{"id":204,"date":"2025-02-13T22:44:59","date_gmt":"2025-02-13T22:44:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/double-angle-half-angle-and-reduction-formulas\/"},"modified":"2025-10-14T19:08:09","modified_gmt":"2025-10-14T19:08:09","slug":"double-angle-half-angle-and-reduction-formulas","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/double-angle-half-angle-and-reduction-formulas\/","title":{"raw":"Double Angle, Half Angle, and Reduction Formulas: Learn It 1","rendered":"Double Angle, Half Angle, and Reduction Formulas: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Use double-angle formulas to find exact values.<\/li>\r\n \t<li style=\"font-weight: 400;\">Use double-angle formulas to verify identities.<\/li>\r\n \t<li style=\"font-weight: 400;\">Use reduction formulas to simplify an expression.<\/li>\r\n \t<li style=\"font-weight: 400;\">Use half-angle formulas to find exact values.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Using Double-Angle Formulas to Find Exact Values<\/h2>\r\nIn the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The <strong>double-angle formulas<\/strong> are a special case of the sum formulas, where [latex]\\alpha =\\beta [\/latex]. Deriving the double-angle formula for sine begins with the sum formula,\r\n<div style=\"text-align: center;\">[latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta [\/latex]<\/div>\r\nIf we let [latex]\\alpha =\\beta =\\theta [\/latex], then we have\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(\\theta +\\theta \\right)&amp;=\\sin \\theta \\cos \\theta +\\cos \\theta \\sin \\theta \\\\ \\sin \\left(2\\theta \\right)&amp;=2\\sin \\theta \\cos \\theta \\end{align}[\/latex]<\/div>\r\nDeriving the double-angle for cosine gives us three options. First, starting from the sum formula, [latex]\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta [\/latex], and letting [latex]\\alpha =\\beta =\\theta [\/latex], we have\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(\\theta +\\theta \\right)&amp;=\\cos \\theta \\cos \\theta -\\sin \\theta \\sin \\theta \\\\ \\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\end{align}[\/latex]<\/div>\r\nUsing the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &amp;=\\left(1-{\\sin }^{2}\\theta \\right)-{\\sin }^{2}\\theta \\\\ &amp;=1 - 2{\\sin }^{2}\\theta\\end{align}[\/latex]<\/div>\r\nThe second interpretation is:\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &amp;={\\cos }^{2}\\theta -\\left(1-{\\cos }^{2}\\theta \\right) \\\\ &amp;=2{\\cos }^{2}\\theta -1\\end{align}[\/latex]<\/div>\r\nSimilarly, to derive the double-angle formula for tangent, replacing [latex]\\alpha =\\beta =\\theta [\/latex] in the sum formula gives\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&amp;=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }\\\\ \\tan \\left(\\theta +\\theta \\right)&amp;=\\frac{\\tan \\theta +\\tan \\theta }{1-\\tan \\theta \\tan \\theta }\\\\ \\tan \\left(2\\theta \\right)&amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/div>\r\n<div><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>double angle formulas<\/h3>\r\nThe double-angle formulas are summarized as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&amp;=2\\sin \\theta \\cos \\theta\\\\\\text{ }\\\\ \\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &amp;=1 - 2{\\sin }^{2}\\theta \\\\ &amp;=2{\\cos }^{2}\\theta -1 \\\\\\text{ }\\\\ \\tan \\left(2\\theta \\right)&amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.\r\n<\/strong>\r\n<ol>\r\n \t<li>Draw a triangle to reflect the given information.<\/li>\r\n \t<li>Determine the correct double-angle formula.<\/li>\r\n \t<li>Substitute values into the formula based on the triangle.