{"id":1395,"date":"2023-06-05T14:51:18","date_gmt":"2023-06-05T14:51:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/chapter\/solutions-for-sum-and-difference-identities\/"},"modified":"2023-06-05T14:51:18","modified_gmt":"2023-06-05T14:51:18","slug":"solutions-for-sum-and-difference-identities","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/chapter\/solutions-for-sum-and-difference-identities\/","title":{"raw":"Solutions 52: Sum and Difference Identities","rendered":"Solutions 52: Sum and Difference Identities"},"content":{"raw":"\n<h2>Solutions to Odd-Numbered Answers<\/h2>\n1.&nbsp;The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures [latex]x[\/latex], the second angle measures [latex]\\frac{\\pi }{2}-x[\/latex]. Then [latex]\\sin x=\\cos \\left(\\frac{\\pi }{2}-x\\right)[\/latex]. The same holds for the other cofunction identities. The key is that the angles are complementary.\n\n3.&nbsp;[latex]\\sin \\left(-x\\right)=-\\sin x[\/latex], so [latex]\\sin x[\/latex] is odd. [latex]\\cos \\left(-x\\right)=\\cos \\left(0-x\\right)=\\cos x[\/latex], so [latex]\\cos x[\/latex] is even.\n\n5.&nbsp;[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]\n\n7.&nbsp;[latex]\\frac{\\sqrt{6}-\\sqrt{2}}{4}[\/latex]\n\n9.&nbsp;[latex]-2-\\sqrt{3}[\/latex]\n\n11.&nbsp;[latex]-\\frac{\\sqrt{2}}{2}\\sin x-\\frac{\\sqrt{2}}{2}\\cos x[\/latex]\n\n13.&nbsp;[latex]-\\frac{1}{2}\\cos x-\\frac{\\sqrt{3}}{2}\\sin x[\/latex]\n\n15.&nbsp;[latex]\\csc \\theta [\/latex]\n\n17.&nbsp;[latex]\\cot x[\/latex]\n\n19.&nbsp;[latex]\\tan \\left(\\frac{x}{10}\\right)[\/latex]\n\n21.&nbsp;[latex]\\sin \\left(a-b\\right)=\\left(\\frac{4}{5}\\right)\\left(\\frac{1}{3}\\right)-\\left(\\frac{3}{5}\\right)\\left(\\frac{2\\sqrt{2}}{3}\\right)=\\frac{4 - 6\\sqrt{2}}{15}[\/latex]\n[latex]\\cos \\left(a+b\\right)=\\left(\\frac{3}{5}\\right)\\left(\\frac{1}{3}\\right)-\\left(\\frac{4}{5}\\right)\\left(\\frac{2\\sqrt{2}}{3}\\right)=\\frac{3 - 8\\sqrt{2}}{15}[\/latex]\n\n23.&nbsp;[latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex]\n\n25.&nbsp;[latex]\\sin x[\/latex]\n<span id=\"fs-id1615462\">\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164052\/CNX_Precalc_Figure_07_02_201.jpg\" alt=\"Graph of y=sin(x) from -2pi to 2pi.\"><\/span>\n\n27.&nbsp;[latex]\\cot \\left(\\frac{\\pi }{6}-x\\right)[\/latex]\n<span id=\"fs-id2233832\">\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164054\/CNX_Precalc_Figure_07_02_203.jpg\" alt=\"Graph of y=cot(pi\/6 - x) from -2pi to pi - in comparison to the usual y=cot(x) graph, this one is reflected across the x-axis and shifted by pi\/6.\"><\/span>\n\n29.&nbsp;[latex]\\cot \\left(\\frac{\\pi }{4}+x\\right)[\/latex]\n<span id=\"fs-id1877975\">\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164057\/CNX_Precalc_Figure_07_02_205.jpg\" alt=\"Graph of y=cot(pi\/4 + x) - in comparison to the usual y=cot(x) graph, this one is shifted by pi\/4. \"><\/span>\n\n31.&nbsp;[latex]\\frac{\\sin x}{\\sqrt{2}}+\\frac{\\cos x}{\\sqrt{2}}[\/latex]\n<span id=\"fs-id2199923\">\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164059\/CNX_Precalc_Figure_07_02_207.jpg\" alt=\"Graph of y = sin(x) \/ rad2 + cos(x) \/ rad2 - it looks like the sin curve shifted by pi\/4.\"><\/span>\n\n&nbsp;\n\n33.&nbsp;They are the same.\n\n35.&nbsp;They are the different, try [latex]g\\left(x\\right)=\\sin \\left(9x\\right)-\\cos \\left(3x\\right)\\sin \\left(6x\\right)[\/latex].\n\n37.&nbsp;They are the same.