{"id":1399,"date":"2023-06-05T14:51:20","date_gmt":"2023-06-05T14:51:20","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/chapter\/solutions-for-sum-to-product-and-product-to-sum-formulas\/"},"modified":"2023-06-05T14:51:20","modified_gmt":"2023-06-05T14:51:20","slug":"solutions-for-sum-to-product-and-product-to-sum-formulas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/chapter\/solutions-for-sum-to-product-and-product-to-sum-formulas\/","title":{"raw":"Solutions 54: Sum-to-Product and Product-to-Sum Formulas","rendered":"Solutions 54: Sum-to-Product and Product-to-Sum Formulas"},"content":{"raw":"\n

Solutions to Odd-Numbered Exercises<\/h2>\n1. Substitute [latex]\\alpha [\/latex] into cosine and [latex]\\beta [\/latex] into sine and evaluate.\n\n3. Answers 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]\n\n5. [latex]8\\left(\\cos \\left(5x\\right)-\\cos \\left(27x\\right)\\right)[\/latex]\n\n7. [latex]\\sin \\left(2x\\right)+\\sin \\left(8x\\right)[\/latex]\n\n9. [latex]\\frac{1}{2}\\left(\\cos \\left(6x\\right)-\\cos \\left(4x\\right)\\right)[\/latex]\n\n11. [latex]2\\cos \\left(5t\\right)\\cos t[\/latex]\n\n13. [latex]2\\cos \\left(7x\\right)[\/latex]\n\n15. [latex]2\\cos \\left(6x\\right)\\cos \\left(3x\\right)[\/latex]\n\n17. [latex]\\frac{1}{4}\\left(1+\\sqrt{3}\\right)[\/latex]\n\n19. [latex]\\frac{1}{4}\\left(\\sqrt{3}-2\\right)[\/latex]\n\n21. [latex]\\frac{1}{4}\\left(\\sqrt{3}-1\\right)[\/latex]\n\n23. [latex]\\cos \\left(80^\\circ \\right)-\\cos \\left(120^\\circ \\right)[\/latex]\n\n25. [latex]\\frac{1}{2}\\left(\\sin \\left(221^\\circ \\right)+\\sin \\left(205^\\circ \\right)\\right)[\/latex]\n\n27. [latex]\\sqrt{2}\\cos \\left(31^\\circ \\right)[\/latex]\n\n29. [latex]2\\cos \\left(66.5^\\circ \\right)\\sin \\left(34.5^\\circ \\right)[\/latex]\n\n31. [latex]2\\sin \\left(-1.5^\\circ \\right)\\cos \\left(0.5^\\circ \\right)[\/latex]\n\n33. [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]\n\n \n\n35. [latex]\\sin x+\\sin \\left(3x\\right)=2\\sin \\left(\\frac{4x}{2}\\right)\\cos \\left(\\frac{-2x}{2}\\right)=[\/latex]\n[latex]2\\sin \\left(2x\\right)\\cos x=2\\left(2\\sin x\\cos x\\right)\\cos x=[\/latex]\n[latex]4\\sin x{\\cos }^{2}x[\/latex]\n\n37. [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]\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]\n\n39. [latex]2\\cos \\left({35}^{\\circ }\\right)\\cos \\left({23}^{\\circ }\\right),\\text{ 1}\\text{.5081}[\/latex]\n\n41. [latex]-2\\sin \\left({33}^{\\circ }\\right)\\sin \\left({11}^{\\circ }\\right),\\text{ }-0.2078[\/latex]\n\n43. [latex]\\frac{1}{2}\\left(\\cos \\left({99}^{\\circ }\\right)-\\cos \\left({71}^{\\circ }\\right)\\right),\\text{ }-0.2410[\/latex]\n\n45. It is an identity.\n\n47. It is not an identity, but [latex]2{\\cos }^{3}x[\/latex] is.\n\n49. [latex]\\tan \\left(3t\\right)[\/latex]\n\n51. [latex]2\\cos \\left(2x\\right)[\/latex]\n\n53. [latex]-\\sin \\left(14x\\right)[\/latex]\n\n55. Start 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\n