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Given that [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex] and [latex]\\theta [\/latex] is in quadrant II, find the following:\r\n<ol>\r\n \t<li>[latex]\\sin \\left(2\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\cos \\left(2\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\tan \\left(2\\theta \\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"477962\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"477962\"]\r\n\r\nIf we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given [latex]\\tan \\theta =-\\frac{3}{4}[\/latex], such that [latex]\\theta [\/latex] is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because [latex]\\theta [\/latex] is in the second quadrant, the adjacent side is on the <em>x<\/em>-axis and is negative. Use the <strong>Pythagorean Theorem<\/strong> to find the length of the hypotenuse:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\left(-4\\right)}^{2}+{\\left(3\\right)}^{2}&amp;={c}^{2}\\\\ 16+9&amp;={c}^{2}\\\\ 25&amp;={c}^{2}\\\\ c&amp;=5\\end{align}[\/latex]<\/p>\r\nNow we can draw a triangle:\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164101\/CNX_Precalc_Figure_07_03_0022.jpg\" alt=\"Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-4,0), and (-4,3). The angle at the origin is theta. The angle formed by the side (-4,3) to (-4,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 5.\" width=\"487\" height=\"251\" \/>\r\n<ol>\r\n \t<li>Let\u2019s begin by writing the double-angle formula for sine.\r\n<div style=\"text-align: center;\">[latex]\\sin \\left(2\\theta \\right)=2\\sin \\theta \\cos \\theta [\/latex]<\/div>\r\nWe see that we to need to find [latex]\\sin \\theta [\/latex] and [latex]\\cos \\theta [\/latex]. Based on Figure 2, we see that the hypotenuse equals 5, so [latex]\\sin \\theta =\\frac{3}{5}[\/latex], and [latex]\\cos \\theta =\u2212\\frac{4}{5}[\/latex]. Substitute these values into the equation, and simplify.\r\nThus,\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&amp;=2\\left(\\frac{3}{5}\\right)\\left(\u2212\\frac{4}{5}\\right) \\\\ &amp;=\u2212\\frac{24}{25} \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>Write the double-angle formula for cosine.\r\n<div style=\"text-align: center;\">[latex]\\cos \\left(2\\theta \\right)={\\cos }^{2}\\theta -{\\sin }^{2}\\theta [\/latex]<\/div>\r\nAgain, substitute the values of the sine and cosine into the equation, and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&amp;={\\left(\u2212\\frac{4}{5}\\right)}^{2}\u2212{\\left(\\frac{3}{5}\\right)}^{2} \\\\ &amp;=\\frac{16}{25}\u2212\\frac{9}{25} \\\\ &amp;=\\frac{7}{25}\\end{align}[\/latex]<\/div><\/li>\r\n \t<li>Write the double-angle formula for tangent.\r\n<div style=\"text-align: center;\">[latex]\\tan \\left(2\\theta \\right)=\\frac{2\\tan \\theta }{1\u2212{\\tan }^{2}\\theta }[\/latex]<\/div>\r\nIn this formula, we need the tangent, which we were given as [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex]. Substitute this value into the equation, and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(2\\theta \\right)&amp;=\\frac{2\\left(-\\frac{3}{4}\\right)}{1-{\\left(-\\frac{3}{4}\\right)}^{2}} \\\\ &amp;=\\frac{-\\frac{3}{2}}{1-\\frac{9}{16}} \\\\ &amp;=-\\frac{3}{2}\\left(\\frac{16}{7}\\right) \\\\ &amp;=-\\frac{24}{7} \\end{align}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">Given [latex]\\sin \\alpha =\\frac{5}{8}[\/latex], with [latex]\\theta [\/latex] in quadrant I, find [latex]\\cos \\left(2\\alpha \\right)[\/latex].[reveal-answer q=\"827677\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"827677\"][latex]\\cos \\left(2\\alpha \\right)=\\frac{7}{32}[\/latex][\/hidden-answer]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]156232[\/ohm_question]<\/section><section class=\"textbox example\" aria-label=\"Example\">Use the double-angle formula for cosine to write [latex]\\cos \\left(6x\\right)[\/latex] in terms of [latex]\\cos \\left(3x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"676542\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"676542\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(6x\\right)&amp;=\\cos \\left(3x+3x\\right) \\\\ &amp;=\\cos 3x\\cos 3x-\\sin 3x\\sin 3x \\\\ &amp;={\\cos }^{2}3x-{\\sin }^{2}3x \\end{align}[\/latex]<\/p>\r\n<strong>Analysis of the Solution<\/strong>\r\n\r\nThis example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li style=\"font-weight: 400;\">Use double-angle formulas to find exact values.