\n\n39.&nbsp;They are the different, try [latex]g\\left(\\theta \\right)=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }[\/latex].\n\n41.&nbsp;They are different, try [latex]g\\left(x\\right)=\\frac{\\tan x-\\tan \\left(2x\\right)}{1+\\tan x\\tan \\left(2x\\right)}[\/latex].\n\n43.&nbsp;[latex]-\\frac{\\sqrt{3}-1}{2\\sqrt{2}},\\text{ or }-0.2588[\/latex]\n\n45.&nbsp;[latex]\\frac{1+\\sqrt{3}}{2\\sqrt{2}}[\/latex], or 0.9659\n\n47.&nbsp;[latex]\\begin{array}{c}\\tan \\left(x+\\frac{\\pi }{4}\\right)=\\\\ \\frac{\\tan x+\\tan \\left(\\frac{\\pi }{4}\\right)}{1-\\tan x\\tan \\left(\\frac{\\pi }{4}\\right)}=\\\\ \\frac{\\tan x+1}{1-\\tan x\\left(1\\right)}=\\frac{\\tan x+1}{1-\\tan x}\\end{array}[\/latex]\n\n&nbsp;\n\n49.&nbsp;[latex]\\begin{array}{c}\\frac{\\cos \\left(a+b\\right)}{\\cos a\\cos b}=\\\\ \\frac{\\cos a\\cos b}{\\cos a\\cos b}-\\frac{\\sin a\\sin b}{\\cos a\\cos b}=1-\\tan a\\tan b\\end{array}[\/latex]\n\n51.&nbsp;[latex]\\begin{array}{c}\\frac{\\cos \\left(x+h\\right)-\\cos x}{h}=\\\\ \\frac{\\cos x\\mathrm{cosh}-\\sin x\\mathrm{sinh}-\\cos x}{h}=\\\\ \\frac{\\cos x\\left(\\mathrm{cosh}-1\\right)-\\sin x\\mathrm{sinh}}{h}=\\cos x\\frac{\\cos h - 1}{h}-\\sin x\\frac{\\sin h}{h}\\end{array}[\/latex]\n\n53.&nbsp;True\n\n55.&nbsp;True. Note that [latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\left(\\pi -\\gamma \\right)[\/latex] and expand the right hand side.\n","rendered":"<h2>Solutions to Odd-Numbered Answers<\/h2>\n<p>1.&nbsp;The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures [latex]x[\/latex], the second angle measures [latex]\\frac{\\pi }{2}-x[\/latex]. Then [latex]\\sin x=\\cos \\left(\\frac{\\pi }{2}-x\\right)[\/latex]. The same holds for the other cofunction identities. The key is that the angles are complementary.<\/p>\n<p>3.&nbsp;[latex]\\sin \\left(-x\\right)=-\\sin x[\/latex], so [latex]\\sin x[\/latex] is odd. [latex]\\cos \\left(-x\\right)=\\cos \\left(0-x\\right)=\\cos x[\/latex], so [latex]\\cos x[\/latex] is even.<\/p>\n<p>5.&nbsp;[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]<\/p>\n<p>7.&nbsp;[latex]\\frac{\\sqrt{6}-\\sqrt{2}}{4}[\/latex]<\/p>\n<p>9.&nbsp;[latex]-2-\\sqrt{3}[\/latex]<\/p>\n<p>11.&nbsp;[latex]-\\frac{\\sqrt{2}}{2}\\sin x-\\frac{\\sqrt{2}}{2}\\cos x[\/latex]<\/p>\n<p>13.&nbsp;[latex]-\\frac{1}{2}\\cos x-\\frac{\\sqrt{3}}{2}\\sin x[\/latex]<\/p>\n<p>15.&nbsp;[latex]\\csc \\theta[\/latex]<\/p>\n<p>17.&nbsp;[latex]\\cot x[\/latex]<\/p>\n<p>19.&nbsp;[latex]\\tan \\left(\\frac{x}{10}\\right)[\/latex]<\/p>\n<p>21.&nbsp;[latex]\\sin \\left(a-b\\right)=\\left(\\frac{4}{5}\\right)\\left(\\frac{1}{3}\\right)-\\left(\\frac{3}{5}\\right)\\left(\\frac{2\\sqrt{2}}{3}\\right)=\\frac{4 - 6\\sqrt{2}}{15}[\/latex]<br \/>\n[latex]\\cos \\left(a+b\\right)=\\left(\\frac{3}{5}\\right)\\left(\\frac{1}{3}\\right)-\\left(\\frac{4}{5}\\right)\\left(\\frac{2\\sqrt{2}}{3}\\right)=\\frac{3 - 8\\sqrt{2}}{15}[\/latex]<\/p>\n<p>23.&nbsp;[latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex]<\/p>\n<p>25.&nbsp;[latex]\\sin x[\/latex]<br \/>\n<span id=\"fs-id1615462\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164052\/CNX_Precalc_Figure_07_02_201.jpg\" alt=\"Graph of y=sin(x) from -2pi to 2pi.\" \/><\/span><\/p>\n<p>27.&nbsp;[latex]\\cot \\left(\\frac{\\pi }{6}-x\\right)[\/latex]<br \/>\n<span id=\"fs-id2233832\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164054\/CNX_Precalc_Figure_07_02_203.jpg\" alt=\"Graph of y=cot(pi\/6 - x) from -2pi to pi - in comparison to the usual y=cot(x) graph, this one is reflected across the x-axis and shifted by pi\/6.