{"id":342,"date":"2026-02-16T20:54:21","date_gmt":"2026-02-16T20:54:21","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/?post_type=chapter&#038;p=342"},"modified":"2026-02-17T23:14:30","modified_gmt":"2026-02-17T23:14:30","slug":"trigonometric-identities-and-equations-get-stronger-answer-key","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/trigonometric-identities-and-equations-get-stronger-answer-key\/","title":{"raw":"Trigonometric Identities and Equations: Get Stronger Answer Key","rendered":"Trigonometric Identities and Equations: Get Stronger Answer Key"},"content":{"raw":"<h1>Simplifying Trigonometric Expressions with Identities<\/h1>\r\n3. When [latex]\\cos t=0[\/latex], then [latex]\\sec t=\\frac{1}{0}[\/latex], which is undefined.\r\n\r\n5.\u00a0[latex]\\sin x[\/latex]\r\n\r\n7.\u00a0[latex]\\sec x[\/latex]\r\n\r\n9.\u00a0[latex]\\csc t[\/latex]\r\n\r\n11.\u00a0[latex]-1[\/latex]\r\n\r\n13.\u00a0[latex]{\\sec }^{2}x[\/latex]\r\n\r\n15.\u00a0[latex]{\\sin }^{2}x+1[\/latex]\r\n\r\n29.\u00a0Answers will vary. Sample proof:\r\n[latex]\\cos x-{\\cos }^{3}x=\\cos x\\left(1-{\\cos }^{2}x\\right)[\/latex]\r\n[latex]=\\cos x{\\sin }^{2}x[\/latex]\r\n\r\n31.\u00a0Answers will vary. Sample proof:\r\n\r\n[latex]\\frac{1+{\\sin }^{2}x}{{\\cos }^{2}x}=\\frac{1}{{\\cos }^{2}x}+\\frac{{\\sin }^{2}x}{{\\cos }^{2}x}={\\sec }^{2}x+{\\tan }^{2}x={\\tan }^{2}x+1+{\\tan }^{2}x=1+2{\\tan }^{2}x[\/latex]\r\n\r\n33.\u00a0Answers will vary. Sample proof:\r\n\r\n[latex]{\\cos }^{2}x-{\\tan }^{2}x=1-{\\sin }^{2}x-\\left({\\sec }^{2}x - 1\\right)=1-{\\sin }^{2}x-{\\sec }^{2}x+1=2-{\\sin }^{2}x-{\\sec }^{2}x[\/latex]\r\n\r\n39.\u00a0Proved with negative and Pythagorean identities\r\n<h1>Sum and Difference Identities<\/h1>\r\n1.\u00a0The 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.\r\n\r\n5.\u00a0[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]\r\n\r\n7.\u00a0[latex]\\frac{\\sqrt{6}-\\sqrt{2}}{4}[\/latex]\r\n\r\n9.\u00a0[latex]-2-\\sqrt{3}[\/latex]\r\n\r\n11.\u00a0[latex]-\\frac{\\sqrt{2}}{2}\\sin x-\\frac{\\sqrt{2}}{2}\\cos x[\/latex]\r\n\r\n13.\u00a0[latex]-\\frac{1}{2}\\cos x-\\frac{\\sqrt{3}}{2}\\sin x[\/latex]\r\n\r\n15.\u00a0[latex]\\csc \\theta [\/latex]\r\n\r\n17.\u00a0[latex]\\cot x[\/latex]\r\n\r\n19.\u00a0[latex]\\tan \\left(\\frac{x}{10}\\right)[\/latex]\r\n\r\n21.\u00a0[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]\r\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]\r\n\r\n23.\u00a0[latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex]\r\n\r\n25.\u00a0[latex]\\sin x[\/latex]\r\n<span id=\"fs-id1615462\">\r\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>\r\n\r\n27.\u00a0[latex]\\cot \\left(\\frac{\\pi }{6}-x\\right)[\/latex]\r\n<span id=\"fs-id2233832\">\r\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><span id=\"fs-id1877975\"><\/span>\r\n\r\n31.\u00a0[latex]\\frac{\\sin x}{\\sqrt{2}}+\\frac{\\cos x}{\\sqrt{2}}[\/latex]\r\n<span id=\"fs-id2199923\">\r\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>\r\n\r\n43.\u00a0[latex]-\\frac{\\sqrt{3}-1}{2\\sqrt{2}},\\text{ or }-0.2588[\/latex]\r\n\r\n45.\u00a0[latex]\\frac{1+\\sqrt{3}}{2\\sqrt{2}}[\/latex], or 0.9659\r\n\r\n47.\u00a0[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]\r\n\r\n&nbsp;\r\n\r\n49.\u00a0[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]\r\n\r\n51.\u00a0[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]\r\n\r\n53.\u00a0True\r\n\r\n55.\u00a0True. Note that [latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\left(\\pi -\\gamma \\right)[\/latex] and expand the right hand side.\r\n<h1>Double-Angle, Half-Angle, and Reduction Formulas<\/h1>\r\n1.\u00a0Use the Pythagorean identities and isolate the squared term.\r\n\r\n5.\u00a0a) [latex]\\frac{3\\sqrt{7}}{32}[\/latex] b) [latex]\\frac{31}{32}[\/latex] c) [latex]\\frac{3\\sqrt{7}}{31}[\/latex]\r\n\r\n7.\u00a0a) [latex]\\frac{\\sqrt{3}}{2}[\/latex] b) [latex]-\\frac{1}{2}[\/latex] c) [latex]-\\sqrt{3}[\/latex]\r\n\r\n15.\u00a0[latex]\\frac{\\sqrt{2-\\sqrt{3}}}{2}[\/latex]\r\n\r\n17.\u00a0[latex]2+\\sqrt{3}[\/latex]\r\n\r\n19.\u00a0[latex]-1-\\sqrt{2}[\/latex]\r\n\r\n21.\u00a0a) [latex]\\frac{3\\sqrt{13}}{13}[\/latex] b) [latex]-\\frac{2\\sqrt{13}}{13}[\/latex] c) [latex]-\\frac{3}{2}[\/latex]\r\n\r\n23.\u00a0a) [latex]\\frac{\\sqrt{10}}{4}[\/latex] b) [latex]\\frac{\\sqrt{6}}{4}[\/latex] c) [latex]\\frac{\\sqrt{15}}{3}[\/latex]\r\n\r\n25.\u00a0[latex]\\frac{120}{169},-\\frac{119}{169},-\\frac{120}{119}[\/latex]\r\n\r\n27.\u00a0[latex]\\frac{2\\sqrt{13}}{13},\\frac{3\\sqrt{13}}{13},\\frac{2}{3}[\/latex]\r\n\r\n29.\u00a0[latex]\\cos \\left({74}^{\\circ }\\right)[\/latex]\r\n\r\n31.