{"id":15687,"date":"2019-09-05T17:47:05","date_gmt":"2019-09-05T17:47:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=15687"},"modified":"2025-02-05T05:23:31","modified_gmt":"2025-02-05T05:23:31","slug":"problem-set-51-solving-trigonometric-equations-with-identities","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/problem-set-51-solving-trigonometric-equations-with-identities\/","title":{"raw":"Problem Set 51: Solving Trigonometric Equations With Identities","rendered":"Problem Set 51: Solving Trigonometric Equations With Identities"},"content":{"raw":"<section id=\"fs-id2347805\" class=\"section-exercises\"><\/section>1. We know [latex]g\\left(x\\right)=\\cos x[\/latex] is an even function, and [latex]f\\left(x\\right)=\\sin x[\/latex] and [latex]h\\left(x\\right)=\\tan x[\/latex] are odd functions. What about [latex]G\\left(x\\right)={\\cos }^{2}x,F\\left(x\\right)={\\sin }^{2}x[\/latex], and [latex]H\\left(x\\right)={\\tan }^{2}x?[\/latex] Are they even, odd, or neither? Why?\r\n\r\n2. Examine the graph of [latex]f\\left(x\\right)=\\sec x[\/latex] on the interval [latex]\\left[-\\pi ,\\pi \\right][\/latex]. How can we tell whether the function is even or odd by only observing the graph of [latex]f\\left(x\\right)=\\sec x?[\/latex]\r\n\r\n3. After examining the reciprocal identity for [latex]\\sec t[\/latex], explain why the function is undefined at certain points.\r\n\r\n4. All of the Pythagorean identities are related. Describe how to manipulate the equations to get from [latex]{\\sin }^{2}t+{\\cos }^{2}t=1[\/latex] to the other forms.\r\n\r\nFor the following exercises, use the fundamental identities to fully simplify the expression.\r\n\r\n5. [latex]\\sin x\\cos x\\sec x[\/latex]\r\n\r\n6.\u00a0[latex]\\sin \\left(-x\\right)\\cos \\left(-x\\right)\\csc \\left(-x\\right)[\/latex]\r\n\r\n7. [latex]\\tan x\\sin x+\\sec x{\\cos }^{2}x[\/latex]\r\n\r\n8.\u00a0[latex]\\csc x+\\cos x\\cot \\left(-x\\right)[\/latex]\r\n\r\n9.\u00a0[latex]\\frac{\\cot t+\\tan t}{\\sec \\left(-t\\right)}[\/latex]\r\n\r\n10.\u00a0[latex]3{\\sin }^{3}t\\csc t+{\\cos }^{2}t+2\\cos \\left(-t\\right)\\cos t[\/latex]\r\n\r\n11. [latex]-\\tan \\left(-x\\right)\\cot \\left(-x\\right)[\/latex]\r\n\r\n12.\u00a0[latex]\\frac{-\\sin \\left(-x\\right)\\cos x\\sec x\\csc x\\tan x}{\\cot x}[\/latex]\r\n\r\n13. [latex]\\frac{1+{\\tan }^{2}\\theta }{{\\csc }^{2}\\theta }+{\\sin }^{2}\\theta +\\frac{1}{{\\sec }^{2}\\theta }[\/latex]\r\n\r\n14.\u00a0[latex]\\left(\\frac{\\tan x}{{\\csc }^{2}x}+\\frac{\\tan x}{{\\sec }^{2}x}\\right)\\left(\\frac{1+\\tan x}{1+\\cot x}\\right)-\\frac{1}{{\\cos }^{2}x}[\/latex]\r\n\r\n15. [latex]\\frac{1-{\\cos }^{2}x}{{\\tan }^{2}x}+2{\\sin }^{2}x[\/latex]\r\n\r\nFor the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression.\r\n\r\n16. [latex]\\frac{\\tan x+\\cot x}{\\csc x};\\cos x[\/latex]\r\n\r\n17. [latex]\\frac{\\sec x+\\csc x}{1+\\tan x};\\sin x[\/latex]\r\n\r\n18.