{"id":17292,"date":"2020-04-19T22:20:50","date_gmt":"2020-04-19T22:20:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/?post_type=chapter&#038;p=17292"},"modified":"2020-05-21T05:27:38","modified_gmt":"2020-05-21T05:27:38","slug":"chapter-8-solutions-to-odd-numbered-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/chapter\/chapter-8-solutions-to-odd-numbered-problems\/","title":{"raw":"Chapter 5 Solutions to Odd-Numbered Problems","rendered":"Chapter 5 Solutions to Odd-Numbered Problems"},"content":{"raw":"<h2>Section 5.1 Solutions<\/h2>\r\n1.\u00a0All three functions, [latex]F,G[\/latex], and [latex]H[\/latex], are even.\r\n\r\nThis is because [latex]F\\left(-x\\right)=\\sin \\left(-x\\right)\\sin \\left(-x\\right)=\\left(-\\sin x\\right)\\left(-\\sin x\\right)={\\sin }^{2}x=F\\left(x\\right),G\\left(-x\\right)=\\cos \\left(-x\\right)\\cos \\left(-x\\right)=\\cos x\\cos x={\\cos }^{2}x=G\\left(x\\right)[\/latex] and [latex]H\\left(-x\\right)=\\tan \\left(-x\\right)\\tan \\left(-x\\right)=\\left(-\\tan x\\right)\\left(-\\tan x\\right)={\\tan }^{2}x=H\\left(x\\right)[\/latex].\r\n\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\n17.\u00a0[latex]\\frac{1}{\\sin x}[\/latex]\r\n\r\n19.\u00a0[latex]\\frac{1}{\\cot x}[\/latex]\r\n\r\n21.\u00a0[latex]\\tan x[\/latex]\r\n\r\n23.\u00a0[latex]-4\\sec x\\tan x[\/latex]\r\n\r\n25.\u00a0[latex]\\pm \\sqrt{\\frac{1}{{\\cot }^{2}x}+1}[\/latex]\r\n\r\n27.\u00a0[latex]\\frac{\\pm \\sqrt{1-{\\sin }^{2}x}}{\\sin x}[\/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\n35.\u00a0False\r\n\r\n37.\u00a0False\r\n\r\n39.\u00a0Proved with negative and Pythagorean identities\r\n\r\n41.\u00a0True\r\n\r\n[latex]3{\\sin }^{2}\\theta +4{\\cos }^{2}\\theta =3{\\sin }^{2}\\theta +3{\\cos }^{2}\\theta +{\\cos }^{2}\\theta =3\\left({\\sin }^{2}\\theta +{\\cos }^{2}\\theta \\right)+{\\cos }^{2}\\theta =3+{\\cos }^{2}\\theta [\/latex]\r\n<h2>Section 5.2 Solutions<\/h2>\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\n3.\u00a0[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.\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>\r\n\r\n29.\u00a0[latex]\\cot \\left(\\frac{\\pi }{4}+x\\right)[\/latex]\r\n<span id=\"fs-id1877975\">\r\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>\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\n&nbsp;\r\n\r\n33.\u00a0They are the same.\r\n\r\n35.\u00a0They are the different, try [latex]g\\left(x\\right)=\\sin \\left(9x\\right)-\\cos \\left(3x\\right)\\sin \\left(6x\\right)[\/latex].\r\n\r\n37.\u00a0They are the same.\r\n\r\n39.\u00a0They are the different, try [latex]g\\left(\\theta \\right)=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }[\/latex].\r\n\r\n41.\u00a0They are different, try [latex]g\\left(x\\right)=\\frac{\\tan x-\\tan \\left(2x\\right)}{1+\\tan x\\tan \\left(2x\\right)}[\/latex].\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<h2>Section 5.3 Solutions<\/h2>\r\n1.\u00a0Use the Pythagorean identities and isolate the squared term.\r\n\r\n3.\u00a0[latex]\\frac{1-\\cos x}{\\sin x},\\frac{\\sin x}{1+\\cos x}[\/latex], multiplying the top and bottom by [latex]\\sqrt{1-\\cos x}[\/latex] and [latex]\\sqrt{1+\\cos x}[\/latex], respectively.\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\n9.\u00a0[latex]\\cos \\theta =-\\frac{2\\sqrt{5}}{5},\\sin \\theta =\\frac{\\sqrt{5}}{5},\\tan \\theta =-\\frac{1}{2},\\csc \\theta =\\sqrt{5},\\sec \\theta =-\\frac{\\sqrt{5}}{2},\\cot \\theta =-2[\/latex]\r\n\r\n11.