<\/li>\n<li style=\"font-weight: 400;\">Use double-angle formulas to verify identities.<\/li>\n<li style=\"font-weight: 400;\">Use reduction formulas to simplify an expression.<\/li>\n<li style=\"font-weight: 400;\">Use half-angle formulas to find exact values.<\/li>\n<\/ul>\n<\/section>\n<h2>Using Double-Angle Formulas to Find Exact Values<\/h2>\n<p>In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The <strong>double-angle formulas<\/strong> are a special case of the sum formulas, where [latex]\\alpha =\\beta[\/latex]. Deriving the double-angle formula for sine begins with the sum formula,<\/p>\n<div style=\"text-align: center;\">[latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta[\/latex]<\/div>\n<p>If we let [latex]\\alpha =\\beta =\\theta[\/latex], then we have<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(\\theta +\\theta \\right)&=\\sin \\theta \\cos \\theta +\\cos \\theta \\sin \\theta \\\\ \\sin \\left(2\\theta \\right)&=2\\sin \\theta \\cos \\theta \\end{align}[\/latex]<\/div>\n<p>Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, [latex]\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta[\/latex], and letting [latex]\\alpha =\\beta =\\theta[\/latex], we have<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(\\theta +\\theta \\right)&=\\cos \\theta \\cos \\theta -\\sin \\theta \\sin \\theta \\\\ \\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\end{align}[\/latex]<\/div>\n<p>Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &=\\left(1-{\\sin }^{2}\\theta \\right)-{\\sin }^{2}\\theta \\\\ &=1 - 2{\\sin }^{2}\\theta\\end{align}[\/latex]<\/div>\n<p>The second interpretation is:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &={\\cos }^{2}\\theta -\\left(1-{\\cos }^{2}\\theta \\right) \\\\ &=2{\\cos }^{2}\\theta -1\\end{align}[\/latex]<\/div>\n<p>Similarly, to derive the double-angle formula for tangent, replacing [latex]\\alpha =\\beta =\\theta[\/latex] in the sum formula gives<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }\\\\ \\tan \\left(\\theta +\\theta \\right)&=\\frac{\\tan \\theta +\\tan \\theta }{1-\\tan \\theta \\tan \\theta }\\\\ \\tan \\left(2\\theta \\right)&=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/div>\n<div>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>double angle formulas<\/h3>\n<p>The double-angle formulas are summarized as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&=2\\sin \\theta \\cos \\theta\\\\\\text{ }\\\\ \\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &=1 - 2{\\sin }^{2}\\theta \\\\ &=2{\\cos }^{2}\\theta -1 \\\\\\text{ }\\\\ \\tan \\left(2\\theta \\right)&=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.<br \/>\n<\/strong><\/p>\n<ol>\n<li>Draw a triangle to reflect the given information.<\/li>\n<li>Determine the correct double-angle formula.<\/li>\n<li>Substitute values into the formula based on the triangle.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/section>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Given that [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex] and [latex]\\theta[\/latex] is in quadrant II, find the following:<\/p>\n<ol>\n<li>[latex]\\sin \\left(2\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\cos \\left(2\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\tan \\left(2\\theta \\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q477962\">Show Solution<\/button><\/p>\n<div id=\"q477962\" class=\"hidden-answer\" style=\"display: none\">\n<p>If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given [latex]\\tan \\theta =-\\frac{3}{4}[\/latex], such that [latex]\\theta[\/latex] is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because [latex]\\theta[\/latex] is in the second quadrant, the adjacent side is on the <em>x<\/em>-axis and is negative. Use the <strong>Pythagorean Theorem<\/strong> to find the length of the hypotenuse:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\left(-4\\right)}^{2}+{\\left(3\\right)}^{2}&={c}^{2}\\\\ 16+9&={c}^{2}\\\\ 25&={c}^{2}\\\\ c&=5\\end{align}[\/latex]<\/p>\n<p>Now we can draw a triangle:<\/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\/27164101\/CNX_Precalc_Figure_07_03_0022.jpg\" alt=\"Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-4,0), and (-4,3). The angle at the origin is theta. The angle formed by the side (-4,3) to (-4,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 5.