\u00a0[latex]\\frac{\\cos x}{1+\\sin x}+\\tan x;\\cos x[\/latex]\r\n\r\n19. [latex]\\frac{1}{\\sin x\\cos x}-\\cot x;\\cot x[\/latex]\r\n\r\n20.\u00a0[latex]\\frac{1}{1-\\cos x}-\\frac{\\cos x}{1+\\cos x};\\csc x[\/latex]\r\n\r\n21. [latex]\\left(\\sec x+\\csc x\\right)\\left(\\sin x+\\cos x\\right)-2-\\cot x;\\tan x[\/latex]\r\n\r\n22.\u00a0[latex]\\frac{1}{\\csc x-\\sin x};\\sec x\\text{ and }\\tan x[\/latex]\r\n\r\n23. [latex]\\frac{1-\\sin x}{1+\\sin x}-\\frac{1+\\sin x}{1-\\sin x};\\sec x\\text{ and }\\tan x[\/latex]\r\n\r\n24.\u00a0[latex]\\tan x;\\sec x[\/latex]\r\n\r\n25. [latex]\\sec x;\\cot x[\/latex]\r\n\r\n26.\u00a0[latex]\\sec x;\\sin x[\/latex]\r\n\r\n27. [latex]\\cot x;\\sin x[\/latex]\r\n\r\n28.\u00a0[latex]\\cot x;\\csc x[\/latex]\r\n\r\nFor the following exercises, verify the identity.\r\n\r\n29. [latex]\\cos x-{\\cos }^{3}x=\\cos x{\\sin }^{2}x[\/latex]\r\n\r\n30. [latex]\\cos x\\left(\\tan x-\\sec \\left(-x\\right)\\right)=\\sin x - 1[\/latex]\r\n\r\n31. [latex]\\frac{1+{\\sin }^{2}x}{{\\cos }^{2}x}=\\frac{1}{{\\cos }^{2}x}+\\frac{{\\sin }^{2}x}{{\\cos }^{2}x}=1+2{\\tan }^{2}x[\/latex]\r\n\r\n32. [latex]{\\left(\\sin x+\\cos x\\right)}^{2}=1+2\\sin x\\cos x[\/latex]\r\n\r\n33. [latex]{\\cos }^{2}x-{\\tan }^{2}x=2-{\\sin }^{2}x-{\\sec }^{2}x[\/latex]\r\n\r\nFor the following exercises, prove or disprove the identity.\r\n\r\n34. [latex]\\frac{1}{1+\\cos x}-\\frac{1}{1-\\cos \\left(-x\\right)}=-2\\cot x\\csc x[\/latex]\r\n\r\n35. [latex]{\\csc }^{2}x\\left(1+{\\sin }^{2}x\\right)={\\cot }^{2}x[\/latex]\r\n\r\n36.\u00a0[latex]\\left(\\frac{{\\sec }^{2}\\left(-x\\right)-{\\tan }^{2}x}{\\tan x}\\right)\\left(\\frac{2+2\\tan x}{2+2\\cot x}\\right)-2{\\sin }^{2}x=\\cos 2x[\/latex]\r\n\r\n37. [latex]\\frac{\\tan x}{\\sec x}\\sin \\left(-x\\right)={\\cos }^{2}x[\/latex]\r\n\r\n38.\u00a0[latex]\\frac{\\sec \\left(-x\\right)}{\\tan x+\\cot x}=-\\sin \\left(-x\\right)[\/latex]\r\n\r\n39. [latex]\\frac{1+\\sin x}{\\cos x}=\\frac{\\cos x}{1+\\sin \\left(-x\\right)}[\/latex]\r\n\r\nFor the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression.\r\n\r\n40. [latex]\\frac{{\\cos }^{2}\\theta -{\\sin }^{2}\\theta }{1-{\\tan }^{2}\\theta }={\\sin }^{2}\\theta [\/latex]\r\n\r\n41. [latex]3{\\sin }^{2}\\theta +4{\\cos }^{2}\\theta =3+{\\cos }^{2}\\theta [\/latex]\r\n\r\n42.\u00a0[latex]\\frac{\\sec \\theta +\\tan \\theta }{\\cot \\theta +\\cos \\theta }={\\sec }^{2}\\theta [\/latex]","rendered":"<section id=\"fs-id2347805\" class=\"section-exercises\"><\/section>\n<p>1. We know [latex]g\\left(x\\right)=\\cos x[\/latex] is an even function, and [latex]f\\left(x\\right)=\\sin x[\/latex] and [latex]h\\left(x\\right)=\\tan x[\/latex] are odd functions. What about [latex]G\\left(x\\right)={\\cos }^{2}x,F\\left(x\\right)={\\sin }^{2}x[\/latex], and [latex]H\\left(x\\right)={\\tan }^{2}x?