\u00a0[latex]\\cos \\left(18x\\right)[\/latex]\r\n\r\n33.\u00a0[latex]3\\sin \\left(10x\\right)[\/latex]\r\n\r\n35.\u00a0[latex]-2\\sin \\left(-x\\right)\\cos \\left(-x\\right)=-2\\left(-\\sin \\left(x\\right)\\cos \\left(x\\right)\\right)=\\sin \\left(2x\\right)[\/latex]\r\n\r\n37.\u00a0[latex]\\begin{array}{l}\\frac{\\sin \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)}{\\tan }^{2}\\theta =\\frac{2\\sin \\left(\\theta \\right)\\cos \\left(\\theta \\right)}{1+{\\cos }^{2}\\theta -{\\sin }^{2}\\theta }{\\tan }^{2}\\theta =\\\\ \\frac{2\\sin \\left(\\theta \\right)\\cos \\left(\\theta \\right)}{2{\\cos }^{2}\\theta }{\\tan }^{2}\\theta =\\frac{\\sin \\left(\\theta \\right)}{\\cos \\theta }{\\tan }^{2}\\theta =\\\\ \\cot \\left(\\theta \\right){\\tan }^{2}\\theta =\\tan \\theta \\end{array}[\/latex]\r\n\r\n39.\u00a0[latex]\\frac{1+\\cos \\left(12x\\right)}{2}[\/latex]\r\n\r\n41.\u00a0[latex]\\frac{3+\\cos \\left(12x\\right)-4\\cos \\left(6x\\right)}{8}[\/latex]\r\n\r\n43.\u00a0[latex]\\frac{2+\\cos \\left(2x\\right)-2\\cos \\left(4x\\right)-\\cos \\left(6x\\right)}{32}[\/latex]\r\n\r\n45.\u00a0[latex]\\frac{3+\\cos \\left(4x\\right)-4\\cos \\left(2x\\right)}{3+\\cos \\left(4x\\right)+4\\cos \\left(2x\\right)}[\/latex]\r\n\r\n47.\u00a0[latex]\\frac{1-\\cos \\left(4x\\right)}{8}[\/latex]\r\n\r\n49.\u00a0[latex]\\frac{3+\\cos \\left(4x\\right)-4\\cos \\left(2x\\right)}{4\\left(\\cos \\left(2x\\right)+1\\right)}[\/latex]\r\n\r\n51.\u00a0[latex]\\frac{\\left(1+\\cos \\left(4x\\right)\\right)\\sin x}{2}[\/latex]\r\n\r\n55.\u00a0[latex]\\frac{2\\tan x}{1+{\\tan }^{2}x}=\\frac{\\frac{2\\sin x}{\\cos x}}{1+\\frac{{\\sin }^{2}x}{{\\cos }^{2}x}}=\\frac{\\frac{2\\sin x}{\\cos x}}{\\frac{{\\cos }^{2}x+{\\sin }^{2}x}{{\\cos }^{2}x}}=[\/latex]\r\n[latex]\\frac{2\\sin x}{\\cos x}.\\frac{{\\cos }^{2}x}{1}=2\\sin x\\cos x=\\sin \\left(2x\\right)[\/latex]\r\n\r\n57.\u00a0[latex]\\frac{2\\sin x\\cos x}{2{\\cos }^{2}x - 1}=\\frac{\\sin \\left(2x\\right)}{\\cos \\left(2x\\right)}=\\tan \\left(2x\\right)[\/latex]\r\n\r\n59.\u00a0[latex]\\begin{array}{l}\\sin \\left(x+2x\\right)=\\sin x\\cos \\left(2x\\right)+\\sin \\left(2x\\right)\\cos x\\hfill \\\\ =\\sin x\\left({\\cos }^{2}x-{\\sin }^{2}x\\right)+2\\sin x\\cos x\\cos x\\hfill \\\\ =\\sin x{\\cos }^{2}x-{\\sin }^{3}x+2\\sin x{\\cos }^{2}x\\hfill \\\\ =3\\sin x{\\cos }^{2}x-{\\sin }^{3}x\\hfill \\end{array}[\/latex]\r\n\r\n61.\u00a0[latex]\\begin{array}{l}\\frac{1+\\cos \\left(2t\\right)}{\\sin \\left(2t\\right)-\\cos t}=\\frac{1+2{\\cos }^{2}t - 1}{2\\sin t\\cos t-\\cos t}\\hfill \\\\ =\\frac{2{\\cos }^{2}t}{\\cos t\\left(2\\sin t - 1\\right)}\\hfill \\\\ =\\frac{2\\cos t}{2\\sin t - 1}\\hfill \\end{array}[\/latex]\r\n<h1>Sum-to-Product and Product-to-Sum Formulas<\/h1>\r\n5.\u00a0[latex]8\\left(\\cos \\left(5x\\right)-\\cos \\left(27x\\right)\\right)[\/latex]\r\n\r\n7.\u00a0[latex]\\sin \\left(2x\\right)+\\sin \\left(8x\\right)[\/latex]\r\n\r\n9.\u00a0[latex]\\frac{1}{2}\\left(\\cos \\left(6x\\right)-\\cos \\left(4x\\right)\\right)[\/latex]\r\n\r\n11.\u00a0[latex]2\\cos \\left(5t\\right)\\cos t[\/latex]\r\n\r\n13.\u00a0[latex]2\\cos \\left(7x\\right)[\/latex]\r\n\r\n15.\u00a0[latex]2\\cos \\left(6x\\right)\\cos \\left(3x\\right)[\/latex]\r\n\r\n17.\u00a0[latex]\\frac{1}{4}\\left(1+\\sqrt{3}\\right)[\/latex]\r\n\r\n19.\u00a0[latex]\\frac{1}{4}\\left(\\sqrt{3}-2\\right)[\/latex]\r\n\r\n21.\u00a0[latex]\\frac{1}{4}\\left(\\sqrt{3}-1\\right)[\/latex]\r\n\r\n23.\u00a0[latex]\\cos \\left(80^\\circ \\right)-\\cos \\left(120^\\circ \\right)[\/latex]\r\n\r\n25.\u00a0[latex]\\frac{1}{2}\\left(\\sin \\left(221^\\circ \\right)+\\sin \\left(205^\\circ \\right)\\right)[\/latex]\r\n\r\n27.\u00a0[latex]\\sqrt{2}\\cos \\left(31^\\circ \\right)[\/latex]\r\n\r\n29.\u00a0[latex]2\\cos \\left(66.5^\\circ \\right)\\sin \\left(34.5^\\circ \\right)[\/latex]\r\n\r\n31.\u00a0[latex]2\\sin \\left(-1.5^\\circ \\right)\\cos \\left(0.5^\\circ \\right)[\/latex]\r\n\r\n33.\u00a0[latex]{2}\\sin \\left({7x}\\right){-2}\\sin{ x}={ 2}\\sin \\left({4x}+{ 3x }\\right)-{ 2 }\\sin\\left({4x } - { 3x }\\right)=\\\\ {2}\\left(\\sin\\left({ 4x }\\right)\\cos\\left({ 3x }\\right)+\\sin\\left({ 3x }\\right)\\cos\\left({ 4x }\\right)\\right)-{ 2 }\\left(\\sin\\left({ 4x }\\right)\\cos\\left({ 3x }\\right)-\\sin \\left({ 3x }\\right)\\cos\\left({ 4x }\\right)\\right)=\\\\{2}\\sin\\left({ 4x }\\right)\\cos\\left({ 3x }\\right)+{2}\\sin\\left({ 3x }\\right)\\cos\\left({ 4x }\\right)-{ 2 }\\sin\\left({ 4x }\\right)\\cos\\left({ 3x }\\right)+{ 2 }\\sin\\left({ 3x }\\right)\\cos\\left({ 4x }\\right)=\\\\{ 4 }\\sin\\left({ 3x }\\right)\\cos\\left({ 4x }\\right)\\\\[\/latex]\r\n\r\n&nbsp;\r\n\r\n35.