\" width=\"487\" height=\"251\" \/><\/p>\n<ol>\n<li>Let\u2019s begin by writing the double-angle formula for sine.\n<div style=\"text-align: center;\">[latex]\\sin \\left(2\\theta \\right)=2\\sin \\theta \\cos \\theta[\/latex]<\/div>\n<p>We see that we to need to find [latex]\\sin \\theta[\/latex] and [latex]\\cos \\theta[\/latex]. Based on Figure 2, we see that the hypotenuse equals 5, so [latex]\\sin \\theta =\\frac{3}{5}[\/latex], and [latex]\\cos \\theta =\u2212\\frac{4}{5}[\/latex]. Substitute these values into the equation, and simplify.<br \/>\nThus,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&=2\\left(\\frac{3}{5}\\right)\\left(\u2212\\frac{4}{5}\\right) \\\\ &=\u2212\\frac{24}{25} \\end{align}[\/latex]<\/div>\n<\/li>\n<li>Write the double-angle formula for cosine.\n<div style=\"text-align: center;\">[latex]\\cos \\left(2\\theta \\right)={\\cos }^{2}\\theta -{\\sin }^{2}\\theta[\/latex]<\/div>\n<p>Again, substitute the values of the sine and cosine into the equation, and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&={\\left(\u2212\\frac{4}{5}\\right)}^{2}\u2212{\\left(\\frac{3}{5}\\right)}^{2} \\\\ &=\\frac{16}{25}\u2212\\frac{9}{25} \\\\ &=\\frac{7}{25}\\end{align}[\/latex]<\/div>\n<\/li>\n<li>Write the double-angle formula for tangent.\n<div style=\"text-align: center;\">[latex]\\tan \\left(2\\theta \\right)=\\frac{2\\tan \\theta }{1\u2212{\\tan }^{2}\\theta }[\/latex]<\/div>\n<p>In this formula, we need the tangent, which we were given as [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex]. Substitute this value into the equation, and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(2\\theta \\right)&=\\frac{2\\left(-\\frac{3}{4}\\right)}{1-{\\left(-\\frac{3}{4}\\right)}^{2}} \\\\ &=\\frac{-\\frac{3}{2}}{1-\\frac{9}{16}} \\\\ &=-\\frac{3}{2}\\left(\\frac{16}{7}\\right) \\\\ &=-\\frac{24}{7} \\end{align}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">Given [latex]\\sin \\alpha =\\frac{5}{8}[\/latex], with [latex]\\theta[\/latex] in quadrant I, find [latex]\\cos \\left(2\\alpha \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q827677\">Show Solution<\/button><\/p>\n<div id=\"q827677\" class=\"hidden-answer\" style=\"display: none\">[latex]\\cos \\left(2\\alpha \\right)=\\frac{7}{32}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm156232\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=156232&theme=lumen&iframe_resize_id=ohm156232&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use the double-angle formula for cosine to write [latex]\\cos \\left(6x\\right)[\/latex] in terms of [latex]\\cos \\left(3x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q676542\">Show Solution<\/button><\/p>\n<div id=\"q676542\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(6x\\right)&=\\cos \\left(3x+3x\\right) \\\\ &=\\cos 3x\\cos 3x-\\sin 3x\\sin 3x \\\\ &={\\cos }^{2}3x-{\\sin }^{2}3x \\end{align}[\/latex]<\/p>\n<p><strong>Analysis of the Solution<\/strong><\/p>\n<p>This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":16,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"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":"original","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\/204"}],"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":8,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/204\/revisions"}],"predecessor-version":[{"id":4677,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/204\/revisions\/4677"}],"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\/204\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=204"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=204"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=204"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=204"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}