[\/latex] Are they even, odd, or neither? Why?<\/p>\n<p>2. Examine the graph of [latex]f\\left(x\\right)=\\sec x[\/latex] on the interval [latex]\\left[-\\pi ,\\pi \\right][\/latex]. How can we tell whether the function is even or odd by only observing the graph of [latex]f\\left(x\\right)=\\sec x?[\/latex]<\/p>\n<p>3. After examining the reciprocal identity for [latex]\\sec t[\/latex], explain why the function is undefined at certain points.<\/p>\n<p>4. All of the Pythagorean identities are related. Describe how to manipulate the equations to get from [latex]{\\sin }^{2}t+{\\cos }^{2}t=1[\/latex] to the other forms.<\/p>\n<p>For the following exercises, use the fundamental identities to fully simplify the expression.<\/p>\n<p>5. [latex]\\sin x\\cos x\\sec x[\/latex]<\/p>\n<p>6.\u00a0[latex]\\sin \\left(-x\\right)\\cos \\left(-x\\right)\\csc \\left(-x\\right)[\/latex]<\/p>\n<p>7. [latex]\\tan x\\sin x+\\sec x{\\cos }^{2}x[\/latex]<\/p>\n<p>8.\u00a0[latex]\\csc x+\\cos x\\cot \\left(-x\\right)[\/latex]<\/p>\n<p>9.\u00a0[latex]\\frac{\\cot t+\\tan t}{\\sec \\left(-t\\right)}[\/latex]<\/p>\n<p>10.\u00a0[latex]3{\\sin }^{3}t\\csc t+{\\cos }^{2}t+2\\cos \\left(-t\\right)\\cos t[\/latex]<\/p>\n<p>11. [latex]-\\tan \\left(-x\\right)\\cot \\left(-x\\right)[\/latex]<\/p>\n<p>12.\u00a0[latex]\\frac{-\\sin \\left(-x\\right)\\cos x\\sec x\\csc x\\tan x}{\\cot x}[\/latex]<\/p>\n<p>13. [latex]\\frac{1+{\\tan }^{2}\\theta }{{\\csc }^{2}\\theta }+{\\sin }^{2}\\theta +\\frac{1}{{\\sec }^{2}\\theta }[\/latex]<\/p>\n<p>14.\u00a0[latex]\\left(\\frac{\\tan x}{{\\csc }^{2}x}+\\frac{\\tan x}{{\\sec }^{2}x}\\right)\\left(\\frac{1+\\tan x}{1+\\cot x}\\right)-\\frac{1}{{\\cos }^{2}x}[\/latex]<\/p>\n<p>15. [latex]\\frac{1-{\\cos }^{2}x}{{\\tan }^{2}x}+2{\\sin }^{2}x[\/latex]<\/p>\n<p>For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression.<\/p>\n<p>16. [latex]\\frac{\\tan x+\\cot x}{\\csc x};\\cos x[\/latex]<\/p>\n<p>17. [latex]\\frac{\\sec x+\\csc x}{1+\\tan x};\\sin x[\/latex]<\/p>\n<p>18.\u00a0[latex]\\frac{\\cos x}{1+\\sin x}+\\tan x;\\cos x[\/latex]<\/p>\n<p>19. [latex]\\frac{1}{\\sin x\\cos x}-\\cot x;\\cot x[\/latex]<\/p>\n<p>20.\u00a0[latex]\\frac{1}{1-\\cos x}-\\frac{\\cos x}{1+\\cos x};\\csc x[\/latex]<\/p>\n<p>21. [latex]\\left(\\sec x+\\csc x\\right)\\left(\\sin x+\\cos x\\right)-2-\\cot x;\\tan x[\/latex]<\/p>\n<p>22.\u00a0[latex]\\frac{1}{\\csc x-\\sin x};\\sec x\\text{ and }\\tan x[\/latex]<\/p>\n<p>23. [latex]\\frac{1-\\sin x}{1+\\sin x}-\\frac{1+\\sin x}{1-\\sin x};\\sec x\\text{ and }\\tan x[\/latex]<\/p>\n<p>24.\u00a0[latex]\\tan x;\\sec x[\/latex]<\/p>\n<p>25. [latex]\\sec x;\\cot x[\/latex]<\/p>\n<p>26.\u00a0[latex]\\sec x;\\sin x[\/latex]<\/p>\n<p>27. [latex]\\cot x;\\sin x[\/latex]<\/p>\n<p>28.\u00a0[latex]\\cot x;\\csc x[\/latex]<\/p>\n<p>For the following exercises, verify the identity.