{"id":3243,"date":"2025-08-15T23:43:46","date_gmt":"2025-08-15T23:43:46","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3243"},"modified":"2025-10-16T17:27:23","modified_gmt":"2025-10-16T17:27:23","slug":"double-angle-half-angle-and-reduction-formulas-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/double-angle-half-angle-and-reduction-formulas-apply-it\/","title":{"raw":"Double Angle, Half Angle, and Reduction Formulas: Apply It","rendered":"Double Angle, Half Angle, and Reduction Formulas: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use double-angle formulas to find exact values.<\/li>\r\n \t<li>Use double-angle formulas to verify identities.<\/li>\r\n \t<li>Use reduction formulas to simplify an expression.<\/li>\r\n \t<li>Use half-angle formulas to find exact values.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Apply Half and Double Angle Identities to Bike Ramps<\/h2>\r\n<section class=\"textbox example\" aria-label=\"Example\">A bicycle ramp is constructed for high-level competition with an angle of [latex]\\theta [\/latex] formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for higher-level competition, what is the measurement of the angle for novice competition?[reveal-answer q=\"685675\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"685675\"]Since the angle for novice competition measures half the steepness of the angle for the high level competition, and [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for high competition, we can find [latex]\\cos \\theta [\/latex] from the right triangle and the Pythagorean theorem so that we can use the half-angle identities.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{3}^{2}+{5}^{2}=34 \\\\ c=\\sqrt{34} \\end{gathered}[\/latex]<\/p>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164106\/CNX_Precalc_Figure_07_03_0042.jpg\" alt=\"Image of a right triangle with sides 3, 5, and rad34. Rad 34 is the hypotenuse, and 3 is the base. The angle formed by the hypotenuse and base is theta. The angle between the side of length 3 and side of length 5 is a right angle.\" width=\"487\" height=\"210\" \/>\r\n\r\nWe see that [latex]\\cos \\theta =\\frac{3}{\\sqrt{34}}=\\frac{3\\sqrt{34}}{34}[\/latex]. We can use the half-angle formula for tangent: [latex]\\tan \\frac{\\theta }{2}=\\sqrt{\\frac{1-\\cos \\theta }{1+\\cos \\theta }}[\/latex]. Since [latex]\\tan \\theta [\/latex] is in the first quadrant, so is [latex]\\tan \\frac{\\theta }{2}[\/latex]. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \\tan \\frac{\\theta }{2}&amp;=\\sqrt{\\frac{1-\\frac{3\\sqrt{34}}{34}}{1+\\frac{3\\sqrt{34}}{34}}} \\\\ &amp;=\\sqrt{\\frac{\\frac{34 - 3\\sqrt{34}}{34}}{\\frac{34+3\\sqrt{34}}{34}}} \\\\ &amp;=\\sqrt{\\frac{34 - 3\\sqrt{34}}{34+3\\sqrt{34}}} \\\\ &amp;\\approx 0.57\\end{align}[\/latex]<\/p>\r\nWe can take the inverse tangent to find the angle: [latex]{\\tan }^{-1}\\left(0.57\\right)\\approx {29.7}^{\\circ }[\/latex]. So the angle of the ramp for novice competition is [latex]\\approx {29.7}^{\\circ }[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">Using the same high-level competition ramp where [latex]\\tan\\theta = \\frac{5}{3}[\/latex], the designers want to create an expert ramp that is twice as steep. What is the angle for the expert competition ramp?<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use double-angle formulas to find exact values.<\/li>\n<li>Use double-angle formulas to verify identities.<\/li>\n<li>Use reduction formulas to simplify an expression.<\/li>\n<li>Use half-angle formulas to find exact values.<\/li>\n<\/ul>\n<\/section>\n<h2>Apply Half and Double Angle Identities to Bike Ramps<\/h2>\n<section class=\"textbox example\" aria-label=\"Example\">A bicycle ramp is constructed for high-level competition with an angle of [latex]\\theta[\/latex] formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for higher-level competition, what is the measurement of the angle for novice competition?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q685675\">Show Solution<\/button><\/p>\n<div id=\"q685675\" class=\"hidden-answer\" style=\"display: none\">Since the angle for novice competition measures half the steepness of the angle for the high level competition, and [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for high competition, we can find [latex]\\cos \\theta[\/latex] from the right triangle and the Pythagorean theorem so that we can use the half-angle identities.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{3}^{2}+{5}^{2}=34 \\\\ c=\\sqrt{34} \\end{gathered}[\/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\/27164106\/CNX_Precalc_Figure_07_03_0042.jpg\" alt=\"Image of a right triangle with sides 3, 5, and rad34. Rad 34 is the hypotenuse, and 3 is the base. The angle formed by the hypotenuse and base is theta. The angle between the side of length 3 and side of length 5 is a right angle.\" width=\"487\" height=\"210\" \/><\/p>\n<p>We see that [latex]\\cos \\theta =\\frac{3}{\\sqrt{34}}=\\frac{3\\sqrt{34}}{34}[\/latex]. We can use the half-angle formula for tangent: [latex]\\tan \\frac{\\theta }{2}=\\sqrt{\\frac{1-\\cos \\theta }{1+\\cos \\theta }}[\/latex]. Since [latex]\\tan \\theta[\/latex] is in the first quadrant, so is [latex]\\tan \\frac{\\theta }{2}[\/latex]. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \\tan \\frac{\\theta }{2}&=\\sqrt{\\frac{1-\\frac{3\\sqrt{34}}{34}}{1+\\frac{3\\sqrt{34}}{34}}} \\\\ &=\\sqrt{\\frac{\\frac{34 - 3\\sqrt{34}}{34}}{\\frac{34+3\\sqrt{34}}{34}}} \\\\ &=\\sqrt{\\frac{34 - 3\\sqrt{34}}{34+3\\sqrt{34}}} \\\\ &\\approx 0.57\\end{align}[\/latex]<\/p>\n<p>We can take the inverse tangent to find the angle: [latex]{\\tan }^{-1}\\left(0.57\\right)\\approx {29.7}^{\\circ }[\/latex]. So the angle of the ramp for novice competition is [latex]\\approx {29.7}^{\\circ }[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">Using the same high-level competition ramp where [latex]\\tan\\theta = \\frac{5}{3}[\/latex], the designers want to create an expert ramp that is twice as steep. What is the angle for the expert competition ramp?<\/section>\n","protected":false},"author":67,"menu_order":20,"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\/3243"}],"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":4,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3243\/revisions"}],"predecessor-version":[{"id":4684,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3243\/revisions\/4684"}],"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\/3243\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3243"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3243"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3243"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3243"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}