\u00a0[latex]2\\sin \\left(\\frac{\\pi }{2}\\right)=2[\/latex]\r\n\r\n13.\u00a0[latex]\\frac{\\sqrt{2-\\sqrt{2}}}{2}[\/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\n53.\u00a0[latex]4\\sin x\\cos x\\left({\\cos }^{2}x-{\\sin }^{2}x\\right)[\/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\r\n63.\u00a0[latex]\\begin{array}{l}\\left({\\cos }^{2}\\left(4x\\right)-{\\sin }^{2}\\left(4x\\right)-\\sin \\left(8x\\right)\\right)\\left({\\cos }^{2}\\left(4x\\right)-{\\sin }^{2}\\left(4x\\right)+\\sin \\left(8x\\right)\\right)=\\hfill \\\\ \\text{ }=\\left(\\cos \\left(8x\\right)-\\sin \\left(8x\\right)\\right)\\left(\\cos \\left(8x\\right)+\\sin \\left(8x\\right)\\right)\\hfill \\\\ \\text{ }={\\cos }^{2}\\left(8x\\right)-{\\sin }^{2}\\left(8x\\right)\\hfill \\\\ \\text{ }=\\cos \\left(16x\\right)\\hfill \\\\ \\hfill \\end{array}[\/latex]\r\n<h2>Section 5.4 Solutions<\/h2>\r\n1. Substitute [latex]\\alpha [\/latex] into cosine and [latex]\\beta [\/latex] into sine and evaluate.\r\n\r\n3.\u00a0Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: [latex]\\frac{\\sin \\left(3x\\right)+\\sin x}{\\cos x}=1[\/latex]. When converting the numerator to a product the equation becomes: [latex]\\frac{2\\sin \\left(2x\\right)\\cos x}{\\cos x}=1\\\\[\/latex]\r\n\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\n45.\u00a0It is an identity.\r\n\r\n47.\u00a0It is not an identity, but [latex]2{\\cos }^{3}x[\/latex] is.\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\n55.\u00a0Start with [latex]\\cos x+\\cos y[\/latex]. Make a substitution and let [latex]x=\\alpha +\\beta [\/latex] and let [latex]y=\\alpha -\\beta [\/latex], so [latex]\\cos x+\\cos y[\/latex] becomes\r\n<p style=\"text-align: center\">[latex]\\cos \\left(\\alpha +\\beta \\right)+\\cos \\left(\\alpha -\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta +\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta =2\\cos \\alpha \\cos \\beta [\/latex]<\/p>\r\nSince [latex]x=\\alpha +\\beta [\/latex] and [latex]y=\\alpha -\\beta [\/latex], we can solve for [latex]\\alpha [\/latex] and [latex]\\beta [\/latex] in terms of <em>x<\/em> and <em>y<\/em> and substitute in for [latex]2\\cos \\alpha \\cos \\beta [\/latex] and get [latex]2\\cos \\left(\\frac{x+y}{2}\\right)\\cos \\left(\\frac{x-y}{2}\\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<h2>Section 5.5 Solutions<\/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\n3.\u00a0If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.\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<h2>Section 5.6 Solutions<\/h2>\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\n15.\u00a0Answers will vary. Sample answer: This function could model temperature changes over the course of one very hot day in Phoenix, Arizona.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164217\/CNX_Precalc_Figure_07_06_202.jpg\" alt=\"Graph of f(x) = -18cos(x*pi\/12) - 5sin(x*pi\/12) + 100 on the interval [0,24]. There is a single peak around 12.\" \/>\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]\r\n\r\n51.\u00a0[latex]y=6{\\left(5\\right)}^{x}+4\\sin \\left(\\frac{\\pi }{2}x\\right)[\/latex]\r\n\r\n53.\u00a0[latex]y=8{\\left(\\frac{1}{2}\\right)}^{x}\\cos \\left(\\frac{\\pi }{2}x\\right)+3[\/latex]","rendered":"<h2>Section 5.1 Solutions<\/h2>\n<p>1.\u00a0All three functions, [latex]F,G[\/latex], and [latex]H[\/latex], are even.