\u00a0[latex]\\sin x+\\sin \\left(3x\\right)=2\\sin \\left(\\frac{4x}{2}\\right)\\cos \\left(\\frac{-2x}{2}\\right)=[\/latex]\r\n[latex]2\\sin \\left(2x\\right)\\cos x=2\\left(2\\sin x\\cos x\\right)\\cos x=[\/latex]\r\n[latex]4\\sin x{\\cos }^{2}x[\/latex]\r\n\r\n37.\u00a0[latex]2\\tan x\\cos \\left(3x\\right)=\\frac{2\\sin x\\cos \\left(3x\\right)}{\\cos x}=\\frac{2\\left(.5\\left(\\sin \\left(4x\\right)-\\sin \\left(2x\\right)\\right)\\right)}{\\cos x}[\/latex]\r\n[latex]=\\frac{1}{\\cos x}\\left(\\sin \\left(4x\\right)-\\sin \\left(2x\\right)\\right)=\\sec x\\left(\\sin \\left(4x\\right)-\\sin \\left(2x\\right)\\right)[\/latex]\r\n\r\n39.\u00a0[latex]2\\cos \\left({35}^{\\circ }\\right)\\cos \\left({23}^{\\circ }\\right),\\text{ 1}\\text{.5081}[\/latex]\r\n\r\n41.\u00a0[latex]-2\\sin \\left({33}^{\\circ }\\right)\\sin \\left({11}^{\\circ }\\right),\\text{ }-0.2078[\/latex]\r\n\r\n43.\u00a0[latex]\\frac{1}{2}\\left(\\cos \\left({99}^{\\circ }\\right)-\\cos \\left({71}^{\\circ }\\right)\\right),\\text{ }-0.2410[\/latex]\r\n\r\n49.\u00a0[latex]\\tan \\left(3t\\right)[\/latex]\r\n\r\n51.\u00a0[latex]2\\cos \\left(2x\\right)[\/latex]\r\n\r\n53.\u00a0[latex]-\\sin \\left(14x\\right)[\/latex]\r\n\r\n57.\u00a0[latex]\\frac{\\cos \\left(3x\\right)+\\cos x}{\\cos \\left(3x\\right)-\\cos x}=\\frac{2\\cos \\left(2x\\right)\\cos x}{-2\\sin \\left(2x\\right)\\sin x}=-\\cot \\left(2x\\right)\\cot x[\/latex]\r\n\r\n59.\u00a0[latex]\\begin{array}{l}\\frac{\\cos \\left(2y\\right)-\\cos \\left(4y\\right)}{\\sin \\left(2y\\right)+\\sin \\left(4y\\right)}=\\frac{-2\\sin \\left(3y\\right)\\sin \\left(-y\\right)}{2\\sin \\left(3y\\right)\\cos y}=\\\\ \\frac{2\\sin \\left(3y\\right)\\sin \\left(y\\right)}{2\\sin \\left(3y\\right)\\cos y}=\\tan y\\end{array}[\/latex]\r\n\r\n61.\u00a0[latex]\\begin{array}{l}\\cos x-\\cos \\left(3x\\right)=-2\\sin \\left(2x\\right)\\sin \\left(-x\\right)=\\\\ 2\\left(2\\sin x\\cos x\\right)\\sin x=4{\\sin }^{2}x\\cos x\\end{array}[\/latex]\r\n\r\n63.\u00a0[latex]\\tan \\left(\\frac{\\pi }{4}-t\\right)=\\frac{\\tan \\left(\\frac{\\pi }{4}\\right)-\\tan t}{1+\\tan \\left(\\frac{\\pi }{4}\\right)\\tan \\left(t\\right)}=\\frac{1-\\tan t}{1+\\tan t}[\/latex]\r\n<h1>Solving Trigonometric Equations<\/h1>\r\n<h2>Solutions to Odd-Numbered Exercises<\/h2>\r\n1.\u00a0There will not always be solutions to trigonometric function equations. For a basic example, [latex]\\cos \\left(x\\right)=-5[\/latex].\r\n\r\n5.\u00a0[latex]\\frac{\\pi }{3},\\frac{2\\pi }{3}[\/latex]\r\n\r\n7.\u00a0[latex]\\frac{3\\pi }{4},\\frac{5\\pi }{4}[\/latex]\r\n\r\n9.\u00a0[latex]\\frac{\\pi }{4},\\frac{5\\pi }{4}[\/latex]\r\n\r\n11.\u00a0[latex]\\frac{\\pi }{4},\\frac{3\\pi }{4},\\frac{5\\pi }{4},\\frac{7\\pi }{4}[\/latex]\r\n\r\n13.\u00a0[latex]\\frac{\\pi }{4},\\frac{7\\pi }{4}[\/latex]\r\n\r\n15.\u00a0[latex]\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex]\r\n\r\n17.\u00a0[latex]\\frac{\\pi }{18},\\frac{5\\pi }{18},\\frac{13\\pi }{18},\\frac{17\\pi }{18},\\frac{25\\pi }{18},\\frac{29\\pi }{18}[\/latex]\r\n\r\n19.\u00a0[latex]\\frac{3\\pi }{12},\\frac{5\\pi }{12},\\frac{11\\pi }{12},\\frac{13\\pi }{12},\\frac{19\\pi }{12},\\frac{21\\pi }{12}[\/latex]\r\n\r\n21.\u00a0[latex]\\frac{1}{6},\\frac{5}{6},\\frac{13}{6},\\frac{17}{6},\\frac{25}{6},\\frac{29}{6},\\frac{37}{6}[\/latex]\r\n\r\n23.\u00a0[latex]0,\\frac{\\pi }{3},\\pi ,\\frac{5\\pi }{3}[\/latex]\r\n\r\n25.\u00a0[latex]\\frac{\\pi }{3},\\pi ,\\frac{5\\pi }{3}[\/latex]\r\n\r\n27.\u00a0[latex]\\frac{\\pi }{3},\\frac{3\\pi }{2},\\frac{5\\pi }{3}[\/latex]\r\n\r\n29.\u00a0[latex]0,\\pi [\/latex]\r\n\r\n31.\u00a0[latex]\\pi -{\\sin }^{-1}\\left(-\\frac{1}{4}\\right),\\frac{7\\pi }{6},\\frac{11\\pi }{6},2\\pi +{\\sin }^{-1}\\left(-\\frac{1}{4}\\right)[\/latex]\r\n\r\n33.\u00a0[latex]\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right),\\frac{\\pi }{3}-\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right),\\frac{2\\pi }{3}+\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right),\\pi -\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right),\\frac{4\\pi }{3}+\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right),\\frac{5\\pi }{3}-\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right)[\/latex]\r\n\r\n35.\u00a0[latex]0[\/latex]\r\n\r\n37.\u00a0[latex]\\frac{\\pi }{6},\\frac{5\\pi }{6},\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex]\r\n\r\n39.\u00a0[latex]\\frac{3\\pi }{2},\\frac{\\pi }{6},\\frac{5\\pi }{6}[\/latex]\r\n\r\n41.\u00a0[latex]0,\\frac{\\pi }{3},\\pi ,\\frac{4\\pi }{3}[\/latex]\r\n\r\n43.\u00a0There are no solutions.\r\n\r\n45.