{"id":1393,"date":"2023-06-05T14:51:17","date_gmt":"2023-06-05T14:51:17","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/chapter\/solutions-for-solving-trigonometric-equations-with-identities\/"},"modified":"2023-06-05T14:51:17","modified_gmt":"2023-06-05T14:51:17","slug":"solutions-for-solving-trigonometric-equations-with-identities","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/chapter\/solutions-for-solving-trigonometric-equations-with-identities\/","title":{"raw":"Solutions 51: Solving Trigonometric Equations with Identities","rendered":"Solutions 51: Solving Trigonometric Equations with Identities"},"content":{"raw":"\n

Solutions to Odd-Numbered Exercises<\/h2>\n1. All three functions, [latex]F,G[\/latex], and [latex]H[\/latex], are even.\n\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].\n\n3. When [latex]\\cos t=0[\/latex], then [latex]\\sec t=\\frac{1}{0}[\/latex], which is undefined.\n\n5. [latex]\\sin x[\/latex]\n\n7. [latex]\\sec x[\/latex]\n\n9. [latex]\\csc t[\/latex]\n\n11. [latex]-1[\/latex]\n\n13. [latex]{\\sec }^{2}x[\/latex]\n\n15. [latex]{\\sin }^{2}x+1[\/latex]\n\n17. [latex]\\frac{1}{\\sin x}[\/latex]\n\n19. [latex]\\frac{1}{\\cot x}[\/latex]\n\n21. [latex]\\tan x[\/latex]\n\n23. [latex]-4\\sec x\\tan x[\/latex]\n\n25. [latex]\\pm \\sqrt{\\frac{1}{{\\cot }^{2}x}+1}[\/latex]\n\n27. [latex]\\frac{\\pm \\sqrt{1-{\\sin }^{2}x}}{\\sin x}[\/latex]\n\n29. Answers will vary. Sample proof:\n[latex]\\cos x-{\\cos }^{3}x=\\cos x\\left(1-{\\cos }^{2}x\\right)[\/latex]\n[latex]=\\cos x{\\sin }^{2}x[\/latex]\n\n31. Answers will vary. Sample proof:\n\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]\n\n33. Answers will vary. Sample proof:\n\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]\n\n35. False\n\n37. False\n\n39. Proved with negative and Pythagorean identities\n\n41. True\n\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]\n","rendered":"

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

1. All three functions, [latex]F,G[\/latex], and [latex]H[\/latex], are even.<\/p>\n

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

3. When [latex]\\cos t=0[\/latex], then [latex]\\sec t=\\frac{1}{0}[\/latex], which is undefined.<\/p>\n

5. [latex]\\sin x[\/latex]<\/p>\n

7. [latex]\\sec x[\/latex]<\/p>\n

9. [latex]\\csc t[\/latex]<\/p>\n

11. [latex]-1[\/latex]<\/p>\n

13. [latex]{\\sec }^{2}x[\/latex]<\/p>\n

15. [latex]{\\sin }^{2}x+1[\/latex]<\/p>\n

17. [latex]\\frac{1}{\\sin x}[\/latex]<\/p>\n

19. [latex]\\frac{1}{\\cot x}[\/latex]<\/p>\n

21. [latex]\\tan x[\/latex]<\/p>\n

23. [latex]-4\\sec x\\tan x[\/latex]<\/p>\n

25. [latex]\\pm \\sqrt{\\frac{1}{{\\cot }^{2}x}+1}[\/latex]<\/p>\n

27. [latex]\\frac{\\pm \\sqrt{1-{\\sin }^{2}x}}{\\sin x}[\/latex]<\/p>\n

29. Answers will vary. Sample proof:
\n[latex]\\cos x-{\\cos }^{3}x=\\cos x\\left(1-{\\cos }^{2}x\\right)[\/latex]
\n[latex]=\\cos x{\\sin }^{2}x[\/latex]<\/p>\n

31. Answers will vary. Sample proof:<\/p>\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]<\/p>\n

33. Answers will vary. Sample proof:<\/p>\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]<\/p>\n

35. False<\/p>\n

37. False<\/p>\n

39. Proved with negative and Pythagorean identities<\/p>\n

41. True<\/p>\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]<\/p>\n\n\t\t\t

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