<\/p>\n<p>This is because [latex]F\\left(-x\\right)=\\sin \\left(-x\\right)\\sin \\left(-x\\right)=\\left(-\\sin x\\right)\\left(-\\sin x\\right)={\\sin }^{2}x=F\\left(x\\right),G\\left(-x\\right)=\\cos \\left(-x\\right)\\cos \\left(-x\\right)=\\cos x\\cos x={\\cos }^{2}x=G\\left(x\\right)[\/latex] and [latex]H\\left(-x\\right)=\\tan \\left(-x\\right)\\tan \\left(-x\\right)=\\left(-\\tan x\\right)\\left(-\\tan x\\right)={\\tan }^{2}x=H\\left(x\\right)[\/latex].<\/p>\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>17.\u00a0[latex]\\frac{1}{\\sin x}[\/latex]<\/p>\n<p>19.\u00a0[latex]\\frac{1}{\\cot x}[\/latex]<\/p>\n<p>21.\u00a0[latex]\\tan x[\/latex]<\/p>\n<p>23.\u00a0[latex]-4\\sec x\\tan x[\/latex]<\/p>\n<p>25.\u00a0[latex]\\pm \\sqrt{\\frac{1}{{\\cot }^{2}x}+1}[\/latex]<\/p>\n<p>27.\u00a0[latex]\\frac{\\pm \\sqrt{1-{\\sin }^{2}x}}{\\sin x}[\/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>35.\u00a0False<\/p>\n<p>37.\u00a0False<\/p>\n<p>39.\u00a0Proved with negative and Pythagorean identities<\/p>\n<p>41.\u00a0True<\/p>\n<p>[latex]3{\\sin }^{2}\\theta +4{\\cos }^{2}\\theta =3{\\sin }^{2}\\theta +3{\\cos }^{2}\\theta +{\\cos }^{2}\\theta =3\\left({\\sin }^{2}\\theta +{\\cos }^{2}\\theta \\right)+{\\cos }^{2}\\theta =3+{\\cos }^{2}\\theta[\/latex]<\/p>\n<h2>Section 5.2 Solutions<\/h2>\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>3.\u00a0[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.\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><\/p>\n<p>29.\u00a0[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.\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>&nbsp;<\/p>\n<p>33.\u00a0They are the same.<\/p>\n<p>35.\u00a0They 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.\u00a0They are the same.<\/p>\n<p>39.\u00a0They are the different, try [latex]g\\left(\\theta \\right)=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }[\/latex].<\/p>\n<p>41.\u00a0They 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.\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<h2>Section 5.3 Solutions<\/h2>\n<p>1.\u00a0Use the Pythagorean identities and isolate the squared term.<\/p>\n<p>3.\u00a0[latex]\\frac{1-\\cos x}{\\sin x},\\frac{\\sin x}{1+\\cos x}[\/latex], multiplying the top and bottom by [latex]\\sqrt{1-\\cos x}[\/latex] and [latex]\\sqrt{1+\\cos x}[\/latex], respectively.<\/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>9.\u00a0[latex]\\cos \\theta =-\\frac{2\\sqrt{5}}{5},\\sin \\theta =\\frac{\\sqrt{5}}{5},\\tan \\theta =-\\frac{1}{2},\\csc \\theta =\\sqrt{5},\\sec \\theta =-\\frac{\\sqrt{5}}{2},\\cot \\theta =-2[\/latex]<\/p>\n<p>11.\u00a0[latex]2\\sin \\left(\\frac{\\pi }{2}\\right)=2[\/latex]<\/p>\n<p>13.\u00a0[latex]\\frac{\\sqrt{2-\\sqrt{2}}}{2}[\/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>53.\u00a0[latex]4\\sin x\\cos x\\left({\\cos }^{2}x-{\\sin }^{2}x\\right)[\/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<p>63.\u00a0[latex]\\begin{array}{l}\\left({\\cos }^{2}\\left(4x\\right)-{\\sin }^{2}\\left(4x\\right)-\\sin \\left(8x\\right)\\right)\\left({\\cos }^{2}\\left(4x\\right)-{\\sin }^{2}\\left(4x\\right)+\\sin \\left(8x\\right)\\right)=\\hfill \\\\ \\text{ }=\\left(\\cos \\left(8x\\right)-\\sin \\left(8x\\right)\\right)\\left(\\cos \\left(8x\\right)+\\sin \\left(8x\\right)\\right)\\hfill \\\\ \\text{ }={\\cos }^{2}\\left(8x\\right)-{\\sin }^{2}\\left(8x\\right)\\hfill \\\\ \\text{ }=\\cos \\left(16x\\right)\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<h2>Section 5.4 Solutions<\/h2>\n<p>1. Substitute [latex]\\alpha[\/latex] into cosine and [latex]\\beta[\/latex] into sine and evaluate.