\u00a0[latex]{\\cos }^{-1}\\left(\\frac{1}{3}\\left(1-\\sqrt{7}\\right)\\right),2\\pi -{\\cos }^{-1}\\left(\\frac{1}{3}\\left(1-\\sqrt{7}\\right)\\right)[\/latex]\r\n\r\n47.\u00a0[latex]{\\tan }^{-1}\\left(\\frac{1}{2}\\left(\\sqrt{29}-5\\right)\\right),\\pi +{\\tan }^{-1}\\left(\\frac{1}{2}\\left(-\\sqrt{29}-5\\right)\\right),\\pi +{\\tan }^{-1}\\left(\\frac{1}{2}\\left(\\sqrt{29}-5\\right)\\right),2\\pi +{\\tan }^{-1}\\left(\\frac{1}{2}\\left(-\\sqrt{29}-5\\right)\\right)[\/latex]\r\n\r\n49.\u00a0There are no solutions.\r\n\r\n51.\u00a0There are no solutions.\r\n\r\n53.\u00a0[latex]0,\\frac{2\\pi }{3},\\frac{4\\pi }{3}[\/latex]\r\n\r\n55.\u00a0[latex]\\frac{\\pi }{4},\\frac{3\\pi }{4},\\frac{5\\pi }{4},\\frac{7\\pi }{4}[\/latex]\r\n\r\n57.\u00a0[latex]{\\sin }^{-1}\\left(\\frac{3}{5}\\right),\\frac{\\pi }{2},\\pi -{\\sin }^{-1}\\left(\\frac{3}{5}\\right),\\frac{3\\pi }{2}[\/latex]\r\n\r\n59.\u00a0[latex]{\\cos }^{-1}\\left(-\\frac{1}{4}\\right),2\\pi -{\\cos }^{-1}\\left(-\\frac{1}{4}\\right)[\/latex]\r\n\r\n61.\u00a0[latex]\\frac{\\pi }{3},{\\cos }^{-1}\\left(-\\frac{3}{4}\\right),2\\pi -{\\cos }^{-1}\\left(-\\frac{3}{4}\\right),\\frac{5\\pi }{3}[\/latex]\r\n\r\n63.\u00a0[latex]{\\cos }^{-1}\\left(\\frac{3}{4}\\right),{\\cos }^{-1}\\left(-\\frac{2}{3}\\right),2\\pi -{\\cos }^{-1}\\left(-\\frac{2}{3}\\right),2\\pi -{\\cos }^{-1}\\left(\\frac{3}{4}\\right)[\/latex]\r\n\r\n65.\u00a0[latex]0,\\frac{\\pi }{2},\\pi ,\\frac{3\\pi }{2}[\/latex]\r\n\r\n67.\u00a0[latex]\\frac{\\pi }{3},{\\cos }^{-1}\\left(-\\frac{1}{4}\\right),2\\pi -{\\cos }^{-1}\\left(-\\frac{1}{4}\\right),\\frac{5\\pi }{3}[\/latex]\r\n\r\n69.\u00a0There are no solutions.\r\n\r\n71.\u00a0[latex]\\pi +{\\tan }^{-1}\\left(-2\\right),\\pi +{\\tan }^{-1}\\left(-\\frac{3}{2}\\right),2\\pi +{\\tan }^{-1}\\left(-2\\right),2\\pi +{\\tan }^{-1}\\left(-\\frac{3}{2}\\right)[\/latex]\r\n\r\n73.\u00a0[latex]2\\pi k+0.2734,2\\pi k+2.8682[\/latex]\r\n\r\n75.\u00a0[latex]\\pi k - 0.3277[\/latex]\r\n\r\n77.\u00a0[latex]0.6694,1.8287,3.8110,4.9703[\/latex]\r\n\r\n79.\u00a0[latex]1.0472,3.1416,5.2360[\/latex]\r\n\r\n81.\u00a0[latex]0.5326,1.7648,3.6742,4.9064[\/latex]\r\n\r\n83.\u00a0[latex]{\\sin }^{-1}\\left(\\frac{1}{4}\\right),\\pi -{\\sin }^{-1}\\left(\\frac{1}{4}\\right),\\frac{3\\pi }{2}[\/latex]\r\n\r\n85.\u00a0[latex]\\frac{\\pi }{2},\\frac{3\\pi }{2}[\/latex]\r\n\r\n87.\u00a0There are no solutions.\r\n\r\n89.\u00a0[latex]0,\\frac{\\pi }{2},\\pi ,\\frac{3\\pi }{2}[\/latex]\r\n\r\n91.\u00a0There are no solutions.\r\n\r\n93.\u00a0[latex]{7.2}^{\\circ }[\/latex]\r\n\r\n95.\u00a0[latex]{5.7}^{\\circ }[\/latex]\r\n\r\n97.\u00a0[latex]{82.4}^{\\circ }[\/latex]\r\n\r\n99.\u00a0[latex]{31.0}^{\\circ }[\/latex]\r\n\r\n101.\u00a0[latex]{88.7}^{\\circ }[\/latex]\r\n\r\n103.\u00a0[latex]{59.0}^{\\circ }[\/latex]\r\n\r\n105.\u00a0[latex]{36.9}^{\\circ }[\/latex]\r\n<h1>Modeling with Trigonometric Equations<\/h1>\r\n1.\u00a0Physical behavior should be periodic, or cyclical.\r\n\r\n3.\u00a0Since cumulative rainfall is always increasing, a sinusoidal function would not be ideal here.\r\n\r\n5.\u00a0[latex]y=-3\\cos \\left(\\frac{\\pi }{6}x\\right)-1[\/latex]\r\n\r\n7.\u00a0[latex]5\\sin \\left(2x\\right)+2[\/latex]\r\n\r\n9.\u00a0[latex]4\\cos \\left(\\frac{x\\pi }{2}\\right)-3[\/latex]\r\n\r\n11.\u00a0[latex]5 - 8\\sin \\left(\\frac{x\\pi }{2}\\right)[\/latex]\r\n\r\n13.\u00a0[latex]\\tan \\left(\\frac{x\\pi }{12}\\right)[\/latex]\r\n\r\n17.\u00a09 years from now\r\n\r\n19.\u00a0[latex]56^\\circ \\text{F}[\/latex]\r\n\r\n21.\u00a0[latex]1.8024[\/latex]\u00a0hours\r\n\r\n23.\u00a04:30\r\n\r\n25.\u00a0From July 8 to October 23\r\n\r\n27.\u00a0From day 19 through day 40\r\n\r\n29.\u00a0Floods: July 24 through October 7. Droughts: February 4 through March 27\r\n\r\n31.\u00a0Amplitude: 11, period: [latex]\\frac{1}{6}[\/latex], frequency: 6 Hz\r\n\r\n33.\u00a0Amplitude: 5, period: [latex]\\frac{1}{30}[\/latex], frequency: 30 Hz\r\n\r\n35.\u00a0[latex]P\\left(t\\right)=-15\\cos \\left(\\frac{\\pi }{6}t\\right)+650+\\frac{55}{6}t[\/latex]\r\n\r\n37.\u00a0[latex]P\\left(t\\right)=-40\\cos \\left(\\frac{\\pi }{6}t\\right)+800{\\left(1.04\\right)}^{t}[\/latex]\r\n\r\n39.\u00a0[latex]D\\left(t\\right)=7{\\left(0.89\\right)}^{t}\\cos \\left(40\\pi t\\right)[\/latex]\r\n\r\n41.\u00a0[latex]D\\left(t\\right)=19{\\left(0.9265\\right)}^{t}\\cos \\left(26\\pi t\\right)[\/latex]\r\n\r\n43.\u00a0[latex]20.1[\/latex] years\r\n\r\n45.\u00a017.8 seconds\r\n\r\n47.\u00a0Spring 2 comes to rest first after 8.0 seconds.\r\n\r\n49.\u00a0500 miles, at [latex]{90}^{\\circ }[\/latex]","rendered":"<h1>Simplifying Trigonometric Expressions with Identities<\/h1>\n<p>3. When [latex]\\cos t=0[\/latex], then [latex]\\sec t=\\frac{1}{0}[\/latex], which is undefined.