[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>\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 x<\/em> and 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].\n\n57. [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]\n\n59. [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]\n\n61. [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]\n\n63. [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]\n","rendered":"

Solutions to Odd-Numbered Exercises<\/h2>\n

1. Substitute [latex]\\alpha[\/latex] into cosine and [latex]\\beta[\/latex] into sine and evaluate.<\/p>\n

3. Answers 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

5. [latex]8\\left(\\cos \\left(5x\\right)-\\cos \\left(27x\\right)\\right)[\/latex]<\/p>\n

7. [latex]\\sin \\left(2x\\right)+\\sin \\left(8x\\right)[\/latex]<\/p>\n

9. [latex]\\frac{1}{2}\\left(\\cos \\left(6x\\right)-\\cos \\left(4x\\right)\\right)[\/latex]<\/p>\n

11. [latex]2\\cos \\left(5t\\right)\\cos t[\/latex]<\/p>\n

13. [latex]2\\cos \\left(7x\\right)[\/latex]<\/p>\n

15. [latex]2\\cos \\left(6x\\right)\\cos \\left(3x\\right)[\/latex]<\/p>\n

17. [latex]\\frac{1}{4}\\left(1+\\sqrt{3}\\right)[\/latex]<\/p>\n

19. [latex]\\frac{1}{4}\\left(\\sqrt{3}-2\\right)[\/latex]<\/p>\n

21. [latex]\\frac{1}{4}\\left(\\sqrt{3}-1\\right)[\/latex]<\/p>\n

23. [latex]\\cos \\left(80^\\circ \\right)-\\cos \\left(120^\\circ \\right)[\/latex]<\/p>\n

25. [latex]\\frac{1}{2}\\left(\\sin \\left(221^\\circ \\right)+\\sin \\left(205^\\circ \\right)\\right)[\/latex]<\/p>\n

27. [latex]\\sqrt{2}\\cos \\left(31^\\circ \\right)[\/latex]<\/p>\n

29. [latex]2\\cos \\left(66.5^\\circ \\right)\\sin \\left(34.5^\\circ \\right)[\/latex]<\/p>\n

31. [latex]2\\sin \\left(-1.5^\\circ \\right)\\cos \\left(0.5^\\circ \\right)[\/latex]<\/p>\n

33. [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>\n

35. [latex]\\sin x+\\sin \\left(3x\\right)=2\\sin \\left(\\frac{4x}{2}\\right)\\cos \\left(\\frac{-2x}{2}\\right)=[\/latex]
\n[latex]2\\sin \\left(2x\\right)\\cos x=2\\left(2\\sin x\\cos x\\right)\\cos x=[\/latex]
\n[latex]4\\sin x{\\cos }^{2}x[\/latex]<\/p>\n

37. [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]
\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

39. [latex]2\\cos \\left({35}^{\\circ }\\right)\\cos \\left({23}^{\\circ }\\right),\\text{ 1}\\text{.5081}[\/latex]<\/p>\n

41. [latex]-2\\sin \\left({33}^{\\circ }\\right)\\sin \\left({11}^{\\circ }\\right),\\text{ }-0.2078[\/latex]<\/p>\n

43. [latex]\\frac{1}{2}\\left(\\cos \\left({99}^{\\circ }\\right)-\\cos \\left({71}^{\\circ }\\right)\\right),\\text{ }-0.2410[\/latex]<\/p>\n

45. It is an identity.<\/p>\n

47. It is not an identity, but [latex]2{\\cos }^{3}x[\/latex] is.<\/p>\n

49. [latex]\\tan \\left(3t\\right)[\/latex]<\/p>\n

51. [latex]2\\cos \\left(2x\\right)[\/latex]<\/p>\n

53. [latex]-\\sin \\left(14x\\right)[\/latex]<\/p>\n

55. Start 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

[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

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 x<\/em> and 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

57. [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

59. [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

61. [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

63. [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\n\t\t\t

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Candela Citations<\/h3>\n\t\t\t\t\t
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