\" \/><\/span><\/p>\n<p>29.&nbsp;[latex]\\cot \\left(\\frac{\\pi }{4}+x\\right)[\/latex]<br \/>\n<span id=\"fs-id1877975\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164057\/CNX_Precalc_Figure_07_02_205.jpg\" alt=\"Graph of y=cot(pi\/4 + x) - in comparison to the usual y=cot(x) graph, this one is shifted by pi\/4.\" \/><\/span><\/p>\n<p>31.&nbsp;[latex]\\frac{\\sin x}{\\sqrt{2}}+\\frac{\\cos x}{\\sqrt{2}}[\/latex]<br \/>\n<span id=\"fs-id2199923\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164059\/CNX_Precalc_Figure_07_02_207.jpg\" alt=\"Graph of y = sin(x) \/ rad2 + cos(x) \/ rad2 - it looks like the sin curve shifted by pi\/4.\" \/><\/span><\/p>\n<p>&nbsp;<\/p>\n<p>33.&nbsp;They are the same.<\/p>\n<p>35.&nbsp;They are the different, try [latex]g\\left(x\\right)=\\sin \\left(9x\\right)-\\cos \\left(3x\\right)\\sin \\left(6x\\right)[\/latex].<\/p>\n<p>37.&nbsp;They are the same.<\/p>\n<p>39.&nbsp;They are the different, try [latex]g\\left(\\theta \\right)=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }[\/latex].<\/p>\n<p>41.&nbsp;They are different, try [latex]g\\left(x\\right)=\\frac{\\tan x-\\tan \\left(2x\\right)}{1+\\tan x\\tan \\left(2x\\right)}[\/latex].<\/p>\n<p>43.&nbsp;[latex]-\\frac{\\sqrt{3}-1}{2\\sqrt{2}},\\text{ or }-0.2588[\/latex]<\/p>\n<p>45.&nbsp;[latex]\\frac{1+\\sqrt{3}}{2\\sqrt{2}}[\/latex], or 0.9659<\/p>\n<p>47.&nbsp;[latex]\\begin{array}{c}\\tan \\left(x+\\frac{\\pi }{4}\\right)=\\\\ \\frac{\\tan x+\\tan \\left(\\frac{\\pi }{4}\\right)}{1-\\tan x\\tan \\left(\\frac{\\pi }{4}\\right)}=\\\\ \\frac{\\tan x+1}{1-\\tan x\\left(1\\right)}=\\frac{\\tan x+1}{1-\\tan x}\\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>49.&nbsp;[latex]\\begin{array}{c}\\frac{\\cos \\left(a+b\\right)}{\\cos a\\cos b}=\\\\ \\frac{\\cos a\\cos b}{\\cos a\\cos b}-\\frac{\\sin a\\sin b}{\\cos a\\cos b}=1-\\tan a\\tan b\\end{array}[\/latex]<\/p>\n<p>51.&nbsp;[latex]\\begin{array}{c}\\frac{\\cos \\left(x+h\\right)-\\cos x}{h}=\\\\ \\frac{\\cos x\\mathrm{cosh}-\\sin x\\mathrm{sinh}-\\cos x}{h}=\\\\ \\frac{\\cos x\\left(\\mathrm{cosh}-1\\right)-\\sin x\\mathrm{sinh}}{h}=\\cos x\\frac{\\cos h - 1}{h}-\\sin x\\frac{\\sin h}{h}\\end{array}[\/latex]<\/p>\n<p>53.&nbsp;True<\/p>\n<p>55.&nbsp;True. Note that [latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\left(\\pi -\\gamma \\right)[\/latex] and expand the right hand side.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1395\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":503070,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1395","chapter","type-chapter","status-publish","hentry"],"part":1385,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters\/1395","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/users\/503070"}],"version-history":[{"count":0,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters\/1395\/revisions"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/parts\/1385"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters\/1395\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/media?parent=1395"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapter-type?post=1395"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/contributor?post=1395"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/license?post=1395"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}