<\/p>\n<p>5.\u00a0[latex]\\sin x[\/latex]<\/p>\n<p>7.\u00a0[latex]\\sec x[\/latex]<\/p>\n<p>9.\u00a0[latex]\\csc t[\/latex]<\/p>\n<p>11.\u00a0[latex]-1[\/latex]<\/p>\n<p>13.\u00a0[latex]{\\sec }^{2}x[\/latex]<\/p>\n<p>15.\u00a0[latex]{\\sin }^{2}x+1[\/latex]<\/p>\n<p>29.\u00a0Answers will vary. Sample proof:<br \/>\n[latex]\\cos x-{\\cos }^{3}x=\\cos x\\left(1-{\\cos }^{2}x\\right)[\/latex]<br \/>\n[latex]=\\cos x{\\sin }^{2}x[\/latex]<\/p>\n<p>31.\u00a0Answers will vary. Sample proof:<\/p>\n<p>[latex]\\frac{1+{\\sin }^{2}x}{{\\cos }^{2}x}=\\frac{1}{{\\cos }^{2}x}+\\frac{{\\sin }^{2}x}{{\\cos }^{2}x}={\\sec }^{2}x+{\\tan }^{2}x={\\tan }^{2}x+1+{\\tan }^{2}x=1+2{\\tan }^{2}x[\/latex]<\/p>\n<p>33.\u00a0Answers will vary. Sample proof:<\/p>\n<p>[latex]{\\cos }^{2}x-{\\tan }^{2}x=1-{\\sin }^{2}x-\\left({\\sec }^{2}x - 1\\right)=1-{\\sin }^{2}x-{\\sec }^{2}x+1=2-{\\sin }^{2}x-{\\sec }^{2}x[\/latex]<\/p>\n<p>39.\u00a0Proved with negative and Pythagorean identities<\/p>\n<h1>Sum and Difference Identities<\/h1>\n<p>1.\u00a0The 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>5.\u00a0[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]<\/p>\n<p>7.\u00a0[latex]\\frac{\\sqrt{6}-\\sqrt{2}}{4}[\/latex]<\/p>\n<p>9.\u00a0[latex]-2-\\sqrt{3}[\/latex]<\/p>\n<p>11.\u00a0[latex]-\\frac{\\sqrt{2}}{2}\\sin x-\\frac{\\sqrt{2}}{2}\\cos x[\/latex]<\/p>\n<p>13.\u00a0[latex]-\\frac{1}{2}\\cos x-\\frac{\\sqrt{3}}{2}\\sin x[\/latex]<\/p>\n<p>15.\u00a0[latex]\\csc \\theta[\/latex]<\/p>\n<p>17.\u00a0[latex]\\cot x[\/latex]<\/p>\n<p>19.\u00a0[latex]\\tan \\left(\\frac{x}{10}\\right)[\/latex]<\/p>\n<p>21.\u00a0[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.\u00a0[latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex]<\/p>\n<p>25.\u00a0[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.\u00a0[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><span id=\"fs-id1877975\"><\/span><\/p>\n<p>31.\u00a0[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>43.\u00a0[latex]-\\frac{\\sqrt{3}-1}{2\\sqrt{2}},\\text{ or }-0.2588[\/latex]<\/p>\n<p>45.\u00a0[latex]\\frac{1+\\sqrt{3}}{2\\sqrt{2}}[\/latex], or 0.9659<\/p>\n<p>47.\u00a0[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.\u00a0[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.\u00a0[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.\u00a0True<\/p>\n<p>55.\u00a0True. Note that [latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\left(\\pi -\\gamma \\right)[\/latex] and expand the right hand side.<\/p>\n<h1>Double-Angle, Half-Angle, and Reduction Formulas<\/h1>\n<p>1.\u00a0Use the Pythagorean identities and isolate the squared term.<\/p>\n<p>5.\u00a0a) [latex]\\frac{3\\sqrt{7}}{32}[\/latex] b) [latex]\\frac{31}{32}[\/latex] c) [latex]\\frac{3\\sqrt{7}}{31}[\/latex]<\/p>\n<p>7.\u00a0a) [latex]\\frac{\\sqrt{3}}{2}[\/latex] b) [latex]-\\frac{1}{2}[\/latex] c) [latex]-\\sqrt{3}[\/latex]<\/p>\n<p>15.\u00a0[latex]\\frac{\\sqrt{2-\\sqrt{3}}}{2}[\/latex]<\/p>\n<p>17.\u00a0[latex]2+\\sqrt{3}[\/latex]<\/p>\n<p>19.\u00a0[latex]-1-\\sqrt{2}[\/latex]<\/p>\n<p>21.\u00a0a) [latex]\\frac{3\\sqrt{13}}{13}[\/latex] b) [latex]-\\frac{2\\sqrt{13}}{13}[\/latex] c) [latex]-\\frac{3}{2}[\/latex]<\/p>\n<p>23.\u00a0a) [latex]\\frac{\\sqrt{10}}{4}[\/latex] b) [latex]\\frac{\\sqrt{6}}{4}[\/latex] c) [latex]\\frac{\\sqrt{15}}{3}[\/latex]<\/p>\n<p>25.\u00a0[latex]\\frac{120}{169},-\\frac{119}{169},-\\frac{120}{119}[\/latex]<\/p>\n<p>27.\u00a0[latex]\\frac{2\\sqrt{13}}{13},\\frac{3\\sqrt{13}}{13},\\frac{2}{3}[\/latex]<\/p>\n<p>29.\u00a0[latex]\\cos \\left({74}^{\\circ }\\right)[\/latex]<\/p>\n<p>31.\u00a0[latex]\\cos \\left(18x\\right)[\/latex]<\/p>\n<p>33.\u00a0[latex]3\\sin \\left(10x\\right)[\/latex]<\/p>\n<p>35.\u00a0[latex]-2\\sin \\left(-x\\right)\\cos \\left(-x\\right)=-2\\left(-\\sin \\left(x\\right)\\cos \\left(x\\right)\\right)=\\sin \\left(2x\\right)[\/latex]<\/p>\n<p>37.\u00a0[latex]\\begin{array}{l}\\frac{\\sin \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)}{\\tan }^{2}\\theta =\\frac{2\\sin \\left(\\theta \\right)\\cos \\left(\\theta \\right)}{1+{\\cos }^{2}\\theta -{\\sin }^{2}\\theta }{\\tan }^{2}\\theta =\\\\ \\frac{2\\sin \\left(\\theta \\right)\\cos \\left(\\theta \\right)}{2{\\cos }^{2}\\theta }{\\tan }^{2}\\theta =\\frac{\\sin \\left(\\theta \\right)}{\\cos \\theta }{\\tan }^{2}\\theta =\\\\ \\cot \\left(\\theta \\right){\\tan }^{2}\\theta =\\tan \\theta \\end{array}[\/latex]<\/p>\n<p>39.\u00a0[latex]\\frac{1+\\cos \\left(12x\\right)}{2}[\/latex]<\/p>\n<p>41.\u00a0[latex]\\frac{3+\\cos \\left(12x\\right)-4\\cos \\left(6x\\right)}{8}[\/latex]<\/p>\n<p>43.