<\/p>\n<p>3.\u00a0Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: [latex]\\frac{\\sin \\left(3x\\right)+\\sin x}{\\cos x}=1[\/latex]. When converting the numerator to a product the equation becomes: [latex]\\frac{2\\sin \\left(2x\\right)\\cos x}{\\cos x}=1\\\\[\/latex]<\/p>\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>45.\u00a0It is an identity.<\/p>\n<p>47.\u00a0It is not an identity, but [latex]2{\\cos }^{3}x[\/latex] is.<\/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>55.\u00a0Start with [latex]\\cos x+\\cos y[\/latex]. Make a substitution and let [latex]x=\\alpha +\\beta[\/latex] and let [latex]y=\\alpha -\\beta[\/latex], so [latex]\\cos x+\\cos y[\/latex] becomes<\/p>\n<p style=\"text-align: center\">[latex]\\cos \\left(\\alpha +\\beta \\right)+\\cos \\left(\\alpha -\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta +\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta =2\\cos \\alpha \\cos \\beta[\/latex]<\/p>\n<p>Since [latex]x=\\alpha +\\beta[\/latex] and [latex]y=\\alpha -\\beta[\/latex], we can solve for [latex]\\alpha[\/latex] and [latex]\\beta[\/latex] in terms of <em>x<\/em> and <em>y<\/em> and substitute in for [latex]2\\cos \\alpha \\cos \\beta[\/latex] and get [latex]2\\cos \\left(\\frac{x+y}{2}\\right)\\cos \\left(\\frac{x-y}{2}\\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<h2>Section 5.5 Solutions<\/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>3.\u00a0If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.<\/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<h2>Section 5.6 Solutions<\/h2>\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>15.\u00a0Answers will vary. Sample answer: This function could model temperature changes over the course of one very hot day in Phoenix, Arizona.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164217\/CNX_Precalc_Figure_07_06_202.jpg\" alt=\"Graph of f(x) = -18cos(x*pi\/12) - 5sin(x*pi\/12) + 100 on the interval [0,24]. There is a single peak around 12.\" \/><\/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<p>51.\u00a0[latex]y=6{\\left(5\\right)}^{x}+4\\sin \\left(\\frac{\\pi }{2}x\\right)[\/latex]<\/p>\n<p>53.\u00a0[latex]y=8{\\left(\\frac{1}{2}\\right)}^{x}\\cos \\left(\\frac{\\pi }{2}x\\right)+3[\/latex]<\/p>\n","protected":false},"author":264444,"menu_order":5,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-17292","chapter","type-chapter","status-publish","hentry"],"part":16602,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/17292","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/users\/264444"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/17292\/revisions"}],"predecessor-version":[{"id":17640,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/17292\/revisions\/17640"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/parts\/16602"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/17292\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/media?parent=17292"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=17292"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/contributor?post=17292"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/license?post=17292"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}