersecting lines L1 and L2 also intersecting the x-axis. The acute angle formed by the intersection of L1 and L2 is theta. The acute angle formed by L2 and the x-axis is theta 1, and the acute angle formed by the x-axis and L1 is theta 2.\" width=\"487\" height=\"289\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q429859\">Show Solution<\/button><\/p>\n<div id=\"q429859\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the difference formula for tangent, this problem does not seem as daunting as it might.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\tan \\theta &=\\tan \\left({\\theta }_{2}-{\\theta }_{1}\\right) \\\\ &=\\frac{\\tan {\\theta }_{2}-\\tan {\\theta }_{1}}{1+\\tan {\\theta }_{1}\\tan {\\theta }_{2}} \\\\ &=\\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><span style=\"text-align: center; font-size: 0.9em;\">\u00a0<\/span><\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164049\/CNX_Precalc_Figure_07_02_0062.jpg\" alt=\"Two right triangles. Both share the same base, 50 feet. The first has a height of 40 ft and hypotenuse S. The second has height 47 ft and hypotenuse R. The height sides of the triangles are overlapping. There is a B degree angle between R and the base, and an a degree angle between the two hypotenuses within the B degree angle.\" width=\"590\" height=\"322\" \/>For a climbing wall, a guy-wire [latex]R[\/latex] is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire [latex]S[\/latex] attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle [latex]\\alpha[\/latex] between the wires.\u00a0<span id=\"fs-id1386435\"><br \/>\n<\/span><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q48078\">Show Solution<\/button><\/p>\n<div id=\"q48078\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let\u2019s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that [latex]\\tan \\beta =\\frac{47}{50}[\/latex], and [latex]\\tan \\left(\\beta -\\alpha \\right)=\\frac{40}{50}=\\frac{4}{5}[\/latex]. We can then use difference formula for tangent.<\/p>\n<p style=\"text-align: center;\">[latex]\\tan \\left(\\beta -\\alpha \\right)=\\frac{\\tan \\beta -\\tan \\alpha }{1+\\tan \\beta \\tan \\alpha }[\/latex]<\/p>\n<p>Now, substituting the values we know into the formula, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\frac{4}{5}=\\frac{\\frac{47}{50}-\\tan \\alpha }{1+\\frac{47}{50}\\tan \\alpha } \\\\ 4\\left(1+\\frac{47}{50}\\tan \\alpha \\right)=5\\left(\\frac{47}{50}-\\tan \\alpha \\right) \\end{gathered}[\/latex]<\/p>\n<p>Use the distributive property, and then simplify the functions.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}4\\left(1\\right)+4\\left(\\frac{47}{50}\\right)\\tan \\alpha =5\\left(\\frac{47}{50}\\right)-5\\tan \\alpha\\\\ 4+3.76\\tan \\alpha =4.7 - 5\\tan \\alpha \\\\ 5\\tan \\alpha +3.76\\tan \\alpha =0.7\\\\ 8.76\\tan \\alpha =0.7\\\\ \\tan \\alpha \\approx 0.07991 \\\\ {\\tan }^{-1}\\left(0.07991\\right)\\approx .079741 \\end{gathered}[\/latex]<\/p>\n<p>Now we can calculate the angle in degrees.<\/p>\n<p style=\"text-align: center;\">[latex]\\alpha \\approx 0.079741\\left(\\frac{180}{\\pi }\\right)\\approx {4.57}^{\\circ }[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Occasionally, when an application appears that includes a right triangle, we may think that solving is a matter of applying the Pythagorean Theorem. That may be partially true, but it depends on what the problem is asking and what information is given.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n","protected":false},"author":67,"menu_order":14,"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\/3239"}],"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":3,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3239\/revisions"}],"predecessor-version":[{"id":4672,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3239\/revisions\/4672"}],"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\/3239\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3239"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3239"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3239"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3239"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}