\u00a0[latex]\\frac{2+\\cos \\left(2x\\right)-2\\cos \\left(4x\\right)-\\cos \\left(6x\\right)}{32}[\/latex]<\/p>\n<p>45.\u00a0[latex]\\frac{3+\\cos \\left(4x\\right)-4\\cos \\left(2x\\right)}{3+\\cos \\left(4x\\right)+4\\cos \\left(2x\\right)}[\/latex]<\/p>\n<p>47.\u00a0[latex]\\frac{1-\\cos \\left(4x\\right)}{8}[\/latex]<\/p>\n<p>49.\u00a0[latex]\\frac{3+\\cos \\left(4x\\right)-4\\cos \\left(2x\\right)}{4\\left(\\cos \\left(2x\\right)+1\\right)}[\/latex]<\/p>\n<p>51.\u00a0[latex]\\frac{\\left(1+\\cos \\left(4x\\right)\\right)\\sin x}{2}[\/latex]<\/p>\n<p>55.\u00a0[latex]\\frac{2\\tan x}{1+{\\tan }^{2}x}=\\frac{\\frac{2\\sin x}{\\cos x}}{1+\\frac{{\\sin }^{2}x}{{\\cos }^{2}x}}=\\frac{\\frac{2\\sin x}{\\cos x}}{\\frac{{\\cos }^{2}x+{\\sin }^{2}x}{{\\cos }^{2}x}}=[\/latex]<br \/>\n[latex]\\frac{2\\sin x}{\\cos x}.\\frac{{\\cos }^{2}x}{1}=2\\sin x\\cos x=\\sin \\left(2x\\right)[\/latex]<\/p>\n<p>57.\u00a0[latex]\\frac{2\\sin x\\cos x}{2{\\cos }^{2}x - 1}=\\frac{\\sin \\left(2x\\right)}{\\cos \\left(2x\\right)}=\\tan \\left(2x\\right)[\/latex]<\/p>\n<p>59.\u00a0[latex]\\begin{array}{l}\\sin \\left(x+2x\\right)=\\sin x\\cos \\left(2x\\right)+\\sin \\left(2x\\right)\\cos x\\hfill \\\\ =\\sin x\\left({\\cos }^{2}x-{\\sin }^{2}x\\right)+2\\sin x\\cos x\\cos x\\hfill \\\\ =\\sin x{\\cos }^{2}x-{\\sin }^{3}x+2\\sin x{\\cos }^{2}x\\hfill \\\\ =3\\sin x{\\cos }^{2}x-{\\sin }^{3}x\\hfill \\end{array}[\/latex]<\/p>\n<p>61.\u00a0[latex]\\begin{array}{l}\\frac{1+\\cos \\left(2t\\right)}{\\sin \\left(2t\\right)-\\cos t}=\\frac{1+2{\\cos }^{2}t - 1}{2\\sin t\\cos t-\\cos t}\\hfill \\\\ =\\frac{2{\\cos }^{2}t}{\\cos t\\left(2\\sin t - 1\\right)}\\hfill \\\\ =\\frac{2\\cos t}{2\\sin t - 1}\\hfill \\end{array}[\/latex]<\/p>\n<h1>Sum-to-Product and Product-to-Sum Formulas<\/h1>\n<p>5.\u00a0[latex]8\\left(\\cos \\left(5x\\right)-\\cos \\left(27x\\right)\\right)[\/latex]<\/p>\n<p>7.\u00a0[latex]\\sin \\left(2x\\right)+\\sin \\left(8x\\right)[\/latex]<\/p>\n<p>9.\u00a0[latex]\\frac{1}{2}\\left(\\cos \\left(6x\\right)-\\cos \\left(4x\\right)\\right)[\/latex]<\/p>\n<p>11.\u00a0[latex]2\\cos \\left(5t\\right)\\cos t[\/latex]<\/p>\n<p>13.\u00a0[latex]2\\cos \\left(7x\\right)[\/latex]<\/p>\n<p>15.\u00a0[latex]2\\cos \\left(6x\\right)\\cos \\left(3x\\right)[\/latex]<\/p>\n<p>17.\u00a0[latex]\\frac{1}{4}\\left(1+\\sqrt{3}\\right)[\/latex]<\/p>\n<p>19.\u00a0[latex]\\frac{1}{4}\\left(\\sqrt{3}-2\\right)[\/latex]<\/p>\n<p>21.\u00a0[latex]\\frac{1}{4}\\left(\\sqrt{3}-1\\right)[\/latex]<\/p>\n<p>23.\u00a0[latex]\\cos \\left(80^\\circ \\right)-\\cos \\left(120^\\circ \\right)[\/latex]<\/p>\n<p>25.\u00a0[latex]\\frac{1}{2}\\left(\\sin \\left(221^\\circ \\right)+\\sin \\left(205^\\circ \\right)\\right)[\/latex]<\/p>\n<p>27.\u00a0[latex]\\sqrt{2}\\cos \\left(31^\\circ \\right)[\/latex]<\/p>\n<p>29.\u00a0[latex]2\\cos \\left(66.5^\\circ \\right)\\sin \\left(34.5^\\circ \\right)[\/latex]<\/p>\n<p>31.\u00a0[latex]2\\sin \\left(-1.5^\\circ \\right)\\cos \\left(0.5^\\circ \\right)[\/latex]<\/p>\n<p>33.\u00a0[latex]{2}\\sin \\left({7x}\\right){-2}\\sin{ x}={ 2}\\sin \\left({4x}+{ 3x }\\right)-{ 2 }\\sin\\left({4x } - { 3x }\\right)=\\\\ {2}\\left(\\sin\\left({ 4x }\\right)\\cos\\left({ 3x }\\right)+\\sin\\left({ 3x }\\right)\\cos\\left({ 4x }\\right)\\right)-{ 2 }\\left(\\sin\\left({ 4x }\\right)\\cos\\left({ 3x }\\right)-\\sin \\left({ 3x }\\right)\\cos\\left({ 4x }\\right)\\right)=\\\\{2}\\sin\\left({ 4x }\\right)\\cos\\left({ 3x }\\right)+{2}\\sin\\left({ 3x }\\right)\\cos\\left({ 4x }\\right)-{ 2 }\\sin\\left({ 4x }\\right)\\cos\\left({ 3x }\\right)+{ 2 }\\sin\\left({ 3x }\\right)\\cos\\left({ 4x }\\right)=\\\\{ 4 }\\sin\\left({ 3x }\\right)\\cos\\left({ 4x }\\right)\\\\[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>35.\u00a0[latex]\\sin x+\\sin \\left(3x\\right)=2\\sin \\left(\\frac{4x}{2}\\right)\\cos \\left(\\frac{-2x}{2}\\right)=[\/latex]<br \/>\n[latex]2\\sin \\left(2x\\right)\\cos x=2\\left(2\\sin x\\cos x\\right)\\cos x=[\/latex]<br \/>\n[latex]4\\sin x{\\cos }^{2}x[\/latex]<\/p>\n<p>37.\u00a0[latex]2\\tan x\\cos \\left(3x\\right)=\\frac{2\\sin x\\cos \\left(3x\\right)}{\\cos x}=\\frac{2\\left(.5\\left(\\sin \\left(4x\\right)-\\sin \\left(2x\\right)\\right)\\right)}{\\cos x}[\/latex]<br \/>\n[latex]=\\frac{1}{\\cos x}\\left(\\sin \\left(4x\\right)-\\sin \\left(2x\\right)\\right)=\\sec x\\left(\\sin \\left(4x\\right)-\\sin \\left(2x\\right)\\right)[\/latex]<\/p>\n<p>39.\u00a0[latex]2\\cos \\left({35}^{\\circ }\\right)\\cos \\left({23}^{\\circ }\\right),\\text{ 1}\\text{.5081}[\/latex]<\/p>\n<p>41.\u00a0[latex]-2\\sin \\left({33}^{\\circ }\\right)\\sin \\left({11}^{\\circ }\\right),\\text{ }-0.2078[\/latex]<\/p>\n<p>43.