<\/p>\n<p>29. [latex]\\cos x-{\\cos }^{3}x=\\cos x{\\sin }^{2}x[\/latex]<\/p>\n<p>30. [latex]\\cos x\\left(\\tan x-\\sec \\left(-x\\right)\\right)=\\sin x - 1[\/latex]<\/p>\n<p>31. [latex]\\frac{1+{\\sin }^{2}x}{{\\cos }^{2}x}=\\frac{1}{{\\cos }^{2}x}+\\frac{{\\sin }^{2}x}{{\\cos }^{2}x}=1+2{\\tan }^{2}x[\/latex]<\/p>\n<p>32. [latex]{\\left(\\sin x+\\cos x\\right)}^{2}=1+2\\sin x\\cos x[\/latex]<\/p>\n<p>33. [latex]{\\cos }^{2}x-{\\tan }^{2}x=2-{\\sin }^{2}x-{\\sec }^{2}x[\/latex]<\/p>\n<p>For the following exercises, prove or disprove the identity.<\/p>\n<p>34. [latex]\\frac{1}{1+\\cos x}-\\frac{1}{1-\\cos \\left(-x\\right)}=-2\\cot x\\csc x[\/latex]<\/p>\n<p>35. [latex]{\\csc }^{2}x\\left(1+{\\sin }^{2}x\\right)={\\cot }^{2}x[\/latex]<\/p>\n<p>36.\u00a0[latex]\\left(\\frac{{\\sec }^{2}\\left(-x\\right)-{\\tan }^{2}x}{\\tan x}\\right)\\left(\\frac{2+2\\tan x}{2+2\\cot x}\\right)-2{\\sin }^{2}x=\\cos 2x[\/latex]<\/p>\n<p>37. [latex]\\frac{\\tan x}{\\sec x}\\sin \\left(-x\\right)={\\cos }^{2}x[\/latex]<\/p>\n<p>38.\u00a0[latex]\\frac{\\sec \\left(-x\\right)}{\\tan x+\\cot x}=-\\sin \\left(-x\\right)[\/latex]<\/p>\n<p>39. [latex]\\frac{1+\\sin x}{\\cos x}=\\frac{\\cos x}{1+\\sin \\left(-x\\right)}[\/latex]<\/p>\n<p>For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression.<\/p>\n<p>40. [latex]\\frac{{\\cos }^{2}\\theta -{\\sin }^{2}\\theta }{1-{\\tan }^{2}\\theta }={\\sin }^{2}\\theta[\/latex]<\/p>\n<p>41. [latex]3{\\sin }^{2}\\theta +4{\\cos }^{2}\\theta =3+{\\cos }^{2}\\theta[\/latex]<\/p>\n<p>42.\u00a0[latex]\\frac{\\sec \\theta +\\tan \\theta }{\\cot \\theta +\\cos \\theta }={\\sec }^{2}\\theta[\/latex]<\/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-15687\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <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>. <strong>License Terms<\/strong>: Download for free at: http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/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":169554,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at: http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-15687","chapter","type-chapter","status-publish","hentry"],"part":14191,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/15687","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/169554"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/15687\/revisions"}],"predecessor-version":[{"id":15688,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/15687\/revisions\/15688"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/14191"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/15687\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=15687"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=15687"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=15687"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=15687"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}