\u00a0[latex]\\frac{1}{2}\\left(\\cos \\left({99}^{\\circ }\\right)-\\cos \\left({71}^{\\circ }\\right)\\right),\\text{ }-0.2410[\/latex]<\/p>\n<p>49.\u00a0[latex]\\tan \\left(3t\\right)[\/latex]<\/p>\n<p>51.\u00a0[latex]2\\cos \\left(2x\\right)[\/latex]<\/p>\n<p>53.\u00a0[latex]-\\sin \\left(14x\\right)[\/latex]<\/p>\n<p>57.\u00a0[latex]\\frac{\\cos \\left(3x\\right)+\\cos x}{\\cos \\left(3x\\right)-\\cos x}=\\frac{2\\cos \\left(2x\\right)\\cos x}{-2\\sin \\left(2x\\right)\\sin x}=-\\cot \\left(2x\\right)\\cot x[\/latex]<\/p>\n<p>59.\u00a0[latex]\\begin{array}{l}\\frac{\\cos \\left(2y\\right)-\\cos \\left(4y\\right)}{\\sin \\left(2y\\right)+\\sin \\left(4y\\right)}=\\frac{-2\\sin \\left(3y\\right)\\sin \\left(-y\\right)}{2\\sin \\left(3y\\right)\\cos y}=\\\\ \\frac{2\\sin \\left(3y\\right)\\sin \\left(y\\right)}{2\\sin \\left(3y\\right)\\cos y}=\\tan y\\end{array}[\/latex]<\/p>\n<p>61.\u00a0[latex]\\begin{array}{l}\\cos x-\\cos \\left(3x\\right)=-2\\sin \\left(2x\\right)\\sin \\left(-x\\right)=\\\\ 2\\left(2\\sin x\\cos x\\right)\\sin x=4{\\sin }^{2}x\\cos x\\end{array}[\/latex]<\/p>\n<p>63.\u00a0[latex]\\tan \\left(\\frac{\\pi }{4}-t\\right)=\\frac{\\tan \\left(\\frac{\\pi }{4}\\right)-\\tan t}{1+\\tan \\left(\\frac{\\pi }{4}\\right)\\tan \\left(t\\right)}=\\frac{1-\\tan t}{1+\\tan t}[\/latex]<\/p>\n<h1>Solving Trigonometric Equations<\/h1>\n<h2>Solutions to Odd-Numbered Exercises<\/h2>\n<p>1.\u00a0There will not always be solutions to trigonometric function equations. For a basic example, [latex]\\cos \\left(x\\right)=-5[\/latex].<\/p>\n<p>5.\u00a0[latex]\\frac{\\pi }{3},\\frac{2\\pi }{3}[\/latex]<\/p>\n<p>7.\u00a0[latex]\\frac{3\\pi }{4},\\frac{5\\pi }{4}[\/latex]<\/p>\n<p>9.\u00a0[latex]\\frac{\\pi }{4},\\frac{5\\pi }{4}[\/latex]<\/p>\n<p>11.\u00a0[latex]\\frac{\\pi }{4},\\frac{3\\pi }{4},\\frac{5\\pi }{4},\\frac{7\\pi }{4}[\/latex]<\/p>\n<p>13.\u00a0[latex]\\frac{\\pi }{4},\\frac{7\\pi }{4}[\/latex]<\/p>\n<p>15.\u00a0[latex]\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex]<\/p>\n<p>17.\u00a0[latex]\\frac{\\pi }{18},\\frac{5\\pi }{18},\\frac{13\\pi }{18},\\frac{17\\pi }{18},\\frac{25\\pi }{18},\\frac{29\\pi }{18}[\/latex]<\/p>\n<p>19.\u00a0[latex]\\frac{3\\pi }{12},\\frac{5\\pi }{12},\\frac{11\\pi }{12},\\frac{13\\pi }{12},\\frac{19\\pi }{12},\\frac{21\\pi }{12}[\/latex]<\/p>\n<p>21.\u00a0[latex]\\frac{1}{6},\\frac{5}{6},\\frac{13}{6},\\frac{17}{6},\\frac{25}{6},\\frac{29}{6},\\frac{37}{6}[\/latex]<\/p>\n<p>23.\u00a0[latex]0,\\frac{\\pi }{3},\\pi ,\\frac{5\\pi }{3}[\/latex]<\/p>\n<p>25.\u00a0[latex]\\frac{\\pi }{3},\\pi ,\\frac{5\\pi }{3}[\/latex]<\/p>\n<p>27.\u00a0[latex]\\frac{\\pi }{3},\\frac{3\\pi }{2},\\frac{5\\pi }{3}[\/latex]<\/p>\n<p>29.\u00a0[latex]0,\\pi[\/latex]<\/p>\n<p>31.\u00a0[latex]\\pi -{\\sin }^{-1}\\left(-\\frac{1}{4}\\right),\\frac{7\\pi }{6},\\frac{11\\pi }{6},2\\pi +{\\sin }^{-1}\\left(-\\frac{1}{4}\\right)[\/latex]<\/p>\n<p>33.\u00a0[latex]\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right),\\frac{\\pi }{3}-\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right),\\frac{2\\pi }{3}+\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right),\\pi -\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right),\\frac{4\\pi }{3}+\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right),\\frac{5\\pi }{3}-\\frac{1}{3}\\left({\\sin }^{-1}\\left(\\frac{9}{10}\\right)\\right)[\/latex]<\/p>\n<p>35.\u00a0[latex]0[\/latex]<\/p>\n<p>37.\u00a0[latex]\\frac{\\pi }{6},\\frac{5\\pi }{6},\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex]<\/p>\n<p>39.\u00a0[latex]\\frac{3\\pi }{2},\\frac{\\pi }{6},\\frac{5\\pi }{6}[\/latex]<\/p>\n<p>41.\u00a0[latex]0,\\frac{\\pi }{3},\\pi ,\\frac{4\\pi }{3}[\/latex]<\/p>\n<p>43.\u00a0There are no solutions.<\/p>\n<p>45.\u00a0[latex]{\\cos }^{-1}\\left(\\frac{1}{3}\\left(1-\\sqrt{7}\\right)\\right),2\\pi -{\\cos }^{-1}\\left(\\frac{1}{3}\\left(1-\\sqrt{7}\\right)\\right)[\/latex]<\/p>\n<p>47.\u00a0[latex]{\\tan }^{-1}\\left(\\frac{1}{2}\\left(\\sqrt{29}-5\\right)\\right),\\pi +{\\tan }^{-1}\\left(\\frac{1}{2}\\left(-\\sqrt{29}-5\\right)\\right),\\pi +{\\tan }^{-1}\\left(\\frac{1}{2}\\left(\\sqrt{29}-5\\right)\\right),2\\pi +{\\tan }^{-1}\\left(\\frac{1}{2}\\left(-\\sqrt{29}-5\\right)\\right)[\/latex]<\/p>\n<p>49.\u00a0There are no solutions.<\/p>\n<p>51.\u00a0There are no solutions.<\/p>\n<p>53.\u00a0[latex]0,\\frac{2\\pi }{3},\\frac{4\\pi }{3}[\/latex]<\/p>\n<p>55.\u00a0[latex]\\frac{\\pi }{4},\\frac{3\\pi }{4},\\frac{5\\pi }{4},\\frac{7\\pi }{4}[\/latex]<\/p>\n<p>57.\u00a0[latex]{\\sin }^{-1}\\left(\\frac{3}{5}\\right),\\frac{\\pi }{2},\\pi -{\\sin }^{-1}\\left(\\frac{3}{5}\\right),\\frac{3\\pi }{2}[\/latex]<\/p>\n<p>59.\u00a0[latex]{\\cos }^{-1}\\left(-\\frac{1}{4}\\right),2\\pi -{\\cos }^{-1}\\left(-\\frac{1}{4}\\right)[\/latex]<\/p>\n<p>61.\u00a0[latex]\\frac{\\pi }{3},{\\cos }^{-1}\\left(-\\frac{3}{4}\\right),2\\pi -{\\cos }^{-1}\\left(-\\frac{3}{4}\\right),\\frac{5\\pi }{3}[\/latex]<\/p>\n<p>63.\u00a0[latex]{\\cos }^{-1}\\left(\\frac{3}{4}\\right),{\\cos }^{-1}\\left(-\\frac{2}{3}\\right),2\\pi -{\\cos }^{-1}\\left(-\\frac{2}{3}\\right),2\\pi -{\\cos }^{-1}\\left(\\frac{3}{4}\\right)[\/latex]<\/p>\n<p>65.\u00a0[latex]0,\\frac{\\pi }{2},\\pi ,\\frac{3\\pi }{2}[\/latex]<\/p>\n<p>67.\u00a0[latex]\\frac{\\pi }{3},{\\cos }^{-1}\\left(-\\frac{1}{4}\\right),2\\pi -{\\cos }^{-1}\\left(-\\frac{1}{4}\\right),\\frac{5\\pi }{3}[\/latex]<\/p>\n<p>69.\u00a0There are no solutions.<\/p>\n<p>71.\u00a0[latex]\\pi +{\\tan }^{-1}\\left(-2\\right),\\pi +{\\tan }^{-1}\\left(-\\frac{3}{2}\\right),2\\pi +{\\tan }^{-1}\\left(-2\\right),2\\pi +{\\tan }^{-1}\\left(-\\frac{3}{2}\\right)[\/latex]<\/p>\n<p>73.\u00a0[latex]2\\pi k+0.2734,2\\pi k+2.8682[\/latex]<\/p>\n<p>75.\u00a0[latex]\\pi k - 0.3277[\/latex]<\/p>\n<p>77.\u00a0[latex]0.6694,1.8287,3.8110,4.9703[\/latex]<\/p>\n<p>79.\u00a0[latex]1.0472,3.1416,5.2360[\/latex]<\/p>\n<p>81.\u00a0[latex]0.5326,1.7648,3.6742,4.9064[\/latex]<\/p>\n<p>83.\u00a0[latex]{\\sin }^{-1}\\left(\\frac{1}{4}\\right),\\pi -{\\sin }^{-1}\\left(\\frac{1}{4}\\right),\\frac{3\\pi }{2}[\/latex]<\/p>\n<p>85.\u00a0[latex]\\frac{\\pi }{2},\\frac{3\\pi }{2}[\/latex]<\/p>\n<p>87.\u00a0There are no solutions.<\/p>\n<p>89.\u00a0[latex]0,\\frac{\\pi }{2},\\pi ,\\frac{3\\pi }{2}[\/latex]<\/p>\n<p>91.\u00a0There are no solutions.<\/p>\n<p>93.\u00a0[latex]{7.2}^{\\circ }[\/latex]<\/p>\n<p>95.\u00a0[latex]{5.7}^{\\circ }[\/latex]<\/p>\n<p>97.\u00a0[latex]{82.4}^{\\circ }[\/latex]<\/p>\n<p>99.\u00a0[latex]{31.0}^{\\circ }[\/latex]<\/p>\n<p>101.\u00a0[latex]{88.7}^{\\circ }[\/latex]<\/p>\n<p>103.\u00a0[latex]{59.0}^{\\circ }[\/latex]<\/p>\n<p>105.\u00a0[latex]{36.9}^{\\circ }[\/latex]<\/p>\n<h1>Modeling with Trigonometric Equations<\/h1>\n<p>1.\u00a0Physical behavior should be periodic, or cyclical.<\/p>\n<p>3.\u00a0Since cumulative rainfall is always increasing, a sinusoidal function would not be ideal here.<\/p>\n<p>5.\u00a0[latex]y=-3\\cos \\left(\\frac{\\pi }{6}x\\right)-1[\/latex]<\/p>\n<p>7.\u00a0[latex]5\\sin \\left(2x\\right)+2[\/latex]<\/p>\n<p>9.\u00a0[latex]4\\cos \\left(\\frac{x\\pi }{2}\\right)-3[\/latex]<\/p>\n<p>11.\u00a0[latex]5 - 8\\sin \\left(\\frac{x\\pi }{2}\\right)[\/latex]<\/p>\n<p>13.\u00a0[latex]\\tan \\left(\\frac{x\\pi }{12}\\right)[\/latex]<\/p>\n<p>17.\u00a09 years from now<\/p>\n<p>19.\u00a0[latex]56^\\circ \\text{F}[\/latex]<\/p>\n<p>21.\u00a0[latex]1.8024[\/latex]\u00a0hours<\/p>\n<p>23.\u00a04:30<\/p>\n<p>25.\u00a0From July 8 to October 23<\/p>\n<p>27.\u00a0From day 19 through day 40<\/p>\n<p>29.\u00a0Floods: July 24 through October 7. Droughts: February 4 through March 27<\/p>\n<p>31.\u00a0Amplitude: 11, period: [latex]\\frac{1}{6}[\/latex], frequency: 6 Hz<\/p>\n<p>33.\u00a0Amplitude: 5, period: [latex]\\frac{1}{30}[\/latex], frequency: 30 Hz<\/p>\n<p>35.\u00a0[latex]P\\left(t\\right)=-15\\cos \\left(\\frac{\\pi }{6}t\\right)+650+\\frac{55}{6}t[\/latex]<\/p>\n<p>37.\u00a0[latex]P\\left(t\\right)=-40\\cos \\left(\\frac{\\pi }{6}t\\right)+800{\\left(1.04\\right)}^{t}[\/latex]<\/p>\n<p>39.\u00a0[latex]D\\left(t\\right)=7{\\left(0.89\\right)}^{t}\\cos \\left(40\\pi t\\right)[\/latex]<\/p>\n<p>41.\u00a0[latex]D\\left(t\\right)=19{\\left(0.9265\\right)}^{t}\\cos \\left(26\\pi t\\right)[\/latex]<\/p>\n<p>43.\u00a0[latex]20.1[\/latex] years<\/p>\n<p>45.\u00a017.8 seconds<\/p>\n<p>47.\u00a0Spring 2 comes to rest first after 8.0 seconds.<\/p>\n<p>49.\u00a0500 miles, at [latex]{90}^{\\circ }[\/latex]<\/p>\n","protected":false},"author":13,"menu_order":15,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":224,"module-header":"- Select Header -","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\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/342"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/342\/revisions"}],"predecessor-version":[{"id":361,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/342\/revisions\/361"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/224"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/342\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=342"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=342"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=342"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=342"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}